Discussion:
Abstract Algebra Broken
Tim Golden BandTech.com
2020-08-31 12:50:10 UTC
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.

https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation

These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.

After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.

To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.

Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.

We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
Mike Terry
2020-08-31 15:10:13 UTC
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]

This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)

Regards,
Mike.
Tim Golden BandTech.com
2020-08-31 15:40:24 UTC
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Mike Terry
2020-08-31 22:09:42 UTC
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...

By a1 X, do you mean:

a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]

Mike.
Lalo Torres
2020-09-01 01:51:09 UTC
" but this operation has not been defined "

Bring to the table just one known or not-known example where this is not defined

For example in square matrices :

a₀·I + a₁·X + a₂·X² + ... + aₙ₋₁·Xⁿ⁻¹ + aₙ·Xⁿ = O

I : https://en.wikipedia.org/wiki/Identity_matrix
O : https://en.wikipedia.org/wiki/Zero_matrix
· : https://en.wikipedia.org/wiki/Scalar_multiplication#Scalar_multiplication_of_matrices
https://en.wikipedia.org/wiki/Matrix_multiplication#Powers_of_a_matrix

" They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers "

tropical geometry is a semiring, but just in case :

Tropical Polynomials by Bryant Mathews (semiring)
https://circles.math.ucla.edu/circles/lib/data/Handout-2121-1848.pdf
Tropical algebra by Amanda Ellis
https://www.math.utah.edu/mathcircle/notes/MathCircleIv2.pdf

I do not know if the conflict can be solved, maybe it can be precisely outlined where is exactly the conflictive zone.

Looks in the following links, maybe, one or two links can be useful :

https://en.wikipedia.org/wiki/Quotient

https://en.wikipedia.org/wiki/Polynomial#Abstract_algebra

https://en.wikipedia.org/wiki/Exponentiation#Generalizations

https://en.wikipedia.org/wiki/Indeterminate_(variable)#Polynomials
https://math.stackexchange.com/questions/495465/what-is-an-indeterminate-in-a-polynomial-ring
https://proofwiki.org/wiki/Definition:Polynomial_Ring/Indeterminate

https://en.wikipedia.org/wiki/Polylogarithmic_function
https://en.wikipedia.org/wiki/Trigonometric_polynomial

https://proofwiki.org/wiki/Definition:Polynomial_Function/Ring
https://proofwiki.org/wiki/Definition:Polynomial_over_Ring
https://proofwiki.org/wiki/Definition:Polynomial
https://proofwiki.org/wiki/Definition:Polynomial_Ring
https://en.wikipedia.org/wiki/Ring_of_polynomial_functions

https://en.wikipedia.org/wiki/Matrix_polynomial
https://en.wikipedia.org/wiki/Polynomial_matrix
https://en.wikipedia.org/wiki/Matrix_ring
https://proofwiki.org/wiki/Ring_of_Square_Matrices_over_Ring_is_Ring

https://commalg.subwiki.org/wiki/Polynomial_ring_over_a_field
https://commalg.subwiki.org/wiki/Polynomial_ring
https://en.wikipedia.org/wiki/Algebra_over_a_field#Associative_algebras_over_rings
https://proofwiki.org/wiki/Definition_of_Polynomial_from_Polynomial_Ring_over_Sequence

https://proofwiki.org/wiki/Polynomial_Ring_of_Sequences_is_Ring
https://proofwiki.org/wiki/Definition:Ring_of_Polynomial_Forms
https://proofwiki.org/wiki/Definition:Ring_(Abstract_Algebra)

https://en.wikipedia.org/wiki/Center_(ring_theory)
https://encyclopediaofmath.org/wiki/Centre_of_a_ring

An Arithmetic for Rooted Trees by Fabrizio Luccio
http://export.arxiv.org/abs/1510.05512

https://en.wikipedia.org/wiki/Binary_operation
https://en.wikipedia.org/wiki/Binary_function
https://en.wikipedia.org/wiki/Partial_function

https://en.wikipedia.org/wiki/Homogeneous_function

https://groupprops.subwiki.org/
https://encyclopediaofmath.org/
https://commalg.subwiki.org/
https://proofwiki.org/
Tim Golden BandTech.com
2020-09-01 11:29:36 UTC
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
X + ...
=
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
a1 X
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up. Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]
Mike.
zelos...@gmail.com
2020-09-01 11:48:29 UTC
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
X + ...
=
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
a1 X
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up. Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]
Mike.
again mate, read the formal definition. There is no sum, it is only historic notation.
Mike Terry
2020-09-01 17:42:25 UTC
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
X + ...
=
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
OK, so you are saying that you mean the product of a1 and X, so clearly
the product operation involved is a binary operation. [It involves two
operands]

There are now two possibilities: you consider a1 to be a polynomial, in
which case we have regular polynomial multiplication, which is properly
defined, or you consider a1 to be a member of the underlying ring.

It seems your issue is with the second approach. Lets focus on
polynomials over R, which is what you discuss below.

I'll assume you're ok with the definition of a polynomial over R. There
are several equivalent approaches to how these are defined, and I could
expand on this, but they all lead the same way, defining addition and
multiplication /of polynomials/.

But now we have something else: a binary operation taking a real number
and a real polynomial. Of course such an operation needs to be properly
defined to have a meaning, as it's not covered by the definition of
polynomial multiplication. Let's write the op as a binary function sm,
so sm: (R x R[X]) ---> R[x]. [R is the set of real numbers, and R[x]
the set of polynomials over R.] I choose the notation "sm" for "scalar
multiplication".

And here is the definition for sm: if a is in R and p is a polynomial
over R, represented as
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
then
sm(a,p) = (a p_0) + (a p_1) x + (a p_2) x^2 + ... + (a p_n) x^n

There is an important point here. When I wrote p = p_0 + p_1 x + p_2
x^2 + ... + p_n x^n above, I am not writing a long sequence of sums and
products on the right hand side (rhs) of the equals! If I were, you
could correctly claim the definition is circular! I am writing a
polynomial specified in whatever notation polynomials have been
previously defined, except admittedly I've just assumed for convenience
they've been introduced as expressions of this form. (Even though there
are other technical ways of introducing them, authors would typically
make a point that they can be represented in this notation, at least for
typographical convenience.)

So sm is well defined, no problem with this approach either.
Effectively, this is considering R[x] to be a module over its base ring
R, and sm is just like multiplying vectors in a vector space by a scalar
in its base field. [A module is akin to a vector space, except it has a
base /ring/ rather than a base /field/.]

And of course, now that we've defined sm, we can go on and prove basic
things like
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n =
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n)

(Here, the + signs on lhs of equals are just part of the notation for a
polynomial p, not an operation sign, while the + signs on rhs ARE the
binary operation acting on R[x].)

And just to add to notation confusion, it is typical tradition to write
scalar multiplication of sm(a, p) simply as a p, as you did initially,
in which case we can also write

sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n) =
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n

but the meaning of rhs here is quite different from meaning of lhs
above: the rhs here IS a long sequence of (polynomial) sums and
(scalar) products.

Does it seem terrible to you that the notation is ambiguous? The point
is that we realise it is ambiguous, but provably no harm is done, as
we've proved that all the different interpretation lead to the same
result, so no harm is done. This is common practice in mathematics,
balancing simplicity of notation against formal syntactical correctness.
Post by Tim Golden BandTech.com
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
a1 X
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
No, there is nothing "undefined" here. You've not properly grasped the
definition of polynomials I think. Perhaps my explanation above will
help, but possibly we need to go back to the basics of how polynomials
are rigorously defined, and the basic operations on them.

I wouldn't mind doing that if you're serious, or alternatively let me
know if you still think something is undefined...
Post by Tim Golden BandTech.com
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
OK, the problem here is either that you missed the definition for the
scalar product, or possibly the author of the text you're using omitted
it for whatever reason. That does not make abstract algebra "wrong" in
any way, and the problem with your OP was that it comes across as a rant.

Well, lets say it was a rant, but for most people if they were trying to
understand a new field of knowledge and didn't follow something, they
would ASK FOR HELP IN EXPLAINING THEIR CONFUSION, rather than rant on
about how the field is a pile of crap etc.. Do you see how the latter
behaviour justifiably invites laughter and (frankly) scorn from readers?

Anyway, I gave the definitions for scalar multiplication above, so
hopefully all is clear now! :)
Post by Tim Golden BandTech.com
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
In this context, when you consider the sum a0 + a1 X, it is understood
that a0 is the /polynomial/ a0 X^0, and it is simply by tradition that
we often write the shorter a0. Similar to the situation above with sm,
there are basic (provable) results which underly this slight abuse of
notation, rendering it harmless.

Specifically, it is shown that the set of polynomials of the form (a0
x^0) together with polynomial addition and multiplication is
/isomorphic/ to the set of real numbers with real number addition and
multiplication. That is, we have the correspondence

a0 x^0 <----> a0

and we show this correspondence respects the operations of addition and
multiplication appropriate for each side of the correspondence.

Example: (3 x^0) + (5 x^0) = (8 X^0) <----> 8 = 3 + 5, and also
3 x^0 <----> 3 and 5 x^0 <----> 5. (Yes, this is as obvious is it
seems!) So algebraicly R and the set of polynomials of form a x^0
behave exactly the same, and we informally identify them together in day
to day use. (This is like we identify the real number 2 with the
natural number 2, although it can be argued they are conceptually
distinct.)
Post by Tim Golden BandTech.com
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up.
I hope I've shown there is no "breakage". At worst there is some minor
abuse of notation going on, which does no harm, and is completely
understood by everybody except you.

It may not be your fault that you missed out on a fuller explanation in
posting a rant) is down to you...

Regards,
Mike.
Post by Tim Golden BandTech.com
Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]
Mike.
Lalo Torres
2020-09-01 20:06:20 UTC
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)

In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Tim Golden BandTech.com
2020-09-02 11:34:59 UTC
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.

Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.

I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Mostowski Collapse
2020-09-02 11:53:59 UTC
"a1 x" is not necessarily multiplicatively, but
additively closed if you include the zero,

with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
validity, since we checked the special case of R[X]:

a1 x + a2 x = (a1 + a2) x

a1 x * a2 x = (a1 * a2) x^2

But if you use a Boolean ring, where x^2 = x, you get,
and its closed:

a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x

Thats what Boole used, x^2 = x, or x^2 - x = 0, or:

x*(x-1) = 0

The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Mostowski Collapse
2020-09-02 11:58:29 UTC
You find x^2 = x here in Boole's work,
just at the beginning:

first principles. page 16
The Mathematical Analysis by George Boole, 1847
http://www.gutenberg.org/ebooks/36884

Have Fun!
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Tim Golden BandTech.com
2020-09-02 12:40:11 UTC
Post by Mostowski Collapse
You find x^2 = x here in Boole's work,
first principles. page 16
The Mathematical Analysis by George Boole, 1847
http://www.gutenberg.org/ebooks/36884
Thanks Mostowski.
It does seem to be a good read.
When a1 is a 'real coefficient' Mostowski... is the language of the polynomial voo-doo?
I hope you are not a witch.
Post by Mostowski Collapse
Have Fun!
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Tim Golden BandTech.com
2020-09-02 12:16:17 UTC
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
Well Mostowski Collapse, a1 x clearly is a product, and that is what I am discussing. Once again somebody skips right over the fundamental construction. There is an operator already in
a1 X .
Is it a binary operator? Does it fit the ring definition? If not is that a notable detail in a subject which goes to this trouble of defining operators? Particularly when a1 is real and those binary operations are defined on reals... do you see what I see? I've just put your nose right on it here. Does it smell?
Post by Mostowski Collapse
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Mostowski Collapse
2020-09-02 12:23:34 UTC
Of course you can topple something. Even
with a board in front of the head you can
topple something.

Just put a toddler into a room with a tower
of toy boxes, and give him a stick. He will
surely make the tower fell down.

A polynomial a1+a2*X+..+an*X^n from a given
a tuple <a1,a2,..,an>.

Not sure whether this can lead to a revolution.

LoL
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
Well Mostowski Collapse, a1 x clearly is a product, and that is what I am discussing. Once again somebody skips right over the fundamental construction. There is an operator already in
a1 X .
Is it a binary operator? Does it fit the ring definition? If not is that a notable detail in a subject which goes to this trouble of defining operators? Particularly when a1 is real and those binary operations are defined on reals... do you see what I see? I've just put your nose right on it here. Does it smell?
Post by Mostowski Collapse
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Mostowski Collapse
2020-09-02 12:25:29 UTC
Corr.:

Just put a blind toddler into a room with a
pinata somewhere, and give him a stick. He will
surely make hit the pinata sooner or later.
Post by Mostowski Collapse
Of course you can topple something. Even
with a board in front of the head you can
topple something.
Just put a toddler into a room with a tower
of toy boxes, and give him a stick. He will
surely make the tower fell down.
A polynomial a1+a2*X+..+an*X^n from a given
a tuple <a1,a2,..,an>.
Not sure whether this can lead to a revolution.
LoL
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
Well Mostowski Collapse, a1 x clearly is a product, and that is what I am discussing. Once again somebody skips right over the fundamental construction. There is an operator already in
a1 X .
Is it a binary operator? Does it fit the ring definition? If not is that a notable detail in a subject which goes to this trouble of defining operators? Particularly when a1 is real and those binary operations are defined on reals... do you see what I see? I've just put your nose right on it here. Does it smell?
Post by Mostowski Collapse
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Tim Golden BandTech.com
2020-09-02 12:46:22 UTC
Post by Mostowski Collapse
Just put a blind toddler into a room with a
pinata somewhere, and give him a stick. He will
surely make hit the pinata sooner or later.
Yes; quite so. Along the way likely some principles will be gained through the stick as well.
As subjects go the toddler stage is down there in the definition of the ring.
Fundamentals; then suddenly up pops a polynomial in abstract X; something familiar yet quite different too.
Is it so wrong to pick apart that polynomial and have a look at what it is made of?
Would it be so wrong to pick apart the real number and see what it is made of?
Mimicry is far more operant amongst mathematicians than is appreciated.
Toddlers too.
Post by Mostowski Collapse
Post by Mostowski Collapse
Of course you can topple something. Even
with a board in front of the head you can
topple something.
Just put a toddler into a room with a tower
of toy boxes, and give him a stick. He will
surely make the tower fell down.
A polynomial a1+a2*X+..+an*X^n from a given
a tuple <a1,a2,..,an>.
Not sure whether this can lead to a revolution.
LoL
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
"a1 x" is not necessarily multiplicatively, but
Well Mostowski Collapse, a1 x clearly is a product, and that is what I am discussing. Once again somebody skips right over the fundamental construction. There is an operator already in
a1 X .
Is it a binary operator? Does it fit the ring definition? If not is that a notable detail in a subject which goes to this trouble of defining operators? Particularly when a1 is real and those binary operations are defined on reals... do you see what I see? I've just put your nose right on it here. Does it smell?
Post by Mostowski Collapse
additively closed if you include the zero,
with the usual operations of polynomial multiplication and
polynomial addition in variable X and coefficients
from lets say R you get. The result is not a general
a1 x + a2 x = (a1 + a2) x
a1 x * a2 x = (a1 * a2) x^2
But if you use a Boolean ring, where x^2 = x, you get,
a1 x * a2 x = (a1 * a2) x^2 = (a1 * a2) x
x*(x-1) = 0
The special case is now Z[X]/(X^2-X). There is nothing
broken. Its only that you have probably learnt logic
and algebra on a cow farm.
Post by Tim Golden BandTech.com
Post by Lalo Torres
Depending on how you assign(or not assign) what names (of different algebraic structures) to what.
In case you were referring to our little ring has psychological problems, concretely, an identity problem,
and, under some circumstances, the ring pretend to be an algebraic structure, higher in the algebraic hierarchy,
compared to what actually is, specifically, through being named differently of its actual name.
And for binary operation...?
Maybe the usage of the name of an algebraic structure in a compounded name(with several words)
In some way you are requesting a meticulous/thorough examination of the formal definition and several specific examples,
while simultaneously requesting the meaning of each word being used...
Sorry, but I am not the one who built the curriculum known as abstract algebra. I have my nose rubbed in it by the status quo.
Now I rub their noses in it. You seem to think that I am playing on words as if to make english language into mathematics.
No. That said we are deep down at a very fundamental level. The subject bothers to carefully define operators, and this is a first.
It is good and clean. A compiler could work by these principles, but when that same compiler is fed
a0 + a1 X + a2 X X
is will certainly tell you that there is an error. There are five type match errors present in that statement.
If this statement is to compile not by a mathematician but by a machine then another operator must be presented, and because the subject goes to the trouble for the first few operators it ought to go to the trouble for this other operator or two.
Let's not forget that this form is developing multidimensional systems and it does so from infinite dimension. No you won't find this language in the books. It's my interpretation. Rather, staying within what is in the books is what I am trying to do here. It is clear that with real coefficients
a1 X
is not a valid construction. This problem then extends onto the entire polynomial. Undefined product operator present. Non-binary operation detected. Compilation failed. The entire curriculum is up for grabs here. This is something that the straight A's will not understand.
I think you are correct that we are dealing in a matter of language, but I am not the one who is cheating within a clearly stated language. I am the one pointing my finger at it; rubbing your noses in it. It stinks. It is a broken thing. The breaks are carefully covered up and propagated, just as Zelos attempts here.
Tim Golden BandTech.com
2020-09-02 11:49:16 UTC
So below here you state:
"At worst there is some minor abuse of notation going on, which does no harm..."
What exactly is the abuse of notation that you are speaking of?
Just a few short lines will do.
Does the abuse you speak of occur when we write
a1 X
where a1 is real and X is not real?
If so then aren't you conceding my point?
Next of course we really ought to consider how minor abuse can lead to major abuse.
For instance if the abuse in
a1 X
is minor then is the abuse in
a0 + a1 X
larger than the abuse in the more minor abusive form?
Post by Mike Terry
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
X + ...
=
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
OK, so you are saying that you mean the product of a1 and X, so clearly
the product operation involved is a binary operation. [It involves two
operands]
There are now two possibilities: you consider a1 to be a polynomial, in
which case we have regular polynomial multiplication, which is properly
defined, or you consider a1 to be a member of the underlying ring.
It seems your issue is with the second approach. Lets focus on
polynomials over R, which is what you discuss below.
I'll assume you're ok with the definition of a polynomial over R. There
are several equivalent approaches to how these are defined, and I could
expand on this, but they all lead the same way, defining addition and
multiplication /of polynomials/.
But now we have something else: a binary operation taking a real number
and a real polynomial. Of course such an operation needs to be properly
defined to have a meaning, as it's not covered by the definition of
polynomial multiplication. Let's write the op as a binary function sm,
so sm: (R x R[X]) ---> R[x]. [R is the set of real numbers, and R[x]
the set of polynomials over R.] I choose the notation "sm" for "scalar
multiplication".
And here is the definition for sm: if a is in R and p is a polynomial
over R, represented as
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
then
sm(a,p) = (a p_0) + (a p_1) x + (a p_2) x^2 + ... + (a p_n) x^n
There is an important point here. When I wrote p = p_0 + p_1 x + p_2
x^2 + ... + p_n x^n above, I am not writing a long sequence of sums and
products on the right hand side (rhs) of the equals! If I were, you
could correctly claim the definition is circular! I am writing a
polynomial specified in whatever notation polynomials have been
previously defined, except admittedly I've just assumed for convenience
they've been introduced as expressions of this form. (Even though there
are other technical ways of introducing them, authors would typically
make a point that they can be represented in this notation, at least for
typographical convenience.)
So sm is well defined, no problem with this approach either.
Effectively, this is considering R[x] to be a module over its base ring
R, and sm is just like multiplying vectors in a vector space by a scalar
in its base field. [A module is akin to a vector space, except it has a
base /ring/ rather than a base /field/.]
And of course, now that we've defined sm, we can go on and prove basic
things like
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n =
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n)
(Here, the + signs on lhs of equals are just part of the notation for a
polynomial p, not an operation sign, while the + signs on rhs ARE the
binary operation acting on R[x].)
And just to add to notation confusion, it is typical tradition to write
scalar multiplication of sm(a, p) simply as a p, as you did initially,
in which case we can also write
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n) =
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
but the meaning of rhs here is quite different from meaning of lhs
above: the rhs here IS a long sequence of (polynomial) sums and
(scalar) products.
Does it seem terrible to you that the notation is ambiguous? The point
is that we realise it is ambiguous, but provably no harm is done, as
we've proved that all the different interpretation lead to the same
result, so no harm is done. This is common practice in mathematics,
balancing simplicity of notation against formal syntactical correctness.
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
a1 X
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
No, there is nothing "undefined" here. You've not properly grasped the
definition of polynomials I think. Perhaps my explanation above will
help, but possibly we need to go back to the basics of how polynomials
are rigorously defined, and the basic operations on them.
I wouldn't mind doing that if you're serious, or alternatively let me
know if you still think something is undefined...
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
OK, the problem here is either that you missed the definition for the
scalar product, or possibly the author of the text you're using omitted
it for whatever reason. That does not make abstract algebra "wrong" in
any way, and the problem with your OP was that it comes across as a rant.
Well, lets say it was a rant, but for most people if they were trying to
understand a new field of knowledge and didn't follow something, they
would ASK FOR HELP IN EXPLAINING THEIR CONFUSION, rather than rant on
about how the field is a pile of crap etc.. Do you see how the latter
behaviour justifiably invites laughter and (frankly) scorn from readers?
Anyway, I gave the definitions for scalar multiplication above, so
hopefully all is clear now! :)
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
In this context, when you consider the sum a0 + a1 X, it is understood
that a0 is the /polynomial/ a0 X^0, and it is simply by tradition that
we often write the shorter a0. Similar to the situation above with sm,
there are basic (provable) results which underly this slight abuse of
notation, rendering it harmless.
Specifically, it is shown that the set of polynomials of the form (a0
x^0) together with polynomial addition and multiplication is
/isomorphic/ to the set of real numbers with real number addition and
multiplication. That is, we have the correspondence
a0 x^0 <----> a0
and we show this correspondence respects the operations of addition and
multiplication appropriate for each side of the correspondence.
Example: (3 x^0) + (5 x^0) = (8 X^0) <----> 8 = 3 + 5, and also
3 x^0 <----> 3 and 5 x^0 <----> 5. (Yes, this is as obvious is it
seems!) So algebraicly R and the set of polynomials of form a x^0
behave exactly the same, and we informally identify them together in day
to day use. (This is like we identify the real number 2 with the
natural number 2, although it can be argued they are conceptually
distinct.)
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up.
I hope I've shown there is no "breakage". At worst there is some minor
abuse of notation going on, which does no harm, and is completely
understood by everybody except you.
It may not be your fault that you missed out on a fuller explanation in
posting a rant) is down to you...
Regards,
Mike.
Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]
Mike.
Mike Terry
2020-09-02 14:40:55 UTC
Post by Tim Golden BandTech.com
"At worst there is some minor abuse of notation going on, which does no harm..."
What exactly is the abuse of notation that you are speaking of?
Just a few short lines will do.
Certain expressions could be formally interpreted in multiple
(equivalent) ways. In an automated proof system, these would need to be
disambiguated, resulting in a reduction of readability, which is why
mathematicians typically allow such things to go on - no harm is done,
Post by Tim Golden BandTech.com
Does the abuse you speak of occur when we write
a1 X
where a1 is real and X is not real?
Post by Tim Golden BandTech.com
If so then aren't you conceding my point?
No, because no harm is done, and abstract algebra is not "broken" as you
were suggesting, and nothing needs to change. Your rant is just that -
a cranky rant, making ridiculous accusations and claims.
Post by Tim Golden BandTech.com
Next of course we really ought to consider how minor abuse can lead to major abuse.
For instance if the abuse in
a1 X
is minor then is the abuse in
a0 + a1 X
larger than the abuse in the more minor abusive form?
No, about the same, and also totally harmless. So nothing has led to
any "major abuse".

