Download 1 Introduction 2 History 3 Irrationality

Nyle Sutton
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π and e
MATH 492
Introduction
I’m sure that most of us have heard of π and e at some point of our mathematical careers.
We all know that these two numbers are irrational, and a lucky few of us may even know that
these two numbers are transcendental. However, how many of us have ever seen a proof of these
facts? Throughout this talk, I will give proofs that π and e have these properties and hopefully
convince you that these magical numbers truly deserve to be ”superstars” in the mathematical
world.
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History
First I will give you a brief timeline of the history of π and e in relation to their introduction
and initial proofs of irrationality and transcendence.
π was introduced in ancient times by the ancient geometers. They knew this mysterious
number as the ”circle constant” because it arose as the ratio between the circumference and
diameter of any circle.
Fast forward to 1683, when Jacob Bernoulli came across e during his research into compound
interest. At this time, e had not received its name, and instead was known as the result of the
P
1
.
infinite series: ∞
n=0
n!
A little over half a century later in 1737, Leonhard Euler, a student of Bernoulli’s younger
brother, gave e its name and showed that it is irrational through the computation of a continued
fraction.
Shortly thereafter in 1761, Johann Heinrich Lambert proved that π is irrational by showing
if x 6= 0 and x ∈ Q, then tan(x) 6 ∈ Q.
Over a century later in 1873, Charles Hermite finally proved that e is transcendental. This
was an important result, as it had been known that transcendental numbers exist, and e was the
first number to be shown to be transcendental, without having been constructed to have that
property.
Finally, in 1882, Ferdinand von Lindemann was able to prove π to be transcendental, which
finally answered the question of whether or not it is possible to ”square the circle”.
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Irrationality
Before we get into the irrationality proofs, we need to prove a lemma which we will then use in
both proofs.
3.1
Preliminary Lemma
Statement We let n ≥ 1 and define a function f (x) =
(1) f (x) is a polynomial of the form f (x) =
(2) For 0 < x < 1, we have 0 < f (x) <
xn (1 − x)n
. Then the following hold:
n!
1 P2n
ci xi , where each ci ∈ Z.
n! i=n
1
.
n!
(3) The derivatives f (k) (0), f (k) (1) ∈ Z, ∀k ≥ 0.
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Nyle Sutton
π and e
MATH 492
Proof of Lemma Part [1] of the lemma is clear by multiplying out the expression of the
function. Part [2] is also clear because 0 < x < 1 implies 0 < xn < 1 and 0 < (1 − x)n < 1.
Hence the numerator of the function is always between 0 and 1, so the inequality holds.
We now consider part [3] of the lemma. First, we notice that from part [1], if k < n or
k!
k > 2n, f (k) (0) = 0. If n ≤ k ≤ 2n, then f (k) (0) = ck , which is an integer. So f (k) (0) is always
n!
an integer.
Now we notice that f (x) = f (1 − x), so f (k) (x) = (−1)k f (k) (1 − x). Therefore,
f (k) (1) = (−1)k f (k) (0), which is an integer. 3.2
e is Irrational
Statement er is irrational ∀r ∈ Q r {0}.
Notes We notice that this statement is stronger than proving e irrational. We also notice that
s
it is sufficient to show es is irrational ∀s ∈ Z+ . Let r = where s ∈ Z+ and t ∈ Z. Then if er
t
is rational, then (er )t = es is rational, and hence the contrapositive holds.
a
as2n+1
Proof by Contradiction Suppose es = for a, b ∈ Z+ . Let n ∈ N such that
< 1. Let
b
n!
F (x) := s2n f (x) − s2n−1 f 0 (x) ± · · · , where f (x) is the function of the Lemma.
We first notice that F (x) satisfies the differential equation F 0 (x) + sF (x) = s2n+1 f (x). We
d sx
also notice that
[e F (x)] = sesx F (x) + esx F 0 (x) = s2n+1 esx f (x). Now define a number
dx
R1
N := b 0 s2n+1 esx f (x)dx. Then N = b [esx F (x)]10 . Evaluating this gives N = aF (1) − bF (0),
which is an integer by part [3] of the Lemma.
as2n+1
bs2n+1 es
=
< 1. Hence N is not
We now notice by part [2] of the Lemma, 0 < N <
n! +
n!
s
an integer, which is a contradiction, and e is irrational ∀s ∈ Z . Remark Letting r = 1, we get that e is irrational.
