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

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!

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.

• Could you perhaps expand this hint, I can't seem to figure this one out. With $x$ factoring into $a*b$ for example I see $x^n = a^n * b^n$. Actually I think I don't understand what this $a_n$ is in the question. Is it just a bunch of different integers? – Display Name Apr 24 '14 at 13:10
• @DisplayName Yes, the $a_i$ are all integers. What I'd do is write $x=m/n$ for some integers (you can assume $m$ and $n$ have no common divisors). Now, can you do some fiddling and turn it into a question about which primes divide either side? Use that if $k$ divides $a + b + c$ and $k$ divides $b$ and $c$, then $k$ must divide $a$, and use the fact that every integer divides zero. – Eric Auld Apr 24 '14 at 14:19

Suppose not, then

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

case $n=2$ is easy
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 $f(x)=x^n-2$.