π和e是无理数的证明

$\pi$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$k(k\in {\mathbb{Z}}^{+},k⩽n)$阶导数为$0\in \mathbb{Z}$
而$f(x)$的$k(k\in {\mathbb{Z}}^{+},k⩾n)$阶导数为

$$f^{(k)}(x)=\frac{c_{k-n}}{n!}\cdot k!+\frac{c_{k-n+1}}{n!}\cdot \frac{k!}{1!}\cdot x+\cdots +\frac{c_n}{n!}\cdot \frac{k!}{(2n-k)!}\cdot x^{2n-k}$$

$$f^{(k)}(0)=\frac{c_{k-n}}{n!}\cdot k!\in {\mathbb{Z}}^{+}(\because k⩾n)$$

$$\therefore \mathrm{\forall }k\in \mathbb{Z},f^{(k)}(0)\in \mathbb{Z}$$

$$\because f(x)=f(\pi -x)\therefore f^{(k)}(\pi )\in \mathbb{Z}$$

考察积分

$${\int }_0^{\pi }f(x)\sin xdx$$

反复运用分部积分法，可得

$$\begin{array}{rl}& {\int }_0^{\pi }f(x)\sin xdx\\ =& {\int }_0^{\pi }f(x)d(-\cos x)\\ =& f(x)(-\cos x){|}_0^{\pi }+{\int }_0^{\pi }\cos x{f}^{\prime }(x)dx\\ =& f(0)+f(\pi )+{\int }_0^{\pi }{f}^{\prime }(x)d(\sin x)\\ =& f(0)+f(\pi )+f^{\prime }(x)\sin x{|}_0^{\pi }-{\int }_0^{\pi }\sin x{f}^{″}(x)dx\\ =& f(0)+f(\pi )-f^{″}(0)-f^{″}(\pi )+\cdots +(-1{)}^n{f}^{(2n)}(0)+(-1{)}^n{f}^{(2n)}(\pi )\in \mathbb{Z}\end{array}$$

另一方面，当$x\in [0,\pi ]$，有

$$0⩽q-px=p(\pi -x)⩽q$$

$$0⩽\frac{p^n(\pi -x{)}^n}{n!}=\frac{x^n(q-px{)}^n}{n!}⩽\frac{{\pi }^n{q}^n}{n!}$$

$$\therefore 0<{\int }_0^{\pi }f(x)\sin xdx⩽{\int }_0^{\pi }f(x)dx<\frac{{\pi }^{n+1}{q}^n}{n!}$$

当$n$充分大时，必有${\pi }^{n+1}{q}^n<n!$，即${\displaystyle \frac{{\pi }^{n+1}{q}^n}{n!}}<1$，那么${\displaystyle {\int }_0^{\pi }f(x)\sin xdx}$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

$$\begin{array}{rl}& p\cdot n!\cdot e\\ =& p\cdot n!\cdot (1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}+\cdots )\\ =& p\cdot n!(1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!})+p[\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots ]\end{array}$$

显然$p\cdot n!(1+1+{\displaystyle \frac{1}{2!}}+{\displaystyle \frac{1}{3!}}+\cdots +{\displaystyle \frac{1}{n!}})$是整数，记

$$M=p[\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots ]$$

$$M<p[\frac{1}{n+1}+\frac{1}{(n+1{)}^2}+\cdots ]=p\cdot {\displaystyle \frac{\frac{1}{n+1}}{1-\frac{1}{n+1}}}=\frac{p}{n}$$

当$n$足够大时，必有$p<n$，即${\displaystyle \frac{p}{n}}<1$,$M$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.

$$\mathbf{Q}.\mathbf{E}.\mathbf{D}$$