3.3
π is Irrational
Statement π 2 is irrational.
Notes Again, we notice this is a stronger statement than what we are setting out to prove. This
is because if a ∈ Q, then a2 ∈ Q.
πan
a
< 1. Let
Proof by Contradiction Suppose π 2 = for a, b ∈ Z+ . Let n ∈ N such that
b
n!
F (x) := bn π 2n f (x) − π 2n−2 f (2) (x) ± · · · , where f (x) is the function of the Lemma.
We first notice that F (x) satisfies the differential equation F (2) (x) + π 2 F (x) = bn π 2n+2 f (x).
d
We also notice that
[F 0 (x)sinπx − πF (x)cosπx] = F (2) (x) + π 2 F (x) sinπx = bn π 2n+2 f (x)sinπx =
dx
1
R1 n
1 0
2 n
π a f (x)sinπx. Define a number N := π 0 a f (x)sinπxdx.. Then N =
F (x)sinπx − F (x)cosπx .
π
0
Evaluating this gives N = F (0) + F (1), which is an integer by part [3] of the Lemma.
πan
We now notice by part [2] of the Lemma, 0 < N <
< 1. Hence N is not an integer,
n!
which is a contradiction, and π 2 is irrational. 2
Nyle Sutton
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π and e
MATH 492
Transcendence
We can now move on to the transcendence proofs. However, we must first define what we mean
by saying a number is transcendental.
4.1
Algebraic Numbers
We define an algebraic number to be a root of a nonzero polynomial with rational coefficients.
1 √
Some examples are 0, 1, −1, , 2, i, 3 + i. I will leave it as an exercise to find the suitable
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polynomials. Before the mid-19th century, mathematicians puzzled over whether or not this
definition is trivial.
In 1844, Joseph Liouville gave a proof that there exist numbers which do not fit this definition.
We call these numbers transcendental numbers.
In 1851, Liouville demonstrated the first concrete
P∞
−k!
example of a transcendental number: k=1 10 . Later papers showed that the set of algebraic
numbers is countably infinite. So not only does there exist transcendental numbers, but there
are a lot more of them than there are algebraic numbers.
4.2
Lindemann-Weierstrass Theorem
In 1882, Ferdinand von Lindemann was able to give proof to the following theorem:
Statement If α1 , . . . , αn are distinct algebraic numbers, then {eα1 , . . . , eαn } forms a linearly
independent set over the algebraic numbers.
I will not show the proof in this talk, but if you would like to see it, go to Wikipedia and
look up the Lindemann- Weierstrass Theorem. There is a good proof of the theorem there.
What is important to note is that this theorem gives us a method to prove certain numbers are
transcendental. We will use this fact to prove a few corollaries about e and π.
4.3
Corollaries
Statement If α is a nonzero algebraic number, then eα is transcendental.
Proof We note that α 6= 0. Thus by the theorem, {1, eα } is linearly independent over the
algebraic numbers. In particular, let β be an algebraic number. Then eα − β 6= 0, and hence
eα 6= β. Remark Letting α = 1, we get that e is transcendental.
Statement π is transcendental.
Proof by Contradiction We first have to note the fact that the algebraic numbers are closed
under multiplication. Suppose π is algebraic. Then iπ is algebraic. Hence by the first corollary, eiπ is transcendental. But eiπ = −1, which is algebraic. This is a contradiction, and π is
transcendental. 4.4
What does transcendence offer?
Up to now, transcendence seems to be a superfluous trait of numbers. However, it gives us a
very powerful method of showing numbers to be irrational. This can be seen by noting that
every rational number is algebraic. Hence every transcendental number is irrational. Thus by
proving e and π to be transcendental, we have shown a second proof of their irrationality.
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Nyle Sutton
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π and e
MATH 492
Open Problems
Even using methods taken from Algebra and using tricks such as defining functions or proving
a number transcendental, irrationality is difficult to prove. As such, there are a lot of numbers
for which we don’t know if they are rational or not. Some examples of these are π + e, π − e,
πe, π e , and ln(π). There is a lot of room for discovery, and finding a reliable method for proving
the irrationality of numbers would be a major find.
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Acknowledgments
1 Aigner, M., & Ziegler, G. M. (2000). Proofs from the book. The Australian Mathematical
Society.
2 Baker, A. (Ed.). (1990). Transcendental number theory. Cambridge University Press.
3 Gel’fond, A. O. (2003). Transcendental and algebraic numbers. Courier Dover Publications.
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