I get the impression you've not properly read what I wrote and
definitely not learned anything from it, so I won't be commenting
further, unless I feel you are trying to genuinely understand something

Mike.
Julio Di Egidio
2020-09-02 15:03:47 UTC
Post by Mike Terry
Post by Tim Golden BandTech.com
"At worst there is some minor abuse of notation going on, which does no harm..."
What exactly is the abuse of notation that you are speaking of?
Just a few short lines will do.
Certain expressions could be formally interpreted in multiple
(equivalent) ways. In an automated proof system, these would need to be
disambiguated, resulting in a reduction of readability, which is why
mathematicians typically allow such things to go on - no harm is done,
Alas, as usual, I find your approach totally misguided:

- Mathematicians give careful definitions, then the notation is
neither ambiguous nor problematic, it just depends on the specific
context.

they better find a different hobby, as maths is full of it: OTOH,
that is certainly not a problem for a machine.

- Moreover, the reality here is that Mr Golden is not complaining
involved. (To dislike the notation, he'd firstly have to agree on
the definitions.)

Bah...

Julio
Tim Golden BandTech.com
2020-09-07 12:21:09 UTC
Post by Mike Terry
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
[snip rest of ridiculous rant]
X + ...
=
Post by Mike Terry
Post by Tim Golden BandTech.com
Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
Regards,
Mike.
Hi Mike.
a1 X
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
OK, so you are saying that you mean the product of a1 and X, so clearly
the product operation involved is a binary operation. [It involves two
operands]
There are now two possibilities: you consider a1 to be a polynomial, in
which case we have regular polynomial multiplication, which is properly
defined, or you consider a1 to be a member of the underlying ring.
It seems your issue is with the second approach. Lets focus on
polynomials over R, which is what you discuss below.
I'll assume you're ok with the definition of a polynomial over R. There
are several equivalent approaches to how these are defined, and I could
expand on this, but they all lead the same way, defining addition and
multiplication /of polynomials/.
But now we have something else: a binary operation taking a real number
and a real polynomial. Of course such an operation needs to be properly
defined to have a meaning, as it's not covered by the definition of
polynomial multiplication. Let's write the op as a binary function sm,
so sm: (R x R[X]) ---> R[x]. [R is the set of real numbers, and R[x]
the set of polynomials over R.] I choose the notation "sm" for "scalar
multiplication".
And here is the definition for sm: if a is in R and p is a polynomial
over R, represented as
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
then
sm(a,p) = (a p_0) + (a p_1) x + (a p_2) x^2 + ... + (a p_n) x^n
There is an important point here. When I wrote p = p_0 + p_1 x + p_2
x^2 + ... + p_n x^n above, I am not writing a long sequence of sums and
products on the right hand side (rhs) of the equals! If I were, you
could correctly claim the definition is circular!
I've got to quote you here:

" When I wrote
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
above, I am not writing a long sequence of sums and products " - Mike Terry

Mike this is extremely weak. If you did not mean to have a sum of products then why would you write a sum of products?
Why would the operators work out as a sum of products? Why would you make use of a 'polynomial with real coefficients'? You have falsified the subject here by denying that the binary operators work within the polynomial, and yet you disown this detail for your own convenience. This sort of wheedling does expose not just minor abuse. Of course this current abuse that I speak of is you attempting some sort of fixup on a failed construction. Obviously for you to go this far suggests that in secret you have gazed upon the black swan. That you would go this far to save the thing it falsifies is a fine indication. I certainly do not claim any circular definition. The very name 'polynomial' means many terms. It existed prior to abstract algebra and possibly because of this it gets less scrutiny by the beloved suppliers of book fees. The ring sum is so strongly implied that for you to deny that the '+' signs in the construction are addition (the ring defined operator) is a sure loser. I just love how others pop in here to make such corrections. By disowning the polynomial you feel that you have corrected the subject... perfect status quo position. Thanks Mike.

Circular definitions are actually far less offensive than constructions which falsify their own axioms. All of your argumentation does not confront the black swan
1.23 X
This is why I've avoided it till now. When someone provides a falsification of a subject the falsification has to be wrong in order for you to have refuted it. This means that you have to work in terms of this expression, which you have not done here. All that you have done, and all that others have done, is provide their own interpretation on the standing subject. It would seem as if (taking your statement above here seriously) that what you are attempting is a claim that the polynomial may not be dismantled. This is a false claim, and your own attempt to deny that the sums are actual ring sums is quite a position for you to land in... all the while admitting 'minor abuse' of the propped system. Again, you have supported my argument by dodging it and further you have landed yourself in a mess of goose shit. You've slipped in it and it is all over your clothing and your face here.

I'd love a link to a text which denies that the polynomial expression in use in abstract algebra is not actually a sum of terms. Thanks Mike. You've really demonstrated what a hack abstract algebra actually is. The quotient and ideal use similar language as you are using here. You are so brilliant that maybe the goose shit will burn off.

May the black swan live on.
Post by Mike Terry
I am writing a
polynomial specified in whatever notation polynomials have been
previously defined, except admittedly I've just assumed for convenience
they've been introduced as expressions of this form. (Even though there
are other technical ways of introducing them, authors would typically
make a point that they can be represented in this notation, at least for
typographical convenience.)
So sm is well defined, no problem with this approach either.
Effectively, this is considering R[x] to be a module over its base ring
R, and sm is just like multiplying vectors in a vector space by a scalar
in its base field. [A module is akin to a vector space, except it has a
base /ring/ rather than a base /field/.]
And of course, now that we've defined sm, we can go on and prove basic
things like
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n =
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n)
(Here, the + signs on lhs of equals are just part of the notation for a
polynomial p, not an operation sign, while the + signs on rhs ARE the
binary operation acting on R[x].)
And just to add to notation confusion, it is typical tradition to write
scalar multiplication of sm(a, p) simply as a p, as you did initially,
in which case we can also write
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n) =
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
but the meaning of rhs here is quite different from meaning of lhs
above: the rhs here IS a long sequence of (polynomial) sums and
(scalar) products.
Does it seem terrible to you that the notation is ambiguous? The point
is that we realise it is ambiguous, but provably no harm is done, as
we've proved that all the different interpretation lead to the same
result, so no harm is done. This is common practice in mathematics,
balancing simplicity of notation against formal syntactical correctness.
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
a1 X
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
No, there is nothing "undefined" here. You've not properly grasped the
definition of polynomials I think. Perhaps my explanation above will
help, but possibly we need to go back to the basics of how polynomials
are rigorously defined, and the basic operations on them.
I wouldn't mind doing that if you're serious, or alternatively let me
know if you still think something is undefined...
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
OK, the problem here is either that you missed the definition for the
scalar product, or possibly the author of the text you're using omitted
it for whatever reason. That does not make abstract algebra "wrong" in
any way, and the problem with your OP was that it comes across as a rant.
Well, lets say it was a rant, but for most people if they were trying to
understand a new field of knowledge and didn't follow something, they
would ASK FOR HELP IN EXPLAINING THEIR CONFUSION, rather than rant on
about how the field is a pile of crap etc.. Do you see how the latter
behaviour justifiably invites laughter and (frankly) scorn from readers?
Anyway, I gave the definitions for scalar multiplication above, so
hopefully all is clear now! :)
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
In this context, when you consider the sum a0 + a1 X, it is understood
that a0 is the /polynomial/ a0 X^0, and it is simply by tradition that
we often write the shorter a0. Similar to the situation above with sm,
there are basic (provable) results which underly this slight abuse of
notation, rendering it harmless.
Specifically, it is shown that the set of polynomials of the form (a0
x^0) together with polynomial addition and multiplication is
/isomorphic/ to the set of real numbers with real number addition and
multiplication. That is, we have the correspondence
a0 x^0 <----> a0
and we show this correspondence respects the operations of addition and
multiplication appropriate for each side of the correspondence.
Example: (3 x^0) + (5 x^0) = (8 X^0) <----> 8 = 3 + 5, and also
3 x^0 <----> 3 and 5 x^0 <----> 5. (Yes, this is as obvious is it
seems!) So algebraicly R and the set of polynomials of form a x^0
behave exactly the same, and we informally identify them together in day
to day use. (This is like we identify the real number 2 with the
natural number 2, although it can be argued they are conceptually
distinct.)
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up.
I hope I've shown there is no "breakage". At worst there is some minor
abuse of notation going on, which does no harm, and is completely
understood by everybody except you.
It may not be your fault that you missed out on a fuller explanation in
posting a rant) is down to you...
Regards,
Mike.
Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]
Mike.
zelos...@gmail.com
2020-09-01 05:40:14 UTC
Post by Tim Golden BandTech.com
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
Because its about rings, not real numbers.
Post by Tim Golden BandTech.com
After the careful construction of the ring from two groups, both operators obeying the closure principle
Not all rings have both operations as groups.
Post by Tim Golden BandTech.com
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
how is that not a polynomial?
Lalo Torres
2020-09-01 07:02:14 UTC
" Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together "

One can think of the ring K[X] as arising from K by adding one new element X that is external to K, commutes with all elements of K, and has no other specific properties. (This may be used for defining polynomial rings.) from https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)

" Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect "

Expand this point. This may have several meaning, specify...

" The sad part is that the weakness of the construction cannot be challenged "

If you are somehow speaking against a construct, you could come up with some alternative, even with a basic and primitive construct.
Otherwise, you leaves us into choose "mid-air with no-ground under our feet" or "our everyday customs". Since the mind is lazy almost
always prefer the last option. Different thing is to provide an alternative.

I guess that a majority of pro-academia depend on their academic status, but this is not a generalization to all.
Not all pro-academia people depend on their status to pay the bread, or a reputation to be defended.
It is just that some people are in a situation, lined up with academia, without much to lose if the wind change the direction...
Lalo Torres
2020-09-01 08:43:46 UTC
On the other hand, I could accept 'X as apples with An as apples',
I could reject 'X as pineapples with An as apples'
And indeed accept 'X as fruits with An as apples'

In which case you could give more examples regarding what you think can be mixed, or not mixed...
Tim Golden BandTech.com
2020-09-03 19:34:52 UTC
Post by Lalo Torres
On the other hand, I could accept 'X as apples with An as apples',
I could reject 'X as pineapples with An as apples'
And indeed accept 'X as fruits with An as apples'
In which case you could give more examples regarding what you think can be mixed, or not mixed...
It is such a simplistic instance that I put out here, and yet none of these people who I can accept have mastered abstract algebra can come down to such a simple level. In this case it is not for me really to say what I think can and cannot be mixed. The requirements have already been formalized and I am working within those formalities on an extremely simple instance. The polynomial construction fails the ring requirements under these conditions; particularly when we specify real coefficients for the still abstract polynomial in X where X has no type constraint whatsoever. I make one simple black swan as
1.23 X
and none here will address it. This expression above offends the ring requirements. It is this simple. Instantiation breaks the long back of abstract algebra.
Mostowski Collapse
2020-09-03 19:45:41 UTC
The instantiation already breaks if you
plug in decimal numbers with two digits.

1.23 * 9.87 = 12.1401

The result has 4 digits and not anymore
two digits. We can say this little example

already Fucked Up Beyond All Recognition
abstract algebra and renders it totally useless.
Post by Tim Golden BandTech.com
Post by Lalo Torres
On the other hand, I could accept 'X as apples with An as apples',
I could reject 'X as pineapples with An as apples'
And indeed accept 'X as fruits with An as apples'
In which case you could give more examples regarding what you think can be mixed, or not mixed...
It is such a simplistic instance that I put out here, and yet none of these people who I can accept have mastered abstract algebra can come down to such a simple level. In this case it is not for me really to say what I think can and cannot be mixed. The requirements have already been formalized and I am working within those formalities on an extremely simple instance. The polynomial construction fails the ring requirements under these conditions; particularly when we specify real coefficients for the still abstract polynomial in X where X has no type constraint whatsoever. I make one simple black swan as
1.23 X
and none here will address it. This expression above offends the ring requirements. It is this simple. Instantiation breaks the long back of abstract algebra.
Mostowski Collapse
2020-09-03 19:49:36 UTC
Or as a friend of mine used to say,
when the ice machine didn't work

since he didn't fill water,
abstract algebra is borked.

https://www.urbandictionary.com/define.php?term=borked

LMAO!
Post by Mostowski Collapse
The instantiation already breaks if you
plug in decimal numbers with two digits.
1.23 * 9.87 = 12.1401
The result has 4 digits and not anymore
two digits. We can say this little example
already Fucked Up Beyond All Recognition
abstract algebra and renders it totally useless.
Post by Tim Golden BandTech.com
Post by Lalo Torres
On the other hand, I could accept 'X as apples with An as apples',
I could reject 'X as pineapples with An as apples'
And indeed accept 'X as fruits with An as apples'
In which case you could give more examples regarding what you think can be mixed, or not mixed...
It is such a simplistic instance that I put out here, and yet none of these people who I can accept have mastered abstract algebra can come down to such a simple level. In this case it is not for me really to say what I think can and cannot be mixed. The requirements have already been formalized and I am working within those formalities on an extremely simple instance. The polynomial construction fails the ring requirements under these conditions; particularly when we specify real coefficients for the still abstract polynomial in X where X has no type constraint whatsoever. I make one simple black swan as
1.23 X
and none here will address it. This expression above offends the ring requirements. It is this simple. Instantiation breaks the long back of abstract algebra.
Tim Golden BandTech.com
2020-09-03 20:01:59 UTC
Post by Mostowski Collapse
The instantiation already breaks if you
plug in decimal numbers with two digits.
1.23 * 9.87 = 12.1401
The result has 4 digits and not anymore
two digits. We can say this little example
already Fucked Up Beyond All Recognition
Well at least I've now learned what FUBAR means.
I can't really agree with your statement here though.
I guess u r far less serious than I had anticipated.
You are on a diversion from a diversion here.
I've stuck your nose right under the black swan's ass,
it left you a nice load on the tip of your nose,
and still you won't wipe it.
Post by Mostowski Collapse
abstract algebra and renders it totally useless.
Post by Lalo Torres
On the other hand, I could accept 'X as apples with An as apples',
I could reject 'X as pineapples with An as apples'
And indeed accept 'X as fruits with An as apples'
In which case you could give more examples regarding what you think can be mixed, or not mixed...
It is such a simplistic instance that I put out here, and yet none of these people who I can accept have mastered abstract algebra can come down to such a simple level. In this case it is not for me really to say what I think can and cannot be mixed. The requirements have already been formalized and I am working within those formalities on an extremely simple instance. The polynomial construction fails the ring requirements under these conditions; particularly when we specify real coefficients for the still abstract polynomial in X where X has no type constraint whatsoever. I make one simple black swan as
1.23 X
and none here will address it. This expression above offends the ring requirements. It is This must be true only mthis simple. Instantiation breaks the long back of abstract algebra.
Mostowski Collapse
2020-09-03 20:21:55 UTC
We can say abstract algebra is plastered
with black swans, possible all dead.
Its very difficult to not step on them
and navigate through this swamp call

abstract algebra. This is because the
oppose, they swallow all the fake science

and text book. Lets make mathematics great
again and drain this swamp.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
The instantiation already breaks if you
plug in decimal numbers with two digits.
1.23 * 9.87 = 12.1401
The result has 4 digits and not anymore
two digits. We can say this little example
already Fucked Up Beyond All Recognition
Well at least I've now learned what FUBAR means.
I can't really agree with your statement here though.
I guess u r far less serious than I had anticipated.
You are on a diversion from a diversion here.
I've stuck your nose right under the black swan's ass,
it left you a nice load on the tip of your nose,
and still you won't wipe it.
Lalo T.
2020-09-03 21:49:00 UTC
I know Tim is completely able to operate the construction and obtain numbers from it.
Also he will not have problems with operating "more composite" objects like:

https://en.wikipedia.org/wiki/Quaternion#Matrix_representations
https://en.wikipedia.org/wiki/Supermatrix
https://en.wikipedia.org/wiki/Split-octonion#Zorn's_vector-matrix_algebra
https://en.wikipedia.org/wiki/Polynomial_matrix
https://en.wikipedia.org/wiki/Zhegalkin_polynomial
https://en.wikipedia.org/wiki/Matrix_polynomial

...or objecs that changing the usual two operations of a structure, like in tropical geometry,
Note that arithmetic in tropical geometry the "addition operation" does not have inverse, hence, a semiring, but it serves to this purpose.
https://circles.math.ucla.edu/circles/lib/data/Handout-2121-1848.pdf (tropical polynomial)
I put this example just for the sake, that you can replace the usual operations of certain algebraic structures usually perceived as the usual addition and product in of numbers,
in this case, this object has two operations, where the "addition operation" is the min operation(or max operation) and the "product operation" is the usual addition.

As Julio said, Tim does not have any problem with notation, and I agree with that.

It is just the case that Tim does not accept the definition (the very contruction).

But let's proceed.

An external binary operation is a binary function from K × S to S. This differs from a binary operation on a set in the sense in that K need not be S; its elements come from outside...
from :
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
https://en.wikipedia.org/wiki/Scalar_multiplication#Interpretation}

Now, suppose we take, for example, matrices of mxm (squares matrices)
https://en.wikipedia.org/wiki/Square_matrix

...being its element real numbers. We endow this matrices with the operation "Entrywise sum"

So far we have (matrices of mxm, entrywise sum '+') and this is an abelian group:

https://proofwiki.org/wiki/Definition:Group
https://proofwiki.org/wiki/Definition:Abelian_Group

We proceed to endow our object with another operation, concretely, matrix multiplication :

https://en.wikipedia.org/wiki/Matrix_multiplication#Definition
https://en.wikipedia.org/wiki/File:Matrix_multiplication_diagram_2.svg

So far we have (matrices of mxm, entrywise sum '+', matrix multiplication '*') and this is a "Unit Ring with zero divisors".

https://mathworld.wolfram.com/UnitRing.html

For the present purpose we are insterested in the "Unit Ring" part.

Instantiating your statement to square matrices :

a1 X <---here a1 is a real number and X a square matrix where its elements are real numbers.

Here we use an specific object (matrices of mxm, entrywise sum '+', scalar multiplication '·') !!!
We say here that our matrices of mxm is provided with a structure of a Vector Space.

a1·X <--it is defined
where symbol ' · ' is the scalar multiplication between a real and a square matrix (not the Dot product)

Now, It happens that in :

a0 I + a1 X + a2 X X + a3 X X X + ...
a0·(I) + a1·(X) + a2·(X*X) + a3·(X*X*X) ...

Tim request, the application of the ring definition :

https://proofwiki.org/wiki/Definition:Ring_(Abstract_Algebra)

Our object is :

(a) (matrices of mxm, entrywise sum '+', matrix multiplication '*')

but in the other definition we have :

" The polynomial ring in X over K is equipped with an addition, a multiplication AND SCALAR MULTIPLICATION that make it a commutative algebra "

this is :

(b) (matrices of mxm, entrywise sum '+', matrix multiplication '*',scalar multiplication '·')

Which indeed, is a structure with three operations, not just two. Note that (b) is algebraically bigger than (a)
("less primitive than")

Certainly Tim rejects (b) as a ring, due to the third operation (scalar multiplication ' · ' )
and also alluding to the ring closure property, concretely, the operation ' · ',
which start from the set of :

"the cartesian product of 'matrices of mxm' with ' with 'real numbers' "

ending in the set of :

'matrices of mxm'

Aₘₓₘ X r ---> Aₘₓₘ (here ' X ' symbol is the cartesian product)

Tim could use the words "perceptual blindness", "ignorance" or "doublethink" on us, since we SEEM to reject his statements.

Note the words "scalar-valued polynomial" in :
https://en.wikipedia.org/wiki/Matrix_polynomial

As a curiosity we can get the same effect in :

(1) λ₀·I + λ₁·X + λ₂·X² + ... + λₙ₋₁·Xⁿ⁻¹ + λₙ·Xⁿ where λ are Scalars, X matrices

(2) A₀*I + A₁*X + A₂*X² + ... + Aₙ₋₁*Xⁿ⁻¹ + Aₙ*Xⁿ where A are Scalar Matrices, and X matrices

https://en.wikipedia.org/wiki/Center_(ring_theory) (in square matrices)

In this context, the scalar multiplication is well-defined, and also an atractive operation to have in a structure...
One likes to "scale" things. Which bring to the table the question of the origin of the scalar multiplication (used as a external operation, also a bigwig/kingpin )
In the case where is it possible to use the "second operation" of a ring to get the same results, equivalent to asking why dont use the "internal" product of a ring if
we get the same effects, why do I have to outsource an operation that my team can do for themself ?

Note in "Buildings, polytopes and tropical convexity" by Annette Werner
https://www.uni-frankfurt.de/50581267/tropical_geometry11.pdf
and look the word "tropical scalar multiplication" in page 2
(semi-module structure)
Mostowski Collapse
2020-09-03 22:44:35 UTC
Doesn't work for multivariante polynomials.
Matrix multiplication is not commutative.

You don't have in general:

/* is not generally valid */

X*Y = Y*X

https://en.wikipedia.org/wiki/Matrix_multiplication#Non-commutativity
Post by Lalo T.
a1 X <---here a1 is a real number and X a square matrix where its elements are real numbers.
Lalo T.
2020-09-03 23:29:27 UTC
Yes, certainly I inject the definition "The polynomial ring, K[X], in X over a field (or, more generally, a COMMUTATIVE ring) K can be defined..."

...and after, I used a non-commutative ring, which is mistake.

Certainly, Tim's initial post on this thread use rings, not commutative rings.
Lalo T.
2020-09-04 02:19:12 UTC
In the post, what I am contending is that the conflict between Tim and everyone else, in this thread, is only apparent.
I think it could be cleared through careful and detailed examination, with specific examples.

