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

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arxiv 2003.06627 v3 pith:25J6JLMO submitted 2020-03-14 math.GM

classification math.GM
keywords sqrtirrationalgeometricproofadaptingbiggercannotgeometrically
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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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