Home > Analysis, Problem Solving > Pi and e are irrational/transcendental

Pi and e are irrational/transcendental


Two of the most used constants in mathematics and mathematical analysis are the numbers \pi and e. These constants, defined in the following, are proven to be irrational. Moreover, these numbers are transcendent, which means that there is no polynomial equation with rational coefficients which have \pi or e as a root.

The proof of these facts will be presented below.

First we prove that e is irrational. Else, suppose e=\frac{a}{b}, with a,b positive integers. This means that n!be=n!a,\ \forall n \geq 0. Since e=\sum_{k=0}^\infty \frac{1}{k!} we have \displaystyle bn!(1+...+\frac{1}{n!})+b(\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...)=n!a. The right hand side is an integer. The left hand side has two terms. The first one is an integer, and the second one is between b/(n+1) and b/n, which is not an integer. Why is that? The first inequality is obvious: a sum of positive terms is greater than one of its terms. For the second inequality, see that b(\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...)<b(\frac{1}{n+1}+\frac{1}{(n+1)^2}+...)=b \frac{1}{n+1-1}=b/n. The contradiction assures us of the irrationality of e.

Using a similar method, we can prove even more, that e^2 is irrational. To do this, assume that e^2=a/b, which is equivalent to be=ae^{-1}. Using the same series expansion for e as before, and the similar expansion 1/e=\sum_{k=0}^\infty (-1)^k \cdot 1/k! we see that n!be=n!ae^{-1} can again be written in the following way: K_b+P_b=K_a+P_a, where K_a,K_b are integers (n! multiplied with the respective parts of the series with denominator less or equal to n!), and P_b and P_a are the “fractional parts” for which we can prove that b/(n+1)

For e^4\in \Bbb{R}\setminus \Bbb{Q}, a similar strategy can be used. See Proofs from the Book. But let’s get to serious business and prove that e^r is irrational for all rational nonzero r. To do this, we need the following

Lemma. For some fixed n \geq 1, let \displaystyle f(x)=\frac{x^n(1-x)^n}{n!}. This function has the following propetrties:
(i) f is a polynomial of the form f(x)=\frac{1}{n!} \sum_{k=0}^{2n} c_ix^i, where c_i are integers.
(ii) For 0<x<1 we have 0<f(x) as^{2n+1}.
We can see that F^\prime(x)=-sF(x)+s^{2n+1}f(x), which leads us to \frac{d}{dx}(e^{sx}F(x))=se^{sx}F(x)+e^{sx}F^\prime(x)=s^{2n+1}e^{sx}f(x). Assume that e^s=a/b with a,b positive integers. Then N=\displaystyle b\int_0^1 s^{2n+1}e^{sx}f(x)dx=b(e^sF(1)-F(0))=aF(1)-bF(0), which is an integer from part (iii) of the lemma.
But the lemma gives us the estimates
\displaystyle 0<N=b\int_0^1 s^{2n+1}e^{sx}f(x)dx<bs^{2n+1}e^{s}\frac{1}{n!}=\frac{as^{2n+1}}{n!} < 1 Contradiction. Therefore e^s is not rational.
If for some nonzero rational r=a/b,\ a>0 we have e^r \in \Bbb{Q}, then (e^{r})^b=e^a is also rational. Contradiction.

The above argument works in showing that \pi^2 is not rational. Assuming \pi^2=a/b, define
F(x)=b^n(\pi^{2n}f(x)-\pi^{2n-2}f^{(2)}(x)+\pi^{2n-4}f^{(4)}(x)-...) which satisfies F^{(2)}(x)=-\pi^2F(x)+b^n\pi^{2n+2}f(x). From the lemma, we get that F(0),F(1) are integers. Differentiating (F^\prime(x)\sin \pi x- \pi F(x)\cos \pi x) we get \pi^2 a^n f(x)\sin \pi x.
Therefore N=\pi \int_0^1 a^n f(x)\sin \pi x d x=F(0)+F(1) is a positive integer.
Part (ii) of the lemma implies that 0< N <\frac{\pi a^n}{n!} which can be made arbitrarily small for n great enough. This contradicts the fact that N is a positive integer. Therefore \pi^2 and then, obviously, \pi are irrational.

A proof of the fact that e and \pi are both transcendental can be found here.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: