# $\sqrt[m]{2}$ is an Irrational Number ## Problem 179

Prove that $\sqrt[m]{2}$ is an irrational number for any integer $m \geq 2$. Add to solve later

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## Hint.

Use ring theory:

1. Consider the polynomial $f(x)=x^m-2$.
2. Apply Eisenstein’s criterion, show that $f(x)$ is irreducible over $\Q$.

## Proof.

Consider the monic polynomial $f(x)=x^m-2$ 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

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