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 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. 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 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 1 infinite series: P1 . 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 2 Q, then tan(x) 62 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". 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 2 . 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) 2 Z; 8k ≥ 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 8r 2 Q r f0g. 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 8s 2 +. Let r = where s 2 + and t 2 . 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 2 +. Let n 2 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 equation 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 8s 2 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 2 Q, then a2 2 Q. a πan Proof by Contradiction Suppose π2 = for a; b 2 +. Let n 2 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. 2 Nyle Sutton π and e MATH 492 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 p 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, 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 P1 −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: α1 αn Statement If α1; : : : ; αn are distinct algebraic numbers, then fe ; : : : ; e g 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, f1; eαg 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. 3 Nyle Sutton π and e MATH 492 5 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, 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 number theory. Cambridge University Press. 3 Gel'fond, A. O. (2003). Transcendental and algebraic numbers.