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

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    @Seirios Sorry, I'm not getting you. A figure maybe?2012-09-23

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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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    @GerryMyerson Yeah, yeah, but playing with [desmos.com](http://www.desmos.com) (graphing calculator) shows that it's pretty likely. (Graph `x+1` and `a^x`, and then adjust the value of `a`.)2015-04-06
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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:

Density in the circle