Two Proofs of the Irrationality of Euler’s Constant

Recently, it was \(\pi\)-day, and my little brother Daniel’s birthday. However, much irreverence to the spirit of \(\pi\)-day celebration, we shall discuss the irrationality of Euler’s constant \(e\). While reading Apostol’s Mathematical Analysis, I came across a proof of the irrationality of \(e\) quite early on, in Theorem 1.11.¹ This proof was discussed much later in Rudin’s Principles of Mathematical Analysis, in the section dedicated to Euler’s constant in the third chapter (on series and sequences).² Although I have previously stated that Rudin’s laconic writing style may not be best suited for me to read, I prefer Rudin’s more straightforward approach in this proof in contrast to Apostol’s proof.

First, we assume the series definition of \(e\): \[e = \sum_{k=0}^{\infty} \frac{1}{k!}\] where \(n!\) is defined to be the factorial of \(n\) and \(0! = 1.\) Immediately, one could notice that since \(n! > 2^{n-1}\) for \(n>2.\) Hence, applying the reciprocal of the previous statement, we see that \[\begin{aligned} e & = 1 + 1 + \frac{1}{2} + \sum_{n=3}^{\infty} \frac{1}{n!} \\ & < 1 + 1 + \frac{1}{2} + \sum_{n=3}^{\infty} \frac{1}{2^{n-1}} \\ & = 1 + 1 + \sum_{n=2}^{\infty} \frac{1}{2^{n-1}} \\ & = 2 + \sum_{n=1}^{\infty} \frac{1}{2^n} \\ & = 2 + 1 = 3. \end{aligned}\]

Using our knowledge of geometric series, we showed that \(e < 3.\) Furthermore, let \((s_n)_{n \in \mathbb{Z}^+}\) be the sequence of partial sums of the series \(e,\) which means \[s_n = \sum_{k = 0}^{n} \frac{1}{k!}.\] Then we find that each step, which could be regarded as an approximation of \(e\) since the series converges to \(e\) (by definition), approximates \(e\) surprisingly well. That is, the error between \(e\) and \(s_n\) decreases very quickly as we increase \(n.\) And since \(s_n\) is a monotone increasing sequence (for each term added to calculate \(e\) is positive) \(e - s_n\) is always positive.³

\[\begin{aligned} e - s_n & = \sum_{k = 0}^{\infty} \frac{1}{k!} - \sum_{k = 0}^{n} \frac{1}{k!} \\ & = \sum_{k = n+1}^{\infty} \frac{1}{k!} \\ & = \frac{1}{(n+1)!} \sum_{k=n+1}^{\infty} \frac{(n+1)!}{k!} \\ & < \frac{1}{(n+1)!} \sum_{k=0}^{\infty} \frac{1}{(n+1)^k} \\ & = \frac{1}{(n+1)!} \frac{1}{1-\frac{1}{n+1}} \\ & = \frac{1}{(n+1)!} \frac{n+1}{n} = \frac{1}{n!n} \end{aligned} \]

Therefore, \(0 < e - s_n < \frac{1}{n!n},\) and the rate at which \(s_n\) converges could be calculated. For instance, \(s_{10}\) approximate \(e\) with an error less that \(10^{-7}.\)

Next, we will take a look at the two proofs that \(e\) is irrational.

Proof in Rudin’s Book

Seeking a contradiction, assume that \(e\) is rational. Then, there must be positive⁴ integers \(p\) and \(q\) such that \(e = p/q.\) But based on our last computation, we know that \[0 < e - s_q < \frac{1}{q!q},\] and thus \[0 < q!e - q!s_q < \frac{1}{q} < 1.\] By our hypothesis that \(e = p/q,\) we see that \(q!e = p(q-1)!.\) And it is clear that \[q!s_q = q!\left(1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{q!}\right) = \sum_{k=0}^q \frac{k!}{q!} \in \mathbb{Z}.\] Therefore, their difference, \(q!e - q!s_q,\) must also be an integer. However, it could then be derived that there exists an integer strictly between \(0\) and \(1,\) leading to a contradiction. We then conclude that \(e\) could not be rational and is thus irrational.

Note that the brevity to this proof relied on computing the error terms and partial sums above.

Proof in Apostol’s Book

Apostol took an interesting maneuver and tries to show that the reciprocal of \(e,\) \(e^{-1},\) is irrational instead. A real number is rational if and only if its reciprocal is also rational. For that, we need to define the function \(e^x\) as a power series in which our definition of \(e\) is recovered as a special case.⁵ \[e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}.\]

As Apostol does not present specific reasoning and justification for his theorem, I will not be presenting it here. It is slightly more complicated but resembles Rudin’s proof in the sense that it obtains a contradiction through the impossibility of an integer strictly between two consecutive integers.

Equivalent Definitions of Euler’s Constant

Another important definition of \(e\) uses limits, which establishes \(e\) as the constant of compound growth at continuous rates, substantiating its importance in many real-world applications.

\[e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\]

Thus, we can define \[t_n = \left(1 + \frac{1}{n}\right)^n,\]

and since we are dealing with a real sequence, we know that \(e\) has to be equal to the limit superior and limit inferior of its partial sum sequence \(s_n.\) Then, it is possible to show that \(t_n\) also approaches \(e\) by comparing their upper and lower limits.

Not only is \(e\) irrational, \(e\) is also a trancedental number! That means \(e\) cannot be the root of a nonzero (finite) polynomial with rational coefficients. This fact, however, is much harder to prove.

1. Tom Mike Apostol, Mathematical Analysis, 2nd ed. (Reading, MA: Addison-Wesley, 1974), 7. ↩

2. Walter Rudin, Principles of Mathematical Analysis (New York u.a., NY: McGraw-Hill u.a., 1964), 64-65. ↩

3. Some of the steps in the following calculation may not be very obvious. It is recommended to verify these steps as an exercise to the reader. They should become obvious once the reader writes out the first few terms. ↩

4. Both integers can be positive because we know that \(e > 0.\) ↩

5. As a special case, we have that \(e = e^1 = \sum_{k=0}^{\infty} 1^k/k!\) ↩