标签：mathbb 无理数 frac cdot 证明 cdots pi

\(\pi\)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.

\[\mathbf{Q}.\mathbf{E} .\mathbf{D} \]

标签：mathbb,无理数,frac,cdot,证明,cdots,pi
From： https://www.cnblogs.com/mdntct/p/17478425.html