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?



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.

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    $\begingroup$ How did you get $\rho_1 < 1$? $\endgroup$ – Isomorphism Sep 6 '15 at 7:37
  • $\begingroup$ 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, $\rho_1 = 1$. $\endgroup$ – templatetypedef Jan 31 '19 at 23:32

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

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    $\begingroup$ 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. $\endgroup$ – Sandeep Silwal Aug 12 '14 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 '14 at 23:25

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