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

Problems and solutions of ring theory in abstract algebra

Problem 179

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

 
LoadingAdd to solve later

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 $m\geq 2$.


LoadingAdd to solve later

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Ring theory
Problems and solutions of ring theory in abstract algebra
The Ideal Generated by a Non-Unit Irreducible Element in a PID is Maximal

Let $R$ be a principal ideal domain (PID). Let $a\in R$ be a non-unit irreducible element. Then show that the...

Close