Proof of Rational Root Theorem

[Paragon]

I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:

Say we have a polynomial:

$$F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$

where all of the coefficients are integers and $$a_{0}$$ does not equal zero.

I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that:

$$F(\frac{p}{q}) = 0$$

p and q are prime numbers. So I want to prove that p is a factor of $$a_{0}$$ and q is a factor of $$a_{n}$$. Next, I insert p/q into F(x):

$$F(\frac{p}{q}) = a_{n}\left(\frac{p}{q}\right)^{n} + a_{n-1}\left(\frac{p}{q}\right)^{n-1} + ... + a_{1}\left(\frac{p}{q}\right) + a_{0} = 0$$

multiplying by $$q^{n}$$ we get:

$$a_{n}p^{n} + a_{n-1}p^{n-1}q + ... + a_{1}p q^{n-1} + a_{0}q^{n} = 0$$

and factorizing out p:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

As p, q are prime numbers, the Greatest Common Devisor of p and q is 1, i.e. GCD(p, q) = 1. And hence, p devides $$a_{0}$$ and is then a factor of $$a_{0}$$.

As the same kind of argument applies to q being a factor of $$a_{n}$$, I thought that this proved the theorem (p/q are roots of F(x)), but no... The problem is in:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer?

 

[Muzza]

I want to find a nice and elegant proof of the Rational Root theorem, but I get stuck. I read some stuff on the Internet, but I have not found a complete proof of the theorem. Here's my try:
Say we have a polynomial:
$$F(x) = \sum ^{n}_{r = 0} a_{n}x^{n} = a_{n}x^{n} + a_{n - 1}x^{n-1} + ... + a_{1}x + a_{0}$$
where all of the coefficients are integers and $$a_{0}$$ does not equal zero.
I want to prove that there is a factor of $$a_{n}$$, call it q, and a factor of $$a_{0}$$, call it p, so that:
$$F(\frac{p}{q}) = 0$$

You won't be able to prove that. Exercise: find a polynomial with integer coefficients which doesn't have any rational roots.

The "rational root" theorem that /I'm/ familiar with states that if p/q is a rational root of F and gcd(p, q) = 1, then q is a factor of a_n and p is a factor of a_0. (Note that this theorem doesn't guarantee that F actually has any rational roots).

how do I prove that: $$a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}$$ is an integer?

A finite sum of integers is clearly an integer itself.

 

[matt grime]

Firstly, as stated your trying to prove something that is false. There is no reason to suppose that an integer polynomial has *any* rational roots.Then there is the issue of what a rational root might look like even if it had one: consider 2x-1, its rational root is 1/2, and not of the form p/q where p and q are primes,

However, I don't see that issue you have is at all an issue: all of the a_r are integers, p and q are integers, and multiplying and adding together integers yields integers.

 

[Paragon]

Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n.

The statement: "A finite sum of integers is clearly an integer itself." hit the nail =)

Thanks guys, this helped a lot!

 

[matt grime]

Indeed, the fact that the polynomial had rational zeros was a presumption. And p, q are factors of a_0 and a_n.

You asserted they were primes. They are at best coprime, or prime to each other.

 

[ila]

and factorizing out p:

$$p (a_{n}p^{n-1} - a_{n-a}p^{n-2} q - ... - a_{1} q^{n-1}) = a_{0} q^{n}$$

And hence, p devides $$a_{0}$$ and is then a factor of $$a_{0}$$.

My maths isn't that strong. I am having trouble with this proof as well. Basically its the logic of saying that p divides a0 and hence is a factor. Why couldn't q be factored out of the equation (instead of p) and then divide the a0*q^n on the RHS by q and just say that q is a factor of a0 (in addition to a factor of q^n).

Lastly, I don't understand why just because p divides $$a_{0}$$ means that it is a factor of a0. The logic seems to be that just because q^n can't be factored by p therefor it implies that a0 can be factored by p.

 

[HallsofIvy]

I would approach it from the other end. IF polynomial p(x) has root m/n, then (x- m/n) is a factor: p(x)= (x- m/n)q(x). So np(x)= (nx- m)q(x)= (nx-m)(a_n-1x^n-1+ ...+ a₀). It should be obvious from that, that the constant term has m as a factor. It is a little harder, because of that "n" in "np(x)", to see that the leading coefficient has m as a factor.