Proposition. Let $\alpha\in\mathbb{R}$. If there is a sequence of integers $a_n,b_n$ such that $0<|b_n\alpha-a_n|\longrightarrow 0^+$ as $n\longrightarrow \infty$, then $\alpha$ is irrational.

How to prove that $\pi$ is irrational using this proposition?

I know several proof of the irrationality of π with complex analysis, but I think in this way is very difficult.

For example to prove the irrationality of $e$ consider

$$0<n!e-n!\left(1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}\right)\le\frac{1}{n}\longrightarrow0^+$$

Any hint would be appreciated.

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    $\begingroup$ Is this an exercise from a book or just something you were curious about? I ask because I would like to know if it's solvable before thinking too much about it. $\endgroup$ – Potato May 9 '14 at 3:21
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    $\begingroup$ @Potato I know several proof of the irrationality of $\pi$ with complex analysis, but I think in this way is very difficult $\endgroup$ – felipeuni May 9 '14 at 3:27
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    $\begingroup$ I don't think there is a known proof like that. The reason is that it can always be reduced into a 'standard' proof of irrationality (like the one shown above for $e$) i.e. assume $x$ rational and show that it gives $0<$integer$<1$ and such a proof is to my knowledge not known - see for example "Proofs from the book" which is a collection of the simplest and most beautiful proofs out there. $\endgroup$ – Winther May 9 '14 at 3:27
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    $\begingroup$ If you are looking for a relatively simple proof of the irrationality of $\pi$ I think Nivens proof is not very difficult. en.wikipedia.org/wiki/… $\endgroup$ – Test123 May 9 '14 at 3:31
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    $\begingroup$ The OP isn't asking for just any proof, clearly he just wants to see it using this technique. Math isn't just about results, it's also about finding new proofs or cool ones. $\endgroup$ – Squirtle May 9 '14 at 3:40

The classical proof that $\pi$ is irrational is due to Lambert and rests on his continued fraction expansion expression for $\tan x$. I include below a proof that a continued fraction expansion is irrational and the application to the irrationality of $\pi$. I copied these from notes that I made for myself some time ago so I hope they are comprehensible, I can also post a proof of the Lambert expansion if you wish. The only references I know are old (very old) there is a two volume algebra book by Chrystal, and an old french calculus book by Bertrand which have information on this material.

In the continued fraction, $$\cfrac{b_1}{a_1 - \cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}} $$ assume that where $1+b_n \leq a_n$ for all $n$, and that we have $1+b_n < a_n$ infinitely often. Then the fraction is irrational. $a_n, b_n$ integers.

Proof Assume that the fraction is rational, say $$\frac{\lambda_1}{\lambda_0}=\cfrac{b_1}{a_1 - \cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}}} $$ where $\lambda_1$ and $\lambda_0$ are positive integers, now since the fraction converges to a number less than one, $\lambda_1 < \lambda_0$ If we set $$\rho_1=\cfrac{b_2}{a_2 - \cfrac{b_3}{a_3 - \cfrac{b_4}{a_4 -\cdots}}} $$ then we have $$\frac{\lambda_1}{\lambda_0}=\frac{b_1}{a_1 -\rho_1}$$ so$$\rho_1=\frac{a_1 \lambda_1 - b_1 \lambda_0}{\lambda_1} < 1$$ So $\rho_1=\frac{\lambda_2}{\lambda_1}$ where $\lambda_2 < \lambda_1$.

Continuing in this way we obtain a strictly decreasing sequence of positive integers, $\lambda_0 > \lambda_1 > \cdots$, a contradiction.

[Lambert] $$\tan x=\cfrac{x}{1- \cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}}$$

This expression leads to the following fundamental result.

Theorem [Lambert] $\pi$ is irrational.

Proof Assume that $\pi$ is rational, then $\frac{\pi}{4}$ is also rational. Let $$\frac{\pi}{4}=\frac{a}{b},$$ and substitute $x=\frac{\pi}{4}$ into Lambert's continued fraction for $\tan x$.

We get \begin{equation*} \begin{split} 1=&\cfrac{\frac{a}{b}}{1- \cfrac{\frac{a^2}{b^2 }}{3-\cfrac{\frac{a^2 }{b^2}}{5-\cfrac{\frac{a^2}{b^2}}{7-\cdots}}}} \\ &=\cfrac{a}{b- \cfrac{a^2}{3b-\cfrac{a^2}{5b-\cfrac{a^2}{7b-\cdots}}}} \\ \end{split} \end{equation*} Now since eventually $nb > a^2 +1$ we have that this expression is irrational, and this is absurd since it is equal to $1$. Therefore $\pi$ is irrational.

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    $\begingroup$ I don't see the connection between your answer and the question. Could you please give some hints, how the proposition from the question above is involved? $\endgroup$ – Markus Scheuer May 16 '14 at 13:13

From the irrationality and uniformity of $\pi$ it follows, that such a sequence exists. Numerically you could easily construct it (look for increasingly long sequences of zeros in eg. the binary representation and use the string before that point as $a_n$ with a $b_n$ of the form $2^m$). Unfortunately that does not help you in proving the irrationality of $\pi$.

From the definition $|b_n \alpha - a_n|\rightarrow 0$ it is obvious, that $\frac{a_n}{b_n}$ is an approximation to $\alpha$. In fact, let $e_n=\left|\alpha - \frac{a_n}{b_n}\right|$ denote the error of the $n$-th approximation, then (for infinitely many $n$)

$$ \begin{align} \left|b_n\alpha - a_n\right| &< \left|b_{n-1}\alpha - a_{n-1}\right| \\ \Rightarrow e_n &= \left|\alpha - \frac{a_n}{b_n}\right| \\ &<\frac{b_{n-1}}{b_n} \left|\alpha - \frac{a_{n-1}}{b_{n-1}}\right| \\ &=\frac{b_{n-1}}{b_n} e_{n-1} \end{align} $$

which implies superlinear convergence of $\frac{a_n}{b_n}\rightarrow\pi$.

Unfortunately I am not aware of a sequence that converges superlinearly to $\pi$. The best I know gives 16 bits per iterations - linearly. Deconstructing such a sequence into $a$ and $b$ we would have $b_n=2^{16}b_{n-1}$ and $e_n = 2^{-16}e_{n-1}$ which clearly would not satisfy the above inequality.

Still, as I only know a small fraction of all the sequences converging to $\pi$ that exist, looking for one that satisfies this necessary condition and checking whether you can deconstruct it into $a_n$ and $b_n$ might be your best bet.


If we assume $ \pi = \lim_{n \to \infty} [\sum_{i=1}^n \sqrt{n^2-(i-1)^2}]{4 \over n^2}$ (which can be obtained by applying a riemann sum over a half circle function $y = \sqrt{R^2-x^2}$, on the interval $[0,r]$), we have a structure like the one you used for $e$, I guess. It isn't an integer, but can be a hint.

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    $\begingroup$ From your answer, you can not get integer sequences $\{a_n\}$ and $\{b_n\}$. So your answer is wrong. $\endgroup$ – xpaul May 12 '14 at 0:22
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    $\begingroup$ It is not a direct answer, but is produces the irrationality by contradicting the division algorithm, and this is similar to what is sought. One can look at the convergents to $\pi$ given by some continued fraction expansion, say Bounker's, and test if they satisfy the requirements. $\endgroup$ – Rene Schipperus May 20 '14 at 12:16

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