Johann Heinrich Lambert
(1728 - 1777)
That's what you call a theorem...
is irrational !!! (and so
is 2 in
fact for Legendre...)
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
irrationality in 1761 , which we will also
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 !
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
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
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
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
Lemma 2 :
We have, for x with its
tangent defined as,
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).
: 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
To simplify things, we will consider tan()=0,
which is what Legendre used to proove that2was
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
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