I am trying to find Lambert's original proof that $\pi$ is irrational. Wikipedia has a little description but it is quite lacking. Can someone direct me to Lambert's original proof or post his proof here?



3 Answers 3


Here is the complete proof:

Part I Derivation of continued fraction for $\tan(x)$

Part II Proof of Irrationality of continued fractions

Part III Proof of Irrationality of $\pi$.

Part I

We show

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

First we treat Lambert's continued fraction expression for the quotient of two power series. Let $F_0$ and $F_1$ be two power series we may assume without loss of generality that they both have nonzero constant term and that in fact both have constant term of $1$.

Define $$F_1-F_0=b_1 x F_2$$ $$F_2-F_1=b_2 x F_3$$ $$F_3-F_2=b_3 x F_4$$ $$F_{n+1}-F_n =b_{n+1} x F_{n+2}$$

where at each step we make the assumption that $F_n$ has a constant term of $1$. In general we would simply get higher powers of $x$ in the equation. Define $$G_n=\frac{F_{n+1}}{F_n}$$ then we have by dividing the last equation by $F_{n+1}$, $$G_n=\frac{1}{1-b_{n+1}xG_{n+1}}$$

therefore $$\frac{F_1}{F_0}=G_0=\cfrac{1}{1- \cfrac{b_1 x}{1-\cfrac{b_ 2 x}{1-\cfrac{b_3 x}{1-\cdots}}}}$$

Now we shall apply this to obtain continued fraction expansions for several functions including $\tan x$.

Let us define the particular power series, \begin{equation*} \begin{split} F_n =&1+ \frac{x}{1! (\gamma +n)} +\frac{x^2}{2! (\gamma +n)(\gamma +n+1)} +\frac{x^3}{3! (\gamma +n)(\gamma +n+1)(\gamma +n+2)} \cdots \\ &1+ \sum\limits_{k=1}^{\infty} \frac{x^k}{k! (\gamma +n)\cdots (\gamma +n+k-1)} \end{split} \end{equation*}

Then we have $$F_{n+1}-F_n =-\frac{x}{(\gamma +n)(\gamma +n+1)} F_{n+2}.$$

Thus with $b_n = -\frac{x}{(\gamma +n)(\gamma +n+1)}$ we get \begin{equation*} \begin{split} \frac{F_1}{F_0}=&\cfrac{1}{1+\cfrac{\frac{x}{(\gamma)(\gamma +1)}}{1+\cfrac{\frac{x}{(\gamma +1)(\gamma +2)}}{1+\cfrac{\frac{x}{(\gamma +2)(\gamma +3)}}{1+\cdots}}}} \\ &=\cfrac{\gamma}{\gamma +\cfrac{x}{(\gamma+1)+\cfrac{x}{(\gamma +2)+\cfrac{x}{(\gamma +3)+\cdots}}}} \end{split} \end{equation*}

Now let us set $\gamma=\frac{1}{2}$ and instead of $x$ write $-\frac{x^2}{4}$

Then $$F_1= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k+1)!}=\frac{\sin x}{x}$$ and

$$F_0= \sum\limits_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}=\cos x.$$

Putting this altogether we get

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

Part II

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.

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. Part III

This expression leads to the following fundamental result.

$\pi$ is irrational.

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.

  • 4
    $\begingroup$ How did you get $\rho_1 < 1$? $\endgroup$ Sep 6, 2015 at 7:37
  • 2
    $\begingroup$ Following up on @Isomorphism's comment, if you assume that $\rho_1 < 1$, why can you then plug in $x = \frac{\pi}{4}$ and apply the theorem? In that case, $\rho_1 = 1$. $\endgroup$ Jan 31, 2019 at 23:32
  • $\begingroup$ It seems to me that the concerns expressed by Isomorphism and @templatetypedef are valid ones. I am posting an answer to address those concerns. $\endgroup$
    – user43208
    Nov 7, 2022 at 5:06

There is a proof here, but i am not sure if it's helpful or not: http://www.pi314.net/eng/lambert.php

Here is a link to the original article: http://www.kuttaka.org/~JHL/L1768b.pdf

  • 1
    $\begingroup$ I don't know what to make of the proof given in your first link. The proof given of lemma 1 does not make any sense. $\endgroup$ Aug 12, 2014 at 22:43
  • $\begingroup$ @Sandeep Silwal Lambert's original proof is long and complicated with respect to some other proofs (as you can see from the original article). All 'Lambert's proofs' i came across have gaps in them (for me at least), since they try to shorten it. Have a look at this: paramanands.blogspot.com.tr/2011/05/… and the next post paramanands.blogspot.com.tr/2011/05/…. Maybe this will help. $\endgroup$
    – Alistair
    Aug 12, 2014 at 23:25

This post is meant to supplement the useful answer of Rene Schipperus, in view of the unanswered comments by Isomorphism and templatetypedef under Schipperus's answer (which seem to me to be valid concerns regarding Part II).

Claim: let $a$ be a positive integer, and let $b_1, b_2, b_3, \ldots$ be a sequence of positive integers, and assume that for all $n$ we have

$$0 < \cfrac{a}{b_n - \cfrac{a}{b_{n+1} - \cfrac{a}{b_{n+2} - \ldots}}} < 1.$$ Then each of these continued fractions is irrational.

Putting aside questions of convergence, the proof was essentially given by Schipperus in his answer. Without loss of generality, we prove the continued fraction in the case $n=1$ is irrational. Suppose it were rational, say $\frac{c_2}{c_1}$ where $c_1, c_2$ are positive integers. Then $c_1 > c_2$ by our assumption in the case $n=1$. Write

$$\frac{c_2}{c_1} = \frac{a}{b_1 - e_2}$$ so that $e_2$ is the continued fraction above in the case $n=2$. By our assumption, we have $0 < e_2 < 1$. Solving for $e_2$, we have

$$e_2 = \frac{c_2b_1 - ac_1}{c_2}$$ so that defining $c_3 = c_2b_1 - ac_1$, we have that $c_3$ is a positive integer less than $c_2$. Proceeding in this way, we produce a strictly decreasing sequence of positive integers $c_1 > c_2 > c_3 > \ldots$, contradiction.

Now we finish off Schipperus's proof as follows. Consider the sequence of numbers

$$\cfrac{a^2}{(2k+1)b - \cfrac{a^2}{(2k+3)b - \cfrac{a^2}{(2k+5)b - \ldots}}}$$ for some large $k$. It is enough to prove that one of these numbers is irrational. But clearly these numbers decrease as $k$ increases, and for large enough $n$ we reach a point where for all $k \geq n$, all these numbers are in the range strictly between $0$ and $1$. This completes the proof.


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