刘维尔数

刘维尔定理：若无理数${\displaystyle \alpha }$是代数数，即整系数${\displaystyle n}$次多项式${\displaystyle f}$的根，那么存在实数${\displaystyle A>0}$，对于所有${\displaystyle p,q\in \mathbb {Z} ,q>0}$有

${\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert >{\frac {A}{q^{n}}}}$

证明：令${\displaystyle M=\max \left\{\left|f'(x)\right||x\in \left[\alpha -1,\alpha +1\right]\right\}}$，记${\displaystyle f}$的其它的不重复的根为 ${\displaystyle \alpha _{1},\alpha _{2},...,\alpha _{m}}$，取这样的A

${\displaystyle 0<A<\min \left(1,{\frac {1}{M}},\left\vert \alpha -\alpha _{1}\right\vert ,\left\vert \alpha -\alpha _{2}\right\vert ,\ldots ,\left\vert \alpha -\alpha _{m}\right\vert \right)}$

如果存在使定理不成立的${\displaystyle p,q}$，就有

${\displaystyle \left\vert \alpha -{\frac {p}{q}}\right\vert \leq {\frac {A}{q^{n}}}\leq A<\min \left(1,{\frac {1}{M}},\left\vert \alpha -\alpha _{1}\right\vert ,\left\vert \alpha -\alpha _{2}\right\vert ,\ldots ,\left\vert \alpha -\alpha _{m}\right\vert \right)}$

那么，${\displaystyle {\frac {p}{q}}\in \left[\alpha -1,\alpha +1\right]\land {\frac {p}{q}}\notin \left\{\alpha _{1},\alpha _{2},...,\alpha _{m}\right\}}$

据拉格朗日中值定理，存在${\displaystyle \alpha }$和${\displaystyle {\frac {p}{q}}}$之间的${\displaystyle x_{0}}$使得

${\displaystyle f(\alpha )-f(p/q)=(\alpha -p/q)\cdot f'(x_{0})}$

有

${\displaystyle \left\vert (\alpha -p/q)\right\vert =\left\vert f(\alpha )-f(p/q)\right\vert /\left\vert f'(x_{0})\right\vert =\left\vert f(p/q)/f'(x_{0})\right\vert \,}$

${\displaystyle f}$是多项式，所以

${\displaystyle \left\vert f(p/q)\right\vert =\left\vert \sum _{i=0}^{n}c_{i}p^{i}q^{-i}\right\vert ={\frac {\left\vert \sum _{i=0}^{n}c_{i}p^{i}q^{n-i}\right\vert }{q^{n}}}\geq {\frac {1}{q^{n}}}}$

由于${\displaystyle \left|f'(x_{0})\right|\leq M}$和${\displaystyle 1/M>A}$

${\displaystyle \left\vert \alpha -p/q\right\vert =\left\vert f(p/q)/f'(x_{0})\right\vert \geq 1/(Mq^{n})>A/q^{n}\geq \left\vert \alpha -p/q\right\vert }$

矛盾。

证明刘维尔数是超越数：有刘维尔数${\displaystyle x}$，它是无理数，如果它是代数数则

${\displaystyle \exists n\in \mathbb {Z} ,A>0\forall p,q\left(\left\vert x-{\frac {p}{q}}\right\vert >{\frac {A}{q^{n}}}\right)}$

取满足${\displaystyle {\frac {1}{2^{r}}}\leq A}$的正整数${\displaystyle r}$，并令${\displaystyle m=r+n}$，存在整数${\displaystyle a,b}$其中 ${\displaystyle b>1}$有

${\displaystyle \left|x-{\frac {a}{b}}\right|<{\frac {1}{b^{m}}}={\frac {1}{b^{r+n}}}={\frac {1}{b^{r}b^{n}}}\leq {\frac {1}{2^{r}}}{\frac {1}{b^{n}}}\leq {\frac {A}{b^{n}}}}$

与上式矛盾。故刘维尔数是超越数。