TITLE: Prove $2^{1/3}$ is irrational.
QUESTION [37 upvotes]: Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer. 

Assume $2^{1/2}$ is irrational. 
$2^{1/3} * 2^{x} = 2^{1/2} \Rightarrow x = 1/6$. 
$2^{1/3} * {2^{1/2}}^{1/3} = 2^{1/2}$.
if $2^{1/2}$ is irrational, then ${2^{1/2}}^{1/3}$ is irrational. 
$2^{1/3} = 2^{1/2} / {2^{1/2}}^{1/3}$. 
$2^{1/3}$ equals an irrational number divided by an irrational number. 
$2^{1/3}$ is an irrational number.

REPLY [87 votes]: I can't resist:  Suppose $2^{\frac{1}{3}}=\frac{n}{m}$.  Then $$2m^3=n^3,$$ or in other words $$m^3+m^3=n^3.$$  But this contradicts Fermats Last Theorem.