# Is there any sans-calculus proof of irrationality of $\pi$?

Is there a proof that will convince someone who doesn't understand calculus, of $\pi$'s irrationality .

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I don't believe anyone has discovered any such proof yet. – Shahab Sep 23 '12 at 9:13
$\pi$ is irrational iff the orbit of $(1,0)$ under rotations of angle $1$ is dense in the circle; so maybe you can convince someone that $\pi$ is irrational by drawing approximations of this orbit (of course, it is not a proof). – Seirios Sep 23 '12 at 9:29
@Seirios Sorry, I'm not getting you. A figure maybe? – user13107 Sep 23 '12 at 9:37

As far as I know, there aren't any simpler proofs. And because pi is transcendental, it doesn't seem like the proof would lend itself well to anything easier than calculus.

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Ok. May be not easier than calculus, but how about something different than calculus? – user13107 Sep 23 '12 at 9:22
My first thought was that you could prove that $\pi$ is transcendental, which would be sufficient, but that explanation would take much longer and would probably be less satisfying for those who don't have the necessary abstract algebra background. Lambert's proof uses continued fractions and essentially concludes that if $\tan(x)$ is rational, then x is irrational. Since $\tan \left( \frac{\pi}{4} \right) = 1$, then $\frac{\pi}{4}$ is irrational. I haven't really looked into it, but it might be more accessible. – JoeDub Sep 23 '12 at 9:40
Cool, thanks for the pointers. Sorry, can't vote up w/o rep. – user13107 Sep 23 '12 at 9:56
In a similar vein, no one seems to know a proof that $\pi$ is not constructible without proving that it is transcendental. See math.stackexchange.com/questions/103786/… – lhf Sep 23 '12 at 11:32
$e$ is transcendental, but if you accept $e=\sum^{\infty}(n!)^{-1}$, you can prove it irrational without Calculus. Also $.1101001000100001000001\dots$ is transcendental, but its irrationality is trivial and no Calculus is needed. – Gerry Myerson Sep 23 '12 at 12:26

I developp my comment: Take the point $x_0=(1,0)$ on the unit circle $S^1$; if $R$ is the rotation of one radian around $(0,0)$, let $x_1=Rx_0$, $x_2=Rx_1$, $x_3=Rx_2$ and so on.

In complex notation, you have $x_n=e^{in}$. If $\pi$ is rational, there exist $p,q>0$ such that $\pi=p/q$ and so $x_{2p}=e^{i2p}=e^{i2q\pi}=x_0$; thus, $\{x_n, n \geq 0\}$ is discrete in $S^1$. Otherwise, $\{x_n,n \geq 0\}$ is dense in $S^1$, ie. it seems to cover the circle.

Consequently, you can convince someone without calculus that $\pi$ is irrational by showing that $\{x_n, \geq 0\}$ seems to cover the circle; of course, it is not a proof.

Here are some iterations:

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