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Home , Charles Hermite, Ferdinand von Lindemann

Nyle Sutton π and e MATH 492

1 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 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.

2 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 as the ” 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 1 infinite : P∞ . 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, 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, was able to prove π to be transcendental, which finally answered the question of whether or not it is possible to ”square the circle”.

3 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 xn(1 − x)n Statement We let n ≥ 1 and define a function f(x) = . Then the following hold: n! 1 (1) f(x) is a polynomial of the form f(x) = P2n c xi, where each c ∈ . n! i=n i i Z 1 (2) For 0 < x < 1, we have 0 < f(x) < . n!

(3) The derivatives f (k)(0), f (k)(1) ∈ Z, ∀k ≥ 0.

1 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) = c , which is an integer. So f (k)(0) is always n! k an integer. Now we notice that f(x) = f(1 − x), so f (k)(x) = (−1)kf (k)(1 − x). Therefore, (k) k (k) f (1) = (−1) f (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 ∈ +. Let r = where s ∈ + and t ∈ . Then if er Z t Z Z is rational, then (er)t = es is rational, and hence the contrapositive holds. a as2n+1 Proof by Contradiction Suppose es = for a, b ∈ +. Let n ∈ such that < 1. Let b Z N n! F (x) := s2nf(x) − s2n−1f 0(x) ± · · · , where f(x) is the function of the Lemma. We first notice that F (x) satisfies the differential F 0(x) + sF (x) = s2n+1f(x). We d also notice that [esxF (x)] = sesxF (x) + esxF 0(x) = s2n+1esxf(x). Now define a number dx R 1 2n+1 sx sx 1 N := b 0 s e f(x)dx. Then N = b [e F (x)]0. Evaluating this gives N = aF (1) − bF (0), which is an integer by part [3] of the Lemma. bs2n+1es as2n+1 We now notice by part [2] of the Lemma, 0 < N < = < 1. Hence N is not 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 π2 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. a πan Proof by Contradiction Suppose π2 = for a, b ∈ +. Let n ∈ such that < 1. Let b Z N n! F (x) := bn π2nf(x) − π2n−2f (2)(x) ± · · · , where f(x) is the function of the Lemma. We first notice that F (x) satisfies the differential equation F (2)(x) + π2F (x) = bnπ2n+2f(x). d We also notice that [F 0(x)sinπx − πF (x)cosπx] = F (2)(x) + π2F (x) sinπx = bnπ2n+2f(x)sinπx = dx  1 2 n R 1 n 1 0 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! 2 which is a contradiction, and π is irrational. 

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4 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 2 polynomials. Before the mid-19th century, mathematicians puzzled over whether or not this definition is trivial. In 1844, 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 : 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: α1 αn Statement If α1, . . . , αn are distinct algebraic numbers, then {e , . . . , e } 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 α e 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 corol- lary, 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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5 Open Problems

Even using methods taken from 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, πe, 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.

6 Acknowledgments

1 Aigner, M., & Ziegler, G. M. (2000). Proofs from the book. The Australian Mathematical Society.

2 Baker, A. (Ed.). (1990). Transcendental . Cambridge University Press.

3 Gel’fond, A. O. (2003). Transcendental and algebraic numbers. Courier Dover Publica- tions.

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