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!
Around
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
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 *a*_{i} and *b*_{i} relative integers.
If *a*_{i}<b_{i}from a
certain *i* value, then *x* is
irrational.
Demonstration :
Let's suppose that already from the *i=1 *value,
we have *a*_{i}<b_{i} which
does not eliminate generalities.
For *iN**, we then
have *b*_{i}-1<b_{i}+<b_{i}+1 and so
as *a*_{i} and *b*_{i} 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 *b*_{i} 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 that*x<1* with *x* rational :
As in the previous study, *p*_{1} has the same properties as *x*,which
means *p*_{1}<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 *P*_{n} et *Q*_{n} 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 that^{2}was
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.
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