8
$\begingroup$

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 !

$\endgroup$
  • 1
    $\begingroup$ I don't havea proof but I read that there is one using continued fractions by Lambert en.wikipedia.org/wiki/… $\endgroup$ – user58512 Jan 26 '13 at 13:00
  • 6
    $\begingroup$ Pretty sure $\tan 0 = 0$. $\endgroup$ – Michael Albanese Jan 26 '13 at 13:00
  • 1
    $\begingroup$ @MichaelAlbanese, ... lol. Any nonzero. $\endgroup$ – user58512 Jan 26 '13 at 13:01
  • $\begingroup$ @MichaelAlbanese , ok ! , add other condition that x = 0 $\endgroup$ – Fawzy Hegab Jan 26 '13 at 13:02
  • $\begingroup$ Here's a blog explaining the proof in his own words paramanands.wordpress.com/2011/04/28/… $\endgroup$ – user58512 Jan 26 '13 at 13:03
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\tan x \notin \mathbb{Q}$.

$\endgroup$
  • $\begingroup$ The wikipedia article does have a nice proof; thanks. I know that Lambert proved the irrationality of $\pi$; did he prove the more general question being asked (and answered) here? $\endgroup$ – Pete L. Clark Dec 5 '13 at 5:44
  • $\begingroup$ I've edited my post to add more relevant details. $\endgroup$ – yoknapatawpha Dec 5 '13 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.