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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.


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