anonymous
  • anonymous
Explain this proof of polynomial congruences by George Andrews. I don't get where it says: "However p (not divide symbol) aₒ, and since the w_i are mutually incongruent, the difference between any two w_i is not divisible by p." How so? https://books.google.com/books?id=eVwvvwZeBf4C&lpg=PA58&pg=PA72#v=onepage&q&f=false
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
It's actually on the next page after you click on the link, but you might wanna read the proof first
anonymous
  • anonymous
I think it's saying, if a is incongruent b mod c, then c does not divide (a-b)
ganeshie8
  • ganeshie8
that is lagrange's theorem, one of the most important theorems in group theory

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anonymous
  • anonymous
oh ok. But can you please explain the statement above? ^^
anonymous
  • anonymous
why doesn't p divide the difference of any of the two w_i? I was thinking it's true by definition
ganeshie8
  • ganeshie8
before explaining that, let me ask you a question from the same proof
anonymous
  • anonymous
ok
anonymous
  • anonymous
The proof use both induction and contradiction at the same time. I'm still trying to figure the general layout of the proof
ganeshie8
  • ganeshie8
did you get below part of the proof |dw:1441691724959:dw|
ganeshie8
  • ganeshie8
why must \(g(x)\) be congruent to \(0\) for "every" integer \(x\) ?
anonymous
  • anonymous
I think I did. But let me explain to see if I really did. But book first proved base cases for n = 0 and n = 1. Now it assumed that f(x) is a polynomial with k degree but with (k+1) mutually incongruent solutions
ganeshie8
  • ganeshie8
are you on windows/mac ?
anonymous
  • anonymous
IS that the induction step btw?
anonymous
  • anonymous
i'm on windows 7
anonymous
  • anonymous
inductive*
ganeshie8
  • ganeshie8
see if you can install teamviewer http://download.teamviewer.com/download/TeamViewer_Setup_en.exe
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
should I choose "basic installation" ?
ganeshie8
  • ganeshie8
just click next keeping whatever is default
anonymous
  • anonymous
installed. Let me enter your id
ganeshie8
  • ganeshie8
click on "Meeting" and enter below id m14-003-342
anonymous
  • anonymous
I think i'm in
anonymous
  • anonymous
I can hear you
anonymous
  • anonymous
Control panel?
anonymous
  • anonymous
I can see your screen now
anonymous
  • anonymous
am I supposed to turn on Microphone setting or something?
anonymous
  • anonymous
uh... how so?
anonymous
  • anonymous
I am supposed to connect a physical micrphone to my computer?
anonymous
  • anonymous
microphone*
ganeshie8
  • ganeshie8
degree of g(x) is less than k
ganeshie8
  • ganeshie8
so \(g(x)\) satisfies induction assumption but \(g(x)\) has \(k\) roots so it must be the case that ALL coefficients of \(g(x)\) are divisible by \(p\)
anonymous
  • anonymous
hold on, let me re-read the part that you posted
anonymous
  • anonymous
the w's are the roots of f(x) right?
ganeshie8
  • ganeshie8
\(w_1, w_2, \ldots, w_k, w_{k+1}\) satisfy \(f(x)\) \(w_1, w_2, \ldots, w_k\) satisfy \(g(x)\)
anonymous
  • anonymous
wait, how do w1, ..., wk satisfy g(x)??
anonymous
  • anonymous
right
anonymous
  • anonymous
oh ok, if you plug any value of w1, ..., wk, in for x, we get: g(x) = f(x), for all x of w1,...wk
anonymous
  • anonymous
right
anonymous
  • anonymous
ok, but what does g(x) = f(x) mean though? Does it mean g(x) ≡ f(x) (mod p) for all x of w1, ... ,wk?
anonymous
  • anonymous
oh ok.
ganeshie8
  • ganeshie8
\(f(x)=x^9 + 2x^3 \) \(g(x) = 8x^9 + 9x^3\) then \(f(x) \equiv g(x) \pmod 7\)
anonymous
  • anonymous
ok, so you're saying [(x^9 + 2x^3) - (8x^9 + 9x^3) ]/7 is an integer for some x
anonymous
  • anonymous
for all x?
