On Tuesday, September 1, 2020 at 1:42:32 PM UTC-4, Mike Terry wrote:
> On 01/09/2020 12:29, Tim Golden BandTech.com wrote:
> > On Monday, August 31, 2020 at 6:09:53 PM UTC-4, Mike Terry wrote:
> >> On 31/08/2020 16:40, Tim Golden BandTech.com wrote:
> >>> On Monday, August 31, 2020 at 11:10:23 AM UTC-4, Mike Terry wrote:
> >>>> [snip rest of ridiculous rant]
> >>>> X + ...
> > =
> >>>> 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
>
> > This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
> OK, so you are saying that you mean the product of a1 and X, so clearly
> the product operation involved is a binary operation. [It involves two
> operands]
>
> There are now two possibilities: you consider a1 to be a polynomial, in
> which case we have regular polynomial multiplication, which is properly
> defined, or you consider a1 to be a member of the underlying ring.
>
> It seems your issue is with the second approach. Lets focus on
> polynomials over R, which is what you discuss below.
>
> I'll assume you're ok with the definition of a polynomial over R. There
> are several equivalent approaches to how these are defined, and I could
> expand on this, but they all lead the same way, defining addition and
> multiplication /of polynomials/.
>
> But now we have something else: a binary operation taking a real number
> and a real polynomial. Of course such an operation needs to be properly
> defined to have a meaning, as it's not covered by the definition of
> polynomial multiplication. Let's write the op as a binary function sm,
> so sm: (R x R[X]) ---> R[x]. [R is the set of real numbers, and R[x]
> the set of polynomials over R.] I choose the notation "sm" for "scalar
> multiplication".
>
> And here is the definition for sm: if a is in R and p is a polynomial
> over R, represented as
> p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
> then
> sm(a,p) = (a p_0) + (a p_1) x + (a p_2) x^2 + ... + (a p_n) x^n
>
> There is an important point here. When I wrote p = p_0 + p_1 x + p_2
> x^2 + ... + p_n x^n above, I am not writing a long sequence of sums and
> products on the right hand side (rhs) of the equals! If I were, you
> could correctly claim the definition is circular!
I've got to quote you here:
" When I wrote
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
above, I am not writing a long sequence of sums and products " - Mike Terry
Mike this is extremely weak. If you did not mean to have a sum of products then why would you write a sum of products?
Why would the operators work out as a sum of products? Why would you make use of a 'polynomial with real coefficients'? You have falsified the subject here by denying that the binary operators work within the polynomial, and yet you disown this detail for your own convenience. This sort of wheedling does expose not just minor abuse. Of course this current abuse that I speak of is you attempting some sort of fixup on a failed construction. Obviously for you to go this far suggests that in secret you have gazed upon the black swan. That you would go this far to save the thing it falsifies is a fine indication. I certainly do not claim any circular definition. The very name 'polynomial' means many terms. It existed prior to abstract algebra and possibly because of this it gets less scrutiny by the beloved suppliers of book fees. The ring sum is so strongly implied that for you to deny that the '+' signs in the construction are addition (the ring defined operator) is a sure loser. I just love how others pop in here to make such corrections. By disowning the polynomial you feel that you have corrected the subject... perfect status quo position. Thanks Mike.
Circular definitions are actually far less offensive than constructions which falsify their own axioms. All of your argumentation does not confront the black swan
1.23 X
This is why I've avoided it till now. When someone provides a falsification of a subject the falsification has to be wrong in order for you to have refuted it. This means that you have to work in terms of this expression, which you have not done here. All that you have done, and all that others have done, is provide their own interpretation on the standing subject. It would seem as if (taking your statement above here seriously) that what you are attempting is a claim that the polynomial may not be dismantled. This is a false claim, and your own attempt to deny that the sums are actual ring sums is quite a position for you to land in... all the while admitting 'minor abuse' of the propped system. Again, you have supported my argument by dodging it and further you have landed yourself in a mess of goose shit. You've slipped in it and it is all over your clothing and your face here.
I'd love a link to a text which denies that the polynomial expression in use in abstract algebra is not actually a sum of terms. Thanks Mike. You've really demonstrated what a hack abstract algebra actually is. The quotient and ideal use similar language as you are using here. You are so brilliant that maybe the goose shit will burn off.
May the black swan live on.
>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
> your studies, but your underlying response to this (your attitude 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.
> >
> >
> >> [i.e. whose X coefficient is 1, and all others zero]
> >> c) something else?
> >> [e.g. maybe X represents multiplication or something!]
> >>
> >> Mike.