$x^2-p=0$, with $p$ prime, have irrational roots.

Question

Unaware that $\sqrt{p}$ is irrational, prove that as $x^2-p=0$ have irrational root for $p$ prime? How would you use the criterion of Eisenstein?

Answer 1

The most common Eisenstein trick is to change $x=y+c$ where c is a constant, and then use Eisenstein's. You need a c that makes the y term divisible by p. Does this help?

Answer 2

Perhaps the following generalization will help see things more clearly. Let $n>1$ be any natural number and $p$ a prime. Let $\alpha$ be the real $n$-th root of $p$. If $\alpha \in \mathbb Q$, then $\alpha$ is a root of $X^n-p$, and thus $(X-\alpha )$ would divide $X^n-p$ (since in general, if $\beta $ is a root of a polynomial $f(X)$, then $(X-\beta )$ divides $f(X)$). Thus, if $\alpha \in \mathbb Q$, then $X^n-p$ is reducible over $ Q$.

Now, the polynomial $X^n-p$ is irreducible by Eisenstein's criterion (directly, using the prime $p$). So, it follows that $X^n-p$ is irreducible over $\mathbb Q$ and thus that $\alpha \notin \mathbb Q$.

Answer 3

There is a very naive solution, without using any criterion at all.

Let $\frac{a}{b}$ be a rational root of the equation $x^2-p$. Then, it follows easily that $b|1$ and $a|p$. But $p$ is a prime. So, $a=p$ or $a=1$. (or $a=-p,-1$).

But it can be easily checked that $1$ or $p$ aren't roots of the given equation. So, the equation must have irrational roots.