In addition, that the Scalar Multiplication operation can be seen, in some instances as a mix of :

https://en.wikipedia.org/wiki/Alien_hand_syndrome
https://www.investopedia.com/terms/o/outsourcing.asp

"Design Decision" or not, the point here is that this last topic has some tentacles towards
Theoretical Physics (topic which I am ignorant of) in terms of the physical significance of maths,
while attempting to isolate the scalar multiplication "landscape".

In any case, my intention with the last topic is not as much, divert the attention of the thread.
In one of the his statements Tim says that something is stinking. I do not agree with him in this regard,
that "a0 X" is ill-formed formula, and I think it is a well-formed formula.

Although as a consecuence, this thread give me the chance to appreciate from another angle the Vector War debates of 150 years ago.
Lalo T.
2020-09-04 03:33:23 UTC
Long story short, I think the smelling thing (stink or aroma) is not the definitions nor all constructions on top of these definitions,
rather, the motivations behind the definitions, which in part, are beyond the scope of Mathematics...
Tim Golden BandTech.com
2020-09-04 12:40:27 UTC
Post by Lalo T.
In the post, what I am contending is that the conflict between Tim and everyone else, in this thread, is only apparent.
I think it could be cleared through careful and detailed examination, with specific examples.
https://en.wikipedia.org/wiki/Alien_hand_syndrome
https://www.investopedia.com/terms/o/outsourcing.asp
"Design Decision" or not, the point here is that this last topic has some tentacles towards
Theoretical Physics (topic which I am ignorant of) in terms of the physical significance of maths,
while attempting to isolate the scalar multiplication "landscape".
In any case, my intention with the last topic is not as much, divert the attention of the thread.
In one of the his statements Tim says that something is stinking. I do not agree with him in this regard,
that "a0 X" is ill-formed formula, and I think it is a well-formed formula.
OK, so Lalo can come to the black swan and state that it is a well-formed formula.
That's nice. At least you have had a visit to the thing and attempted to look at it.
Next part of the puzzle is to actually address whether "a0 X" is a binary operation...
Is it ring behaved?

Thus far you have not invoked any of the language that makes up the ground work of abstract algebra.
And Lalo you must know by now that the X is not real and that a0 is real.
So you are staring the swan in the face. Is it a black swan?
Look again please. The full polynomial is simply a long sum of such individual parts. The conflict worsens here, but that is not the problem. Just one black swan will do in the name of falsification... a detail that none here deny. They simply avoid it altogether. Perhaps usenet is providing a new form: proof by vacuum. The usenet dodge is something very familiar to me. Really I write more for the onlookers. We have to get some kind of entertainment into the thread. To me the best is a mix of strong rhetoric with direct content. But still I think focus is more important here than the branching meandering that is your habit. Whatever: you are free here in an uncensored distributed medium that is something future bots will look over to at least gain an understanding of the human race and its individual inhabitants.

Thank you so much for even going this far. It exposes the others as WIMPS as in the weakly interacting type. None has come as far as you. Even though Mostowski has it on his nose he still cannot smell it. I wouldn't mind hearing about the vector war which no doubt requires the usage of ordered values. This is another area where polysign shine. Also though the lack of need for the cartesian product, whose use in the ring definition is problematic as well. I know that is a bit cryptic so let's go ahead and have a diversion from your own diversions here. In ordinary physical terms we do not generally multiply values on a real line and land with a value on a real line. We instead land such things in units; say meters; whose product then yield square meters. Thus the space of the result is not at all the space of the sources. Within the formal definition of the binary operator we often will see
f : S X S -> S
which is exactly inverted from the ordinary physical sense. You see the abuse of dimensionality by mathematicians is profound and so the interdimensional interpretation has quite a lot of room. Abstract algebra goes on to develop two dimensional numbers from infinite dimensional numbers. Brilliant eh?

I do think it is valid to cast doubt on the accumulation of information that is modern mathematics. And yes, notation does matter. It is as if we have granted humans compiler level integrity by naming them 'mathematician' but it is by no means true. To me this break in the fundamentals is exciting and because we are down in so low and deep any modification at this level is bound to have extensive side effects. Kaboom!
Post by Lalo T.
Although as a consecuence, this thread give me the chance to appreciate from another angle the Vector War debates of 150 years ago.
Peter
2020-09-04 14:09:06 UTC
Post by Tim Golden BandTech.com
Post by Lalo T.
In the post, what I am contending is that the conflict between Tim and everyone else, in this thread, is only apparent.
I think it could be cleared through careful and detailed examination, with specific examples.
https://en.wikipedia.org/wiki/Alien_hand_syndrome
https://www.investopedia.com/terms/o/outsourcing.asp
"Design Decision" or not, the point here is that this last topic has some tentacles towards
Theoretical Physics (topic which I am ignorant of) in terms of the physical significance of maths,
while attempting to isolate the scalar multiplication "landscape".
In any case, my intention with the last topic is not as much, divert the attention of the thread.
In one of the his statements Tim says that something is stinking. I do not agree with him in this regard,
that "a0 X" is ill-formed formula, and I think it is a well-formed formula.
OK, so Lalo can come to the black swan and state that it is a well-formed formula.
That's nice. At least you have had a visit to the thing and attempted to look at it.
Next part of the puzzle is to actually address whether "a0 X" is a binary operation...
No it isn't. In a ring the binary operations are addition and
multiplication. "a0 X" is the result of multiplying the ring element a0
and the ring element X (that X is premultiplied n=by an invisible 1) but
it isn't a binary operation. There is obviously a difference between
binary operations and what results when two elements are combined by the
operation.
Post by Tim Golden BandTech.com
Is it ring behaved?
Thus far you have not invoked any of the language that makes up the ground work of abstract algebra.
And Lalo you must know by now that the X is not real and that a0 is real.
So you are staring the swan in the face. Is it a black swan?
Look again please. The full polynomial is simply a long sum of such individual parts. The conflict worsens here, but that is not the problem. Just one black swan will do in the name of falsification... a detail that none here deny. They simply avoid it altogether. Perhaps usenet is providing a new form: proof by vacuum. The usenet dodge is something very familiar to me. Really I write more for the onlookers. We have to get some kind of entertainment into the thread. To me the best is a mix of strong rhetoric with direct content. But still I think focus is more important here than the branching meandering that is your habit. Whatever: you are free here in an uncensored distributed medium that is something future bots will look over to at least gain an understanding of the human race and its individual inhabitants.
Thank you so much for even going this far. It exposes the others as WIMPS as in the weakly interacting type. None has come as far as you. Even though Mostowski has it on his nose he still cannot smell it. I wouldn't mind hearing about the vector war which no doubt requires the usage of ordered values. This is another area where polysign shine. Also though the lack of need for the cartesian product, whose use in the ring definition is problematic as well. I know that is a bit cryptic so let's go ahead and have a diversion from your own diversions here. In ordinary physical terms we do not generally multiply values on a real line and land with a value on a real line. We instead land such things in units; say meters; whose product then yield square meters. Thus the space of the result is not at all the space of the sources. Within the formal definition of the binary operator we often will see
f : S X S -> S
which is exactly inverted from the ordinary physical sense. You see the abuse of dimensionality by mathematicians is profound and so the interdimensional interpretation has quite a lot of room. Abstract algebra goes on to develop two dimensional numbers from infinite dimensional numbers. Brilliant eh?
I do think it is valid to cast doubt on the accumulation of information that is modern mathematics. And yes, notation does matter. It is as if we have granted humans compiler level integrity by naming them 'mathematician' but it is by no means true. To me this break in the fundamentals is exciting and because we are down in so low and deep any modification at this level is bound to have extensive side effects. Kaboom!
Post by Lalo T.
Although as a consecuence, this thread give me the chance to appreciate from another angle the Vector War debates of 150 years ago.
Tim Golden BandTech.com
2020-09-01 11:49:43 UTC
Post by Lalo Torres
" Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together "
One can think of the ring K[X] as arising from K by adding one new element X that is external to K, commutes with all elements of K, and has no other specific properties. (This may be used for defining polynomial rings.) from https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
Nice find. It is not pretty, but worst of all the notion of binary operators is still in conflict with this construction. I would suggest that this univariate definition is inherently conflicted. It actually breaks the ring definition now formally. Before it was a cover up; now under this definition it is an out and out lie. Still it is an excellent find and does deserve a follow up within this thread. So long as they maintain K as some abstract thing you see it doesn't quite fully fall apart. Upon instantiating the reals with this X in the univariate mode then we break the operators since
a1 X
does not resolve. Thus the operators cannot be binary operators and so the ring definition is explicitly broken. We are in the same position with a slight variation in interpretation, but the failing is the same.
Post by Lalo Torres
" Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect "
Expand this point. This may have several meaning, specify...
" The sad part is that the weakness of the construction cannot be challenged "
If you are somehow speaking against a construct, you could come up with some alternative, even with a basic and primitive construct.
Otherwise, you leaves us into choose "mid-air with no-ground under our feet" or "our everyday customs". Since the mind is lazy almost
always prefer the last option. Different thing is to provide an alternative.
I guess that a majority of pro-academia depend on their academic status, but this is not a generalization to all.
Not all pro-academia people depend on their status to pay the bread, or a reputation to be defended.
It is just that some people are in a situation, lined up with academia, without much to lose if the wind change the direction...
I think I'd rather just remain within the falsification of abstract algebra at the polynomial stage rather than state its replacement.
The falsification is free-standing and needs no such patch.
On academia; well; it makes a good punching bag. Of course some of my language is rhetoric. But also there is content. My belief is that the two stand together to make strong points. Some of the rhetoric stands true without the math content. I don't need to falsify abstract algebra to criticize the academic system, but it helps. I don't need to criticize the academic system to falsify abstract algebra, but the connection is fairly clear. Anyway we are free here to speak our minds even if some go sour. Those who post pure rhetoric without any content are truly suspect. Fortunately Mike has allowed a discussion with content to take place, though his opening was pure rhetoric. We'll see how it goes.
Julio Di Egidio
2020-09-01 12:33:33 UTC
Post by Tim Golden BandTech.com
a1 X
does not resolve. Thus the operators cannot be binary operators and so
the ring definition is explicitly broken. We are in the same position
with a slight variation in interpretation, but the failing is the same.
The "ring definition" is simply irrelevant there. The coefficients come
from a ring, but X, the "indeterminate" in the formation of a polynomial,
*as well as the operations* involved in that structure formation, are
simply *formal terms* standing for the place of the coefficients in a
sequence: there is no evaluation involved.

As for evaluation, we actually first go from the polynomial to a
polynomial function where critical is that the domain of the function
is itself a ring and contains the ring of coefficients, then we do
evaluation of the function, where there is now no problem with taking
the operations involved as operations: on the function domain ring...

HTH,

Julio
Tim Golden BandTech.com
2020-09-02 12:11:11 UTC
Post by Tim Golden BandTech.com
a1 X
does not resolve. Thus the operators cannot be binary operators and so
the ring definition is explicitly broken. We are in the same position
with a slight variation in interpretation, but the failing is the same.
The "ring definition" is simply irrelevant there. The coefficients come
from a ring, but X, the "indeterminate" in the formation of a polynomial,
*as well as the operations* involved in that structure formation, are
simply *formal terms* standing for the place of the coefficients in a
sequence: there is no evaluation involved.
As for evaluation, we actually first go from the polynomial to a
polynomial function where critical is that the domain of the function
is itself a ring and contains the ring of coefficients, then we do
evaluation of the function, where there is now no problem with taking
the operations involved as operations: on the function domain ring...
HTH,
Julio
Hi Julio. I appreciate that you will uphold the standard interpretation fairly well. So let's get on with it. I particularly am discussing polynomials with real coefficients, so a(n) are real valued. This is commonly done within the curriculum. So we start with the acclaimed ring behaved
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
which is not ring behaved since the product is not a binary operation. This is the conflict in its entirety, though the extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined. Clearly we have a real being added to a nonreal. Is this addition a binary operator or isn't it? If it is not a binary operator then should this operator be more explicitly defined within abstract algebra since the subject has bothered to define operators so carefully?

This procedure then extends onto the a2 term and so forth. In effect the entire polynomial is constructed of undefined operators.

Instantiation is often actually a problem in higher maths. Sure we can construct all sorts of fancy ideas and notations on paper. Humans excel at it. Quite often I myself will sit down to design something and it gets complicated. Quite often the final solution is considerably simpler. Instantiation of the high ideas that 'mathematicians' work on often descend down to fairly trivial instances with no fascinating instances actually constructed. I do put abstract algebra here, except that it is more badly broken. Quite oddly it does some work. Well, this is true of humans since we came down out of the trees. We've gotten by and we continue to develop in a progression. Often this progression is prompted by past mistakes. Academia has such a massive and breathless accumulation now... could it be time to topple some of it?
Julio Di Egidio
2020-09-02 13:41:11 UTC
Post by Tim Golden BandTech.com
Post by Tim Golden BandTech.com
a1 X
does not resolve. Thus the operators cannot be binary operators and so
the ring definition is explicitly broken. We are in the same position
with a slight variation in interpretation, but the failing is the same.
The "ring definition" is simply irrelevant there. The coefficients come
from a ring, but X, the "indeterminate" in the formation of a polynomial,
*as well as the operations* involved in that structure formation, are
simply *formal terms* standing for the place of the coefficients in a
sequence: there is no evaluation involved.
As for evaluation, we actually first go from the polynomial to a
polynomial function where critical is that the domain of the function
is itself a ring and contains the ring of coefficients, then we do
evaluation of the function, where there is now no problem with taking
the operations involved as operations: on the function domain ring...
Hi Julio. I appreciate that you will uphold the standard interpretation
fairly well. So let's get on with it. I particularly am discussing
polynomials with real coefficients, so a(n) are real valued. This is
commonly done within the curriculum. So we start with the acclaimed ring
behaved
a0 + a1 X + a2 X X + ...
You keep equivocating on what "ring-behaved" refers to. It's
P[x] = a0+a1*X+... (the whole expression) that is the member of a
ring, a ring of polynomials. And it's the a_i themselves that
are members of a ring, of scalars, e.g. the real numbers. But X
is NOT a member of a ring, and the operations *in the expression*
of P[X] are NOT ring operations, they are just *formal symbols*
that stand for the places of the coefficients in *a sequence*.

And what is a sequence if not a vector? As Wikipedia puts it:
<< The set of polynomials in X_1, ..., X_n, denoted K[X_1, ...,
X_n], is thus *a vector space* (or a free module, if K is a ring)
that has the monomials as a basis. >>

So, as for the ring operations *of polynomials*: << Addition and
scalar multiplication of polynomials are those of a vector space
or free module equipped by a specific basis (here the basis of the
monomials). [...] The multiplication is [...] The verification
of the axioms of an *associative algebra* is straightforward. >>

<https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(multivariate_case)>
Post by Tim Golden BandTech.com
This procedure then extends onto the a2 term and so forth. In effect
the entire polynomial is constructed of undefined operators.
Rather, "+" means different things in different contexts: it's you
who are not taking the definition for serious. And then you might
at best complain that that notation is ambiguous, but, arguably,
it's not even so: that notation is rather conducive to polynomial
functions.

And I am not even at all an expert in that field: (not only I do
apologise in advance for any imprecision in my above statements,)
there may very well be more stringent reasons that justify the
use of exactly that notation. Anyway, even should you dislike
the notation, it is a fact that the maths is just fine.

Indeed, I won't insist: HTH and good luck.

Julio
Peter
2020-09-04 14:03:29 UTC
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.

Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
Tim Golden BandTech.com
2020-09-04 14:43:25 UTC
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.
Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
Come now; you must see that instantiation is always an option. We should replace your word 'any' above with 'every' and then I might agree with your statement partially. It is as it you would deny that
(4.5)(0.1) = 0.45
is not an instance of a ring operation. These are all real values. The reals are acclaimed ring status. So it as if your philosophy states that one must never instantiate a concrete instance in abstract algebra. Now that would be some axiom and it would solidify the status of the subject.

I will falsify your words more carefully in the near future. I do appreciate that you are one of two now who has at least approached the black swan.
Good on you Peter. Yet within your refutation you are somewhat declaring that you can approach the black swan no further. Still, I believe I have enough to falsify you, and that is exactly as it should be on my own language if I am wrong. You are attempting this so I do see this as a potentially productive debate. Thank you for joining in. Best of all with careful language our statements here stand freely and if they are flawed the flaw can be amplified. I will try not to dodge as others do here. Please do provide amplification on my own errors as aggressively as possible. Of course simplicity is helpful, but then too variations matter I think. I've settled onto a concrete instance of
1.23 X
to study and I can easily back off of this but I do suspect this is the simplest of forms to study. There are two operands and one operator right?
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
Peter
2020-09-04 15:11:03 UTC
Post by Tim Golden BandTech.com
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.
Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
Come now; you must see that instantiation is always an option. We should replace your word 'any' above with 'every' and then I might agree with your statement partially. It is as it you would deny that
(4.5)(0.1) = 0.45
is not an instance of a ring operation. These are all real values. The reals are acclaimed ring status. So it as if your philosophy states that one must never instantiate a concrete instance in abstract algebra. Now that would be some axiom and it would solidify the status of the subject.
I will falsify your words more carefully in the near future. I do appreciate that you are one of two now who has at least approached the black swan.
Good on you Peter. Yet within your refutation you are somewhat declaring that you can approach the black swan no further. Still, I believe I have enough to falsify you, and that is exactly as it should be on my own language if I am wrong. You are attempting this so I do see this as a potentially productive debate. Thank you for joining in. Best of all with careful language our statements here stand freely and if they are flawed the flaw can be amplified. I will try not to dodge as others do here. Please do provide amplification on my own errors as aggressively as possible. Of course simplicity is helpful, but then too variations matter I think. I've settled onto a concrete instance of
1.23 X
to study and I can easily back off of this but I do suspect this is the simplest of forms to study. There are two operands and one operator right?
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
a0 +(that's the ring addition) a1 X = a0 + a1 X

is an example of two ring element a0 and a1 X adding to make one ring
element.

If you read a book on algebra you'll see a proof that the set of
polynomials with one indeterminate (here X) and coefficients from a ring
is itself a ring under addition and multiplication of polynomials.
Tim Golden BandTech.com
2020-09-05 15:45:38 UTC
Post by Peter
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.
Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
Come now; you must see that instantiation is always an option. We should replace your word 'any' above with 'every' and then I might agree with your statement partially. It is as it you would deny that
(4.5)(0.1) = 0.45
is not an instance of a ring operation. These are all real values. The reals are acclaimed ring status. So it as if your philosophy states that one must never instantiate a concrete instance in abstract algebra. Now that would be some axiom and it would solidify the status of the subject.
I will falsify your words more carefully in the near future. I do appreciate that you are one of two now who has at least approached the black swan.
Good on you Peter. Yet within your refutation you are somewhat declaring that you can approach the black swan no further. Still, I believe I have enough to falsify you, and that is exactly as it should be on my own language if I am wrong. You are attempting this so I do see this as a potentially productive debate. Thank you for joining in. Best of all with careful language our statements here stand freely and if they are flawed the flaw can be amplified. I will try not to dodge as others do here. Please do provide amplification on my own errors as aggressively as possible. Of course simplicity is helpful, but then too variations matter I think. I've settled onto a concrete instance of
1.23 X
to study and I can easily back off of this but I do suspect this is the simplest of forms to study. There are two operands and one operator right?
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
a0 +(that's the ring addition) a1 X = a0 + a1 X
is an example of two ring element a0 and a1 X adding to make one ring
element.
Then how come you now have three elements with two operators?
This goes in contradiction to the ring definition. It states that two elements will resolve to one element.
I do appreciate what you are saying and I do see that you are obeying the standard interpretation.
Can we please specify real coefficients. Will this be alright with you?
In other words in the expression above a0 and a1 are both real valued.
The notion of an element as a singular thing seems such a fundamental concept and yet possibly this ray of argumentation that you are on here will lead us down into set theory fundamentals. I do suspect that if elements are composed of multiple elements then they are not elementary. In other words your argument above is false. And so in covering this ground you in effect are falsifying abstract algebra under this interpretation.

Of course it starts to feel a bit silly right? We are being such math weenies here. Yet we have whittled down now to a fairly simple and by the way alternative position on elements in set theory. I do still believe that one of the real elements can be
1.23
and that the ring behaviors can still be applied. I guess though that rather than confuse our responses I'll wait for your response here and maybe we can pick up this trail of concrete instantiation as well here off of this subthread. Here is that statement of yours transcribed here:
Your phrase "ring behaved polynomial" is odd.
It suggests to me that you think if R[X] is a ring
then any element of R[X] is a ring, which is potty.
The real numbers constitute a ring, 4.5 doesn't.

I have to admit that I just did a google search on '"ring behaved" polynomial' and get only 54 results so I accept your criticism though I don't see any real abuse of language here. Logically I am finding that the supposed ring behaved polynomial is not ring behaved. I suppose some would simply say 'is a ring' or 'is not a ring' and you are on the side of 'is a ring' and I am on the side of 'is not a ring' But isn't your dialect above really explaining why I use this terminology? The fact that for real values
( 0.01) ( 1.23 ) = 0.00123
is an expression that is consistent with the ring definition and so is ring behaved though clearly these are elements of the set and not the set of reals itself, which is the ring. Oddly enough we are right back to this sticking point of elemental quality within a set. Had I stated that
( 0.01) ( 1.23 ) = #0.00123
where the # indicates a value that is outside of the set of real numbers, then we would say that this expression violates the ring requirements so long as we still kept the declaration of real valued antecedents. I would say then that this is not a ring behaved expression. I do not care to engage is more highfaluting language which no doubt exists for such concepts but lacks instantiable types. We do know the real value well and we do know that the curriculum of abstract algebra does make use of the polynomial with real coefficients. I do point out here the conflict of this construction and particularly claims that these are ring behaved.

