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What is a particularly elegant proof that $e$ is irrational that requires a minimum background in analysis?

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    $\begingroup$ This may be matter of opinion, but I think that "minimal background" and "elegant" are nearly diametrically opposed. Elegant proofs often required really heavy machinery, while elementary proofs often require a great deal of computation. These two things seem to be in opposition, no? $\endgroup$ – Xander Henderson Nov 15 '17 at 0:54
  • $\begingroup$ @XanderHenderson: A proof based on simple ideas (requiring very little mathematical machinery) can be elegant. However it is difficult to find such proofs. $\endgroup$ – Paramanand Singh Nov 15 '17 at 1:01
  • $\begingroup$ Keith Conrad does it in a few lines here: math.uconn.edu/~kconrad/blurbs/analysis/irrational.pdf $\endgroup$ – symplectomorphic Nov 15 '17 at 1:01
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    $\begingroup$ The usual one is already quite essential: if $e$ is rational, then $k!e$ is integer for some $k\ge2$. But then necessarily $\sum_{j=1}^\infty\frac{k!}{(k+j)!}$ is an integer, which is absurd because of the estimate $\frac{k!}{(k+j)!}\le 3^{-j}$. $\endgroup$ – user228113 Nov 15 '17 at 1:05
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Suppose $e=\frac ab$ for $a,b\in \mathbb N$ It follows that $e^{-1}=\frac ba$. But $$\frac ba=e^{-1}=1-\frac 1{1!}+\frac 1{2!}\cdots $$ And, as that is an alternating series for which the terms decrease to $0$ we can write $$0<\big|\frac ba-(1-\cdots + (-1)^{a+1}\frac 1{a!})\big|<\frac 1{(a+1)!}$$ Multiplying by $a!$ gives $$0<\big|b(a-1)!-(a!-\cdots +(-1)^{a+1}\big|<\frac 1{a+1}$$

But the expression inside the absolute value is clearly an integer and there can be no integer between $0$ and $\frac 1{a+1}$

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The following proof is due to J. Sondow, and assuming you're prepared to accept the following definition of $e$:

$$e := \lim_{n \to \infty} \sum_{k=0}^n\frac{1}{k!}$$

the only prerequisite is the nested interval theorem.(A simultaneous result of this proof is that the above limit exists, so the knowledge of $e$'s convergence is not a prerequisite).

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PROOF:

Let $I_1 = [2,3]$ and inductively define a sequence of closed, nested intervals by breaking $I_n$ into into $n+1$ equal length intervals and taking $I_{n+1} = [a_{n+1},b_{n+1}]$ to be the second of these.

Explicitly, $$a_{n} = a_{n-1}+ \frac{1}{n!} \quad \quad b_n = a_n + \frac{1}{n!}$$

It follows that $$a_n = \sum_{k=0}^n \frac{1}{k!} $$

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Let $$c_k:= \frac{n!}{k!} \quad s_n = \sum_{k=0}^n c_k$$ and note that $c_k,s_n \in \mathbb{Z}$ and $$\frac{c_k}{n!} = \frac{1}{k!}$$

so that we can also write:

$$a_n = \sum_{k=0}^n \frac{c_k}{n!} = \frac{1}{n!}\sum_{k=0}^n c_k = \frac{s_n}{n!} \quad \ \quad I_n = [a_n,b_n] = \bigg[\frac{s_n}{n!},\frac{s_n+1}{n!}\bigg]$$

Therefore, $$x \in I_{n+1} \implies \frac{s_n}{n!} < x <\frac{s_n+1}{n!}$$

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Since $c_n$ is an integer, it follows that if $x \in I_{n+1}$ then $x$ cannot be written in the form $x = \frac{m}{n!}$ for any $m \in \mathbb{Z}$

Since any number of the form $\frac{m}{n}$ can be written as $\frac{m(n-1)!}{n!}$, we also have $$x \in I_{n+1} \implies x \neq \frac{m}{n} \space for \space any \space m \in \mathbb{Z}$$

