Liouville (18091882)
What is now proved was once only imagin'd
(William Blake)
All authors credit Liouville with being the discoverer (creator?) of the first example of a transcendental number, but there is
definite confusion
with regard to
dates
. All are agreed on the date (1844), and even on the approximation theorem (again 1844) that he used, but not on the
actual number
that he first proved to be transcendental.
In
seems
that in his 1844 paper he proved there were transcendental numbers by using his
approximation theorem
and
continued fractions
(indeed anyone who knows both can readily create their own private transcendental numbers at will), but, it
seems
, it wasn't until 1851 that he gave the following example (that many authors
state
he gave in 1844), the number now named the
Liouvillian number
, namely the sum of the infinite series
), the decimal number:
0.
11
000
1
00000000000000000
1
0000...
where the
1
's occur at the
, ... places.
Here is one case (perhaps the
only
case) where a beginner may follow with little difficulty a transcendence
proof
. It all depends on a completely simple, but utterly important observation concerning
how well
algebraic numbers may be
approximated
by rational numbers:
Theorem (Liouville, 1844). Let
be any real algebraic number of degree
n
(
), then there is a positive constant
(i.e. the value of
c
depends
only
on
) such that
>
... (i)
for
all
rational numbers
(
p
and
q
integers, with
q
> 0).
Remark 1. Liouville's theorem is also true when
n
= 1, in other words in the case where
is itself a rational number. There is then an obvious modification that needs to be made in the statement of the theorem: instead of "for
all
rational numbers
", one now needs the necessary restriction
. The validity of (i) is then a complete triviality, for suppose
(
a
and
b
integers,
b
> 0) and that
. Then
=
=
, is at least
.
since the previous numerator,
, is
at least
1 (it can't be 0 since otherwise
)
Thus (i) holds with (say)
.
Remark 2. Inequality (i) is completely trivial if
is not close to
, since, in that case, the left hand side of (i) is not small, whereas
has minimum value 1, and is quite small when
q
is large. In short, one should
appreciate
that the inequality (i) is only of
significance
for rationals
that are
close
to
.
(The) Proof of Liouville's theorem (is easy). Let
be the irreducible polynomial with integer coefficients such that
, and let
p
and
q
be integers with
q
> 0. Then, by the mean value theorem,
= (
)*f '(
) ... (ii)
for some
between
and
(where f '(
x
) is the first derivative of f(
x
)).
If
then (i) holds (trivially) with (e.g.)
, whereas if
, then
, and thus f '(
) is bounded above by a constant
c
' whose value depends only on (numbers associated with)
:
f '(
) =
where
H
is the maximum of the absolute values of the coefficients
. Multiplying through (ii) by
, and using
(since f(
x
) is irreducible of least degree 2, and so has no rational roots), gives
=
*f '(
) <
*
c
'
(i) follows immediately.
___________________
I include the following only to illustrate the
idea
behind the proof of Liouville's theorem. I choose an algebraic
of degree 3, one that is a solution of the (irreducible) equation
:
>
restart;
>
f := x > 6*x^3  14*x^2  x + 11;
>
factor(f(x));
>
plot(f(x), x = 1..2);
>
I choose
to be the largest of those solutions (you can see it is
slightly more
than 1.925...)
>
fsolve(f(x) = 0);
>
and a rational number
that is
near
to
(in the following diagram I have chosen
two
rational numbers that are close to
:

, slightly to the left of
, and

, slightly to the right of
Then, close in
near
, the graph of the function is
almost linear
, and thus (whether
be
or
):
(which is a nonzero rational number  with denominator
 and so has minimum value
)
is
almost
equal to
*f '(
)
It should be clear, then, that
is bounded,
from below
, by a constant multiplied by
.
>
with(plots): with(plottools):
pl1 := plot(f(x), x = 1.89..1.955, thickness=2):
pl2 := textplot([1.93,0.03,`alpha`],align=LEFT):
p1 := 19: q1 := 10:
l1 := line([p1/q1, 0], [p1/q1, f(p1/q1)], color=navy, thickness=2):
p2 := 39: q2 := 20:
l2 := line([p2/q2, 0], [p2/q2, f(p2/q2)], color=brown, thickness=2):
pl3 := textplot([1.9,0.06,`p1/q1`],align=ABOVE):
pl4 := textplot([1.9,0.3,`f(p1/q1)`],align=RIGHT):
pl5 := textplot([1.95,0.02,`p2/q2`],align=BELOW):
pl6 := textplot([1.95,0.35,`f(p2/q2)`],align=LEFT):
display([pl1, pl2, l1, l2, pl3, pl4, pl5, pl6]);
Warning, the name changecoords has been redefined
>
Once one is in possession of Liouville's theorem, transcendental numbers simply fall into ones lap (in fact, with Cantorian hindsight, one may construct an uncountable number of such Liouvilletype numbers). Anyone who is familiar with continued fraction expansions of irrational numbers will see immediately that all one has to do  to produce transcendental numbers  is to
define
numbers whose partial quotients grow in size sufficiently rapidly... I think that is what Liouville did in his 1844 paper (I've never read the original paper, and stand to be corrected), whereas it was in a later (1851) paper that he gave (what many seem to think was in the 1844 one) his wellknown decimal example mentioned earlier:
= 0.
11
000
1
00000000000000000
1
0000...
Proof of transcendence of
(the real number
cannot
be algebraic since it possesses rational approximations that are
incompatible
with it being algebraic; the relevant rational approximations are simply the early, (first
m
terms, varying part), of the infinite sum).
Now
is a rational number
with denominator
, and thus
, where the remainder term,
, is
positive
and
less
than
.
(true for all
m
= 1, 2, 3, 4, ... )
That latter inequality is incompatible with
being an algebraic number: for if
were algebraic of degree
n
then one would have
>
, which is clearly impossible for
m
sufficiently large.
Cantor motivated observation. With Cantorian hindsight one may in fact observe that there are an uncountable number of transcendental numbers of Liouville type (meaning ones whose transcendence may be established by the Liouville approach). Simply let {
} be any infinite sequence in which
or 2 for all
n
. There is an uncountable number of such sequences, and, for any one of those sequences, the number
defined by
is transcendental, and no two such numbers are equal (different sequences produce different numbers,
in this case
).
________________
Actually, there is a way of seeing that the above number is transcendental, without knowing anything about the above approximation theorem. Indeed this alternative way could be understood by a
numerate school pupil
with some ability. All that is needed is that one should know 
really know!
 how to multiply decimals together, as I will now illustrate.
Question
. Which do you think is easier to work out:

the
square
of the infinite decimal number
0.01010101
... ?, or

the
square
of the infinite Liouville decimal number
0.11000100
... ?
If you tried to do the first one longhand (as we all learned at school) I think you would soon run into trouble. Try it! Ask your pupils to try it. The 'carrying' gets to be troublesome...
It shouldn't surprise you to know that the squared decimal will be
periodic
: that's simply because the above decimal  which should be seen as the sum of the
infinite geometric series
with initial term
and 'common ratio'
 is, of course, the decimal expansion of the rational number
, and thus the sought, squared decimal, will be whatever is the decimal expansion of
. But what are the digits in that expansion? First, here are the already seen digits in the infinite expansion of
:
>
with(numtheory):
Warning, the protected name order has been redefined and unprotected
>
pdexpand(1/99);
>
That final "[0, 1]" is to be repeated ad infinitum to produce the decimal expansion of
.
Here are a few others before we see the decimal expansion of
:
>
pdexpand(1/7);
which means that
>
pdexpand(1/28);
which means that
(with only the '571428' being repeated)
>
pdexpand(97/26);
which means that
(with only the '307692' being repeated)
And finally, the decimal expansion of
is (anyone like to guess before I reveal it?):
>
pdexpand(1/99^2);
>
There are 198 digits in the repeating cycle. Don't try  by hand!  working out the decimal expansion of the cube of
: there are 19602 digits in the cycle. Anyone familiar with the relevant Number Theory will know it's to do with the fact that
= 19602 (= 2*
), and the above decimal expansion is to do with the fact that
= 198 (= 2*99):
>
order(10, 99^2);
>
and thus that the periodic block of the decimal expansion of
commences with three 0s, followed by:
>
(10^198  1)/99^2;
>
It might initially surprise the unsuspecting that it's much easier to work out  by hand!  the decimal value of the square of the above Liouvillian decimal. And its cube, and its fourth power, and ... . Of course one sees why it's easier... : those huge blocks of 0's between the 1's. That alone  without having to use Liouville's approximation theorem  allows one to argue that the Louville decimal is transcendental, as I now illustrate.
First I
manufacture
a certain quadratic polynomial,
, for which the above Liouville number (
) is
almost
a root (i.e.
is
almost
a solution of
).
This is
how
I arrived at that quadratic (don't worry if you don't follow the continued fraction reasoning behind it): I began by forming the rational number corresponding to the initial part of the Liouville number; in fact simply the third partial sum
, which is a rational number, and so has a terminating continued fraction expansion:
>
alpha3 := add(1/(10)^(n!), n = 1..3);
>
cfrac(alpha3, quotients);
>
Anyone who knows about continued fractions, and knows the effect of a large partial quotient, will spot that '99', and realise why I now define the infinite periodic continued fraction with partial quotients [0, 9, 11, 99, 9, 11, 99, ... ], having initial 0, followed by the infinitely
repeated
block [9, 11, 99]:
>
cf3 := [[0], [9, 11, 99]];
>
beta3 := invcfrac(cf3);
>
Digits := 10:
evalf(beta3); # you see the 'extra' bit:
>
That number, beta3 (which is close to alpha3, which is close to
), is the solution of a quadratic equation, and that equation is recovered by setting
,
,
, etc:
>
(expand((100*x + 4949)^2)  24601601);
The coefficients obviously have gcd = 200:
>
igcd(10000, 989800, 109000);
>
solve(50*x^2 + 4949*x  545 = 0); # the first is L3
>
Now observe how close
x
= alpha3 is to being a solution of the equation
. I will calculate the value of

