Proof of the Irrationality of e

In a previous post I proved that \log_{10} 5 is irrational.  In this post I prove the irrationality of e.

A proof of the irrationality of e must start by defining e.  There are some different ways to do that.  We’ll take e to be the unique positive number b such that \dfrac{d}{dx} b^x = b^x.  (This property itself can be proved from other ways to define e.  See, for example, Fitzpatrick’s Advanced Calculus [1], Section 5.2.)  We’ll also make the assumptions that (1) e < 3 and (2) e^x is an increasing function.

Under this definition of e, the Lagrange Remainder Theorem says that, given x > 0 and n \in \mathbb{N}, there exists c \in (0,x) such that

\displaystyle e^x = 1 + x + \frac{x^2}{2} + \cdots + \frac{x^n}{n!} + \frac{e^c x^{n+1}}{(n+1)!}.

Taking x = 1, we obtain, for some c \in (0,1),

\displaystyle e = 1 + 1 + \frac{1}{2} + \cdots + \frac{1}{n!} + \frac{e^c}{(n+1)!}.

\displaystyle \implies 0 < e - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right] = \frac{e^c }{(n+1)!} \leq \frac{e}{(n+1)!} \leq \frac{3}{(n+1)!}.

Now, suppose e is rational.  Then there exist m_0, n_0 \in \mathbb{N} with e = m_0/n_0.  Therefore,

\displaystyle 0 < \frac{m_0}{n_0} - \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{(n+1)!}.

Multiplying by n!, this becomes

\displaystyle 0 < n! \frac{m_0}{n_0} - n! \left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!} \right] \leq \frac{3}{n+1}.

Since the above inequality is true for all n \in \mathbb{N}, it is true for n = \max \{n_0, 3\}.  For this value of n,  \displaystyle n!\frac{m_0}{n_0} - n!\left[ 2 + \frac{1}{2} + \cdots + \frac{1}{n!}\right] is an integer, and 3/(n+1) \leq 3/4.

This means that there is an integer in the interval (0,3/4).  However, we know that there is no integer in the interval (0,1).  Contradiction; thus e is irrational.

[1] Patrick M. Fitzpatrick, Advanced Calculus, American Mathematical Society, 2006.

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