From this we see that $$x \in \bigcap_{n=1}^\infty I_n \implies x \notin \mathbb{Q}$$

Since the lengths of the intervals $ b_n-a_n = \frac{1}{n!}$ converge to $0$,

we have by the nested interval theorem that

$$\bigcap_{n=1}^\infty I_n = \{ \space \lim_{n \to \infty} a_n \space \} = \{e\}$$

and thus $e \notin \mathbb{Q}$

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Yet another one following the same theme (but it deals directly with $e$, not $e^{-1}$). I am adding this because it avoids the complications of dealing with the reciprocal of $e$ and alternating series. It uses very basic concepts such as factorials and geometric series. It is, in fact, the most accessible proof of irrationality of a transcendental number that I have ever encountered. It's also very easy to remember.

Suppose that $e$ were rational, i.e. $e = \frac ab$ where $a$ and $b$ are coprime positive integers. Clearly $b>1$.

The series expansion of $e$ can be written:

$\displaystyle e = \frac 1{0!} + \frac 1{1!} + \frac 1{2!} + ... + \frac 1{b!} + ...$

(I am making explicitly clear that $b!$ appears as a denominator in that expansion).

Now,

$eb! = \frac {b!}{0!} + \frac {b!}{1!} + \frac {b!}{2!} + ... + \frac {b!}{b!} + \frac{b!}{(b+1)!} + ...$

Note that every term up to and including $\frac {b!}{b!} (= 1)$ is clearly a positive integer. Let this integer be called $N$. Now let us consider the positive sum of the remaining terms (call this $S$).

$\displaystyle eb! = N + \frac{b!}{(b+1)!} + \frac{b!}{(b+2)!}... = N + \frac{1}{b+1} + \frac{1}{(b+1)(b+2)} + ... = N + S$

Compare $S$ to the geometric series $\displaystyle G = \frac{1}{b+1} + \frac 1{(b+1)^2} + \frac 1{(b+1)^3} + ...$.

It is clear that the first term of the geometric series is equal to that of $S$ while the subsequent corresponding terms are obviously larger (since the denominators are smaller).

But $G = \displaystyle \frac {\frac{1}{b+1}}{1-\frac{1}{b+1}} = \frac 1b$

So $0 < S < G < 1$.

Since $eb!( = \frac ab \cdot b! = a(b-1)!$) is clearly a positive integer, we have that $S = a(b-1)! - N$ which is an integer that lies strictly between $0$ and $1$ (contradiction).

Hence the original premise is false, and $e$ is irrational.

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  • $\begingroup$ Do you know who first gave this proof? $\endgroup$ – Shahab Oct 2 '18 at 1:45
  • $\begingroup$ @Shahab I'm not sure but it's such an easy and elegant proof that I'm able to reproduce it from memory after seeing it (or a close variant) only once years back. One of the nicest elementary proofs for the irrationality of a transcendental number. $\endgroup$ – Deepak Oct 4 '18 at 6:10
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If one is aware of the Gauss continued fraction for the (hyperbolic) cotangent function he/she has a straightforward proof. Indeed, $$\forall z\in\left(0,\tfrac{\pi}{2}\right),\qquad \coth(z)=\frac{1}{z}+\frac{z}{3+\frac{z^2}{5+\frac{z^2}{7+\ldots}}}\tag{1}$$ leads to $$ \coth(1)=\frac{e^2-1}{e^2+1}=[1;3,5,7,9,\ldots]\not\in\mathbb{Q}. \tag{2}$$ If $e$ were a rational number, $\coth(1)$ would be a rational number too, but $(2)$ shows that is not the case (all the elements of $\mathbb{Q}^+$ have a finite ordinary continued fraction).

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  • $\begingroup$ May I request an explanation for the downvote? $\endgroup$ – Jack D'Aurizio Nov 15 '17 at 16:59

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