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Prove that if $x$ is a non-zero rational number, then $\tan(x)$ is not a rational number and use this to prove that $\pi$ is not a rational number.

I heard that this was proved two hundred years ago. I need this proof because I want to know the proof of why $\pi$ is not rational.

I need the simplest proof!

thanx !

Michael Albanese
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FNH
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    I don't havea proof but I read that there is one using continued fractions by Lambert http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Lamberts_proof –  Jan 26 '13 at 13:00
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    Pretty sure $\tan 0 = 0$. – Michael Albanese Jan 26 '13 at 13:00
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    @MichaelAlbanese, ... lol. Any nonzero. –  Jan 26 '13 at 13:01
  • @MichaelAlbanese , ok ! , add other condition that x = 0 – FNH Jan 26 '13 at 13:02
  • Here's a blog explaining the proof in his own words http://paramanands.wordpress.com/2011/04/28/continued-fraction-expansion-of-tanx/ –  Jan 26 '13 at 13:03
  • There is a translation of Lamberts paper in "Pi - A source book" - Borwein, Jonathan M.; Borwein, Peter B., Pi, a source book –  Jan 26 '13 at 13:07
  • I've worked out tan(x) as continued fraction here http://math.stackexchange.com/a/288321/58512 - this is the first step of the proof. –  Jan 27 '13 at 20:47

2 Answers2

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Read the excellent book of Ivan Niven, "Irrational Numbers". There is the answer for your interesting question.

Mario De León
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The proof from a few hundred years ago was done by Lambert and Miklós Laczkovich provided a simplified version later on. The Wikipedia page for "Proof that $\pi$ is irrational" provides this proof (in addition to some other discussion).

http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Laczkovich.27s_proof

Edit: Proving the more general statement here hinges upon Claim 3 in Laczkovich's proof. Defining the functions $f_k(x)$ by \begin{equation} f_k(x) = 1 - \frac{x^2}{k} + \frac{x^4}{2!k(k+1)} - \frac{x^6}{3!k(k+1)(k+2)} + \cdots \end{equation} it can be seen (using Taylor series) that \begin{equation} f_{1/2}(x/2) = \cos(x) \end{equation} and \begin{equation} f_{3/2}(x/2) = \frac{\sin(x)}{x} \end{equation} so that \begin{equation} \tan x = x\frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \end{equation}

Taking any $x \in \mathbb{Q} \backslash \{0\}$ we know that $x/2 \in \mathbb{Q} \backslash \{0\}$ and also that $x^2/4 \in \mathbb{Q} \backslash \{0\}$ as well. Then $x/2$ satisfies the hypotheses required by Claim 3.

Using Claim 3 and taking $k = 1/2$, we have \begin{equation} \frac{f_{k+1}(x/2)}{f_k(x/2)} = \frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \notin \mathbb{Q} \end{equation} which then also implies that \begin{equation} \frac{x}{2}\frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \notin \mathbb{Q} \end{equation} Multiplying by 2 then gives $\tan x \notin \mathbb{Q}$.

yoknapatawpha
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