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Prove that that e is irrational.

Recall e $ = \exp(1)$ so 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?

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3 Answers 3

up vote 29 down vote accepted


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.

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how did you get that p+r? –  terrible at math Mar 15 at 19:03
@terribleatmath he merely named those two summations as $p+r$. Great solution, (+1) –  Sabyasachi Mar 15 at 19:07
ah right. can either of you guys explain to me why the first sum is an integer and the second is not? –  terrible at math Mar 15 at 19:09
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. –  Joshua Biderman Mar 15 at 19:19

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 $e$.

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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.

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