Re: Understanding the quotient ring nomenclature

On Jun 18, 11:50 am, Brian Chandler <imaginator...@xxxxxxxxxxxxx>
wrote:
Tim BandTech.com wrote:
On Jun 18, 1:14 am, Brian Chandler <imaginator...@xxxxxxxxxxxxx>
wrote:
Tim BandTech.com wrote:

Aside: Is English your first language?
Yes.

Hmm. Then please stop saying things like "definitions are offended",
because they are meaningless. (Not mistranslations as I thought they
might have been.)

I will stand by this language and reinforce the instance below here.
Suppose a to be an element in the ring A and b to be an element in the
ring B. Now I provide two new values
c = a b
and
d = a + b
These two operators, while they clearly purvey product and sum, have
not been provided. Therefore a claim to have provided these operators
via this construction has offended the ring definition. Definitions
can be offended.

Please we are talking past each other so much and here is a point that
you've rejected. Let's see if this can be resolved. Do you accept this
usage of my language or do you reject it? If you reject it please
point out the exact location of the conflict. I sincerely see none and
so I have no idea how so much rejection can take place on such a
simple point.


You still want your polynomial to be elemental.

I think I have cracked it. By "elemental" you mean this:

[We] want the polynomial to be an element of the set of things we are
talking about.

Elements within the ring terminology are members of the domain under
consideration.

Uh oh. And if the "domain" (bad word) is the "domain" of polynomials,
are the elements not polynomials?

No. The polynomial is expressed on the ring A. The elements a(n) and x
are in the domain A. This A is completely abstract until a more
specific ring instance is applied. The ring principles are highly
universal. This is the only way that such an overarching
generalization can be made.
A does not inherently include the reals nor any other specific ring
instance. It is A as in Abstract. Upon specifying a nonabstract domain
all of the construction members take on that value. For instance if I
were to construct
"Polynomials on the reals"
then both a(n) and x will be real valued. There seems to be a claim
that it is possible to construct
"Polynomials with real coeffients on A[X]"
where a(n) are real yet x will still be in A. This is a misnomer. The
maintenance of two ring types will require a plethora of product and
sum types, especially if those products and sums are to remain within
the ring nomenclature. For instance each time a real a(n) is
multiplied by x^n this product offends the ring definitions of the two
rings specified since neither can accomodate it. Each of the
established operators has a specific domain which it operates in.


<skip more confusion: your arguments are not wrong enough to be
refuted>

Yes, of course we do, otherwise we wouldn't be able to talk about
them. Do you think certain patterns should be ineligible for
investigation? Consider the arithmetic of (indefinitely long, but
finite) bit strings under the operations of XOR and MUL(tiply). Do you
think these form a mathematical structure? Do you think investigation
of this mathematical structure is valid (should be allowed)? Do you
see why I choose this particular example?

If you provide accurate sum and product operations on the "bit string"
that are consistent with the ring terminology then they would form a
ring.

Oh, really? Well whaddya think "XOR" and "MUL(tiply)" are, if not well-
defined operations on bit strings? Write x^y for x XOR y, and p * q
for p MUL q, and check against the ring axioms, which you can read on
Wikipedia. Here's a start:

0 = bit string ...0000000
1 = bit string ...0000001
Check: ^ is commutative: p^q = q^p (well-known programming fact)
Check: * is associative: (a * b) * c = a * (b * c)
etc.
Check distributive law: (a ^ b) * c = (a^c) * (b^c) ... bit harder so
just choose values
a=10111011 b=101100 c=1011 (or simpler!) and calculate test case by
hand

Now for the big question: this is (take it from me, or prove it to
yourself, as you wish) a ring. The "things" to which we are applying
the ring operations are the bit strings. Question: the ring is a set,
and the elements of this set are:

(a) the bit strings
(b) 0 and 1
(c) the natural numbers
(d) polynomials over GF2
(e) hydrogen and stupidity

Take your time. Pick one (1) answer only to be the Correct Answer.

a) the bit strings. When selecting one member of the set we will
always get a bit string.

"In mathematics, an element or member of a set is any one of the
distinct objects that make up that set."
- http://en.wikipedia.org/wiki/Element_(mathematics)


As I said to Leland I'm getting tired of this topic. As far as I can
tell you all have serious mimicry issues.

"Mimicry"? You mean you think we all copy each other? Or at least that
we all have secret meetings to make sure we all say things you are
sure are wrong? Hmm.

Brian Chandler

.

Relevant Pages

  • Re: Understanding the quotient ring nomenclature
    ... are the elements not polynomials? ... The polynomial is expressed on the ring A. ... the ring operations are the bit strings. ... provided you note that this is not the naturals with the "usual" ...
    (sci.math)
  • Re: Understanding the quotient ring nomenclature
    ... Hmm. ... Elements within the ring terminology are members of the domain under ... are the elements not polynomials? ... finite) bit strings under the operations of XOR and MUL. ...
    (sci.math)
  • Re: Tilted fore and aft neck humbucker
    ... admit I don't know enough to reference the sound. ... The neck pup is quite ... reversing the ring so the screw side is the one closer to the strings. ...
    (rec.music.makers.guitar.jazz)
  • Re: Technical problem
    ... open strings so that they ring for their correct time value. ... How do you damp open strings? ... I'm letting those open strings ring out where there is a rest all the time. ...
    (rec.music.classical.guitar)
  • Re: Understanding the quotient ring nomenclature
    ... When it is named "the ring A" then we see that this domain has been ... But the 4.3 here is not in the reals. ... a and x and so the polynomial form can be built up yet what domain it ... polynomials isn't even as large as you can go. ...
    (sci.math)