Prove $2^{1/3}$ is irrational. 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. 


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*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.


Let $x=2^{1/3}$ be a rational $\frac{p}{q}$ where $p$ and $q$ are natural numbers having no common factors.
Then $x^3 = 2$, and $x = \frac{x^3}{x^2} = \frac{2}{(2^{1/3})^2} = \frac{2}{\big(\frac{p}{q}\big)^2}$
Hence $x = \frac{p}{q} = \frac{2q^2}{p^2}$.
Since $\frac{p}{q}$ is in its lowest terms, then the second denominator $p^2$ is a multiple of $q$, which is a contradiction unless $q=1$.
But if $q=1$, then $x=2^{1/3}=p$, a natural number, so $x$ is a natural number as well. But $x^3=2$, and no natural number is equal to 2 when cubed. Hence, we have a contradiction, and so $x$ must be irrational, as required.
