$\sqrt[m]{2}$ is an Irrational Number
Problem 179
Prove that $\sqrt[m]{2}$ is an irrational number for any integer $m \geq 2$.
Contents
Hint.

Use ring theory:
 Consider the polynomial $f(x)=x^m2$.
 Apply Eisenstein’s criterion, show that $f(x)$ is irreducible over $\Q$.
Proof.
Consider the monic polynomial $f(x)=x^m2$ in $\Z[x]$.
The constant term is divisible by the prime $2$ and not divisible by $2^2$.
Thus, by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible over the rational numbers $\Q$.
In particular, it does not have a degree $1$ factor.
If $\sqrt[m]{2}$ is rational, then $x\sqrt[m]{2}\in Q[x]$ is a degree $1$ factor of $f(x)$ and this cannot happen.
Therefore, $\sqrt[m]{2}$ is an irrational number for any integer $m\geq 2$.
Add to solve later