Peter you are the strongest respondent to this thread. I appreciate that you are taking this on and that you are taking me seriously. In effect you expose better than I the weakness of the other responses here. The informational concepts ought to be free-standing regardless of these human interactions that involve personal slights and geese crapping on people's noses and so forth. Please feel free to throw in a few yourself since you are on usenet here. Poor Mostowski still can't smell it and it's been days. I don't think he sees it either. Just let it dry and scrape it off Mostowski. Any day now.
Post by Peter
If you read a book on algebra you'll see a proof that the set of
polynomials with one indeterminate (here X) and coefficients from a ring
is itself a ring under addition and multiplication of polynomials.
Peter
2020-09-06 16:40:55 UTC
Post by Tim Golden BandTech.com
Post by Peter
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.
Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
Come now; you must see that instantiation is always an option. We should replace your word 'any' above with 'every' and then I might agree with your statement partially. It is as it you would deny that
(4.5)(0.1) = 0.45
is not an instance of a ring operation. These are all real values. The reals are acclaimed ring status. So it as if your philosophy states that one must never instantiate a concrete instance in abstract algebra. Now that would be some axiom and it would solidify the status of the subject.
I will falsify your words more carefully in the near future. I do appreciate that you are one of two now who has at least approached the black swan.
Good on you Peter. Yet within your refutation you are somewhat declaring that you can approach the black swan no further. Still, I believe I have enough to falsify you, and that is exactly as it should be on my own language if I am wrong. You are attempting this so I do see this as a potentially productive debate. Thank you for joining in. Best of all with careful language our statements here stand freely and if they are flawed the flaw can be amplified. I will try not to dodge as others do here. Please do provide amplification on my own errors as aggressively as possible. Of course simplicity is helpful, but then too variations matter I think. I've settled onto a concrete instance of
1.23 X
to study and I can easily back off of this but I do suspect this is the simplest of forms to study. There are two operands and one operator right?
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
a0 +(that's the ring addition) a1 X = a0 + a1 X
is an example of two ring element a0 and a1 X adding to make one ring
element.
Then how come you now have three elements with two operators?
This goes in contradiction to the ring definition. It states that two elements will resolve to one element.
I do appreciate what you are saying and I do see that you are obeying the standard interpretation.
Can we please specify real coefficients. Will this be alright with you?
Yes. I would it be alright with you for you to stop using meaningless
phrases such as

the ring definition [...] states that two elements will
resolve to one element.
Post by Tim Golden BandTech.com
In other words in the expression above a0 and a1 are both real valued.
The notion of an element as a singular thing seems such a fundamental concept and yet possibly this ray of argumentation that you are on here will lead us down into set theory fundamentals. I do suspect that if elements are composed of multiple elements then they are not elementary. In other words your argument above is false. And so in covering this ground you in effect are falsifying abstract algebra under this interpretation.
Of course it starts to feel a bit silly right? We are being such math weenies here. Yet we have whittled down now to a fairly simple and by the way alternative position on elements in set theory. I do still believe that one of the real elements can be
1.23
Your phrase "ring behaved polynomial" is odd.
It suggests to me that you think if R[X] is a ring
then any element of R[X] is a ring, which is potty.
The real numbers constitute a ring, 4.5 doesn't.
I have to admit that I just did a google search on '"ring behaved" polynomial' and get only 54 results so I accept your criticism though I don't see any real abuse of language here. Logically I am finding that the supposed ring behaved polynomial is not ring behaved. I suppose some would simply say 'is a ring' or 'is not a ring' and you are on the side of 'is a ring' and I am on the side of 'is not a ring' But isn't your dialect above really explaining why I use this terminology? The fact that for real values
( 0.01) ( 1.23 ) = 0.00123
is an expression that is consistent with the ring definition and so is ring behaved
There you go again. How about sticking to the language which one finds
in algebra texts? Have you ever read one?
Post by Tim Golden BandTech.com
though clearly these are elements of the set and not the set of reals itself, which is the ring. Oddly enough we are right back to this sticking point of elemental quality within a set. Had I stated that
( 0.01) ( 1.23 ) = #0.00123
where the # indicates a value that is outside of the set of real numbers, then we would say that this expression violates the ring requirements so long as we still kept the declaration of real valued antecedents. I would say then that this is not a ring behaved expression. I do not care to engage is more highfaluting language which no doubt exists for such concepts but lacks instantiable types. We do know the real value well and we do know that the curriculum of abstract algebra does make use of the polynomial with real coefficients. I do point out here the conflict of this construction and particularly claims that these are ring behaved.
Peter you are the strongest respondent to this thread. I appreciate that you are taking this on and that you are taking me seriously.
What makes you think that?
Post by Tim Golden BandTech.com
In effect you expose better than I the weakness of the other responses here. The informational concepts ought to be free-standing regardless of these human interactions that involve personal slights and geese crapping on people's noses and so forth. Please feel free to throw in a few yourself since you are on usenet here. Poor Mostowski still can't smell it and it's been days. I don't think he sees it either. Just let it dry and scrape it off Mostowski. Any day now.
Post by Peter
If you read a book on algebra you'll see a proof that the set of
polynomials with one indeterminate (here X) and coefficients from a ring
is itself a ring under addition and multiplication of polynomials.
Tim Golden BandTech.com
2020-09-06 20:25:43 UTC
Post by Tim Golden BandTech.com
Post by Peter
[...] I particularly am discussing polynomials with real
coefficients, so a(n) are real valued. This is commonly done within
the curriculum. So we> a(n) are real valued. This is commonly done
a0 + a1 X + a2 X X + ...
Now we choose zero values for a(n) except we'll
keep a1 variable. Thus the ring behaved polynomial condenses to
a1 X
That is one polynomial in the ring R[X]. Since R[X] *is* a ring it will
also contain (a1)^2 XX.
Your phrase "ring behaved polynomial" is odd. It suggests to me that
you think if R[X] is a ring then any element of R[X] is a ring, which is
potty. The real numbers constitute a ring, 4.5 doesn't.
Come now; you must see that instantiation is always an option. We should replace your word 'any' above with 'every' and then I might agree with your statement partially. It is as it you would deny that
(4.5)(0.1) = 0.45
is not an instance of a ring operation. These are all real values. The reals are acclaimed ring status. So it as if your philosophy states that one must never instantiate a concrete instance in abstract algebra. Now that would be some axiom and it would solidify the status of the subject.
I will falsify your words more carefully in the near future. I do appreciate that you are one of two now who has at least approached the black swan.
Good on you Peter. Yet within your refutation you are somewhat declaring that you can approach the black swan no further. Still, I believe I have enough to falsify you, and that is exactly as it should be on my own language if I am wrong. You are attempting this so I do see this as a potentially productive debate. Thank you for joining in. Best of all with careful language our statements here stand freely and if they are flawed the flaw can be amplified. I will try not to dodge as others do here. Please do provide amplification on my own errors as aggressively as possible. Of course simplicity is helpful, but then too variations matter I think. I've settled onto a concrete instance of
1.23 X
to study and I can easily back off of this but I do suspect this is the simplest of forms to study. There are two operands and one operator right?
which is not ring behaved since the product is not a binary
operation. This is the conflict in its entirety, though the
extensions go to considering
a0 + a1 X
since this addition operation is not binary and so is undefined.
a0 +(that's the ring addition) a1 X = a0 + a1 X
is an example of two ring element a0 and a1 X adding to make one ring
element.
Then how come you now have three elements with two operators?
This goes in contradiction to the ring definition. It states that two elements will resolve to one element.
I do appreciate what you are saying and I do see that you are obeying the standard interpretation.
Can we please specify real coefficients. Will this be alright with you?
Yes. I would it be alright with you for you to stop using meaningless
phrases such as
the ring definition [...] states that two elements will
resolve to one element.
Post by Tim Golden BandTech.com
In other words in the expression above a0 and a1 are both real valued.
The notion of an element as a singular thing seems such a fundamental concept and yet possibly this ray of argumentation that you are on here will lead us down into set theory fundamentals. I do suspect that if elements are composed of multiple elements then they are not elementary. In other words your argument above is false. And so in covering this ground you in effect are falsifying abstract algebra under this interpretation.
Of course it starts to feel a bit silly right? We are being such math weenies here. Yet we have whittled down now to a fairly simple and by the way alternative position on elements in set theory. I do still believe that one of the real elements can be
1.23
Your phrase "ring behaved polynomial" is odd.
It suggests to me that you think if R[X] is a ring
then any element of R[X] is a ring, which is potty.
The real numbers constitute a ring, 4.5 doesn't.
I have to admit that I just did a google search on '"ring behaved" polynomial' and get only 54 results so I accept your criticism though I don't see any real abuse of language here. Logically I am finding that the supposed ring behaved polynomial is not ring behaved. I suppose some would simply say 'is a ring' or 'is not a ring' and you are on the side of 'is a ring' and I am on the side of 'is not a ring' But isn't your dialect above really explaining why I use this terminology? The fact that for real values
( 0.01) ( 1.23 ) = 0.00123
is an expression that is consistent with the ring definition and so is ring behaved
There you go again. How about sticking to the language which one finds
in algebra texts? Have you ever read one?
We are not in algebra per se. We are in the curriculum of abstract algebra. I have no idea how you could come to disagree with this language particularly when you find no actual fault with the language. I have linked to numerous online references here; not as numerous as Lalo though. It is almost as if you want to deny that the real numbers do fit the ring definition. We are simply multiplying two real values above. Why are you in denial of this simplistic instance? You need to understand Peter that abstract algebra is radically different from ordinary algebra that we learn in high school. Your request that I stay within the language of some abstract algebra text will inherently deny falsification won't it? They are in the business of propagating this topic. Possibly you are misunderstanding the nature of the polynomial within AA. It is not at all the polynomial that we learn in high school algebra where x takes real values. That polynomial in real x is fine. The AA version is not so fine. So again, when I write:
( 0.01) ( 1.23 ) = 0.00123
where all these values are real numbers we have a concrete instance of multiplication that satisfies the ring definition. These are elements in the reals that I've written above. Again I will source for you the definition of a binary operator, which is required by ring definitions:
https://en.wikipedia.org/wiki/Binary_operation
and particularly
https://en.wikipedia.org/wiki/Closure_(mathematics)
which isn't really strongly stressed on wikipedia, but is in many renditions incuded right in the ring requirements. Possibly of importance for you to understand what is going on:
"The closure axiom is already implied by the condition that +/• be a binary operation. Some authors therefore omit this axiom. "
- https://en.wikipedia.org/wiki/Ring_(mathematics)#ref_c
By definition the product
1.23 X
simply cannot satisfy the closure requirement. It is a one line falsification. Yes, I can expound and take a few alternate forms, but really this crux is about all that there is to my argument. This black swan is so simple. I show you the black swan and you run away and hide in a cage. I will not lock the door. But it is up to you to step out and face the black swan.

We have to allow for the long form sum of the polynomial
a + b + c + d + e + f + ...
and witness that while there are umpteen operators and terms present in this expression it is implied that they all satisfy the closure requirement; otherwise they would break the ring requirements. In order to dismantle the polynomial; something that the AA texts fail to do; we simply need to study a few of these terms. One of these terms really is a fine instance of the black swan that falsifies the abstract algebra polynomial with real coefficients. Once again up on your head with its ass ready to dump on you:
1.23 X

s x
where s is sign and x is magnitude. What results? Quite a bit, and it actually overlaps with abstract algebra. But you see because I dismantled the real value and found its substructure; its elemental nature; the real number itself is no longer fundamental for me. Indeed the complex numbers come right along as three-signed numbers with no additional rules required. The real numbers are no more fundamental than are the complex numbers. They exist side by side as members in a family of number systems
P1, P2, P3, P4, ...
These systems inherently demand their dimensional geometry without orthogonality and without any need of the Cartesian product. They stand freely and primitively without any need for additional language. Through polysign numbers I was put onto abstract algebra. The bizarre stage of the polynomial development and its awkward quotient and ideal is not at all something to be proud of. Some texts even go into defense mode prior to bringing these details up. It is very clear that through all of the contorted language the polynomial is an infinite dimensional construction and by forcing modulo behavior on it reduced dimension is gotten. In effect they celebrate building a two dimensional complex value out of an infinite dimensional series. Great. All the while sitting on top of the pristine ring definition... which they completely obliterated when they built the polynomial. They are all telling you lies. The entire field is a bunch of PhDs looking for financial compensation. The room to publish must expand exponentially right along with the human population. This is the academic system that has brought you abstract algebra and which will continue to uphold it as if it has great integrity. To what degree is the mathematics community authentically brainwashed and to what degree are the highest performing mimics inherently susceptible to brainwashing? This level of philosophy I actually do arrive at through my work. We are humans doing mathematics. We cannot write ourselves a free ticket out of humanity. The very separation of the subject matter out into physics, philosophy, and mathematics; the queen; may merely be a matter of making room for more PhDs. That procedure has gone on and on now. These divides are false. These subjects are one. After all of your escapism into mathematics when you step outside the door and witness the world in amazement and ponder the weakness of the human condition then you will have arrived at a truth that is far beneath the subject at hand. We are the greatest mimics on the planet, and as a result we struggle very hard to find the truth.
Post by Tim Golden BandTech.com
though clearly these are elements of the set and not the set of reals itself, which is the ring. Oddly enough we are right back to this sticking point of elemental quality within a set. Had I stated that
( 0.01) ( 1.23 ) = #0.00123
where the # indicates a value that is outside of the set of real numbers, then we would say that this expression violates the ring requirements so long as we still kept the declaration of real valued antecedents. I would say then that this is not a ring behaved expression. I do not care to engage is more highfaluting language which no doubt exists for such concepts but lacks instantiable types. We do know the real value well and we do know that the curriculum of abstract algebra does make use of the polynomial with real coefficients. I do point out here the conflict of this construction and particularly claims that these are ring behaved.
Peter you are the strongest respondent to this thread. I appreciate that you are taking this on and that you are taking me seriously.
What makes you think that?
Post by Tim Golden BandTech.com
In effect you expose better than I the weakness of the other responses here. The informational concepts ought to be free-standing regardless of these human interactions that involve personal slights and geese crapping on people's noses and so forth. Please feel free to throw in a few yourself since you are on usenet here. Poor Mostowski still can't smell it and it's been days. I don't think he sees it either. Just let it dry and scrape it off Mostowski. Any day now.
Post by Peter
If you read a book on algebra you'll see a proof that the set of
polynomials with one indeterminate (here X) and coefficients from a ring
is itself a ring under addition and multiplication of polynomials.
Lalo T.
2020-09-06 21:01:29 UTC
In the context of your initial thesis, it is not possible ( that you, with compiler integrity quality standard,
pointing to facts, and we also pointing to facts) the existence of opposition. If it is possible, then the playground is not accurate.

What I mean by "Language Issues" is, for example, in :
https://en.wikipedia.org/wiki/Ring_(mathematics)#Definition
in the phrase "A ring is a set R" the symbol 'R' is not for real numbers (just an example)

I take for granted that you already distinguish operations of the rings from how its elements are
constituted. Either way, that is not the core issue.

I can not use the three-word forbidden concept starting with "Exte... ".
Otherwise I would get caught in your "clopenness" bait and I would disrespect

I'm requesting your examination of three points :

(1) The polynomial ring example (non-evaluated, not mixing)

(2) The matricial example (evaluated, mixing and not mixing)

(3) The context where you originally pick your model example (???)

Let's focus more on (3)

" How is it that I am the only one here who does go to the trouble of this sort of language? "

I guess that you pick your model example from somewhere. This does not appear out of the blue.
Tim, you mentioned that through Polysigns you saw this issue. Bring to the table the EXACT
POINT that inspire your questioning, an the discussion will move towards there...
Also, you must point out, why do I can, or can not, mix reals in your set, and details related.

An arbitrary question would be : Is the case that while examining the relationship between
RxC and P4, you were requesting your R-BO-C, without mentioning the forbidden concept ?

https://en.wikipedia.org/wiki/Module_(mathematics)
note " A module over a ring is a generalization of the notion of VECTOR SPACE over a field "
https://en.wikipedia.org/wiki/Vector_space#History
https://en.wikipedia.org/wiki/Linear_map

"As for a resolution as you are working towards above; that is good, but as to the initial falsification"
Presumably those two points are different. Maybe the other topic if just off-topic. Maybe yes, maybe no...
"Can we dispense with scalar multiplication when it is possible ?" is presumably off-topic.

Well, frisky reptilians or not, I have not stated that the refutation of your falsification have been performed.
Current options :

(a) thesis is true

(b) thesis is false

(c) non-falsifiable in a boring way

(d) non-falsifiable in an interesting way

I would not further define (c) and (d), or I 'll get into trouble.

"In the philosophy of science, falsifiability or refutability is the capacity for a statement, theory or
hypothesis to be contradicted by evidence. For example, the statement "All swans are white" is falsifiable
because one can observe that black swans exist." ( from : https://en.wikipedia.org/wiki/Falsifiability )

It would be not accurate on my side to say "I don't agree with you Tim".
I would like to say that "I do not agree with you and you are not accurate".
But first, to disagree (or agree) I have check the place where the complaint is originated.
Maybe the place has not issues and all is fine, ok, but the thing is the place is under examination.
So far, suspicion on (d). The only certainty that I have, is that something is going on...

These "two" oddities (although of different nature) pretty near of each other seem to indicate
the presence of a wandering Strange Loop in the surrounding area. Before pressing more on the
last point, state your new conditions, or I will give up and I ll be absorved by the "proof by vacuum".
...and wing-beaten
Peter
2020-09-01 11:24:10 UTC
Post by Tim Golden BandTech.com
After the careful construction of the ring from two groups,
You seem not to know what a ring is.
Tim Golden BandTech.com
2020-09-01 11:51:52 UTC
Post by Peter
Post by Tim Golden BandTech.com
After the careful construction of the ring from two groups,
You seem not to know what a ring is.
U R A lamb so long as you post no content.
Awaiting Falsification.
zelos...@gmail.com
2020-09-01 11:27:32 UTC
Post by Tim Golden BandTech.com
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
It distinctly disobeys the closure requirement
You might want to learn the more formal definition because the whole shebang about X is an artifact of history more than anything formal in abstract algebra. let (G,0, + ) be a commutative monoid, adn R a ring, then we define the polynomials to be R[G] to be the set {x e R^G: n e G & x(n)=0 for almost all.}, define addition as (x+y)(n)=x(n)+y(n) and multiplication as (x*y)(n)=sum_{i=0}^n x(i)*y(m), where i+m=n.

So it works perfectly fine, it is closed and satisfies the ring axioms.

the whole thing about X and all is just an artifact of history, and in our usual case, G is the additive monoid of natural numbers.
Tim Golden BandTech.com
2020-09-01 12:11:23 UTC
Post by ***@gmail.com
Post by Tim Golden BandTech.com
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
It distinctly disobeys the closure requirement
You might want to learn the more formal definition because the whole shebang about X is an artifact of history more than anything formal in abstract algebra. let (G,0, + ) be a commutative monoid, adn R a ring, then we define the polynomials to be R[G] to be the set {x e R^G: n e G & x(n)=0 for almost all.}, define addition as (x+y)(n)=x(n)+y(n) and multiplication as (x*y)(n)=sum_{i=0}^n x(i)*y(m), where i+m=n.
So it works perfectly fine, it is closed and satisfies the ring axioms.
the whole thing about X and all is just an artifact of history, and in our usual case, G is the additive monoid of natural numbers.
Sorry, no, the ring definition does not include an additional argument (n). The ring operators are quite simple and well established. If effect here you are also actually admitting that my falsification does hold, since you have not falsified it. You have failed to address closure and the notion of a binary operator. It looks as if you have established a trinary operator, but you've waived your hands and raised the complexity in order to avoid the conflict. How about we look within your definition at
a1 X
and see what sort of operation this is. Shall we say that n = 1 here? What sort of operator is this? Then let's have a look at
a0 + a1 X
You see the problem remains. Are you saying that abstract algebra is historically conflicted? That the works that do not use your notation are bad?
In effect then I suppose we might be in agreement. I'm not really so sure, but I would like to understand the distinction that you make better. It sounds a bit like when complex numbers stopped using sqrt(-1) to develop i and just went ahead and defined a two dimensional product and sum. It so happens that my own fix to these things develops the complex numbers with the same rules that develop the real number and so all of these constructions are needless as the system works in general dimension. The modulo behaved systems already have their kernel within the sign of the real value. It just needs generalization. But that is polysign and this is a falsification of abstract algebra. You've pivoted, but in your pivot you are admitting the weakness that most deny. I wish you could expound and bring it back down to the simplest of instances that I've put under your nose here. I really doubt that more fancy language is going to explain away the problem. I do think that this is an open problem. All problems are ultimately. They have to be. And this failing is a fine instance of why this is the case.
zelos...@gmail.com
2020-09-02 05:29:13 UTC
Post by Tim Golden BandTech.com
Sorry, no, the ring definition does not include an additional argument (n)
I didn't say that rings do, I said for that specific one :)
Post by Tim Golden BandTech.com
The ring operators are quite simple and well established
It all depends on the ring and "simplicity" again depends on teh ring, we use a different definition for polynomial rings. There is no "the operations" for rings. Only that there is AN operation we call addition and AN operation that we cann multiplication.
Post by Tim Golden BandTech.com
f effect here you are also actually admitting that my falsification does hold, since you have not falsified it
I point out that your complain is based on a misunderstanding because in formal definition, the whole X is nothing but notation.
Post by Tim Golden BandTech.com
You have failed to address closure and the notion of a binary operator.
I actually did adress it, I gave you the two binary operators for polynomial, defined them and you can see they are closed.
Post by Tim Golden BandTech.com
It looks as if you have established a trinary operator, but you've waived your hands and raised the complexity in order to avoid the conflict
I haven't made a ternary one, you just don't understand how it is formalyl done.
Post by Tim Golden BandTech.com
and see what sort of operation this is.
Thats no operation, it is notation. A polynomial is a function of |R^|N such that for almost all n e |N, p(n)=0.
Post by Tim Golden BandTech.com
You see the problem remains.
As I pointed out, thats not the formal definition, I gave you the fucking formal definition. Why did you not read it?
Post by Tim Golden BandTech.com
Are you saying that abstract algebra is historically conflicted? That the works that do not use your notation are bad?
What I am saying is that the formal definition of polynomials do not deal with X, unknowns or anything of the like. the usage of X is an artifact of history where it started.
Post by Tim Golden BandTech.com
It sounds a bit like when complex numbers stopped using sqrt(-1) to develop i and just went ahead and defined a two dimensional product and sum.
Complex numbers didn't have a proper formal construction initially, but they eventually did get it.
Post by Tim Golden BandTech.com
You've pivoted, but in your pivot you are admitting the weakness that most deny.
There is no weakness here, there is only you not understanding the differens between notation based on history vs formal abstract algebra definition/construction.
Post by Tim Golden BandTech.com
I really doubt that more fancy language is going to explain away the problem.
I use precise correct language, nothing "fancy". However you calling it fancy does indicate you do not know what it is saying.
Post by Tim Golden BandTech.com
I do think that this is an open problem. All problems are ultimately. They have to be. And this failing is a fine instance of why this is the case.
You might think it but there isn't.

"a_i X, what is the operation?" is not a problem question because it is just NOTATION, there is no operation there. Nothing being done. It is just a different way of writing (0,a_i,0,0,0,...)
which is the formal construction.
Tim Golden BandTech.com
2020-09-02 11:13:28 UTC
Post by ***@gmail.com
Post by Tim Golden BandTech.com
Sorry, no, the ring definition does not include an additional argument (n)
I didn't say that rings do, I said for that specific one :)
Post by Tim Golden BandTech.com
The ring operators are quite simple and well established
It all depends on the ring and "simplicity" again depends on teh ring, we use a different definition for polynomial rings. There is no "the operations" for rings. Only that there is AN operation we call addition and AN operation that we cann multiplication.
Sorry, but 'the operations' are very carefully defined within the ring definition. They are definitely binary operators, and when we have algebraically behaved systems they typically fit it quite well. These operators are explicitly defined to take two elements from a set and yield an element in that same set.

