# 无理数

${\displaystyle \mathbb {N} \subseteq \mathbb {Z} \subseteq \mathbb {Q} \subseteq \mathbb {R} \subseteq \mathbb {C} }$

## 举例

1. ${\displaystyle {\sqrt {3}}=1.73205080\cdots }$
2. ${\displaystyle \log _{10}3=0.47712125\cdots }$
3. ${\displaystyle e=2.71828182845904523536\cdots }$
4. ${\displaystyle \sin {45^{\circ }}={\frac {\sqrt {2}}{2}}=0.70710678\cdots }$
5. 圆周率=3.14159265358979323846264338327950288419716939937510582097494459230781640628620899...

## 性质

• 无理数加或减无理数不一定得无理数，例如:${\displaystyle \log _{10}2+\log _{10}5=\log _{10}10=1}$
• 无理数乘不等于0的有理数必得无理数。
• 无理数的平方根立方根等次方根必得无理数。

## 不知是否是无理数的数

${\displaystyle \pi +e\,}$${\displaystyle \pi -e\,}$等，事实上，对于任何非零整数${\displaystyle m\,}$${\displaystyle n\,}$，不知道${\displaystyle m\pi +ne\,}$是否无理数。

## 无理化作连分数的表达式

${\displaystyle x^{2}=c\qquad (c>0)}$

${\displaystyle \rho ^{2}

{\displaystyle {\begin{aligned}x^{2}\ -\!\rho ^{2}&=c\ -\!\rho ^{2}\\(x\ -\!\rho )(x\ +\!\rho )&=c\ -\!\rho ^{2}\\x\ -\!\rho &={\frac {c\ -\!\rho ^{2}}{\rho \ +\!x}}\\x&=\rho \ +\!{\frac {c\ -\!\rho ^{2}}{\rho \ +\!x}}\\&=\rho \ +\!{\cfrac {c\ -\!\rho ^{2}}{\rho \ +\!\left(\rho \ +\!{\cfrac {c\ -\!\rho ^{2}}{\rho \ +\!x}}\right)}}\\&=\rho \ +\!{\cfrac {c\ -\!\rho ^{2}}{2\rho \ +\!{\cfrac {c\ -\!\rho ^{2}}{2\rho \ +\!{\cfrac {c\ -\!\rho ^{2}}{\ddots \,}}}}}}={\sqrt {c}}\,\end{aligned}}}

## 一些无理数的证明

### 证明${\displaystyle {\sqrt {2}}+{\sqrt {3}}}$是无理数

${\displaystyle 5+2{\sqrt {6}}=p^{2}\Rightarrow {\sqrt {6}}={\frac {p^{2}-5}{2}}}$

### 证明${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}}$是无理数

${\displaystyle 10+2{\sqrt {6}}+2{\sqrt {10}}+2{\sqrt {15}}=p^{2}\Rightarrow {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}={\frac {p^{2}-10}{2}}}$

${\displaystyle 31+10{\sqrt {6}}+6{\sqrt {10}}+4{\sqrt {15}}={\frac {(p^{2}-10)^{2}}{4}}}$

${\displaystyle 3{\sqrt {6}}+{\sqrt {10}}+2({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}})={\frac {{\frac {(p^{2}-10)^{2}}{4}}-31}{2}}}$

${\displaystyle \Rightarrow 3{\sqrt {6}}+{\sqrt {10}}={\frac {{\frac {(p^{2}-10)^{2}}{4}}-31}{2}}-2({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}})}$

${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p}$

${\displaystyle \Rightarrow {\sqrt {2}}+{\sqrt {3}}=p-{\sqrt {5}}}$，两边平方：

${\displaystyle \Rightarrow ({\sqrt {2}}+{\sqrt {3}})^{2}=(p-{\sqrt {5}})^{2}}$

${\displaystyle \Rightarrow 5+2{\sqrt {6}}=p^{2}+5-2p{\sqrt {5}}}$

${\displaystyle \Rightarrow 2({\sqrt {6}}+p{\sqrt {5}})=p^{2}}$

### 证明${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}+{\sqrt {7}}}$是无理数

${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}+{\sqrt {7}}=p}$

${\displaystyle \Rightarrow {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p-{\sqrt {7}}}$，两边平方得到：

${\displaystyle \Rightarrow 10+2{\sqrt {6}}+2{\sqrt {10}}+2{\sqrt {15}}=p^{2}+7-2p{\sqrt {7}}}$

${\displaystyle \Rightarrow {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}={\frac {p^{2}}{2}}-{\frac {3}{2}}}$，得到${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}}$为一有理数

${\displaystyle \Rightarrow {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}={\frac {p^{2}}{2}}-{\frac {3}{2}}-p{\sqrt {7}}}$，两边继续平方：

${\displaystyle \Rightarrow \left({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}\right)^{2}=\left(p^{2}-{\frac {3}{2}}-p{\sqrt {7}}\right)^{2}}$

${\displaystyle \Rightarrow \left({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}\right)^{2}=\left[\left(p^{2}-{\frac {3}{2}}\right)-p{\sqrt {7}}\right]^{2}}$

${\displaystyle \Rightarrow 31+2{\sqrt {60}}+2{\sqrt {90}}+2{\sqrt {150}}=\left(p^{2}-{\frac {3}{2}}\right)^{2}+(-p{\sqrt {7}})^{2}-2\times {p}{\sqrt {7}}\times \left(p^{2}-{\frac {3}{2}}\right)}$

${\displaystyle \Rightarrow 31+4{\sqrt {15}}+6{\sqrt {10}}+10{\sqrt {6}}=\left(p^{2}-{\frac {3}{2}}\right)^{2}+7p^{2}-p(2p^{2}-3){\sqrt {7}}}$

${\displaystyle \Rightarrow 2{\sqrt {10}}+6{\sqrt {6}}+p(2p^{2}-3){\sqrt {7}}=\left(p^{2}-{\frac {3}{2}}\right)^{2}+7p^{2}-4\left({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}\right)-31}$

${\displaystyle {\sqrt {10}}+6{\sqrt {6}}+p(2p^{2}-3){\sqrt {7}}=q=\left(p^{2}-{\frac {3}{2}}\right)^{2}+7p^{2}-4\left({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}\right)-31}$${\displaystyle q}$亦为有理数