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)}+...). 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 we have $0.
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 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.