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

enzotib
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user13107
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  • I don't believe anyone has discovered any such proof yet. – Shahab Sep 23 '12 at 09:13
  • $\pi$ 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 09:29
  • @Seirios Sorry, I'm not getting you. A figure maybe? – user13107 Sep 23 '12 at 09:37

2 Answers2

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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.

JoeDub
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  • Ok. May be not easier than calculus, but how about something *different* than calculus? – user13107 Sep 23 '12 at 09:22
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    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](http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Lambert.27s_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 09:40
  • Cool, thanks for the pointers. Sorry, can't vote up w/o rep. – user13107 Sep 23 '12 at 09:56
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    In a similar vein, no one seems to know a proof that $\pi$ is not constructible without proving that it is transcendental. See http://math.stackexchange.com/questions/103786/direct-proof-that-pi-is-not-constructible – 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
  • @GerryMyerson isn't that representation of $e$ built on calculus? infinite (specifically taylor) series seems to be directly from calculus...unless of course the OP meant modulo integral/differential calculus... – Coffee_Table Jul 21 '14 at 22:33
  • @Coffee, I don't think I was responding directly to the original question, but to some comment, possibly to some comment that has since been deleted. Taylor series certainly presupposes Calculus, but one can do a lot with infinite series without any Calculus (though you do need limits). As for $e$, it is possible to take that infinite series as the definition of $e$, and if you take that approach then no Calculus is needed to prove irrationality. – Gerry Myerson Jul 22 '14 at 00:03
  • @GerryMyerson I hear that. I guess I just always associate certain studies of the "infinite"-i.e., series and limits-with calculus, but I guess there's no reason for that to be set in stone, and it is my own bias. Also, I just saw the date on your comment! It just happens to be that I recently became interested in this very question and my googling led me directly here... – Coffee_Table Jul 22 '14 at 02:23
  • @GerryMyerson You can define $e$ without calculus, if you want: _The unique number $e$ such that $e^x\ge x+1$ for all $x$._ (This definition is surprisingly powerful. It lets you prove that $x^x$ has a minimum at $e^{-1}$, for example, and it lets you prove that $1-\frac12+\frac13-\dotsb=\ln2$. This is all without using calculus. EDIT: Well, technically, you just get to prove that $\ln2-\frac1n<1-\frac12+\dotsb\frac1n<\ln2+\frac1n$, but the Squeeze Theorem implies the infinite version immediately. I suppose that the Squeeze Theorem is _kinda_ calculus-y, but not too calculus-y.) – Akiva Weinberger Apr 03 '15 at 11:16
  • @colum, OK, though somehow you have to prove that there exists a unique number such that $\forall x\, e^x\ge x+1$. – Gerry Myerson Apr 06 '15 at 10:47
  • @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`.) – Akiva Weinberger Apr 06 '15 at 11:49
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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

Seirios
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