Problem 37: Irrationality of e ($\u2713$ $\u2713$) 1997 Paper III

For each positive integer $n\phantom{\rule{0.3em}{0ex}}$, let

$$\begin{array}{llll}\hfill {a}_{n}& =\frac{1}{n+1}+\frac{1}{\left(n+1\right)\left(n+2\right)}+\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}+\cdots \phantom{\rule{0.3em}{0ex}},\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {b}_{n}& =\frac{1}{n+1}+\frac{1}{{\left(n+1\right)}^{2}}+\frac{1}{{\left(n+1\right)}^{3}}+\cdots \phantom{\rule{0.3em}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$- (i)
- Show that ${b}_{n}=1\u2215n\phantom{\rule{0.3em}{0ex}}$.
- (ii)
- Deduce that $0<{a}_{n}<1\u2215n\phantom{\rule{0.3em}{0ex}}$.
- (iii)
- Show that ${a}_{n}=n!\phantom{\rule{0.3em}{0ex}}e-\left[n!\phantom{\rule{0.3em}{0ex}}e\right]\phantom{\rule{0.3em}{0ex}}$, where $\left[x\right]$ denotes the integer part of $x\phantom{\rule{0.3em}{0ex}}$.
- (iv)
- Hence show that e is irrational.

Comments

Each part of this looks horrendously difficult, but it doesn’t take much thought to see what is going on. If you haven’t come across the concept of integer part some examples should make it clear: $\left[21.25\right]=21\phantom{\rule{0.3em}{0ex}}$, $\left[\pi \right]=3\phantom{\rule{0.3em}{0ex}}$, $\left[1.999999\right]=1$ and $\left[2\right]=2\phantom{\rule{0.3em}{0ex}}$.

For part (iii), you need to know the series for the number e, which is $\sum _{n=0}^{\infty}1\u2215n!\phantom{\rule{0.3em}{0ex}}$.

If you are stumped by the last part, just remember the deﬁnition of the word irrational: $x$ is rational if and only if it can be expressed in the form $p\u2215q$ where $p$ and $q$ are integers; if $x$ is not rational, it is irrational. Then look for a proof by contradiction (‘Suppose that $x$ is rational ...’).

The ﬁrst proof that e is irrational was ﬁrst given by Euler (1736). He also named the number e, though he didn’t ‘invent’ it; e is the basis for natural logarithms and as such was used implicitly by John Napier in 1614. Euler’s proof involved the use of continuous fractions. He wrote e as

$$2+\frac{1}{1+\frac{1}{1+\frac{1}{4+\frac{1}{\cdots}}}}$$ |

the numbers in the denominators forming the inﬁnite pattern $1,2,1,1,4,1,1,6,1,1,8\dots $. It is known that any such continuous fraction represents an irrational number.

In general, it is not easy to show that numbers are irrational. Johann Lambert showed that $\pi $ is irrational in 1760, but the simple continued fraction for $\pi $ is not known. It is still not known if $\pi +\mathrm{e}$ is irrational.

The proof that e is irrational given in this question is based on a proof by
Fourier^{27}
in about 1815.

Solution to problem 37

(i) Although the series for ${b}_{n}$ does not at ﬁrst sight look tractable, it is in fact just a geometric progression: the ﬁrst term is $1\u2215\left(n+1\right)$ and the common ratio is also $1\u2215\left(n+1\right)\phantom{\rule{0.3em}{0ex}}$. Thus

(ii) Each term (after the ﬁrst) of ${a}_{n}$ is less than the corresponding term in ${b}_{n}$, so ${a}_{n}<{b}_{n}=1\u2215n\phantom{\rule{0.3em}{0ex}}$.

(iii) Multiplying the series for $\mathrm{e}$ by $n!$ gives

$$n!\mathrm{e}=n!+n!+\frac{1}{2}n!+\cdots +1+{a}_{n}$$and the result follows because ${a}_{n}<1\phantom{\rule{0.3em}{0ex}}$ and all the other terms on the right hand side of the above equation are integers.

(iv) We use proof by contradiction. Suppose that there exist integers $k$ and $m$ such that $\mathrm{e}=k\u2215m$. Then $m!\mathrm{e}$ is certainly an integer. But if $m!\mathrm{e}$ is an integer then $\left[m!\mathrm{e}\right]=m!\mathrm{e}$, which contradicts the result of part (iii) since we know that ${a}_{m}\ne 0$ (it is obvious from the deﬁnition that ${a}_{m}>0$).

Post-mortem

Well, that was short!

There were, however, two key steps which were not difficult in themselves but were not easy to ﬁnd in the context of this question.

The ﬁrst was to recognise that ${b}_{n}$ is, for each $n$, a simple geometric progression. If it had been presented in the form $b=r+{r}^{2}+{r}^{3}+\cdots \phantom{\rule{0.3em}{0ex}}$ it would have been immediately recognisable. Somehow, the fact that the ratio $r$ is given in terms of $n$ — or, even worse, $n+1$ — makes it difficult to spot; not least because $n$ is ﬁxed for each series instead of labelling the terms of an individual series.

The second key step, in part (iv), was to assume that $e=k\u2215m$ and show that $m!\phantom{\rule{0.3em}{0ex}}e$, not $me$, cannot be an integer. It seems so wasteful when it would suffice to show that $me$ is not an integer. We seem to be making a great deal of extra work for ourselves and that proﬂigacy, unusual in mathematics, might well have thrown you off the scent despite the clear signal from part (iii).

I don’t regard the idea of using proof by contradiction as a difficult step in part (iv). An
irrational number is by deﬁnition a number that is not rational, so the idea of a proof by
contradiction should jump out at you. Indeed, the standard example, the proof of the irrationality of
$\sqrt{2}$, is by
contradiction.^{28}

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