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
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
2020-09-01 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
2020-09-01 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.