I should serve her until Sunday :) :) :)

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- Oct 31st 2006, 11:07 PMsbsiteHelp with Assignment
I should serve her until Sunday :) :) :)

- Nov 1st 2006, 03:56 AMCaptainBlack
Problem 1 is fairly easy. Suppose there in a non-integral rational solution

X=a/b where b!=1, and a and b coprime.

Then multiply the equation through by b^(n-1). The first term is then not

an integer, and all the other terms are integers, and the sum is zeros, which

is impossible as the sum cannot be an integer.

Therefore if there is a rational solution it is integral.

RonL - Nov 1st 2006, 07:52 AMThePerfectHacker
Problem 2 is also easy.

The leading coeffienct is 1 and the last coefficent is 1. By the rational roots test a solution is either 1 or -1.

If the solution is 1 then:

$\displaystyle 1^n+p_{n-1}1^{n-1}+...+p_11^n+1=2+p_1+...+p_{n-1}=0$

But that cannot be.

If the solution is -1 then:

$\displaystyle 1+p_1(-1)+p_2(-1)^2+...+p_{n-1}(-1)^{n-1}=0$

Which is it not.

Thus there are no solutions. - Nov 2nd 2006, 12:51 AMsbsiteLittle request???
- Nov 2nd 2006, 01:36 AMsbsite
- Nov 2nd 2006, 05:11 AMThePerfectHacker
Talk normally. And stop with the smiley abuse.

-=USER WARNED=-

~~~

I got that because 1 or -1 are solutions.

If I substitute -1 into that polynomial

Then I get (if you look at the powers only):

$\displaystyle (-1)^n+(-1)^{n-1}+...+(-1)^2+(-1)^1$

That is alternating,...

$\displaystyle 1,-1,1,-1,...$ - Nov 2nd 2006, 05:22 AMCaptainBlack
Suppose $\displaystyle \exists\ a,b \in \mathbb{Z}$ such that $\displaystyle gcd(a,b)=1$, and that $\displaystyle X=a/b$ is a root of:

$\displaystyle

X^n + p_{n-1}X^{n-1}+\ ...\ +p_1X +1=0

$

Then:

$\displaystyle

b^{n-1}X^n =-[p_{n-1}b^{n-1}X^{n-1}+\ ...\ +p_1b^{n-1}X +b^{n-1}]

$

Now:

$\displaystyle

-[p_{n-1}b^{n-1}X^{n-1}+\ ...\ +p_1b^{n-1}X +b^{n-1}]\in \mathbb{Z}

$

but:

$\displaystyle

b^{n-1}X^n \notin \mathbb{Z}

$

a contradiction.

Therefor no such $\displaystyle a, b$ exist and so if the given equation has a rational root that root must be integral.

RonL