at
x
= alpha3, first to 7 decimal places, and then

at
x
= alpha3, to 8 decimal places
and observe
proximity to 545
:
>
evalf(50*alpha3^2 + 4949*alpha3, 7);
evalf(50*alpha3^2 + 4949*alpha3, 8);
>
Thus
x
= alpha3 is
almost
a solution of the equation
, and we reasonably ask:
could
be
a solution?
Without using Liouville's approximation theorem we may easily see that it isn't. I'm not going to
'dot every i and cross every t'
, but rather give pointers and invite you to ponder on your own.
What one wants to do is to somehow calculate
and
see
that it isn't 0.
Note first of all that Maple can't do that
exact
calculation for us (though it
can
do many infinite sums):
>
restart;
alpha := sum(1/10^(n!), n = 1..infinity); # 'add' won't 'work'
>
50*alpha^2 + 4949*alpha  545;
>
But one can calculate, or rather
see
the
essential
part of the decimal expansion of
by hand, and in the process
see
that
. The essential point is that it's actually quite easy to square

much easier
than squaring the earlier challenge decimal .0101010101...  in the sense that one can
see
what its decimal expansion
is
(this is difficult to type up, since I want vertical holds on the place values in the good oldfashioned way of multiplying and 'carrying'). For each line you should keep in mind where the next '1' is occurs, and the next, and the next, ... :
.11000100000000000000000100000000... multiplied by
.11000100000000000000000100000000... is
.011000100000000000000000100000000... plus
.0011000100000000000000000100000000... plus
.00000011000100000000000000000100000000... plus
.00000000000000000000000011000100000000000000000
1
00000000... plus
ad infinitum
, which adds up to
= .01210022000100000000000022000200000000000000000
1
00000000... , thus
= .6050110000500000000000110001000000000000000000
50
0000000...
= 544.3949490000000000000049490000000000000000000000000000... , which adds up to
544.9999600000500000000049600001000000000000000000
50
0000...
and what I am attempting to draw your eye's attention to is the occurence of that
50
, which is going to occur over and over again (as a consequence of which it is obvious that
cannot
be equal to 545, a whole number). I invite you to think about
why
that kind of thing happens, not just in
this
particular case but in the general case. The above kind of numerical play is only to give one a feeling for this, but you need to start thinking about the question: where, in
place
, do we get 1s occuring in the decimal expansion of (not just)
(but
), and what then if the effect of multiplying by coefficients, and adding, ... Can we end up getting '0' after it's all totted up?
The numerical play has a more formal aspect to it (which should be preceded by play), and that formal aspect is the
infinite
version of (e.g.)
,
namely
+
+
+
ad infinitum
so as to see where 1s occur in
(each of those contributes a '
1
' to the decimal expansion) + the 'cross terms', each of which contributes a '
2
' to the decimal, and to see how isolated those
1
s are in terms of their
neighbouring nonzero
digits.
I will end with a lovely example from Conway & Guy's
The Book of Numbers
. They observe that the Liouville number is
almost
a solution of the 6th degree polynomial equation
:
>
Digits := 24: alp := evalf(add(1/(10)^(n!), n = 1..4)):
10*alp^6  75*alp^3  190*alp;
>
They don't tell their readers how that polynomial was arrived at, and I leave it to my reader to speculate.