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arXiv · 2003.06627

A geometric proof that $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{7}$ are irrational

Ricardo A. Podestá
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Abstract

We show geometrically that $\sqrt n$ is irrational for $n=3,5,7$ by adapting Tennenbaum's geometric proof that $\sqrt 2$ is irrational. We also show that this method cannot be used to prove the irrationality of $\sqrt n$ for a bigger $n$.

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Ricardo A. Podestá. 2020-03-14. A geometric proof that $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{7}$ are irrational. https://arxiv.org/abs/2003.06627

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