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Section 6.4 Irrationality of \(e\)

In addition to Corollary 6.1.9, there is another method for proving irrationality which involves this rather simple-looking lemma.

We prove this using the contrapositive. Suppose \(x\) is rational. Then \(x=\frac{a}{b}\) for some integers \(a\) and \(b\) with \(b >0\text{.}\) Hence if \(\frac{p}{q}\) is any other rational number not equal to \(x\text{,}\)

\begin{equation*} \left|x-\frac{p}{q}\right| =\left|\frac{a}{b}-\frac{p}{q}\right| =\left|\frac{aq-pb}{bq}\right| =\frac{|aq-pb|}{bq}\text{.} \end{equation*}

Note that \(aq-pb\) is an integer. Moreover \(aq-pb\neq0\text{,}\) since otherwise we would have \(x = p/q\text{.}\) Hence, \(|aq-pb|\geq 1\text{,}\) and if we set

\begin{equation*} c=\frac{1}{b}\text{,} \end{equation*}

then we have that for any rational number \(\frac{p}{q}\text{,}\)

\begin{equation*} \left|x-\frac{p}{q}\right| \geq \frac{|aq-pb|}{bq}\geq \frac{1}{bq}=\frac{c}{q}\text{.} \end{equation*}

We have shown that, supposing \(x\) is rational, there exists a \(c>0\) such that for all rational numbers \(\frac{p}{q}\) we have \(\left|x-\frac{p}{q}\right| \geq \frac{c}{q}\) and we are done.

As a corollary, we get the following:

Let \(c>0\text{.}\) By assumption, there is an \(N\) so that \(n> N\) implies

\begin{equation*} \left|x-\frac{p_{n}}{q_{n}}\right|q_n\lt c \end{equation*}

or equivalently

\begin{equation*} \left|x-\frac{p_{n}}{q_{n}}\right| \lt \frac{c}{q_n}\text{.} \end{equation*}

Thus, for all \(c>0\text{,}\) we can find \(\frac{p_n}{q_n}\) rational so that (6.4.1) holds. Thus \(x\) is irrational.

Let \(x = e-1\text{.}\) We aim to show that for every \(c>0\text{,}\) we can find a rational number \(\frac{p}{q}\) so that (6.4.1) holds. Lemma 6.4.1 will then imply that \(e-1\text{,}\) and hence \(e\text{,}\) is irrational.

So let \(c>0\text{.}\) Let \(N\in\mathbb{N}\text{.}\) Then

\begin{equation*} x=\sum_{n=1}^{\infty} \frac{1}{n!} =\sum_{n=1}^{N}\frac{1}{n!} + \sum_{n=N+1}^{\infty}\frac{1}{n!} \end{equation*}

Note that

\begin{align*} \sum_{n=1}^{N}\frac{1}{n!} \amp =1+\frac{1}{2!}+\frac{1}{3!}+\cdots + \frac{1}{N!}\\ \amp = \frac{N!}{N!}+\frac{N!/2!}{N!}+\frac{N!/3!}{N!}+\cdots + \frac{1}{N!}\\ \amp = \frac{N!+\frac{N!}{2!}+\frac{N!}{3!}+\cdots + 1}{N!} = \frac{p}{N!} \end{align*}

where \(p\) is an integer since \(N!/k!\) is an integer for \(k=0,1,...,N\text{.}\) We will now show, for \(N\) large enough, that

\begin{equation*} \left| x-\frac{p}{N!}\right|\lt \frac{c}{N!}\text{.} \end{equation*}

We know exactly what \(x-\frac{p}{N!}\) is:

\begin{equation*} x-\frac{p}{N!} =x-\sum_{n=1}^{N}\frac{1}{n!} =\sum_{n=N+1}^{\infty}\frac{1}{n!} =\frac{1}{(N+1)!}\sum_{n=N+1}^{\infty}\frac{(N+1)!}{n!} \end{equation*}

If we can show that there is some constant \(C\) such that

\begin{equation*} \sum_{n=N+1}^{\infty}\frac{(N+1)!}{n!} \leq C \end{equation*}

for all \(N\text{,}\) we are done, since then

\begin{equation*} \left|x-\frac{p}{N!}\right| =\frac{1}{(N+1)!}\sum_{n=N+1}^{\infty}\frac{(N+1)!}{n!} \leq \frac{C}{(N+1)!}=\frac{C/(N+1)}{N!} \end{equation*}

and so if \(\frac{C}{N+1}\lt c\) (which happens if we pick \(N>\frac{C}{c}-1\)), the above gives \(\left|x-\frac{p}{N!}\right|\lt \frac{c}{N!}\text{.}\)

For the missing step, we have that for \(n \geq N+1\text{,}\)

\begin{equation*} \frac{(N+1)!}{n!} = \frac{1}{{n\choose N+1}} \frac{1}{(n-N-1)!} \leq \frac{1}{(n-N-1)!} \end{equation*}

since the binomial coefficients are all nonnegative integers, and we have already seen that

\begin{equation*} \sum_{n = N+1}^\infty \frac{1}{(n-N-1)!} = \sum_{n=0}^\infty\frac{1}{n!} \end{equation*}

converges.

To recap, for every \(c>0\text{,}\) we have shown that we can pick \(N\) so that \(\left|x-\frac{p}{N!}\right|\lt \frac{c}{N!}\text{,}\) which implies that \(e\) is irrational by Lemma 6.4.1.