Prove that e is an irrational number.

Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then

$$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$


$$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$


$$ b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a. $$

Where can I go from here?


5 Answers 5



We first show that $2<\mathrm{e}<3$ (see below), and hence $\mathrm{e}$ is not an integer.

Next, following up OP's thought, assuming $\mathrm{e}=a/b$, we multiply by $b!$ and we obtain $$ \sum_{k=0}^\infty \frac{b!}{k!}=a\cdot (b-1)! \tag{1} $$ The right hand side of $(1)$ is an integer.

The left hand side of $(1)$ is of the form $$ \sum_{k=0}^b \frac{b!}{k!}+\sum_{k=b+1}^\infty \frac{b!}{k!}= p+r. $$ Note that $p=\sum_{k=0}^b \frac{b!}{k!}$ is an integer, while $$ 0<r=\sum_{k=b+1}^\infty \frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+\cdots<\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{1}{b}<1. $$

Note. The fact that $\mathrm{e}\in (2,3)$ can be derived from the inequalities $$ \left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e}<\left(1+\frac{1}{n}\right)^{\!n+1}, $$ for $n=1$ for the left inequality and $n=5$ for the right inequality.

  • $\begingroup$ how did you get that p+r? $\endgroup$ Mar 15, 2014 at 19:03
  • 3
    $\begingroup$ @terribleatmath he merely named those two summations as $p+r$. Great solution, (+1) $\endgroup$
    – Guy
    Mar 15, 2014 at 19:07
  • 1
    $\begingroup$ ah right. can either of you guys explain to me why the first sum is an integer and the second is not? $\endgroup$ Mar 15, 2014 at 19:09
  • 2
    $\begingroup$ In the first sum, $k\leq b$ so $k!|b!$, so each term is an integer, so the sum is an integer. In the second sum, $k>b$, and it can be shown inductively that the sum is less than 1. $\endgroup$ Mar 15, 2014 at 19:19
  • $\begingroup$ Are there any other proofs that wouldn't be based on the same idea of multiplying by $b!$ ? Because most proofs I could find are just some variations of that. $\endgroup$ Jul 16, 2017 at 21:09

I like the following mild variant, which is less popular. Equivalently, we prove that $e^{-1}$ is irrational. Suppose to the contrary that $e^{-1}=\frac{m}{n}$ where $m$ and $n$ are integers with $n\gt 0$. We have $$e^{-1}=\sum_{k=0}^{n}\frac{(-1)^k}{k!}+\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}.$$ Multiply through by $n!$. We get that $$n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}$$ must be an integer.

However, by the reasoning that leads to the Alternating Series Test, we have $$0\lt \left|n!\sum_{n+1}^\infty \frac{(-1)^k}{k!}\right|\lt \frac{n!}{(n+1)!}\lt 1.$$

The (small) advantage is that the estimation of the tail is easier than when we use the series expansion of $e$.

  • 1
    $\begingroup$ I like this proof too! $\endgroup$
    – SiXUlm
    Feb 4, 2016 at 19:55

Here is yet another one, which is one of my favorite irrationality/transcendence proofs :

The confluent hypergeometreic series

$$_{0}F_{1}(k; z) = \sum_{n = 0}^\infty \frac1{(k)_n} \frac{z^n}{n!}$$

Satisfies the more-or-less easily verifiable identity

$$_0F_1(k-1;z) - {}_0F_1(k; z) = \frac{z}{k(k-1)}{}_0F_1(k+1;z)$$

Iterating this, one ends up with the continued fraction

$$\frac{{}_0F_1(k+1;z)}{k{}_0F_1(k;z)} = \frac1{k+\cfrac{z}{k+1+\cfrac{z}{k+2+\cfrac{z}{k+3+\cdots}}}}$$

Now note that $_0F_1(3/2;x^2/4) = \cosh(x)$ and $x \,{}_0F_1(3/2;x^2/4) = \sinh(x)$, hence applying the above one has the pretty well-known continued fraction

$$\tanh(x) = \cfrac{x/2}{\frac{1}{2} + \cfrac{\frac{x^2}{4}}{\frac{3}{2} + \cfrac{\frac{x^2}{4}}{\frac{5}{2} + \cfrac{\frac{x^2}{4}}{\frac{7}{2} + \cdots}}}} = \cfrac{x}{1 + \cfrac{x^2}{3 + \cfrac{x^2}{5 + \cfrac{x^2}{7 + \cdots}}}}$$

Note, however, that $$\tanh(x) = \frac{\exp(x) - \exp(-x)}{\exp(x) + \exp(-x)}$$ but since the continued fraction above is not finite, $\tanh(1)$ is not rational. Hence $e$ is not rational either.


