Does someone know a proof (books, articles) that $\pi$ is not a quadratic irrational?
The proof should not use that $\pi$ is transcendental.
Any hints would be appreciated.
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8$\begingroup$ Niven's proof, which is widely available, shows that $\pi^2$ is irrational (which isn't quite what you want, but it's a start). $\endgroup$– Gerry MyersonAug 18, 2013 at 8:11
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3$\begingroup$ Shameless self-reference : math.stackexchange.com/questions/713467/…. As a simple continued fraction is periodic if and only if it is a CF of some irrational quadratic, this proof essentially proved that $\pi$ is not a quadratic irrational. $\endgroup$– Balarka SenAug 9, 2014 at 8:15
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$\begingroup$ @Balarksa Sen ,Sorry, but all I see in your link is a proof that the CF for $\tanh (1)$ is not periodic. which only implies that $\tanh (1)$ and $e$ are not quadratic over $Q$. Did I miss something? $\endgroup$– DanielWainfleetMar 9, 2016 at 6:54
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1$\begingroup$ Lambert proved $\pi ^2$ is irrational , by use of $\tan (x) = \frac{x}{1-\frac {x^2}{3-\frac{x^2}{5-\frac{x^2}{...}}}}$ $\endgroup$– user217174Mar 11, 2016 at 16:05
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$\begingroup$ Would the proof that $\pi$ is trancendental be sufficient? It doesn't use the fact, it proves it $\endgroup$– Yuriy SMar 15, 2016 at 10:03
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1 Answer
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As mentioned in the comments Lambert proved that $\pi$ is irrational (http://www.pi314.net/eng/lambert.php). Lambert showed that it's irrational by first demonstrating that the continued fraction expansion holds: $$\tan(x) = \cfrac{x}{1 - \cfrac{x^2}{3 - \cfrac{x^2}{5 - \cfrac{x^2}{7 - {}\ddots}}}}.$$ He then proved that if $x$ is non-zero and a rational number then the expression must be irrational. Consequently because $\tan(\frac {\pi}{4}) = 1$, it follows that $\frac {\pi}{4}$ is irrational and as a result that $\pi$ is irrational. (Refer to the Wikipedia article on the proof of $\pi$ being irrational by Lambert - Lambert's Proof)
Laczkovich's Proof is a simplification of Lambert's proof (refer here for more information).
I found a few resources detailing numerous proofs of $\pi$ being irrational, none of them utilising $\pi$ as transcendental. The Wikipedia article lists 6 different proofs (including Lambert's and Laczkovich's). There was also a question asked on this math forum about understanding the following proof of $\pi$ being irrational making use of polynomials and calculus (A simple proof that $\pi$ is irrational).
I hope this helps.
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3$\begingroup$ This doesn't answer the question... OP wants to know if there's a simple proof that $\pi$ isn't a quadratic irrational, i.e., an irrational root of a quadratic with rational coefficients. $\endgroup$– mjqxxxxMar 14, 2016 at 16:59