"In mathematics, a ring is one of the fundamental algebraic structures used in abstract algebra. It consists of a set equipped with two binary operations that generalize the arithmetic operations of addition and multiplication. " - https://en.wikipedia.org/wiki/Ring_(mathematics)

I have no idea why you would wheedle here. Indeed without your ordered notation you are essentially admitting that my claim is accurate. Furthermore the ordered notation has not replaced the traditional notation. Possibly it is in transition, but by no means am I discussing antiquated notation. That you push for this interpretation is indicative of the accuracy of my falsification, which continues to go unaddressed within your own interpretation. Therefor I have to place your own language, no matter how authentically you have written it, into the category of a dodge rather than a direct attack onto my falsification. In taking this step you have essentially built support for my argument though it is indirect.

As we discuss things at this fundamental level of detail to claim that something is 'just notation' is not at all a fair defense. If our notation is conflicted then an attack on notation is of importance. This is after all mathematics that we are discussing right? I do agree generally with your interpretation of the meaning of X. In my own language I am happy to see it as a dimensional placeholder. To then implement it into the ring operators as
a0 + a1 X + a2 X X + a3 X X X + ...
ought to be as offensive to you as it is to me. In effect you are admitting this though it is as if you whisper it here, all the while saying 'NO' with plenty of lung.
Post by ***@gmail.com
Post by Tim Golden BandTech.com
f effect here you are also actually admitting that my falsification does hold, since you have not falsified it
I point out that your complain is based on a misunderstanding because in formal definition, the whole X is nothing but notation.
Post by Tim Golden BandTech.com
You have failed to address closure and the notion of a binary operator.
I actually did adress it, I gave you the two binary operators for polynomial, defined them and you can see they are closed.
Post by Tim Golden BandTech.com
It looks as if you have established a trinary operator, but you've waived your hands and raised the complexity in order to avoid the conflict
I haven't made a ternary one, you just don't understand how it is formalyl done.
Post by Tim Golden BandTech.com
and see what sort of operation this is.
Thats no operation, it is notation. A polynomial is a function of |R^|N such that for almost all n e |N, p(n)=0.
Post by Tim Golden BandTech.com
You see the problem remains.
As I pointed out, thats not the formal definition, I gave you the fucking formal definition. Why did you not read it?
Post by Tim Golden BandTech.com
Are you saying that abstract algebra is historically conflicted? That the works that do not use your notation are bad?
What I am saying is that the formal definition of polynomials do not deal with X, unknowns or anything of the like. the usage of X is an artifact of history where it started.
Post by Tim Golden BandTech.com
It sounds a bit like when complex numbers stopped using sqrt(-1) to develop i and just went ahead and defined a two dimensional product and sum.
Complex numbers didn't have a proper formal construction initially, but they eventually did get it.
Post by Tim Golden BandTech.com
You've pivoted, but in your pivot you are admitting the weakness that most deny.
There is no weakness here, there is only you not understanding the differens between notation based on history vs formal abstract algebra definition/construction.
Post by Tim Golden BandTech.com
I really doubt that more fancy language is going to explain away the problem.
I use precise correct language, nothing "fancy". However you calling it fancy does indicate you do not know what it is saying.
Post by Tim Golden BandTech.com
I do think that this is an open problem. All problems are ultimately. They have to be. And this failing is a fine instance of why this is the case.
You might think it but there isn't.
"a_i X, what is the operation?" is not a problem question because it is just NOTATION, there is no operation there. Nothing being done. It is just a different way of writing (0,a_i,0,0,0,...)
which is the formal construction.
zelos...@gmail.com
2020-09-02 13:40:38 UTC
Post by Tim Golden BandTech.com
Sorry, but 'the operations' are very carefully defined within the ring definition
No, they aren't. They are just stated to have the distirbutive relation and that addition is commutative.
Post by Tim Golden BandTech.com
They are definitely binary operators
Yes, but what operation it is does not matter as long as its binary and sates said qualities.
Post by Tim Golden BandTech.com
These operators are explicitly defined to take two elements from a set and yield an element in that same set.
Yes, but one ring has two operands and another has two different ones.
Post by Tim Golden BandTech.com
Furthermore the ordered notation has not replaced the traditional notation.
It is a sequence or infinite cartesian product which are the same essentially. And yes, it hasn't replaced it due to history, so fucking what?
Post by Tim Golden BandTech.com
Possibly it is in transition, but by no means am I discussing antiquated notation.
Post by Tim Golden BandTech.com
That you push for this interpretation is indicative of the accuracy of my falsification
No, it is demonstrating you are complaining about notation, not formal construction which makes your point moot.
Post by Tim Golden BandTech.com
Therefor I have to place your own language, no matter how authentically you have written it, into the category of a dodge rather than a direct attack onto my falsification. In taking this step you have essentially built support for my argument though it is indirect.
No, it is you being ignorant.

Post by Tim Golden BandTech.com
As we discuss things at this fundamental level of detail to claim that something is 'just notation' is not at all a fair defense.
When your complaint is about ntoation, saying it is just notation IS a method to show your complaint is invalid.
Post by Tim Golden BandTech.com
If our notation is conflicted then an attack on notation is of importance.
Notation is important but this is a historical artifact and it doesn't matter ultimately if you have some intelligence.
Post by Tim Golden BandTech.com
This is after all mathematics that we are discussing right? I do agree generally with your interpretation of the meaning of X. In my own language I am happy to see it as a dimensional placeholder. To then implement it into the ring operators as
a0 + a1 X + a2 X X + a3 X X X + ...
Post by Tim Golden BandTech.com
ought to be as offensive to you as it is to me. In effect you are admitting this though it is as if you whisper it here, all the while saying 'NO' with plenty of lung.
It is just notation, there is nothing offensive about it.
Ross A. Finlayson
2020-09-02 15:59:30 UTC
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.

I'm sure the modular arithmetic and abstract algebra is familiar to everybody.

that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.

that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.

In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.

"To formalize this argument

we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.

This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.

Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.

The entire polynomials fails under their own formalities. "

Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)

I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.

Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)

So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.

Just more, though - not different.

Thanks, it's OK.
Mostowski Collapse
2020-09-02 16:10:54 UTC
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.

Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.

I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-02 16:21:51 UTC
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it

must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.

Take disjunction and negation defined as:

x v y := x + y - x*y

~x := 1 - x

Then take this tautology:

x v ~x = x + (1 - x) - x*(1 - x)

= 1 - x + x^2

= 1 - x + x

= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Lalo T.
2020-09-02 18:19:58 UTC
Ok, manage the email thing. I 'am curious to how far the reasoning can be extended.

With a bit of luck this will not be a lumpy rug
http://rugs.droogkast.com/lumpy-rug/

You seem more centered in the Polynomial Ring concept and Ring of Polynomial Functions concept (in regarding to the Ring concept)

https://en.wikipedia.org/wiki/Scalar_multiplication#Interpretation
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
https://en.wikipedia.org/wiki/External_(mathematics)

https://commalg.subwiki.org/wiki/Polynomial_ring
https://en.wikipedia.org/wiki/Polynomial#Polynomial_functions
https://en.wikipedia.org/wiki/Ring_of_polynomial_functions

https://encyclopediaofmath.org/wiki/Unital_ring

I will try to do another approach (in the case of matrices)

1. a₀·1 + a₁·x + a₂·x² + ... + aₙ₋₁·xⁿ⁻¹ + aₙ·xⁿ = 0

2. A₀·1 + A₁·x + A₂·x² + ... + Aₙ₋₁·xⁿ⁻¹ + Aₙ·xⁿ = O

3. a₀·I + a₁·X + a₂·X² + ... + aₙ₋₁·Xⁿ⁻¹ + aₙ·Xⁿ = O

4. A₀*I + A₁*X + A₂*X² + ... + Aₙ₋₁*Xⁿ⁻¹ + Aₙ*Xⁿ = O

https://en.wikipedia.org/wiki/Scalar_multiplication#Scalar_multiplication_of_matrices
https://en.wikipedia.org/wiki/Diagonal_matrix#Scalar_matrix
https://en.wikipedia.org/wiki/Center_(ring_theory)

Do you believe, looking points (a) and (b), that the scalar multiplication of matrices is non-primitive as operation, and the scalar multiplication of matrices is, indeed, equivalent to use the usual matrix multiplication with "scalar matrices" ? (as if the scalar product were a non-essential operation (since you consider it as an alien operation) )

(a) λ₀·I + λ₁·X + λ₂·X² + ... + λₙ₋₁·Xⁿ⁻¹ + λₙ·Xⁿ = O where '·' is scalar product

equivalent to :

(b) A₀*I + A₁*X + A₂*X² + ... + Aₙ₋₁*Xⁿ⁻¹ + Aₙ*Xⁿ = O where '*' is matrix product

???

Note that instead of use the scalar product ' λ·I = A '
to define our "Scalar Matrices" A (the coefficients in (b) ), we directly define the elements of the Scalar Matrices A in (b)

A = [aₕₖ]ₘₓₘ :
a matrix A of dimensions mxm (m times m), subscript h indicates the row and subscript k indicates the column

where an element aₕₖ = λ if h = k , and aₕₖ = 0 if h =/= k

Would it be fine if we just kick out the scalar product in the case of matrices, and only use matrix multiplication ?

(...And in this respect, the same question for any structure that use scalar multiplication? )
Lalo T.
2020-09-02 21:30:44 UTC
https://hsm.stackexchange.com/questions/11235/who-started-calling-the-matrix-multiplication-multiplication
The vector algebra war: a historical perspective https://arxiv.org/abs/1509.00501v2

Tim, if you requesting a revision of the concept of scalar, with respect to his interaction/relationship with other algebraic structures...

https://en.wikipedia.org/wiki/Scalar_(mathematics)#Etymology
https://en.wikipedia.org/wiki/Scalar_multiplication
https://en.wikipedia.org/wiki/Scaling_(geometry)

...we will end up re-examining and inspecting stuff from mid-19th century or before.

I suppose you think in the context of an abstract X, and the interaction a2 with X² in 'a2 X X'
You indeed will not consider too problematic if we say ' a2*X*X ', but you will consider non well-formulated the statement ' a2·(X*X) '
If somehow, is always the case, or in a number of cases (since the '·' operation seems to have an independent status) that for some
unknown reason, the '·' operation can be reduced to merely the '*' operation. Would the conflict disappear under your quality standard of compiler integrity ?

cheers
Mostowski Collapse
2020-09-03 10:29:20 UTC
Boole also introduced, it can be justified:

∀P^A == A[P/0]*A[P/1]

Its a bold step since Boolean algebras can have more than
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-03 10:31:35 UTC
It also shows that the fundamental theorem
does not hold in Boolean algebras in general.

For a binary Boolean algebra this here has
two roots x=0 and x=1:

x*(1-x) = 0

But in a Boolean algebra with more than two
values, it has much more roots.
Post by Mostowski Collapse
∀P^A == A[P/0]*A[P/1]
Its a bold step since Boolean algebras can have more than
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Ross A. Finlayson
2020-09-04 15:56:34 UTC
Post by Mostowski Collapse
It also shows that the fundamental theorem
does not hold in Boolean algebras in general.
For a binary Boolean algebra this here has
x*(1-x) = 0
But in a Boolean algebra with more than two
values, it has much more roots.
Post by Mostowski Collapse
∀P^A == A[P/0]*A[P/1]
Its a bold step since Boolean algebras can have more than
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
"It has much more roots."
Lalo T.
2020-09-04 23:13:13 UTC
"...polynomial evaluated at .."

Nice to see that the issue has some leverage. Ok Tim, let me add to the pile...

I wonder if this debate can be effectively solved in the realm of Abstract Algebra when Vector Spaces are already present.
My bet is that this conflict can not be solved (or dissolved) in this context (solutions mainly using abstract algebra).
Maybe resorting to other foundational areas. I'm open to be corrected, in case this issue can effectively be solved inside
Abstract Algebra, in which case, I will retract my words.

I bet that our apparent black swan, it is being victim of an ultra-deformulation mechanism, originated by the bullets of an exotic handgun
being used by a frisky reptilian, with the unwanted effect of a misbehavior on the side of the photons lying in the vicinity of our swan.
Of course, this situation being on top of the misbehaviour of our already bad-tempered swan. I think that will be enough some strikes with the mace,
to our frisky friend, not too rough (exercise some tough love) to indicate that we will not tolerate pranks at those levels. Not too hard also,
because animals have rights, and, undercover agents of pro-animal organizations are attentive to this kind of situations, ready to jump...

https://www.flickr.com/photos/elsie/3181428441/
https://www.dailymail.co.uk/news/article-2154839/Black-swan-stands-gatecrashing-group-600-white-ones-ancient-swannery.html

Events with swans have been reported: https://www.youtube.com/watch?v=2OJA2Tw5bzg

Since my intention now is to state that it is not the case that :

(1) "the falsification of abstract algebra" is true

(2) "this conflict is solvable in the usual abstract algebra" is true

(3) "a0 I + a1 X + a2 X X + .. is ill-formed " is true

( where it is not my intention : https://en.wikipedia.org/wiki/Minority_Report_(film) )

...then my own position here would be stepping out of Post-Scalar-Multiplication Abstract Algebra, for the sake of preserving
the structural coherence of the building, or alternatively, look for some Scalar-Free building, in order to make the refutability, or even the formulation of your statement impossible,
although, at an ridiculously expensive price, of course, since I state that the conflict is pointing to scalars and vector spaces.

If I am quoting well, Roger B. refer to "indexed vs ordered".

For the sake of diversity on this dispute over the ring, here is missing the words of wisdom of a true veteran in rings, namely, "The Lord of the Rings", to share his grasp of the issue,
and of course, give his/her two cents on it.

Consider write, in the future, an epic novel about swans, rings and frisky reptilians, and allocate part of the profits to pro-animal organizations (in case you wish).

If we were is a maze, in a competition where only one can get out, and weapons are provided, before proceed to the combat, better check first if the walls can be broken, with the possibililty
of finding a tunnel.

While restricting to the three concepts that you use in your initial post, to some extent I give you the benefit of doubt, but not, at the "Abstract Algebra Falsification" part...
Tim Golden BandTech.com
2020-09-05 14:59:40 UTC
Post by Lalo T.
"...polynomial evaluated at .."
Nice to see that the issue has some leverage. Ok Tim, let me add to the pile...
I wonder if this debate can be effectively solved in the realm of Abstract Algebra when Vector Spaces are already present.
My bet is that this conflict can not be solved (or dissolved) in this context (solutions mainly using abstract algebra).
Maybe resorting to other foundational areas. I'm open to be corrected, in case this issue can effectively be solved inside
Abstract Algebra, in which case, I will retract my words.
I bet that our apparent black swan, it is being victim of an ultra-deformulation mechanism, originated by the bullets of an exotic handgun
being used by a frisky reptilian, with the unwanted effect of a misbehavior on the side of the photons lying in the vicinity of our swan.
Of course, this situation being on top of the misbehaviour of our already bad-tempered swan. I think that will be enough some strikes with the mace,
to our frisky friend, not too rough (exercise some tough love) to indicate that we will not tolerate pranks at those levels. Not too hard also,
because animals have rights, and, undercover agents of pro-animal organizations are attentive to this kind of situations, ready to jump...
https://www.wikihow.com/images/9/9a/Color-Step-7-32.jpg
http://pixeljoint.com/files/icons/full/bicho_lanza.gif
https://www.flickr.com/photos/elsie/3181428441/
https://www.dailymail.co.uk/news/article-2154839/Black-swan-stands-gatecrashing-group-600-white-ones-ancient-swannery.html
Events with swans have been reported: https://www.youtube.com/watch?v=2OJA2Tw5bzg
(1) "the falsification of abstract algebra" is true
(2) "this conflict is solvable in the usual abstract algebra" is true
(3) "a0 I + a1 X + a2 X X + .. is ill-formed " is true
( where it is not my intention : https://en.wikipedia.org/wiki/Minority_Report_(film) )
...then my own position here would be stepping out of Post-Scalar-Multiplication Abstract Algebra, for the sake of preserving
the structural coherence of the building, or alternatively, look for some Scalar-Free building, in order to make the refutability, or even the formulation of your statement impossible,
although, at an ridiculously expensive price, of course, since I state that the conflict is pointing to scalars and vector spaces.
If I am quoting well, Roger B. refer to "indexed vs ordered".
For the sake of diversity on this dispute over the ring, here is missing the words of wisdom of a true veteran in rings, namely, "The Lord of the Rings", to share his grasp of the issue,
and of course, give his/her two cents on it.
Consider write, in the future, an epic novel about swans, rings and frisky reptilians, and allocate part of the profits to pro-animal organizations (in case you wish).
If we were is a maze, in a competition where only one can get out, and weapons are provided, before proceed to the combat, better check first if the walls can be broken, with the possibililty
of finding a tunnel.
While restricting to the three concepts that you use in your initial post, to some extent I give you the benefit of doubt, but not, at the "Abstract Algebra Falsification" part...
Nice piece Lalo. I see some here are like
If I could just find one for Mostowski...
I can't stand all the reptilean theories going around, but it is well proven that we mammals do still have reptillian brain components. Supposedly the fight or flee instinct is embedded there. The amygdala reaction...

It does seem to beg the question of what if Tim is accurate here... what are the consequences? This math was an attempt at a pristine formulation. Clearly it is not pristine. That much is already admitted by others here though they can only whisper it under their breath. Minor abuse; historical notation; these are not my own terms. My terms are considerably stronger. Falsification as a practice seems quite foreign to these users of sci.math. Can it be done so quickly and so shortly as uttering " the polynomial with real coefficients 1.23 X cannot be ring behaved"

Obviously coverage of this claim will have to consider what the elements of the operator are. The ring requires elements in the same set and a singular result which belongs in that same set. If this operator does not fit this requirement then it obviously has offended the ring definition. How is it that I am the only one here who does go to the trouble of this sort of language? The rest are lost up in the high lands of the infinite polynomial. Is it really wise what they are doing? When their infinite construction is composed of a sum that is obeying (supposedly) the ring definition and whose products (supposedly) obey the ring definition? No. We go to the elemental stage and study these things. A conflict is found and the dodge begins all over again. As for a resolution as you are working towards above; that is good, but as to the initial falsification: this part is so important that it has to be established on its own terms and in the simplest language possible. Of course then things do fall apart and we can expand out the attack, but the initial attack has to be down in there in the fundamental and kept quite simple I think. As far as I can tell we can firstly falsify the multiplicative operator with
1.23 X
and then we can go on to falsify the additive operator with
0.01 + 1.23 X
and from there we pretty well have collapsed the usage of the polynomial with real coefficients as a farce which has been pushed onto numerous disbelieving students who for the sake of an A must mimic on and on and on no matter how bad the basis is; at their own cost to boot. Even just the text book fee alone makes it a bad deal. But then pay the guy who shoves this crap down their throats... well, you might say; that sounds a bit harsh. But if I am correct than all of this rhetoric holds up just fine and the guy who came before that guy who shoved the white goose shit down his throat and so on. Ahhh... back to a time when possibly this subject was treated as an open problem... should we remain there at treating these sorts of systems as an open problem then I would say that the maturity of the subject at hand could be established. Instead it is treated as a closed book. One must preserve the book...
Tim Golden BandTech.com
2020-09-03 13:22:33 UTC
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.

Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.

I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.

I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.

Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Julio Di Egidio
2020-09-03 14:48:00 UTC
Post by Tim Golden BandTech.com
I am amazed at the amount of dodge taking place on this thread.
Can anyone take this on directly? Apparently not, since if they
could have they would have. When somebody provides a falsification
you have to return a refutation. In other words I should have made
a mistake in my work here. No one has stepped up and pointed out an
error in my work. Instead they just go on back to the curriculum as
if I posted a message on needing some help understanding the subject.
years now, so let me be more blatant: you are the typical self-deluded
crank in permanent denial, and that's all there is to it. -- That said,
whether you need help or not I'll rather let you decide for yourself.

*Plonk*

Julio
Mostowski Collapse
2020-09-03 14:51:19 UTC
This here is not a polynomial:

a(0) + a(1) X a(2) X X + a(3) X X X + ...

Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.

https://en.wikipedia.org/wiki/Degree_of_a_polynomial

What you indicated was possibly the ring of
see here:

https://en.wikipedia.org/wiki/Formal_power_series

That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are

easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.

Mono-mials are closed over multiplication:

aX^n * bX^m = (a*b)X^(n+m)

The degree behaves as follows:

deg(aX^n) = n

deg(aX^m) = m

deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m

Requiring the monomials over a fixed degree k
are closed, would require that:

deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k

Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].

So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,

when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-03 15:03:02 UTC
If you have more constraints, more polynomials
or monomials get closed concerning multiplication.
This is easy to see, if you have a constraint:

p(X) = a0 + a1 X + ... + an X^n = 0 (*)

Then you can already reduce every polynomial
with degree larger or equal than n. This is
easy to see, simply note that a constraint (*)

as above also implies, provided the coefficients
can divide:

-a0 - a1 X - ... - an-1 X^(n-1)
X^n = -------------------------------
an

a0 a1 an-1
= - --- - --- X - ... - ---- X^(n-1) (**)
an an an

So when ever you have a polynomial of degree
larger or equal than n you can reduce it by one
degree and so on, until it has a degree

below n. This polynomial rewriting possibly
doesn't work for formal power series. Not
sure. But you see a difference between

polynomials and formal power series.
Things get also more complicated in multi-
variante polynomials such as R[X][Y],

because the coefficients from R[X] do
not divide anymore.
Post by Tim Golden BandTech.com
a(0) + a(1) X a(2) X X + a(3) X X X + ...
Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.
https://en.wikipedia.org/wiki/Degree_of_a_polynomial
What you indicated was possibly the ring of
https://en.wikipedia.org/wiki/Formal_power_series
That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are
easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.
aX^n * bX^m = (a*b)X^(n+m)
deg(aX^n) = n
deg(aX^m) = m
deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m
Requiring the monomials over a fixed degree k
deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k
Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].
So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,
when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-03 15:08:32 UTC
In case you are interesed in multivariate
polynomials, you might like this book:

Using Algebraic Geometry
von David A. Cox , John Little , Donal O'Shea (Autor)
https://www.amazon.de/dp/0387207066
Post by Mostowski Collapse
If you have more constraints, more polynomials
or monomials get closed concerning multiplication.
p(X) = a0 + a1 X + ... + an X^n = 0 (*)
Then you can already reduce every polynomial
with degree larger or equal than n. This is
easy to see, simply note that a constraint (*)
as above also implies, provided the coefficients
-a0 - a1 X - ... - an-1 X^(n-1)
X^n = -------------------------------
an
a0 a1 an-1
= - --- - --- X - ... - ---- X^(n-1) (**)
an an an
So when ever you have a polynomial of degree
larger or equal than n you can reduce it by one
degree and so on, until it has a degree
below n. This polynomial rewriting possibly
doesn't work for formal power series. Not
sure. But you see a difference between
polynomials and formal power series.
Things get also more complicated in multi-
variante polynomials such as R[X][Y],
because the coefficients from R[X] do
not divide anymore.
Post by Tim Golden BandTech.com
a(0) + a(1) X a(2) X X + a(3) X X X + ...
Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.
https://en.wikipedia.org/wiki/Degree_of_a_polynomial
What you indicated was possibly the ring of
https://en.wikipedia.org/wiki/Formal_power_series
That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are
easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.
aX^n * bX^m = (a*b)X^(n+m)
deg(aX^n) = n
deg(aX^m) = m
deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m
Requiring the monomials over a fixed degree k
deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k
Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].
So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,
when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-03 15:21:47 UTC
Here is an example of a formal power series
that you cannot reduce for example versus
X^2-X=0. Its simply 1/(1-X) formally:

1
----- = 1 + X + X^2 + X^3 + ...
1 - X

If you reduce with X^2-X=0 respectively
X^2=X, you get:

= 1 + X + X + X + ...

