Show that $x_0$ must be an integer. Conclude that $\sqrt[n]{2}$ is irrational for every $n \geq 2$

Question

I have a problem in my workbook that is as follows:

Let $f = x^n + a_{n-1}x^{n-1}+\dots+a_1x+a_0 = 0 $ with $a_i \in \mathbb{Z}$. Suppose there exists a rational number $x_0$ with $f(x_0) = 0$. Show that $x_0$ must be an integer. Conclude that $\sqrt[n]{2}$ is irrational for every $n \geq 2$.

My problem: I don't really understand why I have $f$ equalling all that, and I assume that it is $f(x)$ equalling all that rather. I imagine I can prove that last statement about the nth root without the first part, but obviously I a meant to progress this problem using the fact first proved. If I could have some direction on this one it would be greatly appreciated!

Answer 1

You are correct; it should say $f(x)$. Often times books are loose in their terminology. A hint I would give you is to say that you should use that every integer has a unique factorization into primes.

Answer 2

Suppose not, then

$$ 2^{\frac{1}{n}} = \frac{p}{q} \iff 2 = ( \frac{p}{q})^n \iff 2 q^n = p^n \iff q^n + q^n = p^n$$

contradicts fermat;s last theorem

case $n=2$ is easy

Answer 3

The first part is a direct consequence of the Gauss lemma (and one can be more specific that any rational root of $f$ must in fact be a divisor of $a_0$), apply it to $f(x)=x^n-2$.