While it may not lead to a better formal proof than other approaches, I find the following quite intuitive. It also compares nicely to Euler's original proof using the continued fraction representation of$~e$.

It is easy to see that every rational number $\alpha\in(0,1)_\Bbb Q$ can be uniquely expressed as $$\alpha = \frac{c_1}{1!}+\cfrac{c_2}{2!}+\cdots+\cfrac{c_k}{k!} \quad\text{with $k>1$, integers $0\leq c_i<i$ for $i=0,\ldots,k$, and $c_k\neq0$.} $$ One can interpret this as the expression of $\alpha$ in the fractional counterpart of the factorial number system. Writing this as $\alpha=(c_1,c_2,\ldots,c_k)_!$ ones has the familiar property of finite decimal representations (without integer part), that the order of two such numbers is given by lexicographic comparison of their representations $(c_1,c_2,\ldots,c_k)$.

Now $e-2$ is the limit of the rational numbers of the form $(0,1,\ldots,1)_!$ as the number of digits $1$ goes to infinity (the "$-2$" removes to two integral terms $\frac1{0!}$ and $\frac1{1!}$). Clearly this cannot converge to any number with a finite representation $(c_1,c_2,\ldots,c_k)_!$.

If one would want to turn this into a formal argument, one could argue as follows. Once the "digit" in position $k+1$ has become $1$, all further numbers in the sequence lie in the closed interval between the $k+1$ digit numbers $(0,1,\ldots,1,1)_!$ and $(0,1,\ldots,1,2)_!$, which interval does not contain any number with a $k$-digit representation $(c_1,c_2,\ldots,c_k)_!$, and in particular the sequence cannot converge to such a number.


Reading Courant-Robbins I was surprised to see that proving the irrationality of $$ e = \sum_{n=0}^{\infty} \frac{1}{n!} = 2 + \frac{1}{2} + \frac{1}{2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \dots $$ isn't harder than proving the irrationality of $\sqrt{2}$, just a bit more laborious (I fear that with $\pi$ things go in a very different way...). I found Courant-Robbins proof development very concise, so I developed the proof in 15 numbered steps (with the purpose to be more explicit and clear). In the first part of the proof (step $1\to 4$) we'll show that surely the first ten digit of $e$ are $2.718281828$ (for the second part of the proof it would be sufficient to prove that $e$ is not natural, but since we're playing, better do a good estimate). In the second part (step $5\to 15$) we'll focus on irrationality of $e$.

  • 1 ) We start proving a formula that we will use two times (step 4 and step 14): if $k$ is a natural number bigger than one, we have $$ \sum_{n=0}^{\infty} \frac{1}{k^n} = \frac{1}{1-\frac{1}{k}} $$ To prove this, let's consider the sum $$ G_n = q^0 + q^1 + q^2 + \ldots + q^n $$ Multiplying both member by $q$ we get $$ q G_n = q^1 + q^2 + q^3 + \ldots + q^{n+1} $$ and subtracting we get (suppose $q \neq 1$) $$ G_n = \frac{1-q^{n+1}}{1-q} $$ If $q=\frac{1}{k}$ with $k$ natural bigger than one, and if the sum goes to $n \to \infty$, the second term in the numerator goes to zero, qed.

  • 2 ) Summing first 13 terms of $\sum \frac{1}{n!}$ (i.e. from $n=0$ to $n=12$ inclusive) we find $$ \sum_{n=0}^{12} \frac{1}{n!} = \frac{260412269}{95800320}$$ that is about $2.71828182829$.

