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praxer
 one year ago
help!
praxer
 one year ago
help!

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praxer
 one year ago
Best ResponseYou've already chosen the best response.1why to check integer roots of a polynomial modulo is used. I am bit confused with such notion. Anyone please explain the use of modulo precisely in proving that polynomial has no integer roots.

praxer
 one year ago
Best ResponseYou've already chosen the best response.1wait! is it about the constant term being divisible by 2. I get this ?!! But why 2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.02 has nothing to do with the problem since 1) the coefficients are integers and 2) you're looking for integer roots, you are allowed to use modular arithmetic here

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0consider a quick example : \[P(x) = x^22x +1 = 0 \] suppose \(n\) is an integer root, then we have : \[P(n) = n^22n+1 = 0\] we are allowed to take mod both sides because we're dealing with integers (n and all coefficients are integers)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0taking mod 2 both sides you get \[ n^22n+1 \equiv 0 \pmod{2}\] which is same as \[ n^2+1 \equiv 0 \pmod{2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0computations get simpler in modular arithmetic, observe that middle 2n term has just disappeared in mod2!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0they are picking mod2 because it is simpler

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0It doesn't matter which modulus, \(m\), you pick it will be true everywhere.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so is below statement clear ? If \(n\) is an integer root of a polynomial \(P(x)\) with integer coeffcients, then \(P(n) \equiv 0 \pmod{m}\).

praxer
 one year ago
Best ResponseYou've already chosen the best response.1yes! now for the question. Can the same technique be used. Like f(x)=(xa)(xb)(xc)(xd)r(x)+5 and.......?????

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what same technique ? do you have a solution ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0The actual problem turns out to be much simpler, we don't need fancy modular arithmetic etc...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Lets first look at a rough sketch of \(f(x)\) since \(f(a)=f(b)=f(c)=f(d)=5\), the graph looks something like : dw:1438956400888:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Pull that graph down by \(5\) units and call it \(g(x)\). where does \(g(x)\) cut the x axis ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, so it is easy to see that \(a,b,c,d\) are the roots of the polynomial \(g(x)\) (pulled down by \(5\) units version of \(f(x)\))

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so \(g(x)\) will be of form : \[g(x) = (xa)(xb)(xc)(xd)h(x)\] where \(h(x)\) is some other polynomial with integer coeffcients

praxer
 one year ago
Best ResponseYou've already chosen the best response.1f(x)=(x−a)(x−b)(x−c)(x−d)h(x)+5 ???

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Yes, rearrange and get : \[f(x)5 = (xa)(xb)(xc)(xd)h(x)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0suppose there exists an integer \(k\) such that \(f(k) = 8\), plugin \(x = k\) in above equation : \[f(k)5 = (ka)(kb)(kc)(kd)h(k)\\~\\85= (ka)(kb)(kc)(kd)h(k)\\~\\ 3=(ka)(kb)(kc)(kd)h(k)\\\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Notice that the right hand side is a product of \(4\) integers but the left hand side is a prime number what can you say ?

praxer
 one year ago
Best ResponseYou've already chosen the best response.1CONTRADICTION AND THERE EXIST NO INTEGER K

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0why is it a contradiction ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0we can have : \(3 = 1*1*1*3\) right ?

praxer
 one year ago
Best ResponseYou've already chosen the best response.1cause primes have only factors

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0primes have only 2 "distinct" positive factors

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0as you saw, we can write \(3\) as product of \(4\) factors : \(1*1*1*3\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so thats not really a contradiction

praxer
 one year ago
Best ResponseYou've already chosen the best response.1okay then what will it be called?? Btw I have a doubt. May I ??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I'm saying that is not a reason for contradiction

praxer
 one year ago
Best ResponseYou've already chosen the best response.1first the why it was used by induction on n and second the use of inductive assumtion

praxer
 one year ago
Best ResponseYou've already chosen the best response.11. Assume is fixed and use induction on 2.hence by the inductive assumption In many proof writing, inductive is being used I know the theory but induction andsolve the +2 question. But many times it is confusing

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0what do you know about induction ?

praxer
 one year ago
Best ResponseYou've already chosen the best response.1well frankly speaking I know that it is like true for n=1 and then if the given condition is true for n=k+1 whenever it is true for n=k then it is true for all natural numbers.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0why is it so ? do you have a proof for that statement ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Prove this : If : 1) The statement is true for \(n=1\) 2) The statement is true for \(n=k+1\), whenever it is true for \(n=k\) Then : the statement is true for all integers greater than \(1\) too.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0not now, when you're free, go through that proof of "why/how induction works" and try to understand :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Its not hard but you just need to see it once and understand

praxer
 one year ago
Best ResponseYou've already chosen the best response.1okay! one thing what do you understand from the page like using the term induction on n. I want to have a normal idea of what others feel when they read that line. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Notice that there are several variables in that problem : \(k,n\) etc you need to clear on what variable you're picking to "induct on".

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0They chose \(n\) to induct on

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0but you may pick \(k\) to induct on too : 1) show that the statement is true for \(k=1\) 2) show that the statement is true for \(k=t+1\), whenever it is true for \(k=t\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0** Notice that there are several variables in that problem : \(k,n\) etc you need to \(\color{red}{\text{be}}\) clear on what variable you're picking to "induct on".
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