ganeshie8
  • ganeshie8
[(x^9 + 2x^3) - (8x^9 + 9x^3) ]/7 is an integer for "every" x
anonymous
  • anonymous
let me simplify it using wolfram alpha to see. Hold on
anonymous
  • anonymous
oh, ok ^^
anonymous
  • anonymous
by assumption?
ganeshie8
  • ganeshie8
Induction assumption : every polynomial of degree \(n\) less than \(k\) has at most \(n\) incongruent solutions mod prime \(p\) \(g(x)\) satisfies induction assumption since degree is less than \(k\) also \(g(x)\) has \(k\) incongruent solutoins so it must be the case that ALL coefficients of \(g(x)\) are divisible by \(p\), why ?
anonymous
  • anonymous
what do yo mean by coefficients of g(x)? you mean the constants in front of the x's?
anonymous
  • anonymous
so you're saying g(x) congruent f(x) congruent 0 because the coefficients of g(x) are divisible by p?
ganeshie8
  • ganeshie8
g(x) congruent 0 for all x because the coefficients of g(x) are divisible by p
anonymous
  • anonymous
oh oh i see. You're saying g(x) congruent 0 (mod p) because the coefficients of g(x) are divisiple by p
anonymous
  • anonymous
uhm... I have no idea :D
anonymous
  • anonymous
you're asking why all the coeffient of g(x) are divisible by p
anonymous
  • anonymous
right
anonymous
  • anonymous
wait why does p | a_n ?
ganeshie8
  • ganeshie8
Induction assumption : If the degree of polynomial is \(n\) and if it is less than \(k\), then it has at most \(n\) incongruent solutions mod prime \(p\) provided \(p\nmid a_n\)
anonymous
  • anonymous
rigt
anonymous
  • anonymous
contrapostive?
ganeshie8
  • ganeshie8
consdier below conditional : If a quadrilateral has four right angles, and if all sides are equal, then it is a square
anonymous
  • anonymous
must be a rectangle?
anonymous
  • anonymous
oh
anonymous
  • anonymous
I think i'm starting to see what you're trying to get at.
ganeshie8
  • ganeshie8
there is some quadrilateral that you know that has four right angles and you also happen to know that it is a square what can you say about its sides
anonymous
  • anonymous
the sides must be equal?
ganeshie8
  • ganeshie8
you know that the degree of g(x) is less than k and you also happen to know that g(x) has "k" incongruent solutions what can you say about its leading coefficient ?
ganeshie8
  • ganeshie8
Induction assumption : If the degree of polynomial is \(n\) and if it is less than \(k\), then it has at most \(n\) incongruent solutions mod prime \(p\) provided \(p\nmid a_n\)
anonymous
  • anonymous
that p divides the leading coefficient. But let me confirm that IF A and B, then C Suppose A and ~C, then ~B right?
ganeshie8
  • ganeshie8
\((A\land B) \implies C \\\iff\\ (A\land \neg C) \implies \neg B\) ?
anonymous
  • anonymous
so what I said is correct? about A,B,C
anonymous
  • anonymous
so about the assumption: A = p divides leading coefficient B = f(x) has n degree C = f(x) has at most n incongruent solutions
anonymous
  • anonymous
p does not*
ganeshie8
  • ganeshie8
so about the assumption: A = p does not divide the leading coefficient B = g(x) has degree n which is less than k C = g(x) has at most n incongruent solutions
anonymous
  • anonymous
right g(x)
anonymous
  • anonymous
ok, so since g(x) has k-1 (which less than k), but it has k incongruent solutions, therefor P must divide the leading coefficient!
anonymous
  • anonymous
sorry, k-1 degree*
anonymous
  • anonymous
i don't know :D
ganeshie8
  • ganeshie8
since \(p\mid a_k\), \(g(x) \equiv 0x^{k}+a_{k-1}x^{k-1}+\cdots + a_1x+a_0 \\\equiv a_{k-1}x^{k-1}+\cdots + a_1x+a_0 \pmod{p}\)
anonymous
  • anonymous
since p | a_k, a_k congrent 0 (mod p) right
anonymous
  • anonymous
i'm still puzzled at why you replaced a_k with 0
anonymous
  • anonymous
why's that? what modulo law was used?
anonymous
  • anonymous
Sorry if I'm slow. Going from integer to polynomial is big jump for me XD
anonymous
  • anonymous
@ganeshie8

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