= 1 + (1 + 1 + 1 + ...) X

So you get a coefficient that tends to
infinity, but such coefficients are not
found in R.
Post by Mostowski Collapse
In case you are interesed in multivariate
Using Algebraic Geometry
von David A. Cox , John Little , Donal O'Shea (Autor)
https://www.amazon.de/dp/0387207066
Post by Mostowski Collapse
If you have more constraints, more polynomials
or monomials get closed concerning multiplication.
p(X) = a0 + a1 X + ... + an X^n = 0 (*)
Then you can already reduce every polynomial
with degree larger or equal than n. This is
easy to see, simply note that a constraint (*)
as above also implies, provided the coefficients
-a0 - a1 X - ... - an-1 X^(n-1)
X^n = -------------------------------
an
a0 a1 an-1
= - --- - --- X - ... - ---- X^(n-1) (**)
an an an
So when ever you have a polynomial of degree
larger or equal than n you can reduce it by one
degree and so on, until it has a degree
below n. This polynomial rewriting possibly
doesn't work for formal power series. Not
sure. But you see a difference between
polynomials and formal power series.
Things get also more complicated in multi-
variante polynomials such as R[X][Y],
because the coefficients from R[X] do
not divide anymore.
Post by Tim Golden BandTech.com
a(0) + a(1) X a(2) X X + a(3) X X X + ...
Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.
https://en.wikipedia.org/wiki/Degree_of_a_polynomial
What you indicated was possibly the ring of
https://en.wikipedia.org/wiki/Formal_power_series
That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are
easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.
aX^n * bX^m = (a*b)X^(n+m)
deg(aX^n) = n
deg(aX^m) = m
deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m
Requiring the monomials over a fixed degree k
deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k
Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].
So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,
when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Mostowski Collapse
2020-09-03 15:25:58 UTC
You could also view it as the observation that
for polynomials p(x) the substitution p(1)

always exists, whereas for formal power series f(x)
the substitution f(1) need not exist.

LoL
Post by Mostowski Collapse
Here is an example of a formal power series
that you cannot reduce for example versus
1
----- = 1 + X + X^2 + X^3 + ...
1 - X
If you reduce with X^2-X=0 respectively
= 1 + X + X + X + ...
= 1 + (1 + 1 + 1 + ...) X
So you get a coefficient that tends to
infinity, but such coefficients are not
found in R.
Post by Mostowski Collapse
In case you are interesed in multivariate
Using Algebraic Geometry
von David A. Cox , John Little , Donal O'Shea (Autor)
https://www.amazon.de/dp/0387207066
Post by Mostowski Collapse
If you have more constraints, more polynomials
or monomials get closed concerning multiplication.
p(X) = a0 + a1 X + ... + an X^n = 0 (*)
Then you can already reduce every polynomial
with degree larger or equal than n. This is
easy to see, simply note that a constraint (*)
as above also implies, provided the coefficients
-a0 - a1 X - ... - an-1 X^(n-1)
X^n = -------------------------------
an
a0 a1 an-1
= - --- - --- X - ... - ---- X^(n-1) (**)
an an an
So when ever you have a polynomial of degree
larger or equal than n you can reduce it by one
degree and so on, until it has a degree
below n. This polynomial rewriting possibly
doesn't work for formal power series. Not
sure. But you see a difference between
polynomials and formal power series.
Things get also more complicated in multi-
variante polynomials such as R[X][Y],
because the coefficients from R[X] do
not divide anymore.
Post by Tim Golden BandTech.com
a(0) + a(1) X a(2) X X + a(3) X X X + ...
Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.
https://en.wikipedia.org/wiki/Degree_of_a_polynomial
What you indicated was possibly the ring of
https://en.wikipedia.org/wiki/Formal_power_series
That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are
easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.
aX^n * bX^m = (a*b)X^(n+m)
deg(aX^n) = n
deg(aX^m) = m
deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m
Requiring the monomials over a fixed degree k
deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k
Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].
So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,
when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
Tim Golden BandTech.com
2020-09-03 19:57:41 UTC
If you could dodge the black swan any harder Mostowski I'd have to call you out.
A simple question for you:
Is
1.23 X
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
The value 1.23 is a real number. Is this valid?

If I could simplify down any farther I would, but I'm afraid I've reached the bottom here. Next though of course I'd ask you if
0.12 + 1.23 X
is a valid polynomial expression within abstract algebra? Of course again these values are real except for the X which has no actual set definition at all. Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof.

None here seem capable of even discussing the act of falsification. Mimicry rules this day.

Two black swans for Mostowski. It seems you see no swans at all.
Post by Mostowski Collapse
You could also view it as the observation that
for polynomials p(x) the substitution p(1)
always exists, whereas for formal power series f(x)
the substitution f(1) need not exist.
LoL
Post by Mostowski Collapse
Here is an example of a formal power series
that you cannot reduce for example versus
1
----- = 1 + X + X^2 + X^3 + ...
1 - X
If you reduce with X^2-X=0 respectively
= 1 + X + X + X + ...
= 1 + (1 + 1 + 1 + ...) X
So you get a coefficient that tends to
infinity, but such coefficients are not
found in R.
Post by Mostowski Collapse
In case you are interesed in multivariate
Using Algebraic Geometry
von David A. Cox , John Little , Donal O'Shea (Autor)
https://www.amazon.de/dp/0387207066
Post by Mostowski Collapse
If you have more constraints, more polynomials
or monomials get closed concerning multiplication.
p(X) = a0 + a1 X + ... + an X^n = 0 (*)
Then you can already reduce every polynomial
with degree larger or equal than n. This is
easy to see, simply note that a constraint (*)
as above also implies, provided the coefficients
-a0 - a1 X - ... - an-1 X^(n-1)
X^n = -------------------------------
an
a0 a1 an-1
= - --- - --- X - ... - ---- X^(n-1) (**)
an an an
So when ever you have a polynomial of degree
larger or equal than n you can reduce it by one
degree and so on, until it has a degree
below n. This polynomial rewriting possibly
doesn't work for formal power series. Not
sure. But you see a difference between
polynomials and formal power series.
Things get also more complicated in multi-
variante polynomials such as R[X][Y],
because the coefficients from R[X] do
not divide anymore.
a(0) + a(1) X a(2) X X + a(3) X X X + ...
Polynomials from R[X] are finite. If p(X) is a polynomial,
then the function deg(p(X)) is defined and
it has a value from the natural numbers.
https://en.wikipedia.org/wiki/Degree_of_a_polynomial
What you indicated was possibly the ring of
https://en.wikipedia.org/wiki/Formal_power_series
That monomials in a fixed degree are not closed
for multiplication was already noted. The meaning
of the word mono-mial and poly-nomial are
easy to understand. A mono-mial has a single
non-zero coefficient. A poly-nomial can have
multiple non-zero coefficients.
aX^n * bX^m = (a*b)X^(n+m)
deg(aX^n) = n
deg(aX^m) = m
deg(aX^n * bX^m) = deg((a*b)X^(n+m)) = n+m
Requiring the monomials over a fixed degree k
deg(aX^k * bX^k) = deg((a*b)X^(2k)) = 2k = k
Which is only possible for k=0. With the monomials
of degree k=0, respectively the polynomials of
degree k=0 you can recover the domain R of R[X].
So R[X] embeds R through polynomials of
degree k=0. Thats what is usually closed over
multiplication when the variables are free,
when we do not have some constraint like X^2-X=0,
as in a Boolean polynomial.
Post by Tim Golden BandTech.com
Post by Mostowski Collapse
A Boolean polynomials with no roots, is a tautology.
If it has no roots, i.e. never becomes zero, it
must be always 1, hence it is a tautology. That it
also reduces algebraically to 1, is quite a feat.
x v y := x + y - x*y
~x := 1 - x
Well, I do perk up a little bit when you mention negation here, but since you define it using a minus its original form is arguably still present.
Still for boolean systems it squares with logic, yet that zero would be the inverse of one does not take much extensive meaning beyond that.
Doesn't this detail really expose the uniqueness of the boolean system? All that notation for such a simple form... It really doesn't seem right.
Still, I have weighed in on it, and now I'd like you to weigh in on this thread Mostowski Collapse.
Perhaps just one more variation should be played out on the polynomial of abstract algebra. The polynomial
a(0) + a(1) X a(2) X X + a(3) X X X + ...
is claimed to be ring behaved for all a(n). Normally mathematicians stay in this infinite form for quite a bit longer before they do any work. The sum and the product of multiple polynomials expose that they still fit this polynomial form (though the product and its indexing is fishy). At this point they feel that they have established the ring behavior of the expression. At some later point they will introduce the polynomial ring with real coefficients. Thus the form that I speak of is in use in the curriculum.
I'm stretching this out a bit since everyone here wants to dodge the simplest detail. A falsification is a falsification, no matter how much icing you throw on top of the cake. Still I can go to a simpler form and that is the variation that I will try here. We've selected real coefficients for the polynomial and still believe it to be ring behaved. Now we select the real coefficients. This would be the first concrete instance of a polynomial; a stage which the ordinary curriculum skips right over; no different than many 'high' mathematics courses do of their own constructions, if indeed any instance can ever be found. So let's really cement that polynomial to
0 + 1.23 X + 0 X X + 0 X X X + 0...
which is a fairly simple instance and now the supposed ring expression becomes
1.23 X
where 1.23 is a real value and X is not a real value. Therefore this first concrete instance of the polynomial form is not ring behaved, as it conflicts with the ring operator definitions. It fails to meet the closure requirement under product. This is a formal falsification. The failure of others here to meet this falsification head on is unacceptable. The math is not general if it cannot withstand instantiation.
I am amazed at the amount of dodge taking place on this thread. Can anyone take this on directly? Apparently not, since if they could have they would have. When somebody provides a falsification you have to return a refutation. In other words I should have made a mistake in my work here. No one has stepped up and pointed out an error in my work. Instead they just go on back to the curriculum as if I posted a message on needing some help understanding the subject.
Falsification methods... I need only provide one black swan to falsify the statement that all swans are white. Somehow you all fail to observe this effect, and this then again brings us into a broader statement on philosophy and the modern human. Could it be that at your computers sipping your coffee you are staring at the black swan in disbelief? Who here can utter a statement about the black swan? None so far except myself. Why this is so; a very fascinating system that we are caught up in. This is sci.math and we are on an uncensored medium. All here are self-censoring. If they witness the black swan even they will have gone too far. Put your hands up before your eyes and ask yourself if you are doing math. Or are you merely a good mimic?
Post by Mostowski Collapse
x v ~x = x + (1 - x) - x*(1 - x)
= 1 - x + x^2
= 1 - x + x
= 1
Post by Mostowski Collapse
In Boolean polynomials, interestingly the fundamental
theorem of algebra, still holds somehow.
Because a Boolean polynomial in one variable with no roots,
i.e. that never becomes zero, reduces to 1.
I guess thats why Boole was so excited.
Post by Ross A. Finlayson
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
It's great, if you found something mathematics owes, i.e. that it gives.
I'm sure the modular arithmetic and abstract algebra is familiar to everybody.
that you know about for example machine arithmetic and polysign
numbers, of course that abstract algebra besides cancellation or
opposition (division, the reciprocal, or inverse) then is linear then
that in the non-linear, as what results in the vertical or pole,
that it's OK and algebra does carry over, boxing up the terms
and circling the others, as what arrive in the results in the
evaluation, of what are the terms, of the abstract algebra.
that you found something that needs fixing - that what you have
in mind you also know that for all the intents and purposes of
abstract algebra, it will be OK in those terms.
In the operator space there then using for example polysign numbers
for points with algebra's pointing to their neighbors, or about moments,
when you find abstract algebra all broken it's usually in terms of
simple matters of scale - that the variables weren't in terms.
"To formalize this argument
we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one.
This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle.
Now likewise
each term in the polynomial can be attacked and
after all they are a sum which
in the ring definition requires
that they all be in one set together.
The entire polynomials fails under their own formalities. "
Now, I actually agree here though for defining "the entire polynomials fails", as,
as together that of course it's read in the un-usual way "the, entire polynomial, 's",
that "fails" only means "doesn't hold up". (Like you say, in the terms.)
I.e. the polynomials entirely hold up the fundamental theorem
of algebra about the existence of roots.
Then initial terms and such moments of characteristics,
that is for example all assigned to numerical methods
then what of course adding numerical methods to non-linear
or non-parametric modeling, often multiplies instead of
cancels, the error term. (The other term after the approximation
the first term of which in the numerical method is correct.)
So, you'd have to write that out some more,
to make it so logically all the conditions arrive
at what you say.
Just more, though - not different.
Thanks, it's OK.
zelos...@gmail.com
2020-09-04 05:25:10 UTC
Post by Tim Golden BandTech.com
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
it is a notation of it, yes.
Post by Tim Golden BandTech.com
The value 1.23 is a real number. Is this valid?
Yes
Post by Tim Golden BandTech.com
is a valid polynomial expression within abstract algebra?
Yes
Post by Tim Golden BandTech.com
Of course again these values are real except for the X which has no actual set definition at all.
As I pointed out already, the X is a historical NOTATION that we use, it means n=1 for our polynomial sequence as I defined earlier. just as X^2 means n=2, etc
Post by Tim Golden BandTech.com
Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof
Complainin about notation, that I already showed you is just notation, is not a disproof, it is however proof of your ignorance.
Lalo T.
2020-09-04 05:54:26 UTC
https://proofwiki.org/wiki/Definition:Scalar
https://proofwiki.org/wiki/Definition:Scalar_Ring

https://mathworld.wolfram.com/ScalarMultiplication.html
https://en.wikipedia.org/wiki/Scalar_multiplication
https://encyclopediaofmath.org/wiki/Scalar

https://en.wikipedia.org/wiki/Classical_Hamiltonian_quaternions
Tim Golden BandTech.com
2020-09-04 11:46:11 UTC
Post by ***@gmail.com
Post by Tim Golden BandTech.com
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
it is a notation of it, yes.
Post by Tim Golden BandTech.com
The value 1.23 is a real number. Is this valid?
Yes
Post by Tim Golden BandTech.com
is a valid polynomial expression within abstract algebra?
Yes
Post by Tim Golden BandTech.com
Of course again these values are real except for the X which has no actual set definition at all.
As I pointed out already, the X is a historical NOTATION that we use, it means n=1 for our polynomial sequence as I defined earlier. just as X^2 means n=2, etc
Post by Tim Golden BandTech.com
Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof
Complainin about notation, that I already showed you is just notation, is not a disproof, it is however proof of your ignorance.
Zelos I am not the one who only just wrote out formalized operators. This is supposedly key to this topic. More fundamental really than the polynomial. The polynomial is claimed to be ring behaved, but my assertion here proves it not to be ring behaved.

It is as if you deny my ability to dismantle the polynomial in the way that I do and yet the way that I do is just as it is constructed. It is a long series of sums. This is how you can come to claim its ring behavior and yet when we study the individual terms we land in this conclusion that these expressions are not ring behaved. Their internal structure is broken.

And now again you pivot back to claiming that this is 'historical' notation but didn't we already cover this above? This is not antiquated notation. It is still in use. It is as if you can see the conflict; the black swan. Only one is needed sir. That there are many and that there are still white swans is not of concern to the process of falsification. One black swan will do:
1.23 X
is not ring behaved. It cannot be. It goes in direct contradiction to the closure requirement of the binary operator. Thus just as soon as the abstracted algebraics goes to use their new construction with 'real coefficients' they broke away from their own rules. Your own inability Zelos to come to speak about the black swan above and admit that it is not ring behaved is proven by the elongated and diversionary terms on which you speak in this thread. This must be about the fifth go around now. I do appreciate you staying on but it could be helpful couldn't it if you could come to facing the music of the quacker that is biting you in the but? Poor Mostowski has it on the nose. You are caught running away. You must face the black swan to halt the attack.
Mostowski Collapse
2020-09-04 12:44:33 UTC
We can only say that the axioms do not
so easily satisfy a submodel property.

If you have N ⊆ M, and M satisfies a
closure property

∀x,y(x e M /\ y e M => x*y e M)

then its not automatic that N also
satisfies a closure property

∀x,y(x e N /\ y e N => x*y e N)

So only some of the N ⊆ M are also
closed. And among the closed only

some satisfy the original axioms.
In Grouptheory such a Group has a special
name, its called Subgroup:

https://en.wikipedia.org/wiki/Subgroup

You can also call N ⊆ M of a group M,
which is not a subgroup a black swan.
But this is a little overdramatic.
Post by Tim Golden BandTech.com
Post by ***@gmail.com
Post by Tim Golden BandTech.com
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
it is a notation of it, yes.
Post by Tim Golden BandTech.com
The value 1.23 is a real number. Is this valid?
Yes
Post by Tim Golden BandTech.com
is a valid polynomial expression within abstract algebra?
Yes
Post by Tim Golden BandTech.com
Of course again these values are real except for the X which has no actual set definition at all.
As I pointed out already, the X is a historical NOTATION that we use, it means n=1 for our polynomial sequence as I defined earlier. just as X^2 means n=2, etc
Post by Tim Golden BandTech.com
Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof
Complainin about notation, that I already showed you is just notation, is not a disproof, it is however proof of your ignorance.
Zelos I am not the one who only just wrote out formalized operators. This is supposedly key to this topic. More fundamental really than the polynomial. The polynomial is claimed to be ring behaved, but my assertion here proves it not to be ring behaved.
It is as if you deny my ability to dismantle the polynomial in the way that I do and yet the way that I do is just as it is constructed. It is a long series of sums. This is how you can come to claim its ring behavior and yet when we study the individual terms we land in this conclusion that these expressions are not ring behaved. Their internal structure is broken.
1.23 X
is not ring behaved. It cannot be. It goes in direct contradiction to the closure requirement of the binary operator. Thus just as soon as the abstracted algebraics goes to use their new construction with 'real coefficients' they broke away from their own rules. Your own inability Zelos to come to speak about the black swan above and admit that it is not ring behaved is proven by the elongated and diversionary terms on which you speak in this thread. This must be about the fifth go around now. I do appreciate you staying on but it could be helpful couldn't it if you could come to facing the music of the quacker that is biting you in the but? Poor Mostowski has it on the nose. You are caught running away. You must face the black swan to halt the attack.
Mostowski Collapse
2020-09-04 12:53:05 UTC
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).

