= Solution
The <exponential function> series gives <Euler's number> $e=\sum_{k=0}^{\infty}1/k!$. Suppose $e=a/b$ with <integers> $a,b$ and $b>0$. Choose $n\geq\max(b,2)$. Since $b$ divides $n!$, the number
$$
N=n!\left(e-\sum_{k=0}^n\frac1{k!}\right)
$$
is an <integer>. But its positive tail satisfies
$$
0<N=\sum_{j=1}^{\infty}\frac{n!}{(n+j)!}
<\sum_{j=1}^{\infty}\frac1{(n+1)^j}=\frac1n<1.
$$
The inequality is strict because the factors after the first exceed $n+1$ when $j\geq2$. No <integer> lies strictly between zero and one. This proves the <irrationality of e>:
$$
\boxed{e
otin\mathbb Q.}
$$
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