\(\pi\)和\(e\)是无理数的证明
证明\(\pi\)是无理数
用反证法，假设
\[\pi=\frac{q}{p}
\]
\[p,q \in \mathbb{Z}^+
\]
构造函数
\[f(x)=\frac{x^n(q-px)^n}{n!}=\frac{p^n x^n(\pi-x)^n}{n!}
\]
设\(f(x)\)展开后变为
\[f(x)=\frac{c_0}{n!}x^n+\frac{c_1}{n!}x^{n+1}+\cdots+\frac{c_n}{n!}x^{2n}
\]
容易发现\(f(x)\)在\(0\)处的\(k(k \in \mathbb{Z}^+,k\leqslant n)\)阶导数为\(0\in \mathbb{Z}\)
而\(f(x)\)的\(k(k \in \mathbb{Z}^+,k\geqslant n)\)阶导数为
\[f^{(k)}(x)=\frac{c_{k-n}}{n!}\cdot k!+\frac{c_{k-n+1}}{n!}\cdot \frac{k!}{1!}\cdot x+\cdots+\frac{c_n}{n!}\cdot\frac{k!}{(2n-k)!}\cdot x^{2n-k}
\]
\[f^{(k)}(0)=\frac{c_{k-n}}{n!}\cdot k!\in \mathbb{Z}^+(\because k \geqslant n)
\]
\[\therefore \forall k \in \mathbb{Z},f^{(k)}(0) \in \mathbb{Z}
\]
\[\because f(x)=f(\pi-x)\text{ } \text{ } \therefore f^{(k)}(\pi)\in \mathbb{Z}
\]
考察积分
\[\int_{0}^{\pi}f(x)\sin{x}\,dx
\]
反复运用分部积分法，可得
\[\begin{aligned}
&\int_{0}^{\pi}f(x)\sin{x}\,dx \cr
=&\int_{0}^{\pi}f(x)\,d(-\cos{x}) \cr
=&f(x)(- \cos{x})|_{0}^{\pi}+\int_{0}^{\pi}\cos{x}f'(x)\,dx \cr
=&f(0)+f(\pi)+\int_{0}^{\pi}f'(x)\,d(\sin{x}) \cr
=&f(0)+f(\pi)+f'(x)\sin{x}|_{0}^{\pi}-\int_{0}^{\pi}\sin{x}f''(x)\,dx \cr
=&f(0)+f(\pi)-f''(0)-f''(\pi)+\cdots+(-1)^n f^{(2n)}(0)+(-1)^n f^{(2n)} (\pi)\in \mathbb{Z}
\end{aligned}
\]
另一方面，当\(x\in [0,\pi]\)，有
\[0 \leqslant q-px=p(\pi-x) \leqslant q
\]
\[0 \leqslant \frac{p^n(\pi-x)^n}{n!}=\frac{x^n(q-p x)^n}{n!}\leqslant \frac{\pi^n q^n}{n!}
\]
\[\therefore 0<\int_{0}^{\pi}f(x)\sin{x}\,dx \leqslant \int_{0}^{\pi}f(x)\,dx<\frac{\pi^{n+1}q^n}{n!}
\]
当\(n\)充分大时，必有\(\pi^{n+1}q^n<n!\)，即\(\dfrac{\pi^{n+1}q^n}{n!}<1\)，那么\(\displaystyle\int_{0}^{\pi}f(x)\sin{x}\,dx\)不是整数，与先前它是整数的结论矛盾.
所以\(\pi\)不是有理数，是无理数.
\[\mathbf{Q}.\mathbf{E} .\mathbf{D}
\]
证明\(e\)是无理数
同样也用反证法，假设
\[e=\frac{q}{p}
\]
\[p,q \in \mathbb{Z}^+
\]
根据泰勒展开，知道
\[e=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}+\cdots
\]
将等式\(e=\frac{q}{p}\)两边乘以\(p\cdot n!\)，得到
\[p \cdot n!\cdot e=q\cdot n!\in \mathbb{Z}
\]
而
\[\begin{aligned}
&p \cdot n!\cdot e\\
=&p \cdot n!\cdot \left(1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}+\cdots \right)\\
=&p\cdot n!\left(1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}\right)+p\left[\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots\right]\\
\end{aligned}
\]
显然\(p\cdot n!\left(1+1+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots+\dfrac{1}{n!}\right)\)是整数，记
\[M=p\left[\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots\right]\]
\[M<p\left[\frac{1}{n+1}+\frac{1}{(n+1)^2}+\cdots\right]=p\cdot \dfrac{\cfrac{1}{n+1}}{1-\cfrac{1}{n+1}}=\frac{p}{n}
\]
当\(n\)足够大时，必有\(p<n\)，即\(\dfrac{p}{n}<1\),\(M\)不是整数，与上面推出的\(p \cdot n!\cdot e\in \mathbb{Z}\)矛盾.
所以\(e\)不是有理数，是无理数.
\[\mathbf{Q}.\mathbf{E} .\mathbf{D}
\]