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## Prove that $$e$$ is an irrational number

### Solution

To start, let’s expand the exponential function
$\mathrm{e}^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}$
$\Rightarrow e=\sum_{k=0}^{\infty} \frac{1}{k !}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots$
Assume that e is rational
$e=\frac{m}{n}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots \quad \text { for two coprime integers } m \text { and } n$
$\frac{m}{n}=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots+\frac{1}{n !}+\frac{1}{(n+1) !}+\frac{1}{(n+2) !}+\cdots$
$n ! \frac{m}{n}=n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\frac{n !}{3 !}+\cdots+\frac{n !}{n !}+\frac{n !}{(n+1) !}+\frac{n !}{(n+2) !}+\cdots$
\begin{aligned} &\mathrm{A}=n ! \frac{m}{n}=(n-1) ! m \in \mathbb{N} \\ &\mathrm{B}=n !+\frac{n !}{1 !}+\frac{n !}{2 !}+\frac{n !}{3 !}+\cdots+\frac{n !}{n !} \in \mathbb{N} \\ &\mathrm{C}=\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots \end{aligned}
$\begin{gathered} \text { For } n \geqslant 1 \\ \frac{1}{n+1} \leqslant \frac{1}{2} \\ \frac{1}{(m+1)(n+2)}<\frac{1}{2^{2}} \\ \frac{1}{(n+1)(n+2)(n+3)}<\frac{1}{2^{3}} \\ \vdots \end{gathered}$
$\Rightarrow\mathrm{C}<\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots=\left(\frac{1}{2}\right)\left(\frac{1}{1-\frac{1}{2}}\right)=1$
$0<c<1 \Rightarrow C \notin \mathbb{N}$
$\text { Therefore }$
$\huge \mathrm{e} \notin \mathbb{Q}$
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Home -> Solved problems -> e irrational number