We have the identity $$\tan x=\frac{x}{1+\underset{n=1}{\overset{\infty}{\mathrm K}} \frac{-x^2}{2n+1}}.$$ From the Wikipedia article on Proof that π is irrational:
[...] Lambert proved that if x is non-zero and rational then this expression must be irrational.
But how can we prove Lambert's assertion? I couldn't find any resource containing the proof.
On
I couldn't find any resource containing the proof.
The Wikipedia page says that Lambert's proof may be found in
Lambert, Johann Heinrich (2004) [1768], "Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques", in Berggren, Lennart; Borwein, Jonathan M.; Borwein, Peter B. (eds.), Pi, a source book (3rd ed.), New York: Springer-Verlag, pp. 129–140, ISBN 0-387-20571-3
Suppose that $x=\frac{a}{b}$ for $a,b\in\mathbb{N}$. Then $$\tan\frac{a}{b}=\frac{a}{b+\underset{n=1}{\overset{\infty}{\mathrm K}} \frac{-a^2}{(2n+1)b}}.$$ Let $3b\gt a^2$ (this can be assumed without loss of generality, if $3b\le a^2$ then for large enough $n$ it's true that $(2n+1)b\gt a^2$), which means that $$0\lt \frac{a^2}{3b+\underset{n=2}{\overset{\infty}{\mathrm K}} \tfrac{-a^2}{(2n+1)b}}\lt 1.$$ Now if $$\frac{a^2}{3b+\underset{n=2}{\overset{\infty}{\mathrm K}} \tfrac{-a^2}{(2n+1)b}}=\frac{c_2}{c_1}$$ for some $c_1,c_2\in\mathbb{N}$ and $0\lt\frac{c_2}{c_1}\lt 1$, then $c_1\gt c_2$. Furthermore, $$0\lt\frac{c_{3}}{c_{2}}=\frac{a^2}{5b+\underset{n=3}{\overset{\infty}{\mathrm K}} \tfrac{-a^2}{(2n+1)b}}\lt 1,$$ $c_{3}\in\mathbb{N}$ and $c_2\gt c_3$, and so on: $$c_1\gt c_2\gt c_3\gt\cdots\gt 0.$$ This is a contradiction since $c_{k\gt 0}$ are natural numbers, so $$\frac{a^2}{3b+\underset{n=2}{\overset{\infty}{\mathrm K}} \tfrac{-a^2}{(2n+1)b}}\ne\frac{c_2}{c_1}\implies \tan\frac{a}{b}\notin\mathbb{Q}.$$ Thus we arrived at the conclusion that $$x\in\mathbb{Q}\setminus\{0\}\implies \tan x\notin\mathbb{Q}.$$