Cube Root of 2 is Irrational

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Theorem

$\sqrt [3] 2$ is irrational.


Proof

Aiming for a contradiction, suppose that $\sqrt [3] 2$ is rational.

Then:

\(\ds \sqrt [3] 2\) \(=\) \(\ds \frac p q\) for some integer $p$ and $q$
\(\ds \leadsto \ \ \) \(\ds p^3\) \(=\) \(\ds q^3 + q^3\)

which contradicts Fermat's Last Theorem.

$\blacksquare$