# A Classic Joke Proof

While this may stretch the stated point of this blog, I’m ultimately the one in control, and I’ll decide what gets posted here. And I think it’s really funny and would be fun to write up.

I am nowhere near the first person who has ever told this joke.

$\textbf{Claim:}$ For all $n > 2$, $\sqrt[n]{2}\not\in\mathbb{Q}$.

$\textbf{Proof:}$ For real numbers $a, b \in \mathbb{R}$, such that $b \neq 0$, suppose that $(\frac{a}{b})^n = 2$. We have that $a^n = 2b^n = b^n + b^n$.

By Fermat’s last theorem, we have $a, b$ are not both integers, and hence that $\frac{a}{b}$ cannot be rational (it obviously implies $a, b$ aren’t both integers, but how does it imply that they aren’t both rational? Think on this…).

Unfortunately, I have not yet read and understood Wiles’ proof of FLT, so I do not know if this is a circular argument.