All formulas
Num. approx.
Misc. math.
Delirium !

Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013

Pi-Day in
Pi Day Countdown
Home Version history Guestbook Who I am Some pictures (fr) Acknowledgements Last modifications Contact

Cette page en français This page in English

Johann Heinrich Lambert
(1728 - 1777)

That's what you call a theorem...

is irrational !!! (and so is 2 in fact for Legendre...)

Biographical notes

Johann Lambert came from a modest background. A self-taught person, he left school when he was 12 years old, and developed an imaginative and complete mind. Some people actually say that Frederick the Great asked him one day which science he was the best at. Very humbly, Lambert replied: "All" !
Working on the beguinnings of the non-Euclidian geometries, but also interested in philosophy and physics, Lambert is famous for demonstrating Pi's irrationality in 1761 , which we will also do!


In fact, Pi 's irrationality is an expected result but also very useful, because it's almost the only one that can give us information about Pi 's decimal places: These aren't periodic !
Lambert actually demonstrated the following theorem : if x#0 is rational, then tan(x) is irrational.
Moreover tan(/4)=1 therefore /4 and thus are irrationnal !


Lambert's demonstration (1761) is a bit complicated, but let's try and summarize it!
In fact, the other proofs I found on the net use another method, always the same one... (see Links)
So, lets vary our pleasures!

The principle is to find a tan(x) development which has particular characteristics.

Lemma 1 :

Let's consider the continuous fraction x, convergent and unlimited :
with ai and bi relative integers.
If ai<bifrom a certain i value, then x is irrational.

Demonstration :

Let's suppose that already from the i=1 value, we have ai<bi which does not eliminate generalities.

For iN*, we then have bi-1<bi+<bi+1 and so as ai and bi are integers separated from at least one unit, we then have : .
The extra term compared with the initial hypothesis has its absolute value smaller than 1 so it can't change the sign of the fraction (because bi is an integer).
This indicates to us that the sign has not changed and so we can conclude that has the same sign as . Moreover, its absolute value is smaller than 1 according to the above inequality.

In a similar way, we obtain that has the same sign as and has its absolute value smaller than 1.

By immediate descending recurrence, we can in fact write that x is of the sign of ,
and of a module less than 1 (x1).

For x=1, the development isn't interesting to study because of a very particular type......
So let's suppose thatx<1 with x rational :

As in the previous study, p1 has the same properties as x,which means p1<1 and so we can conclude r< p.
But then ! By repeating this process we construct an infinite series of fractions, their numerators being integers with a strictly decreasing module, which is absurd !
We must therefore conclude that x is irrational.

Lemma 2 :

We have, for x with its tangent defined as,

Demonstration :

We use for this the sin and cos development:

If we write tan(x) under the form:
and in the same way, we can then write : . By repeating this process we therefore build : .

By almost immediate recurrence we then have

Reciprocally, we must also verify (which I do not want to do !) that the fraction we get effectively converges towards tan(x). The principle is not exactly the same as the one exposed for the demonstration of the continuous fraction of Lord Brounker. With the notes of this last site, in our case, we must show that the reductions Pn et Qn converge uniformly respectively towards sin(x) and cos(x).

Lambert's theorem : is irrational

For this last result, I will not use the traditional considered by Lambert. In fact I have still not understood how this result can be used while we have excluded the x=1 case in the initial demonstration (but I must be stupid...).

To simplify things, we will consider tan()=0, which is what Legendre used to proove that2was irrational.
by taking x=, we then have:
Let's then take note that (2k+1)>2 as soon as k=5 and so on account of the 1, 3/2 lemma, and so 2 is irational (that's Legendre's theorem). Finally, we also have : is irrational.

back to home page