We can now say:

White Swan: S is subgroup and submonoid

Yellow Swan: S is subgroup but not submonoid

Blue Swan: S is not subgroup but submonoid

Black Swan: S is neither subgroup nor submonoid

Example R = polynomials, S = {a1 X e R}. Thats a
yellow swan, and not a black swan.
Post by Mostowski Collapse
We can only say that the axioms do not
so easily satisfy a submodel property.
If you have N ⊆ M, and M satisfies a
closure property
∀x,y(x e M /\ y e M => x*y e M)
then its not automatic that N also
satisfies a closure property
∀x,y(x e N /\ y e N => x*y e N)
So only some of the N ⊆ M are also
closed. And among the closed only
some satisfy the original axioms.
In Grouptheory such a Group has a special
https://en.wikipedia.org/wiki/Subgroup
You can also call N ⊆ M of a group M,
which is not a subgroup a black swan.
But this is a little overdramatic.
Post by Tim Golden BandTech.com
Post by ***@gmail.com
Post by Tim Golden BandTech.com
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
it is a notation of it, yes.
Post by Tim Golden BandTech.com
The value 1.23 is a real number. Is this valid?
Yes
Post by Tim Golden BandTech.com
is a valid polynomial expression within abstract algebra?
Yes
Post by Tim Golden BandTech.com
Of course again these values are real except for the X which has no actual set definition at all.
As I pointed out already, the X is a historical NOTATION that we use, it means n=1 for our polynomial sequence as I defined earlier. just as X^2 means n=2, etc
Post by Tim Golden BandTech.com
Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof
Complainin about notation, that I already showed you is just notation, is not a disproof, it is however proof of your ignorance.
Zelos I am not the one who only just wrote out formalized operators. This is supposedly key to this topic. More fundamental really than the polynomial. The polynomial is claimed to be ring behaved, but my assertion here proves it not to be ring behaved.
It is as if you deny my ability to dismantle the polynomial in the way that I do and yet the way that I do is just as it is constructed. It is a long series of sums. This is how you can come to claim its ring behavior and yet when we study the individual terms we land in this conclusion that these expressions are not ring behaved. Their internal structure is broken.
1.23 X
is not ring behaved. It cannot be. It goes in direct contradiction to the closure requirement of the binary operator. Thus just as soon as the abstracted algebraics goes to use their new construction with 'real coefficients' they broke away from their own rules. Your own inability Zelos to come to speak about the black swan above and admit that it is not ring behaved is proven by the elongated and diversionary terms on which you speak in this thread. This must be about the fifth go around now. I do appreciate you staying on but it could be helpful couldn't it if you could come to facing the music of the quacker that is biting you in the but? Poor Mostowski has it on the nose. You are caught running away. You must face the black swan to halt the attack.
Contrarian
2020-09-04 13:10:12 UTC
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
I think you mean "Golden swan". But then again, gold and black can be the same... (https://en.wikipedia.org/wiki/The_dress)
Post by Mostowski Collapse
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
Tim Golden BandTech.com
2020-09-04 13:46:13 UTC
Post by Contrarian
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
I think you mean "Golden swan". But then again, gold and black can be the same... (https://en.wikipedia.org/wiki/The_dress)
Post by Mostowski Collapse
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
This is all a dodge. The instantiated black swan is
1.23 X
which is not ring behaved but as an instance of a polynomial with real coefficients which is claimed to be ring behaved. This conflict is direct. It is not broad. It is not general. It is extremely specific. Just one of these will do in the name of falsification and the more you dodge it the more you step in it.
< Hmmm? What? Oh, sorry, I didn't see that line that says "1.23 X" and all the text you repetitively write around it.
< Now you claim to have dismantled the polynomial.
< Hmmm. I must be dreaming. Nope. Back to the usual...
I mean, really, it is as if these high faluting people cannot read.
Can't deal with the simplest of terms.
Dodge what I have put under their noses for the umpteenth time.
I work with a hammer and a chisel while you all pull out your swiss army knives...
I suppose even your best blade on that thing is dull. And good luck sharpening it.
And you paid how much for that thing?
Peter
2020-09-04 14:05:33 UTC
Post by Tim Golden BandTech.com
Post by Contrarian
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
I think you mean "Golden swan". But then again, gold and black can be the same... (https://en.wikipedia.org/wiki/The_dress)
Post by Mostowski Collapse
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
This is all a dodge. The instantiated black swan is
1.23 X
which is not ring behaved but as an instance of a polynomial with real coefficients which is claimed to be ring behaved. This
You need to say what you mean by "ring behaved". Does it mean anything
other than "in an element of R[X]"?

conflict is direct. It is not broad. It is not general. It is extremely
specific. Just one of these will do in the name of falsification and
the more you dodge it the more you step in it.
Post by Tim Golden BandTech.com
< Hmmm? What? Oh, sorry, I didn't see that line that says "1.23 X" and all the text you repetitively write around it.
< Now you claim to have dismantled the polynomial.
< Hmmm. I must be dreaming. Nope. Back to the usual...
I mean, really, it is as if these high faluting people cannot read.
Can't deal with the simplest of terms.
Dodge what I have put under their noses for the umpteenth time.
I work with a hammer and a chisel while you all pull out your swiss army knives...
I suppose even your best blade on that thing is dull. And good luck sharpening it.
And you paid how much for that thing?
Mostowski Collapse
2020-09-04 23:35:39 UTC
#algebragate #fakepolynomials #mathcrisis
Post by Contrarian
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
I think you mean "Golden swan". But then again, gold and black can be the same... (https://en.wikipedia.org/wiki/The_dress)
Post by Mostowski Collapse
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
Mostowski Collapse
2020-09-04 23:39:55 UTC
#ringmeltdown #groupsnotwell #closurematters
Post by Mostowski Collapse
#algebragate #fakepolynomials #mathcrisis
Post by Contrarian
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
I think you mean "Golden swan". But then again, gold and black can be the same... (https://en.wikipedia.org/wiki/The_dress)
Post by Mostowski Collapse
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
Lalo T.
2020-09-05 09:13:04 UTC
I wrote :

" If we were is a maze, in a competition where only one can get out, and weapons are provided, before proceed to the combat, better check first if the walls can be broken, with the possibililty
of finding a tunnel. "

I err, I meant :

" If we were iN a maze, in a competition where only one can get out, and weapons are provided, before proceed to the combat, WE better check first if the walls can be broken, with the possibililty
of finding a tunnel.. and find a way out".

Still I think that there are some language issues....
Tim Golden BandTech.com
2020-09-05 15:58:41 UTC
Post by Mostowski Collapse
A subring of a ring (R, +, ∗, 0, 1) is a subset S
of R that is both a subgroup of (R, +, 0) and a
submonoid of (R, ∗, 1).
White Swan: S is subgroup and submonoid
Yellow Swan: S is subgroup but not submonoid
Blue Swan: S is not subgroup but submonoid
Black Swan: S is neither subgroup nor submonoid
Example R = polynomials, S = {a1 X e R}. Thats a
yellow swan, and not a black swan.
If I am to take you seriously here, then this line of argumentation really should lead back onto the polynomial with real coefficients which is where I constructed the black swan
1.23 X
We can make another black swan
0.1 + 1.23 X
and another and another, but one is enough isn't it?

Earlier I was thinking that you are spamming this thread, but I guess maybe you are a deep and slow thinker. You better look out because these birds are quite fast when they attack. I think there is still something on your nose Mostowski.
Post by Mostowski Collapse
Post by Mostowski Collapse
We can only say that the axioms do not
so easily satisfy a submodel property.
If you have N ⊆ M, and M satisfies a
closure property
∀x,y(x e M /\ y e M => x*y e M)
then its not automatic that N also
satisfies a closure property
∀x,y(x e N /\ y e N => x*y e N)
So only some of the N ⊆ M are also
closed. And among the closed only
some satisfy the original axioms.
In Grouptheory such a Group has a special
https://en.wikipedia.org/wiki/Subgroup
You can also call N ⊆ M of a group M,
which is not a subgroup a black swan.
But this is a little overdramatic.
Post by Tim Golden BandTech.com
Post by ***@gmail.com
Post by Tim Golden BandTech.com
an instance of a polynomial with real coefficients? The X here of course is the abstract form of abstract algebra.
it is a notation of it, yes.
Post by Tim Golden BandTech.com
The value 1.23 is a real number. Is this valid?
Yes
Post by Tim Golden BandTech.com
is a valid polynomial expression within abstract algebra?
Yes
Post by Tim Golden BandTech.com
Of course again these values are real except for the X which has no actual set definition at all.
As I pointed out already, the X is a historical NOTATION that we use, it means n=1 for our polynomial sequence as I defined earlier. just as X^2 means n=2, etc
Post by Tim Golden BandTech.com
Why should all here deny such simple forms? Are they too simple? Are they beneath you? Does the terminology 'fundamental' have anything to do with simplicity? Possibly you are a bot and merely work out variations on the higher side of things. If so I really think you should try and compile this low level disproof
Complainin about notation, that I already showed you is just notation, is not a disproof, it is however proof of your ignorance.
Zelos I am not the one who only just wrote out formalized operators. This is supposedly key to this topic. More fundamental really than the polynomial. The polynomial is claimed to be ring behaved, but my assertion here proves it not to be ring behaved.
It is as if you deny my ability to dismantle the polynomial in the way that I do and yet the way that I do is just as it is constructed. It is a long series of sums. This is how you can come to claim its ring behavior and yet when we study the individual terms we land in this conclusion that these expressions are not ring behaved. Their internal structure is broken.
1.23 X
is not ring behaved. It cannot be. It goes in direct contradiction to the closure requirement of the binary operator. Thus just as soon as the abstracted algebraics goes to use their new construction with 'real coefficients' they broke away from their own rules. Your own inability Zelos to come to speak about the black swan above and admit that it is not ring behaved is proven by the elongated and diversionary terms on which you speak in this thread. This must be about the fifth go around now. I do appreciate you staying on but it could be helpful couldn't it if you could come to facing the music of the quacker that is biting you in the but? Poor Mostowski has it on the nose. You are caught running away. You must face the black swan to halt the attack.
Lalo T.
2020-09-06 09:08:55 UTC
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Since it is not EVALUATED you could name it :
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.

"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"

The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +

P = p₀ + p₁x + p₂x² + ...

Q = q₀ + q₁x + q₂x² + ...

(1) addition(P,Q) = P ＋ Q

(2) multiplication(P,Q) = P ✖ Q

( ) scalar_multiplication(t,P) = t·P <--- !!!

https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring

"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "

in the same section :

"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."

more recreational links to the pile :
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does

In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations

This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
Tim Golden BandTech.com
2020-09-06 13:33:34 UTC
Post by Lalo T.
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.
"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"
The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +
P = p₀ + p₁x + p₂x² + ...
Q = q₀ + q₁x + q₂x² + ...
(1) addition(P,Q) = P ＋ Q
(2) multiplication(P,Q) = P ✖ Q
( ) scalar_multiplication(t,P) = t·P <--- !!!
https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring
"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "
"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does
In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
Nice finds here. I suspect these are trivial though. To what degree will
K X S -> S
imply that K is a subset of S? Pretty sure everything else will be noncomputable and thus contradict the statement. This is the odd nature of these supposedly mathematical statements: they have no decent instances to speak of other that trivialities. This same problem exists in abstract algebra, but a magician's cloth is far more extensively in use. It is as if they've gone to the trouble to wipe fingerprints in order to 'prove' the system of AA.

I do like your attempt at drawing a barrier at EVALUATION.
It seems to me that if you would get the AA people to insert the requirement of non-evaluation into their math that we will be done.
I do not think though that you have actually falsified my position here but rather supported it.
Still it is another interpretation that does seem relevant.

If you choose to draw a distinction between the operators in use within the polynomial expression and those used to add and multiply those polynomials you land in the same conundrum. This attempt also is weak in that it forms yet another criticism on the topic of AA... having gone to the trouble of formalizing ring operators how on Earth could you introduce two fresh operators and not formally identify them? So again we have an attack on AA that stands freely without the introduction of any new technology. I'm pretty sure that most here would not take this avenue and will claim that there are just two ring operators in use within the polynomial with real coefficients. If there are more then the subject is a sham for avoiding them completely.
Post by Lalo T.
This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
No Lalo. You are dodging the black swan. And as you close with you didn't even try to catch the thing here. I still do see something in your meandering style. You do manage to fill in some gaps. Here is a nice piece with an entire chapter on the negative sign that I just bumped into. His language is in the open style that I recommend all take up.
https://www.gutenberg.org/files/39088/39088-pdf.pdf
Peter
2020-09-06 13:47:07 UTC
Post by Tim Golden BandTech.com
Post by Lalo T.
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.
"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"
The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +
P = p₀ + p₁x + p₂x² + ...
Q = q₀ + q₁x + q₂x² + ...
(1) addition(P,Q) = P ＋ Q
(2) multiplication(P,Q) = P ✖ Q
( ) scalar_multiplication(t,P) = t·P <--- !!!
https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring
"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "
"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does
In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
Nice finds here. I suspect these are trivial though. To what degree will
K X S -> S
imply that K is a subset of S?
No, and if it where the operation would not be external. Since you're
using X for the indeterminate in a polynomial, may I suggest that you
use x for Cartesian product?
Post by Tim Golden BandTech.com
Pretty sure everything else will be noncomputable
Computability has got nothing to do with your problem.
Post by Tim Golden BandTech.com
and thus contradict the statement. This is the odd nature of these supposedly mathematical statements: they have no decent instances to speak of other that trivialities. This same problem exists in abstract algebra, but a magician's cloth is far more extensively in use. It is as if they've gone to the trouble to wipe fingerprints in order to 'prove' the system of AA.
I do like your attempt at drawing a barrier at EVALUATION.
It seems to me that if you would get the AA people to insert the requirement of non-evaluation into their math that we will be done.
I do not think though that you have actually falsified my position here but rather supported it.
Still it is another interpretation that does seem relevant.
If you choose to draw a distinction between the operators in use within the polynomial expression and those used to add and multiply those polynomials you land in the same conundrum. This attempt also is weak in that it forms yet another criticism on the topic of AA... having gone to the trouble of formalizing ring operators how on Earth could you introduce two fresh operators and not formally identify them? So again we have an attack on AA that stands freely without the introduction of any new technology. I'm pretty sure that most here would not take this avenue and will claim that there are just two ring operators in use within the polynomial with real coefficients. If there are more then the subject is a sham for avoiding them completely.
Post by Lalo T.
This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
No Lalo. You are dodging the black swan. And as you close with you didn't even try to catch the thing here. I still do see something in your meandering style. You do manage to fill in some gaps. Here is a nice piece with an entire chapter on the negative sign that I just bumped into. His language is in the open style that I recommend all take up.
https://www.gutenberg.org/files/39088/39088-pdf.pdf
Tim Golden BandTech.com
2020-09-06 14:35:46 UTC
Post by Tim Golden BandTech.com
Post by Lalo T.
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.
"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"
The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +
P = p₀ + p₁x + p₂x² + ...
Q = q₀ + q₁x + q₂x² + ...
(1) addition(P,Q) = P ＋ Q
(2) multiplication(P,Q) = P ✖ Q
( ) scalar_multiplication(t,P) = t·P <--- !!!
https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring
"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "
"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does
In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
Nice finds here. I suspect these are trivial though. To what degree will
K X S -> S
imply that K is a subset of S?
No, and if it where the operation would not be external. Since you're
using X for the indeterminate in a polynomial, may I suggest that you
use x for Cartesian product?
Clearly you feel badly cornered in a cage not of my making. The black swan chased you in and there you sit.
This here is merely a side track. It is not the formality that is the crux of the thread. Lalo is after the fix to the problem even while he denies that the problem exists. You Peter sit caged from your own earlier track of argumentation in a position where you feel locked from responding to the simplest of requests such as granting that the polynomial with real coefficients is fair game within this discussion while that has been stipulated from the beginning. This lack of concession points out to me your own inauthentic behavior at this point of the debate. I doubt we will ever here an utterance on this from you. Dodge the swan and dodge this concern? I'm fairly certain that your attempt is a flop at this point. You can eat white goose flops for lunch today. Tomorrow is another day.
Post by Tim Golden BandTech.com
Pretty sure everything else will be noncomputable
Computability has got nothing to do with your problem.
Post by Tim Golden BandTech.com
and thus contradict the statement. This is the odd nature of these supposedly mathematical statements: they have no decent instances to speak of other that trivialities. This same problem exists in abstract algebra, but a magician's cloth is far more extensively in use. It is as if they've gone to the trouble to wipe fingerprints in order to 'prove' the system of AA.
I do like your attempt at drawing a barrier at EVALUATION.
It seems to me that if you would get the AA people to insert the requirement of non-evaluation into their math that we will be done.
I do not think though that you have actually falsified my position here but rather supported it.
Still it is another interpretation that does seem relevant.
If you choose to draw a distinction between the operators in use within the polynomial expression and those used to add and multiply those polynomials you land in the same conundrum. This attempt also is weak in that it forms yet another criticism on the topic of AA... having gone to the trouble of formalizing ring operators how on Earth could you introduce two fresh operators and not formally identify them? So again we have an attack on AA that stands freely without the introduction of any new technology. I'm pretty sure that most here would not take this avenue and will claim that there are just two ring operators in use within the polynomial with real coefficients. If there are more then the subject is a sham for avoiding them completely.
Post by Lalo T.
This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
No Lalo. You are dodging the black swan. And as you close with you didn't even try to catch the thing here. I still do see something in your meandering style. You do manage to fill in some gaps. Here is a nice piece with an entire chapter on the negative sign that I just bumped into. His language is in the open style that I recommend all take up.
https://www.gutenberg.org/files/39088/39088-pdf.pdf
Peter
2020-09-06 14:43:17 UTC
Post by Tim Golden BandTech.com
Post by Tim Golden BandTech.com
Post by Lalo T.
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.
"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"
The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +
P = p₀ + p₁x + p₂x² + ...
Q = q₀ + q₁x + q₂x² + ...
(1) addition(P,Q) = P ＋ Q
(2) multiplication(P,Q) = P ✖ Q
( ) scalar_multiplication(t,P) = t·P <--- !!!
https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring
"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "
"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does
In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
Nice finds here. I suspect these are trivial though. To what degree will
K X S -> S
imply that K is a subset of S?
No, and if it where the operation would not be external. Since you're
using X for the indeterminate in a polynomial, may I suggest that you
use x for Cartesian product?
Clearly you feel badly cornered in a cage not of my making. The black swan chased you in and there you sit.
This here is merely a side track. It is not the formality that is the crux of the thread. Lalo is after the fix to the problem even while he denies that the problem exists. You Peter sit caged from your own earlier track of argumentation in a position where you feel locked from responding to the simplest of requests such as granting that the polynomial with real coefficients is fair game within this discussion while that has been stipulated from the beginning. This lack of concession points out to me your own inauthentic behavior at this point of the debate. I doubt we will ever here an utterance on this from you. Dodge the swan and dodge this concern? I'm fairly certain that your attempt is a flop at this point. You can eat white goose flops for lunch today. Tomorrow is another day.
I don't know what the black swan is that you are referring to.
Post by Tim Golden BandTech.com
Post by Tim Golden BandTech.com
Pretty sure everything else will be noncomputable
Computability has got nothing to do with your problem.
Post by Tim Golden BandTech.com
and thus contradict the statement. This is the odd nature of these supposedly mathematical statements: they have no decent instances to speak of other that trivialities. This same problem exists in abstract algebra, but a magician's cloth is far more extensively in use. It is as if they've gone to the trouble to wipe fingerprints in order to 'prove' the system of AA.
I do like your attempt at drawing a barrier at EVALUATION.
It seems to me that if you would get the AA people to insert the requirement of non-evaluation into their math that we will be done.
I do not think though that you have actually falsified my position here but rather supported it.
Still it is another interpretation that does seem relevant.
If you choose to draw a distinction between the operators in use within the polynomial expression and those used to add and multiply those polynomials you land in the same conundrum. This attempt also is weak in that it forms yet another criticism on the topic of AA... having gone to the trouble of formalizing ring operators how on Earth could you introduce two fresh operators and not formally identify them? So again we have an attack on AA that stands freely without the introduction of any new technology. I'm pretty sure that most here would not take this avenue and will claim that there are just two ring operators in use within the polynomial with real coefficients. If there are more then the subject is a sham for avoiding them completely.
Post by Lalo T.
This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
No Lalo. You are dodging the black swan. And as you close with you didn't even try to catch the thing here. I still do see something in your meandering style. You do manage to fill in some gaps. Here is a nice piece with an entire chapter on the negative sign that I just bumped into. His language is in the open style that I recommend all take up.
https://www.gutenberg.org/files/39088/39088-pdf.pdf
Tim Golden BandTech.com
2020-09-06 15:52:36 UTC
Post by Peter
Post by Tim Golden BandTech.com
Post by Tim Golden BandTech.com
Post by Lalo T.
Several user have provided respect https://en.wikipedia.org/wiki/Polynomial_ring
They have stated that evaluated is different compared to the above concept.
Arithmetic of " 'X' marks with attached superscript integers" polynomial
and use the symbol '?' instead of the symbol 'X', if you want.
"However, here, X has not any value (other than itself), and cannot vary, being a constant in the polynomial ring"
The symbols '＋' '✖' '·' will be used for the operations, note that ＋ is bigger than +
P = p₀ + p₁x + p₂x² + ...
Q = q₀ + q₁x + q₂x² + ...
(1) addition(P,Q) = P ＋ Q
(2) multiplication(P,Q) = P ✖ Q
( ) scalar_multiplication(t,P) = t·P <--- !!!
https://proofwiki.org/wiki/Definition:Polynomial_Ring
in the section https://en.wikipedia.org/wiki/Polynomial_ring#Definition_(univariate_case)
https://infogalactic.com/info/Polynomial_ring
"The polynomial ring in X over K is equipped with an addition, a multiplication and a scalar multiplication that make it a commutative algebra. "
"The scalar multiplication is the special case of the multiplication where p = p0 is reduced to its constant term (the term that is independent of X); that is..."
https://www.womansworld.com/posts/entertainment/my-wife-and-my-mother-in-law-166601
https://matrix.fandom.com/wiki/Black_cat
https://en.wikipedia.org/wiki/The_purpose_of_a_system_is_what_it_does
In any case, the conflict seems to lies in the 'external binary operation'.
https://en.wikipedia.org/wiki/External_(mathematics)#Generalizations
https://en.wikipedia.org/wiki/Binary_operation#External_binary_operations
Nice finds here. I suspect these are trivial though. To what degree will
K X S -> S
imply that K is a subset of S?
No, and if it where the operation would not be external. Since you're
using X for the indeterminate in a polynomial, may I suggest that you
use x for Cartesian product?
Clearly you feel badly cornered in a cage not of my making. The black swan chased you in and there you sit.
This here is merely a side track. It is not the formality that is the crux of the thread. Lalo is after the fix to the problem even while he denies that the problem exists. You Peter sit caged from your own earlier track of argumentation in a position where you feel locked from responding to the simplest of requests such as granting that the polynomial with real coefficients is fair game within this discussion while that has been stipulated from the beginning. This lack of concession points out to me your own inauthentic behavior at this point of the debate. I doubt we will ever here an utterance on this from you. Dodge the swan and dodge this concern? I'm fairly certain that your attempt is a flop at this point. You can eat white goose flops for lunch today. Tomorrow is another day.
I don't know what the black swan is that you are referring to.
Post by Tim Golden BandTech.com
Post by Tim Golden BandTech.com
Pretty sure everything else will be noncomputable
Computability has got nothing to do with your problem.
Post by Tim Golden BandTech.com
and thus contradict the statement. This is the odd nature of these supposedly mathematical statements: they have no decent instances to speak of other that trivialities. This same problem exists in abstract algebra, but a magician's cloth is far more extensively in use. It is as if they've gone to the trouble to wipe fingerprints in order to 'prove' the system of AA.
I do like your attempt at drawing a barrier at EVALUATION.
It seems to me that if you would get the AA people to insert the requirement of non-evaluation into their math that we will be done.
I do not think though that you have actually falsified my position here but rather supported it.
Still it is another interpretation that does seem relevant.
If you choose to draw a distinction between the operators in use within the polynomial expression and those used to add and multiply those polynomials you land in the same conundrum. This attempt also is weak in that it forms yet another criticism on the topic of AA... having gone to the trouble of formalizing ring operators how on Earth could you introduce two fresh operators and not formally identify them? So again we have an attack on AA that stands freely without the introduction of any new technology. I'm pretty sure that most here would not take this avenue and will claim that there are just two ring operators in use within the polynomial with real coefficients. If there are more then the subject is a sham for avoiding them completely.
Post by Lalo T.
This is, if someone provides Tim a satisfactory answer (regarding his initial post)
he will be satisfied and will be the end of the whole(as other user put it) shebang.
The requirement is, merely, provide an 'external binary operation'-free type of answer, nothing more than that.
That is, say to Tim, mathematically, "That all is ok, and nothing is happening here"
(without mentioning the aforementioned forbidden word, of course)
No Lalo. You are dodging the black swan. And as you close with you didn't even try to catch the thing here. I still do see something in your meandering style. You do manage to fill in some gaps. Here is a nice piece with an entire chapter on the negative sign that I just bumped into. His language is in the open style that I recommend all take up.
https://www.gutenberg.org/files/39088/39088-pdf.pdf
https://www.gutenberg.org/files/33571/33571-h/33571-h.htm#THE_MAGIC_SWAN

Good luck Peter.
Peter
2020-09-06 16:07:02 UTC
Post by Tim Golden BandTech.com
Post by Peter
I don't know what the black swan is that you are referring to.
Post by Peter
Post by Tim Golden BandTech.com
Pretty sure everything else will be noncomputable
Computability has got nothing to do with your problem.
https://www.gutenberg.org/files/33571/33571-h/33571-h.htm#THE_MAGIC_SWAN
Good luck Peter.
There is no mention of algebra or polynomials in that. Maybe it would
benefit you more to read an algebra text book than fairy tales.
zelos...@gmail.com
2020-09-07 07:26:18 UTC
Post by Tim Golden BandTech.com
Zelos I am not the one who only just wrote out formalized operators. This is supposedly key to this topic. More fundamental really than the polynomial. The polynomial is claimed to be ring behaved, but my assertion here proves it not to be ring behaved.
False, your claims are all about notation. If you do it in proper fashion as with the sequences of almost all are 0s, then you'll see it is a ring.
Post by Tim Golden BandTech.com
It is as if you deny my ability to dismantle the polynomial in the way that I do and yet the way that I do is just as it is constructed. It is a long series of sums. This is how you can come to claim its ring behavior and yet when we study the individual terms we land in this conclusion that these expressions are not ring behaved. Their internal structure is broken.
It works perfectly fine as a ring, I proved it in my bachelor thesis on polynomials :)

And no, it is not a series, it is not a sum, those are just ancient notation.
Post by Tim Golden BandTech.com
This is not antiquated notation. It is still in use
It is, and it is no issue that it is because most people are smarter than you and realise it is just ntoation.
Post by Tim Golden BandTech.com
is not ring behaved
It is an element, an element of a ring is not a ring.
Post by Tim Golden BandTech.com
It goes in direct contradiction to the closure requirement of the binary operator
Not at all, multiply it by any polynomial and you get a polynomial, add any polynomial to it and you get a polynomial, ergo it is closed under both operations.
Post by Tim Golden BandTech.com
Your own inability Zelos to come to speak about the black swan above and admit that it is not ring behaved is proven by the elongated and diversionary terms on which you speak in this thread
There is not much to say because 1.23X means little here. You ahve not shown that it multiplied by some polynomial is not a polynomial.
Lalo T.
2020-09-07 09:01:31 UTC
...even in :
https://en.wikipedia.org/wiki/Algebra_representation

"...is after the fix to the problem even while he denies that the problem exists"

https://en.wikipedia.org/wiki/Talk:Binary_operation

https://hsm.stackexchange.com/questions/11235/who-started-calling-the-matrix-multiplication-multiplication

https://math.stackexchange.com/questions/1348273/external-operation-binary-and-unary-perhaps

The issue loosely make me remember the topic "dimensional numbers"
concretely, like the strange T.n.p of the user Socratis, who contend "There are no dimensionless numbers"
(not necessarily that system, but systems with the same thesis)

in https://en.wikipedia.org/wiki/Vector_space#Definition
note : " Compatibility of scalar multiplication with field multiplication "
https://en.wikipedia.org/wiki/Semigroup_action
https://en.wikipedia.org/wiki/Group_action

https://en.wikipedia.org/wiki/Homogeneous_function

" However, there is still no other terminology available for an 'external monoid' for which this terminology gives us a concise expression. Above all else, this is a reason this term should be of use in the mathematical community."
https://en.wikipedia.org/wiki/External_(mathematics)
https://en.wikipedia.org/wiki/Talk:External_(mathematics)

Contrarian already put an optical effect, and other alredy put some bits, let me add another

The suspicion is that is impossible to conclude/obtain your example from an abstract algebra reasoning.
I tried to obtain your model example but without success.

