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.
$\blacksquare$