praxer
  • praxer
help!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
praxer
  • praxer
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praxer
  • praxer
@SithsAndGiggles
ganeshie8
  • ganeshie8
Hey!

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praxer
  • 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
  • ganeshie8
which notation ?
praxer
  • praxer
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praxer
  • praxer
wait! is it about the constant term being divisible by 2. I get this ?!! But why 2
ganeshie8
  • 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
  • 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
  • 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
  • ganeshie8
computations get simpler in modular arithmetic, observe that middle -2n term has just disappeared in mod2!
ganeshie8
  • ganeshie8
they are picking mod2 because it is simpler
ganeshie8
  • ganeshie8
It doesn't matter which modulus, \(m\), you pick it will be true everywhere.
ganeshie8
  • 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
  • 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
  • ganeshie8
what same technique ? do you have a solution ?
ganeshie8
  • ganeshie8
The actual problem turns out to be much simpler, we don't need fancy modular arithmetic etc...
praxer
  • praxer
okay ! :)
ganeshie8
  • 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
  • ganeshie8
Pull that graph down by \(5\) units and call it \(g(x)\). where does \(g(x)\) cut the x axis ?
praxer
  • praxer
a,b,c,d
ganeshie8
  • 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
  • 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
  • praxer
f(x)=(x−a)(x−b)(x−c)(x−d)h(x)+5 ???
ganeshie8
  • ganeshie8
Yes, rearrange and get : \[f(x)-5 = (x-a)(x-b)(x-c)(x-d)h(x)\]
ganeshie8
  • 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
  • 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
  • praxer
CONTRADICTION AND THERE EXIST NO INTEGER K
ganeshie8
  • ganeshie8
why is it a contradiction ?
ganeshie8
  • ganeshie8
we can have : \(3 = 1*1*1*3\) right ?
praxer
  • praxer
cause primes have only factors
praxer
  • praxer
only factors
ganeshie8
  • ganeshie8
so what
praxer
  • praxer
only 2
ganeshie8
  • ganeshie8
primes have only 2 "distinct" positive factors
ganeshie8
  • ganeshie8
as you saw, we can write \(3\) as product of \(4\) factors : \(1*1*1*3\)
ganeshie8
  • ganeshie8
so thats not really a contradiction
praxer
  • praxer
okay then what will it be called?? Btw I have a doubt. May I ??
ganeshie8
  • ganeshie8
I'm saying that is not a reason for contradiction
ganeshie8
  • ganeshie8
sure, ask
praxer
  • praxer
first the why it was used by induction on n and second the use of inductive assumtion
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praxer
  • 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
  • ganeshie8
what do you know about induction ?
praxer
  • 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
  • ganeshie8
why is it so ? do you have a proof for that statement ?
ganeshie8
  • 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
  • ganeshie8
not now, when you're free, go through that proof of "why/how induction works" and try to understand :)
ganeshie8
  • ganeshie8
Its not hard but you just need to see it once and understand
praxer
  • 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
  • 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
  • ganeshie8
They chose \(n\) to induct on
ganeshie8
  • 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
  • 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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