Peter

2020-10-17 16:18:27 UTC

We are not in algebra per se. We are in the curriculum of

abstract algebra. I have no idea how you could come to

disagree with this language particularly when you find no

actual fault with the language. I have linked to numerous

online references here; not as numerous as Lalo though. It is

almost as if you want to deny that the real numbers do fit

the ring definition.

Consider the following. Letabstract algebra. I have no idea how you could come to

disagree with this language particularly when you find no

actual fault with the language. I have linked to numerous

online references here; not as numerous as Lalo though. It is

almost as if you want to deny that the real numbers do fit

the ring definition.

a_0, a_1, a_2, ... b_0, b_1, b_2, ...

be (not necessarily distinct) sequences of real numbers subject

to the requirements that only finitely many members are

non-zero.

Define the _sum_ of them to be

(a_1 + b_1), (a_2 + b_2, (a_3 + b_3), ...

And the _product_ of them to be

sum_{k=0}^infty sum_{n=0}^k a_n b_{k-n}.

I claim that the set of such sequences under sum and addition

as defined is a ring. Do you agree?

how the polynomial products and sums work out to another polynomial;

but I say no because those sums which build the polynomial are built

via the ring operator, and the products in the terms of the

polynomial are built with that ring product, and they do not fit the

ring definition. This is not a yes/no situation. The ring of

polynomials with real coefficients is not ring behaved.

That ring is, but for notation, the same as the ring of

polynomials of one indeterminate over the real numbers denote

R[X] where R is the real numbers. The element

a_0 + a_1 X + a_2 X^2 +...

of R[X] corresponds to

a_0, a_1, a_2, ...

Do you understand that?

polynomials of one indeterminate over the real numbers denote

R[X] where R is the real numbers. The element

a_0 + a_1 X + a_2 X^2 +...

of R[X] corresponds to

a_0, a_1, a_2, ...

Do you understand that?

the outer level the ring behavior is upheld but it is this method

of dismantling the polynomial which yields the conflict. Each

polynomial is a sum of terms and those terms clearly must obey the

ring definition too. Each term contains products and those products

must obey the ring definition. Otherwise there is a conflict. The

conflict is best exposed on the polynomial with real coefficients

and so to specify an instance

a1 X and witness that this product is not ring behaved;

I've asked you twice before what "ring behaved" means. Would youlike to say now? If you wish, answer with respect to a1 X rather

than generally (though generally would be best).

. But Peter if all

that we do is cover the same ground again that will not be

meaningful. You have had a chance. If all that you mean to do is

repeat the same defense then why bother? This is a big deal, so it is

worth trying again. But it is relevant to take a variation from the

prior course. In fact I am happy to expound for I think I did bump

into the reason that I use this terminology. The Ring definition is

not a concrete thing. Whereas in the definition of real numbers we

can actually name a real number 1.23 and I can comfortably say 'is

real' the ring definition really only defines requirements for

operators, and those operators perform actions. They have behavior.

They take two elements and yield one element. But furthermore there

is nothing concrete about them. Even the real value I would say is

ring behaved. To say that the real number 'is a ring' has poor

grammatical value.

Bugger its 'grammatical value', it's downright false to say that thethat we do is cover the same ground again that will not be

meaningful. You have had a chance. If all that you mean to do is

repeat the same defense then why bother? This is a big deal, so it is

worth trying again. But it is relevant to take a variation from the

prior course. In fact I am happy to expound for I think I did bump

into the reason that I use this terminology. The Ring definition is

not a concrete thing. Whereas in the definition of real numbers we

can actually name a real number 1.23 and I can comfortably say 'is

real' the ring definition really only defines requirements for

operators, and those operators perform actions. They have behavior.