Going for broke, the bet is, like the Professor Kokichi Sugihara :

https://thekidshouldseethis.com/post/professor-kokichi-sugihara-creates-his-mind-blowing-illusions-with-math

you built an Abstract Algebraic Optical Illusion (an impossible construction)

in order to get a mathematical RFC to outgrow "External Binary Operations"

Hence, options (a), (b) and (c) dismissed.
The strange loop is in "External Binary operation" (the house where the paradox inhabits and, together with the critique of the method)

https://en.wikipedia.org/wiki/Strange_loop
https://en.wikipedia.org/wiki/Koan
Tim Golden BandTech.com
2020-09-07 15:24:55 UTC
Post by Lalo T.
https://en.wikipedia.org/wiki/Algebra_representation
"...is after the fix to the problem even while he denies that the problem exists"
https://en.wikipedia.org/wiki/Talk:Binary_operation
https://hsm.stackexchange.com/questions/11235/who-started-calling-the-matrix-multiplication-multiplication
https://math.stackexchange.com/questions/1348273/external-operation-binary-and-unary-perhaps
The issue loosely make me remember the topic "dimensional numbers"
concretely, like the strange T.n.p of the user Socratis, who contend "There are no dimensionless numbers"
(not necessarily that system, but systems with the same thesis)
in https://en.wikipedia.org/wiki/Vector_space#Definition
note : " Compatibility of scalar multiplication with field multiplication "
https://en.wikipedia.org/wiki/Semigroup_action
https://en.wikipedia.org/wiki/Group_action
https://en.wikipedia.org/wiki/Homogeneous_function
" However, there is still no other terminology available for an 'external monoid' for which this terminology gives us a concise expression. Above all else, this is a reason this term should be of use in the mathematical community."
https://en.wikipedia.org/wiki/External_(mathematics)
https://en.wikipedia.org/wiki/Talk:External_(mathematics)
Contrarian already put an optical effect, and other alredy put some bits, let me add another
The suspicion is that is impossible to conclude/obtain your example from an abstract algebra reasoning.
I tried to obtain your model example but without success.
https://thekidshouldseethis.com/post/professor-kokichi-sugihara-creates-his-mind-blowing-illusions-with-math
you built an Abstract Algebraic Optical Illusion (an impossible construction)
in order to get a mathematical RFC to outgrow "External Binary Operations"
Hence, options (a), (b) and (c) dismissed.
The strange loop is in "External Binary operation" (the house where the paradox inhabits and, together with the critique of the method)
https://en.wikipedia.org/wiki/Strange_loop
https://en.wikipedia.org/wiki/Koan
The strange loop does seem interesting, but they've almost demanded that it be paradoxical. Circular axioms are fine IMO if they have consequences. Typically they are able to take on many forms as a result, and attempts to reduce their redundancy might fail beyond a certain minimal rule set. If they support themselves will they support instantiation? Do we even have set theory yet? are there elements within a strange loop? We have to be granted the freedom to construct. Could we in fact normalize their weird prefix by simply admitting that our ordinary numerical radix ten representation of the real numbers fits this loop theory? The minimal requirement is one that we all would like to take... but how many things can be constructed? It's pretty clear that mostly we are caught up in mimicing past constructions; myself included. Part of the way forward is by dismantling those past constructions. Taking freedom from them. Working out variations on them. When consequential details emerge from this method then we ought not to dismiss the results. Rather we ought to propagate those results. When I witness the stupendously detailed nature of abstract algebra and see how simply breaking open the real number can do the job then I am dumbfounded that any would insist that polysign is not remarkable.

My remarks on the human condition I will stand by. They are many and they continue to congeal. The fraud is within us as much as it is from outside. We are social animals. Mathematicians may fit a very particular extreme within this group, and academia ensures the winnowing up of the best mimics. In this age it is more important to add onto the ever burgeoning accumulation. The abuse of dimension that is taking place does deserve ridicule. Possibly it could source a conversation on whether dimension builds from high to low, but mathematics has already built high dimension from down low. It's just been off by one. The mangled thing which is abstract algebra in no way is pristine. Your external binary operation does not answer any problem. It probably doesn't even carry a valid instance. As far as I can tell the most basic of product relations amongst elements of differing sets is one of preservation. For instance when I write
s x
where s is discrete sign and x is continuous magnitude these two different elemental forms married together in a product relation but they do not evaluate. They are the yield of a new set. Notation does matter, and the usage of sign for both an operator and for a value goes undiscussed within such 'minorly abusive' topics as abstract algebra. Well what does it matter if there is no consequence? Why is it that things work while they are still demonstrably conflicted? The trouble is that like polysign numbers there may be a construct sitting beneath our noses that nobody has bothered to build yet. This is the sad nature of the human condition. I can posit polysign as such an instance. Likely there are more. All of us are schooled on similar curricula and will likely reject this thing when it is presented; no different than these here cannot see through the AA system. This condition of the human form including and especially math types is cause for pause. We are so near to a fundamental breakthrough now. Maybe its in our genes. Why should we grant the human the ability to derive the truth in a linguistic form if it has only developed a rudimentary system thus far? The FOX P2 gene... a blessed curse.
bassam karzeddin
2020-09-05 10:20:58 UTC
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
So good of you that you raised a very important issue about the broken algebra basically with its fundamental theorem of absolute stupidity

But one should be aware of the imosibility for those vast majorities of common academic mainstreams (not only in mathematics) to realize anything even with hundreds of simple short proofs since they were brainwashed long ago of early ages and well trained to be true stubborn imbeciles even much worse than their preceding generations

One mustn't expect anything good from true imbeciles and real cranks who are globally in plenty and very available everywhere on this planet

The problem was never the truer superior mathematics with that unsociable creatures, but very deep and so many incurable human mental diseases with over mental inherited retardation

If you can write a book about the main problem, then you may be able to help the new uprising generation not to fall at all as easy prey to all those mentally disordered historical figures as (KUNT, GODEL, IMBECILE, COUNTER, ABEL, Galois, Ruffini, Cardano, Euler, Dedikined, Cauchy, ...., etc) With millions of dwarf followers after them...

Again the main problem was never the truer and superior mathematics itself, but truly about human first reaction towards higher knowledge

Good luck

Bassam King Karzeddin
Tim Golden BandTech.com
2020-09-05 13:57:36 UTC
Post by bassam karzeddin
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
So good of you that you raised a very important issue about the broken algebra basically with its fundamental theorem of absolute stupidity
But one should be aware of the imosibility for those vast majorities of common academic mainstreams (not only in mathematics) to realize anything even with hundreds of simple short proofs since they were brainwashed long ago of early ages and well trained to be true stubborn imbeciles even much worse than their preceding generations
One mustn't expect anything good from true imbeciles and real cranks who are globally in plenty and very available everywhere on this planet
The problem was never the truer superior mathematics with that unsociable creatures, but very deep and so many incurable human mental diseases with over mental inherited retardation
If you can write a book about the main problem, then you may be able to help the new uprising generation not to fall at all as easy prey to all those mentally disordered historical figures as (KUNT, GODEL, IMBECILE, COUNTER, ABEL, Galois, Ruffini, Cardano, Euler, Dedikined, Cauchy, ...., etc) With millions of dwarf followers after them...
Again the main problem was never the truer and superior mathematics itself, but truly about human first reaction towards higher knowledge
Good luck
Bassam King Karzeddin
So nice of you to add your two cents on this thread King Bassam. Still you've dodged the content rather thoroughly. I do wish to hear your own interpretation on
1.23 X
and whether this operation that is in use by the AA types has any validity. Given that they've gone to the trouble of carefully defining the operations of sum and product within the ring definition how then can they turn around and throw 'real coefficients' onto an undefined X without getting themselves into this jam? It is ring behaved and it is not ring behaved. Thus the subject carries its own conflicts to all users of this curriculum. It seems so peculiar in that they were so convincing in going about the ring definition in the first place. How dare them by the way snatch the name 'ring' for the formal definition of operators. It's not wise. That could be interesting to learn about its evolution. A ring clearly has rotational aspects and so does the polynomial but one is continuous whereas the other is quite discrete and really until they wrap that polynomial around with their fowl language of 'ideal' and 'quotient' the thing just shoots you off into higher dimension; no ring at all; not a clean crisp tone; not a torus; not even any rotation there. Ring their necks? Like a bird's? This is what they'd like to do to my black swan. So few can even approach the thing. I'll just make another one. They are all over the place really. They just refuse to see them.
Peter
2020-09-06 10:10:23 UTC
Tim Golden BandTech.com wrote:

. I do wish to hear your own interpretation on
Post by Tim Golden BandTech.com
1.23 X
and whether this operation that is in use by the AA types has any validity.
What operation are you referring to? The operation of pre multiplying X
by 1.23?
Tim Golden BandTech.com
2020-09-06 12:43:04 UTC
. I do wish to hear your own interpretation on
Post by Tim Golden BandTech.com
1.23 X
and whether this operation that is in use by the AA types has any validity.
What operation are you referring to? The operation of pre multiplying X
by 1.23?
Uh-oh. Peter as far as I know nobody has ever required 'pre multiplying' within the field of abstract algebra. In fact because of the algebraic behavior this is a clear sign that you have gone wrong. If the expression
a b c
implied that you must pre-multiply a by b before you multiply the result by c then we would be in a non-commutative realm. This is not the realm that we are dealing in here. There is but one operator present in the expression
1.23 X
so this further points to corruption in your thinking. How could we 'pre multiply' when there is just one multiplication taking place? Certainly because the usual good algebraic behaviors are claimed to be present within the ring of polynomials with real coefficients you are descending into a realm where the black swan is found. At some point you may have to admit that you witness the thing. This breakage from the status quo position may feel quite uncomfortable, and the extensions of it really are remarkable. Still I must welcome you already for you do see. At least if you do not yet then your words here do. I don't mean to cause a mental breakdown. That is not at all what I am after. I would like rather that we consider this topic open. The old position may be the best that could be done with the tools at hand. It may work fairly well. Pythagoras did quite some good work. Newton too. We are in modernity by no means existing in perfection and this is a fine instance that proves it. Will the mathematician or the social animal win here? Here the straight A's have no idea. They are the ones in power. This place here has a purity than no journal will ever carry. You and I can say anything here. It is even distributed media. Gee, could it be that such a medium ought to be suppressed and privatized? Spoiled perhaps?

You have failed to answer whether you are willing to grant the a(n) real values within our discussion. I most certainly am discussing this particular type of polynomial as it strengthens my argument. By the operator rules laid out so carefully within abstract algebra the above construction is not valid, and yet I have built it as an instance of the polynomial with real coefficients. This conflict is not my own. It is AA's own conflict. It is the black swan. Again, just one black swan need be presented to falsify the claim that all swans are white. I am not attempting to introduce any new concept such as "pre multimplication" into the topic at hand. As such this conflict stands freely from within the subject itself. It is the internal structure of the polynomial that is under attack here. This is a point which clearly goes avoided in AA.

There are in fact two interpretations in terms of dismantling the polynomial. Initially on this thread I specified that the real valued a(n) are zero with the exception of a1. I do think this is the most straight-forward interpretation, but there is another readily present that may be more convincing for some. The polynomial is a long sum. As such we can apply the ring principles to this long sum. In other words because the polynomial is a ring we can invoke the binary nature of its operators directly to isolate out
a1 X
from it since this term is in sum with other terms, all of which belong to the same set by the closure requirement. Now, having specified real coefficients we can study this term in isolation and ask whether its operator is ring behaved no different than when we instantiated a polynomial with mostly zero coefficients. Possibly this helps round out the analysis for some. Though it does lean on the ring behavior of the polynomial to prove that its terms lacks ring behavior this is still the same black swan. It's more like biting your butt rather than crapping on your nose.

Really Peter I'd rather see some rhetoric than you ducking the black swan. Be careful where you step. Honestly I do not expect you as an adherent to simply change your position at the prompt of my text here. That is not realistic. The best you can do is to allow your own interpretation to take place. If it is pre-multiplication then so be it, but this is terminology outside of the existing supposedly pristine subject. I think I might have to step back your position into the dodge camp though, but you'd be much farther along than say the 'minor abuse' position and certainly much farther than the 'historical language' position. As for Mostowski I'm afraid he's still rubbing his nose off. We each have to take the math as a burden onto ourselves, for if we do not then we are practicing biblical mathematics. There is no guarantee that the greats who came before us had compiler level integrity. If any have it it is us; not them. Integrity within all human power structures is problematic and it is supposedly mathematicians who get over it. They wrote themselves a free ticket away from it right? Well...
Peter
2020-09-06 12:55:54 UTC
Post by Tim Golden BandTech.com
. I do wish to hear your own interpretation on
1.23 X and whether this operation that is in use by the AA types
has any validity.
What operation are you referring to? The operation of pre
multiplying X by 1.23?
Uh-oh. Peter as far as I know nobody has ever required 'pre
multiplying' within the field of abstract algebra. In fact because of
the algebraic behavior this is a clear sign that you have gone wrong.
If the expression a b c implied that you must pre-multiply a by b
You seem not to know what pre-multiply means. In abc, b is
pre-multiplied by a, not vice versa.
Post by Tim Golden BandTech.com
before you multiply the result by c then we would be in a
non-commutative realm. This is
bassam karzeddin
2020-09-06 10:24:40 UTC
Post by Tim Golden BandTech.com
Post by bassam karzeddin
Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
https://en.wikipedia.org/wiki/Closure_(mathematics)
https://en.wikipedia.org/wiki/Ring_(mathematics)
https://en.wikipedia.org/wiki/Binary_operation
These links all fit together. The ring construction is at the base of the curriculum of abstract algebra. It is treating the operators with extreme care. They are so carefully defined that they don't even require the usage of the terms 'multiply' and 'add' though that is exactly what these operators do in the real numbers.
After the careful construction of the ring from two groups, both operators obeying the closure principle, the polynomial is introduced with an 'X' that is not real, and then those polynomials are simply assigned real coefficients, but this operation has not been defined, for the multiplication of a real by a non-real value X is undefined within their own operator theory that was just so carefully constructed. The term
a1 X
is not a binary operation. It distinctly disobeys the closure requirement. Thus abstract algebra has only managed to rebuild a conundrum that even DesCartes trailed off on in his book Rules For The Direction Of The Mind. The sad part is that the weakness of the construction cannot be challenged. Your own impedance is a fine instance of a subconscious power that doctrine holds over us. We are en mass hypnotized. This obvious blunder within Abstract Algebra is a fine instance. It has nothing to do with polysign. Well, at least this argument holds with or without polysign. I just happen to have seen it through polysign, which brought me to Abstract Algebra. Still it is true that the ultimate behavior of the polynomial that they are after is the modulo effect, which is painfully constructed though the 'quotient' and the 'ideal' which to this day do not make sense to me. I suspect that the cause of the contaminated language is the flaws which they dodge. I have bumped into an online AA book that alludes to it but it has been some time.
To formalize this argument we should really take the acclaimed ring behaved
a0 + a1 X + a2 XX + ...
then assign an = 0 when n not equal to one. This then leaves us the ring behaved
a1 X
which is not ring behaved because it offends the closure principle. Now likewise each term in the polynomial can be attacked and after all they are a sum which in the ring definition requires that they all be in one set together. The entire polynomials fails under their own formalities.
Your own denial of this is fine, for this is also true of anyone else who has bumped into this falsification. Clearly the extent of the problem if I am correct does develop a philosophical statement on the modern human and its culture. The burden of credibility; particularly in a subject such as mathematics; the main burden of the modern student is of mimicry; something which humans excel at. Mimic the complexities of this subject and you can achieve an A grade and a future within academia. Deny the credibility of this subject and you will be shunned out of the system. Even within our own heads this bad dog relation is playing out as we are social animals and so the biblical nature of academia is established not only from without but from within. The developments of academia are more a method of fitting more and more PhDs than it is a pursuit of the truth. Such a branch as abstract algebra with its imperfections goes unchallenged in this arena for good reason. As one PhD steps on another PhDs subject as false the whole system will collapse its extraordinary accumulation. Should it collapse down to truth? I doubt it. I think we are more like monkeys in the tree still trying to figure out fundamental principles. We are only part way and we've landed ourselves in this morass.
We need to maintain these subjects as alive rather than as dead and pickled to perfection. Our past leaders should not be placed up on a mantle as unapproachable. The burden of their work is on us.
So good of you that you raised a very important issue about the broken algebra basically with its fundamental theorem of absolute stupidity
But one should be aware of the imosibility for those vast majorities of common academic mainstreams (not only in mathematics) to realize anything even with hundreds of simple short proofs since they were brainwashed long ago of early ages and well trained to be true stubborn imbeciles even much worse than their preceding generations
One mustn't expect anything good from true imbeciles and real cranks who are globally in plenty and very available everywhere on this planet
The problem was never the truer superior mathematics with that unsociable creatures, but very deep and so many incurable human mental diseases with over mental inherited retardation
If you can write a book about the main problem, then you may be able to help the new uprising generation not to fall at all as easy prey to all those mentally disordered historical figures as (KUNT, GODEL, IMBECILE, COUNTER, ABEL, Galois, Ruffini, Cardano, Euler, Dedikined, Cauchy, ...., etc) With millions of dwarf followers after them...
Again the main problem was never the truer and superior mathematics itself, but truly about human first reaction towards higher knowledge
Good luck
Bassam King Karzeddin
So nice of you to add your two cents on this thread King Bassam. Still you've dodged the content rather thoroughly. I do wish to hear your own interpretation on
1.23 X
and whether this operation that is in use by the AA types has any validity. Given that they've gone to the trouble of carefully defining the operations of sum and product within the ring definition how then can they turn around and throw 'real coefficients' onto an undefined X without getting themselves into this jam? It is ring behaved and it is not ring behaved. Thus the subject carries its own conflicts to all users of this curriculum. It seems so peculiar in that they were so convincing in going about the ring definition in the first place. How dare them by the way snatch the name 'ring' for the formal definition of operators. It's not wise. That could be interesting to learn about its evolution. A ring clearly has rotational aspects and so does the polynomial but one is continuous whereas the other is quite discrete and really until they wrap that polynomial around with their fowl language of 'ideal' and 'quotient' the thing just shoots you off into higher dimension; no ring at all; not a clean crisp tone; not a torus; not even any rotation there. Ring their necks? Like a bird's? This is what they'd like to do to my black swan. So few can even approach the thing. I'll just make another one. They are all over the place really. They just refuse to see them.
*********************************************
Actually, I prefer to deal with much older fundamental and historical common global mistakes to relize the broken algebra in a much easier way than those described as modern set theories of those like (Godel, Hilbert, Dedekind, Cauchy, Galois, ..., etc)

The most three famous historical problems of the Greeks, where nobody seems to ever understand them very correctly up to this date ... nor do they want to understand them for many human reasons that are irrelevant to any fact

Many detailed posts I wrote about them to get the basic idea about what is truly a real number is?

Where the simplest insolvable polynomial is (x^3 = 2), which has no roots at all since it is basically the truer form of doubling the cube problem which is impossible by all the means ... FOR SURE

And I understand that those problems would always creat contradictions in any modern mathematics like those you are well-describing

But I always prefer to deal with the first roots of the problem and never with branched and wrong mathematics

And yes, algebra was good enough until modern mathematicians spoiled it completely with their rotten old refuted beliefs about real numbers (that are strictly described as non-constructible numbers)

In short, every wrong mathematics is springing up from untrue understanding about what is the real number?

Where ALL the published issues in any mathematical sites about serious problems are sourced from this fact (no matter, however smart they are trying to present their problems as if their own actual problems)

All problems in mathematics revolve about the fact of the real number

BKK