# 無理數

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

## 举例

1. ${\displaystyle {\sqrt {3}}}$＝1.73205080…
2. ${\displaystyle \log _{10}}$3＝0.47712125…
3. e＝2.71828182845904523536…
4. sin 45°＝${\textstyle {\frac {\sqrt {2}}{2}}}$＝0.70710678…
5. π＝3.141592653589793238462…

## 性质

• 无理数加或减无理数不一定得无理数，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。

## 不知是否是無理數的數

π＋e、π－e等，事实上，對于任何非零整數${\displaystyle m\,}$${\displaystyle n\,}$，不知道${\displaystyle m\pi +ne\,}$是否無理數。

## 無理化作連分數的表達式

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

${\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}}}

## 無理數之證

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

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

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

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

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

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

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

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

${\displaystyle \Rightarrow {\sqrt {2}}+{\sqrt {3}}=p-{\sqrt {5}}}$，兩邊平方：

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

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

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

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

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

${\displaystyle \Rightarrow {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p-{\sqrt {7}}}$，兩邊平方得

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

${\displaystyle \Rightarrow {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}={\frac {p^{2}}{2}}-{\frac {3}{2}}}$，得到${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}}$為一有理數

${\displaystyle \Rightarrow {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}={\frac {p^{2}}{2}}-{\frac {3}{2}}-p{\sqrt {7}}}$，兩邊繼續平方：

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

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

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

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

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

${\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}$亦為有理數