# proving irrationality of cube-root of 2 using an specific method?

One of the methods mentioned in A Course of Pure Mathematics for proving irrationality of $\sqrt{2}$ is the following:

Suppose, if possible, that $\frac{p}{q}$ is a positive fraction, in its lowest terms, such that $\left(\frac{p}{q}\right)^{2} = 2$ or $p^{2} = 2q^{2}$. It is easy to see that this involves $(2q - p)^{2} = 2(p - q)^{2}$; and so $\frac{2q - p}{p - q}$ is another fraction having the same property. But clearly $q < p < 2q$, and so $p - q < q$. Hence there is another fraction equal to $\frac{p}{q}$ and having a smaller denominator, which contradicts the assumption that $\frac{p}{q}$ is in its lowest terms.

This Method is similar to this one mentioned in Wikipedia, named "Proof by infinite descent, not involving factoring".

My question is,can this method be used for proving irrationality of $\sqrt[\leftroot{-2}\uproot{2}3]{2}$?

Yes, something similar can be made to work with an extra step. Suppose $\frac pq$ is $\sqrt[3]2$ in lowest terms. Then $\frac{p^2}{q^2}$ must be $\sqrt[3]4$ in lowest terms. Now we can check that $x=\frac{2q^2-p^2}{pq-q^2}$ also satisfies $x^3=4$. This has a smaller (but still positive) denominator than $\frac{p^2}{q^2}$, contradiction.
[edit to answer lhf's question] Set $y=\sqrt[3]2$, so that $2=y^3$. Then $2-y^2=y^3-y^2=y^2(y-1)$, and so $\Big(\frac{2-y^2}{y-1}\Big)^3=y^6=4$. Setting $y=p/q$ and multiplying top and bottom by $q^2$ gives the expresion for $x$ above. I think this generalises to $\sqrt[n]2$ by converting all cubes to $n$th powers and squares to $(n-1)$th powers, so you get a contradiction to $2^{(n-1)/n}$ being in lowest terms.
• It'd be nice if you explained how you found $x$ and how this could be generalized to say $\sqrt[n]{2}$. – lhf Feb 2 '17 at 9:09
• thanks you for also covering the generalized part. what I like about this proof is the fact that it follows from $p < 2q$. – user2838619 Feb 2 '17 at 11:14