Tom Niedzielski Presents
EXTREMELY RATIONAL
PROOFS OF IRRATIONALITY
Clever Techniques and Profound Ideas
Topics to be Covered







Joseph Liouville
Proof: Irrationality of e
Proof: Irrationality of e2
Charles Hermite
Hermite’s Helpful Lemma
Proof: Irrationality of π2
Summary
Joseph Liouville (1822-1901)
Life:



Father was a captain under Napoleon
Taught at many different institutions
Public failures hurt him deeply
Achievements:


Most famous for number theory and
Differential Equations
Major works in transcendentals
The Irrational Number e

1
e is. e = 
, which
is
n
n 0 n!
x
x
formula e =  . It’s
First, we’d better go over what
derived from the more general
n 0 n!
used in many branches of mathematics, especially
when describing exponential growth or decay.
The easiest way to discover that e is irrational is to
assume it is rational, and then look for a contradiction.
The Irrational Number e
So, assume e = a/b for integers a,b>0. To eliminate
the RHS fraction, we multiply by n!b on both sides.
Now we see n!be = n!a. n!a = an integer, as we’re
only multiplying integers. n!be is more complicated.
n
1
*
,
k  0 k!

1
* k
,
 n 1 k!
It breaks into bn!
an integer, plus bn!
1
1
b
(

which = n  1 (n  1)(n  2)  ...).
Fortunately, we can narrow down the range this
converges to for large values of n.
The Irrational Number e
We compare b( n 1 1  (n  1)(1n  2)  ...) to a few other values:
b
b
b
b
b
b
b


 ... 


 ... 
2
3
n  1 n  1 (n  1)(n  2)
n  1 (n  1) (n  1)
n
1
1

 ...)
n  1 (n  1)(n  2)
This shows that
is trapped between the
nearest rational values, and thus cannot be an integer.
Of course, this means that the entire left hand side is
not an integer. Since the right hand side is, the two
cannot be equal. This is our contradiction.
Therefore, e is irrational.
b(
Is
2
e
Also Irrational?
The previous techniques can be extended to e2.
Again, we assume e2 = a/b. We then rewrite the
equation as be = ae-1, and multiply by n! to get
n!be = n!ae-1. As we discovered last time, n!be is just
above an integer, coming from the right as n->∞.
Let’s look at the other side. It also breaks down into
integer and non-integer parts.
Is
2
e
Also Irrational?
The non-integer part is
1
1
1
n+1


 ...).
(-1) n!a(
(n  1)! (n  2)! (n  3)!
In the case of even n,
a
1
1
1
1
n+1

(
1




...
-(a/n) < (-1) n!a((n 1)! (n  2)! (n  3)! ) < n  1 n ) < 0.
Notice that for increasing even n, this value becomes a
small negative number, making the RHS slightly lower
than an integer. Since the LHS is slightly above an
integer, they can’t be equal. This is our contradiction.
Charles Hermite (1822-1901)
Life:



Born to a merchant family
Deformity kept him out of wars, schools
Taught from 1847 to 1897
Achievements:



Proved e transcendental in 1873
Solved the general quintic in 1858
Discovered the Hermitian Forms
A Very Useful Lemma
x n (1  x) n
n!
For some fixed n ≥ 1, let f(x) =
This function has three useful properties:
1 2n
1.
f(x) is a polynomial of the form f(x) = n!  ci xi
i n
1
2.
When 0 < x < 1, 0 < f(x) < n!
3.
For all k ≥ 0, the derivatives f(k)(0) and f(k)(1) are
integers.
The third property results from the fact that f(k)(0 or 1)
(where k ≥ 0) goes to 0 unless n ≤ k ≤ 2n, for
which f(k)(0 or 1) is an integer.
A Somewhat Simple Proof for π2
As in our proofs of e and e2, we begin by assuming
the opposite of what we want to prove.
Assume π2 = a/b for integers a,b > 0.
This step is a little different: We now define the new
function F(x) = bn(π2nf(x) - π2n-2f(2)(x) + π2n-4f(4)(x) …)
F(x) has a useful feature:
F’’(x) = -π2F(x) + bnπ2n+2 f(x)
We’ll use this soon.
A Somewhat Simple Proof for π2
Then we take a useful derivative:
d
[F’(x)sin(πx) – πF(x)cos(πx)] = F’’(x)sin(πx) +
dx
πF’(x)cos(πx) - πF’(x)cos(πx) + π2F(x)sin(πx) =
sin(πx)(F’’(x) + π2F(x)) = bnπ2n+2 f(x)sin(πx) =
bn(an/bn) π2f(x)sin(πx) = anπ2f(x)sin(πx).
Now we’re going to use this
information to define
1
another function: N = a n 0 f ( x) sin(x)dx . This in turn =
1
[ F ' ( x) sin(x)  F ( x) cos( x)]10 = F(0) + F(1) = an integer.

A Somewhat Simple Proof for π2
We also know that N is positive, since it is the integral
of a positive region.
a n
Also, we know that N < n! . This is because for 0 < x
<1, 0 < f(x) < 1/n!, and sin(πx) ≤ 1. Notice that
large n will make na! < 1. Since no integers lie
between 0 and 1, N can’t be an integer. N can’t both
be and not be an integer, so we’ve found our
contradiction.
n
A Somewhat Simple Proof for π2
Therefore, π2 is not a rational number.
Note: When we work with irrationality proofs, though
x irrational ≠> x2 irrational, x irrational => x is
irrational.
We can use this and the fact we just proved to
conclude that since π2 is irrational, π is irrational.
In Summary



We have shown that e, e2, π, and π2 are all
irrational numbers.
In each of our cases, proof by contradiction
provided a clean result.
As mathematics advances, it is often beneficial to
revisit old ideas to see if they can be handled
better.
Sources
1.
2.
3.
4.
Proofs from The Book, Chapter 6 (Proofs)
http://www-groups.dcs.stand.ac.uk/~history/Biographies/Hermite.html
(Hermite’s Biography)
http://www-groups.dcs.stand.ac.uk/~history/Biographies/Liouville.html
(Liouville’s Biography)
Dr Biebighauser (Advice on Design & Content)