Help with Assignment

**sbsite** · October 31st 2006, 11:07 PM

I should serve her until Sunday

**CaptainBlack** · November 1st 2006, 03:56 AM

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

**ThePerfectHacker** · November 1st 2006, 07:52 AM

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:
$1^n+p_{n-1}1^{n-1}+...+p_11^n+1=2+p_1+...+p_{n-1}=0$
But that cannot be.

If the solution is -1 then:
$1+p_1(-1)+p_2(-1)^2+...+p_{n-1}(-1)^{n-1}=0$
Which is it not.

Thus there are no solutions.

**sbsite** · November 2nd 2006, 12:51 AM

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

can you writh that in matematics .
i have problems with my english/???

**sbsite** · November 2nd 2006, 01:36 AM

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:
$1^n+p_{n-1}1^{n-1}+...+p_11^n+1=2+p_1+...+p_{n-1}=0$
But that cannot be.

If the solution is -1 then:
$1+p_1(-1)+p_2(-1)^2+...+p_{n-1}(-1)^{n-1}=0$
Which is it not.

Thus there are no solutions.

**ThePerfectHacker** · November 2nd 2006, 05:11 AM

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):
$(-1)^n+(-1)^{n-1}+...+(-1)^2+(-1)^1$
That is alternating,...
$1,-1,1,-1,...$

**CaptainBlack** · November 2nd 2006, 05:22 AM

Originally Posted by sbsite

can you writh that in matematics .
i have problems with my english/???

Suppose $\exists\ a,b \in \mathbb{Z}$ such that $gcd(a,b)=1$ , and that $X=a/b$ is a root of:

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

Then:

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

Now:

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

but:

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

a contradiction.

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

RonL

Thread: Help with Assignment

LinkBack

Thread Tools

Display

Help with Assignment

Little request???

Similar Math Help Forum Discussions

assignment help?

assignment help

assignment

hi everyone, i have an assignment due and i really need help

need help with assignment:

Search Tags