  • 3 ) The difference $\xi$ between this estimate and the true value of $e$ is $$ \xi = \frac{1}{13!} + \frac{1}{14!} + \frac{1}{15!} + \ldots = \frac{1}{13!} + \frac{1}{13!\cdot 14} + \frac{1}{13!\cdot 14 \cdot 15} + \ldots $$ We can overvalue in this way $$ \xi < \frac{1}{13!} + \frac{1}{13!\cdot 13} + \frac{1}{13!\cdot 13^2} + \ldots = \frac{1}{13!} \left( 1 + \frac{1}{13} + \frac{1}{13^2} + \ldots \right) $$ So $$ \xi < \frac{1}{13!} \sum_{n=0}^{\infty} \frac{1}{13^n} $$

  • 4 ) Using the formula proved in step 1 we have $$ \xi < \frac{1}{13!} \cdot \frac{1}{1-\frac{1}{13}} = \frac{1}{12! \cdot 12} = \frac{1}{5748019200} \approx 1,7 \cdot 10^{-10} $$ This means that the error due to the fact we stopped the sum to thirteenth term can influence at most the tenth digit after the dot (adding $0.00000000017$ to $2.71828182829$ we cannot change the first ten digit of the number). The first part of proof is ended: $e$ "starts" with $2.718281828$.

  • 5 ) Let's prove the irrationality of $e$ by supposing, by absurd, that it is rational: $$ e = \frac{p}{q} \qquad p,q \in \mathbb{N} $$

  • 6 ) We already proved that $e \approx 2.7$. In particular, $e \notin \mathbb{N}$, so $$ q \ge 2 $$

  • 7 ) Let's substitute in $e=\frac{p}{q}$ the definition of $e$ and let's multiply two member of this equation by $q!$ (on the right, I simplified $\frac{q!}{q}$ to $(q-1)!$) $$ \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \right) q! = p (q-1) ! $$

  • 8 ) Let's rewrite the first member in a different way $$ \begin{array}{c} \displaystyle \left( 2 + \frac{1}{2!} + \frac{1}{3!} + \ldots \right. \\ \displaystyle \left. + \frac{1}{(q-2) !} + \frac{1}{(q-1) !} + \frac{1}{q!} + \frac{1}{(q+1) !} + \frac{1}{(q+2) !} + \ldots \right) q! \end{array} $$ Doing the product we have $$ 2 q! + 3 \cdot 4 \cdot \ldots \cdot q + 4 \cdot 5 \cdot \ldots \cdot q + \ldots + (q-1) \cdot q + q + 1 + \frac{1}{q+1} + \frac{1}{(q+1) (q+2) } + \ldots $$

  • 9 ) Let's define the number $A$, $B$ and $C$ in this way $$ A = p (q-1) ! $$ $$ B = 2 q! + 3 \cdot 4 \cdot \ldots \cdot q + 4 \cdot 5 \cdot \ldots \cdot q + \ldots + (q-1) \cdot q + q + 1 $$ $$ C = \frac{1}{q+1} + \frac{1}{(q+1) (q+2) } + \ldots $$

  • 10 ) Using the number $A$, $B$ and $C$, and exploiting step 8, the equation of step 7 can be written in this way $$ B + C = A $$

  • 11 ) $A$ e $B$ are clearly integer by definition, so if we prove that $C$ is not integer we have an absurdity starting by the hypothesis of rationality of $e$ and the proof is ended.

  • 12 ) The terms of the serie $\frac{1}{q+1} + \frac{1}{(q+1) (q+2) } + \ldots$ must be smaller (with the only exception of the first term) to the terms of the serie $$ \frac{1}{q+1} + \frac{1}{(q+1) ^2} + \ldots = \sum_{n=1}^{\infty} \frac{1}{(q+1) ^n} = \sum_{n=0}^{\infty} \frac{1}{(q+1) ^n} - 1 $$ So we certainly have $$ C < \sum_{n=0}^{\infty} \frac{1}{(q+1) ^n} - 1 $$

  • 13 ) The serie $\sum_{n=0}^{\infty} \frac{1}{(q+1) ^n}$ take the greatest possible value when the natural number $q$ take the smallest possible value. So we certainly have ( $q=1$ is excluded, as showed in step 6) $$ C < \sum_{n=0}^{\infty} \frac{1}{3^n} - 1 $$

  • 14 ) Exploiting the theorem proved in the step 1, the last inequality becomes $$ C < \frac{1}{1-\frac{1}{3}} - 1 = \frac{1}{2} $$

  • 15 ) We proved that $C<\frac{1}{2}$, so $C \notin \mathbb{N}$. Observing the step 11 we see that the proof is ended: $e$ is irrational.


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