# Help with Assignment

• Oct 31st 2006, 11:07 PM
sbsite
Help with Assignment
I should serve her until Sunday :) :) :)
• Nov 1st 2006, 03:56 AM
CaptainBlack
Quote:

Originally Posted by sbsite
I should serve her until Sunday :) :) :)

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 AM
ThePerfectHacker
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 AM
sbsite
Little request???
Quote:

Originally Posted by CaptainBlack
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

:) :) :) :) :D :D :D :D
can you writh that in matematics .
i have problems with my english/???
• Nov 2nd 2006, 01:36 AM
sbsite
Quote:

Originally Posted by ThePerfectHacker
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.

:) :) :) :) :) :) :) :) :cool: :cool: :cool: :cool: :cool: :cool:
• Nov 2nd 2006, 05:11 AM
ThePerfectHacker
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 AM
CaptainBlack
Quote:

Originally Posted by sbsite
:) :) :) :) :D :D :D :D
can you writh that in matematics .
i have problems with my english/???

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:

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

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