help!

- praxer

help!

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- katieb

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- praxer

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- praxer

@SithsAndGiggles

- ganeshie8

Hey!

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## More answers

- praxer

why 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.

- ganeshie8

which notation ?

- praxer

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- praxer

wait! is it about the constant term being divisible by 2. I get this ?!! But why 2

- ganeshie8

2 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

consider a quick example :
\[P(x) = x^2-2x +1 = 0 \]
suppose \(n\) is an integer root, then we have :
\[P(n) = n^2-2n+1 = 0\]
we are allowed to take mod both sides because we're dealing with integers (n and all coefficients are integers)

- ganeshie8

taking mod 2 both sides you get
\[ n^2-2n+1 \equiv 0 \pmod{2}\]
which is same as
\[ n^2+1 \equiv 0 \pmod{2}\]

- ganeshie8

computations get simpler in modular arithmetic,
observe that middle -2n term has just disappeared in mod2!

- ganeshie8

they are picking mod2 because it is simpler

- ganeshie8

It doesn't matter which modulus, \(m\), you pick
it will be true everywhere.

- ganeshie8

so 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

yes! now for the question. Can the same technique be used. Like f(x)=(x-a)(x-b)(x-c)(x-d)r(x)+5 and.......?????

- ganeshie8

what same technique ? do you have a solution ?

- ganeshie8

The actual problem turns out to be much simpler, we don't need fancy modular arithmetic etc...

- praxer

okay ! :)

- ganeshie8

Lets 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

Pull that graph down by \(5\) units and call it \(g(x)\).
where does \(g(x)\) cut the x axis ?

- praxer

a,b,c,d

- ganeshie8

Yes, 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

so \(g(x)\) will be of form :
\[g(x) = (x-a)(x-b)(x-c)(x-d)h(x)\]
where \(h(x)\) is some other polynomial with integer coeffcients

- praxer

f(x)=(x−a)(x−b)(x−c)(x−d)h(x)+5 ???

- ganeshie8

Yes, rearrange and get :
\[f(x)-5 = (x-a)(x-b)(x-c)(x-d)h(x)\]

- ganeshie8

suppose there exists an integer \(k\) such that \(f(k) = 8\),
plugin \(x = k\) in above equation :
\[f(k)-5 = (k-a)(k-b)(k-c)(k-d)h(k)\\~\\8-5= (k-a)(k-b)(k-c)(k-d)h(k)\\~\\ 3=(k-a)(k-b)(k-c)(k-d)h(k)\\\]

- ganeshie8

Notice 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

CONTRADICTION AND
THERE EXIST NO INTEGER K

- ganeshie8

why is it a contradiction ?

- ganeshie8

we can have :
\(3 = 1*1*1*3\)
right ?

- praxer

cause primes have only factors

- praxer

only factors

- ganeshie8

so what

- praxer

only 2

- ganeshie8

primes have only 2 "distinct" positive factors

- ganeshie8

as you saw, we can write \(3\) as product of \(4\) factors : \(1*1*1*3\)

- ganeshie8

so thats not really a contradiction

- praxer

okay then what will it be called?? Btw I have a doubt. May I ??

- ganeshie8

I'm saying that is not a reason for contradiction

- ganeshie8

sure, ask

- praxer

first the why it was used by induction on n and second the use of inductive assumtion

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- praxer

1. 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

what do you know about induction ?

- praxer

well 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

why is it so ? do you have a proof for that statement ?

- ganeshie8

Prove 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

not now, when you're free, go through that proof of "why/how induction works" and try to understand :)

- ganeshie8

Its not hard but you just need to see it once and understand

- praxer

okay! 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

Notice 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

They chose \(n\) to induct on

- ganeshie8

but 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

**
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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