They take two elements and yield one element. But furthermore there

is nothing concrete about them. Even the real value I would say is

ring behaved. To say that the real number 'is a ring' has poor

grammatical value.

real number 'is a ring'. A ring is three things (S,+,*) where S is a

set and + and * are both binary operations defined on that set. Those

operations are required to satisfy certain conditions (such as

associativity) that you can look up. An example is S being the set of

real numbers and + and * being the usual arithmetical operations. To

wrote of one element of S as being a ring is silly

Sorry

To say that the real number is ring behaved is

much more natural. Both I believe mean to obey the rules of the ring

definition

The definition of a ring refers to a triple, above written as (S,+,*).much more natural. Both I believe mean to obey the rules of the ring

definition

It is silly to write of an element of S as obeying the rules of the ring

definition. Here's an analogy: London is defined to be the capital of

the UK (yes, I know there are other Londons but for my purposes it's

London as defined by me that I'm writing about.) Does John Smith who

lives on Eaton Square (a place in London) obey the rules of the

definition of London? No. Is that a problem? No.

You mention 1.23 X. In terms of the 'sequences of real numbers subject

to the requirements that only finitely many members are non-zero'

definition of polynomial, that is 0, 1.23, 0, 0, .... It is also the

product of 1.23, 0, 0, 0, ... and 0, 1, 0, 0, ... (Which is no more

remarkable than six being both 6 and the product of 2 and 3.)

, but the definition does not describe a set of elements;

it describes what a set of elements can do. It defines their

behavior.

I don't know what 'resolve' means in this context. Nor am I sure I knowit describes what a set of elements can do. It defines their

behavior.

that this term is in sum with other terms that will not obey the

ring definition; and of course we need just one such polynomial

to fail in order to have a falsification; one black swan proves

that not all swans are white and so we can go a bit further with

our specificity of that black swan 1.23 X which does not obey the

ring definition. This is as concrete an instance as I can muster

without breaking any of AA's rules. So there you have it: a ring

behaved polynomial that is not ring behaved. It is nice of you to

make another attempt here. For a while you were the only one who

could face the black swan. Others here have devolved into

statements like That product is not a product The sum is not a

sum Whereas the X is nearly nothing at all Were operators only

just defined in this subject Peter? The ring definition comes

some pages before the introduction of the polynomial in AA right?

Should a subject which cares to define operators use them more

carefully? Do elements contain operators? If they do then those

operators ought to resolve right? This is exactly what the AA

ring definition; and of course we need just one such polynomial

to fail in order to have a falsification; one black swan proves

that not all swans are white and so we can go a bit further with

our specificity of that black swan 1.23 X which does not obey the

ring definition. This is as concrete an instance as I can muster

without breaking any of AA's rules. So there you have it: a ring

behaved polynomial that is not ring behaved. It is nice of you to

make another attempt here. For a while you were the only one who

could face the black swan. Others here have devolved into

statements like That product is not a product The sum is not a

sum Whereas the X is nearly nothing at all Were operators only

just defined in this subject Peter? The ring definition comes

some pages before the introduction of the polynomial in AA right?

Should a subject which cares to define operators use them more

carefully? Do elements contain operators? If they do then those

operators ought to resolve right? This is exactly what the AA

what you mean by elements containing operators. Are you taking about

the way something might be written? E.g., here's an element of the ring

of integers: 6, and it equals 2*3, and '2*3' contains the symbol '*'. To

say, instead, '6 contains *' would be potty.

polynomial does not do.

only just defined in this subject Peter? The ring definition comes

some pages before the introduction of the polynomial in AA right?

Should a subject which cares to define operators use them more

carefully? Do elements contain operators? If they do then those

operators ought to resolve right? This is exactly what the AA

polynomial does not do. Now you are dodging the black swan; something

that you used to be unafraid of. I have already conceded to you that

products of polynomials yield polynomials and sums of polynomials

yield polynomials, but my argument is on the contents of those

polynomials. I dismantle the polynomial to expose the conflict.

--

When, once, reference was made to a statesman almost universally

recognized as one of the villains of this century, in order to

induce him to a negative judgment, he replied: "My situation is

so different from his, that it is not for me to pass judgment".

Ernst Specker on Paul Bernays

When, once, reference was made to a statesman almost universally

recognized as one of the villains of this century, in order to

induce him to a negative judgment, he replied: "My situation is

so different from his, that it is not for me to pass judgment".

Ernst Specker on Paul Bernays