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anonymous
 one year ago
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
anonymous
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's actually on the next page after you click on the link, but you might wanna read the proof first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it's saying, if a is incongruent b mod c, then c does not divide (ab)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that is lagrange's theorem, one of the most important theorems in group theory

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. But can you please explain the statement above? ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why doesn't p divide the difference of any of the two w_i? I was thinking it's true by definition

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0before explaining that, let me ask you a question from the same proof

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The proof use both induction and contradiction at the same time. I'm still trying to figure the general layout of the proof

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0did you get below part of the proof dw:1441691724959:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0why must \(g(x)\) be congruent to \(0\) for "every" integer \(x\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0are you on windows/mac ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0IS that the induction step btw?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0see if you can install teamviewer http://download.teamviewer.com/download/TeamViewer_Setup_en.exe

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should I choose "basic installation" ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0just click next keeping whatever is default

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0installed. Let me enter your id

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0click on "Meeting" and enter below id m14003342

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can see your screen now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0am I supposed to turn on Microphone setting or something?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am supposed to connect a physical micrphone to my computer?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0degree of g(x) is less than k

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so \(g(x)\) 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
 one year ago
Best ResponseYou've already chosen the best response.0hold on, let me reread the part that you posted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the w's are the roots of f(x) right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(w_1, w_2, \ldots, w_k, w_{k+1}\) satisfy \(f(x)\) \(w_1, w_2, \ldots, w_k\) satisfy \(g(x)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait, how do w1, ..., wk satisfy g(x)??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh 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
 one year ago
Best ResponseYou've already chosen the best response.0ok, but what does g(x) = f(x) mean though? Does it mean g(x) ≡ f(x) (mod p) for all x of w1, ... ,wk?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(f(x)=x^9 + 2x^3 \) \(g(x) = 8x^9 + 9x^3\) then \(f(x) \equiv g(x) \pmod 7\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so you're saying [(x^9 + 2x^3)  (8x^9 + 9x^3) ]/7 is an integer for some x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0[(x^9 + 2x^3)  (8x^9 + 9x^3) ]/7 is an integer for "every" x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me simplify it using wolfram alpha to see. Hold on

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Induction 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
 one year ago
Best ResponseYou've already chosen the best response.0what do yo mean by coefficients of g(x)? you mean the constants in front of the x's?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you're saying g(x) congruent f(x) congruent 0 because the coefficients of g(x) are divisible by p?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0g(x) congruent 0 for all x because the coefficients of g(x) are divisible by p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh oh i see. You're saying g(x) congruent 0 (mod p) because the coefficients of g(x) are divisiple by p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhm... I have no idea :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're asking why all the coeffient of g(x) are divisible by p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait why does p  a_n ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Induction 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 \(p\nmid a_n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0consdier below conditional : If a quadrilateral has four right angles, and if all sides are equal, then it is a square

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0must be a rectangle?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think i'm starting to see what you're trying to get at.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0there 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
 one year ago
Best ResponseYou've already chosen the best response.0the sides must be equal?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0Induction 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
 one year ago
Best ResponseYou've already chosen the best response.0that p divides the leading coefficient. But let me confirm that IF A and B, then C Suppose A and ~C, then ~B right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\((A\land B) \implies C \\\iff\\ (A\land \neg C) \implies \neg B\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what I said is correct? about A,B,C

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so about the assumption: A = p divides leading coefficient B = f(x) has n degree C = f(x) has at most n incongruent solutions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0ok, so since g(x) has k1 (which less than k), but it has k incongruent solutions, therefor P must divide the leading coefficient!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0since \(p\mid a_k\), \(g(x) \equiv 0x^{k}+a_{k1}x^{k1}+\cdots + a_1x+a_0 \\\equiv a_{k1}x^{k1}+\cdots + a_1x+a_0 \pmod{p}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since p  a_k, a_k congrent 0 (mod p) right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm still puzzled at why you replaced a_k with 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why's that? what modulo law was used?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry if I'm slow. Going from integer to polynomial is big jump for me XD
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