無理數

证：

假设${\displaystyle {\sqrt {2}}}$是有理数，并且令${\displaystyle {\sqrt {2}}={\frac {p}{q}}}$，${\displaystyle {\frac {p}{q}}}$是最简分数。

两边平方，得到${\displaystyle 2={\frac {p^{2}}{q^{2}}}}$。将此式改写为${\displaystyle 2q^{2}=p^{2}}$，可见${\displaystyle p^{2}}$为偶数。

因为平方运算保持奇偶性，所以${\displaystyle p}$只能为偶数。设${\displaystyle p=2p_{1}}$，其中${\displaystyle p_{1}}$为整数。

代入可得${\displaystyle q^{2}=2p_{1}^{2}}$。同理可得${\displaystyle q}$亦为偶数。

这与${\displaystyle {\frac {p}{q}}}$为最简分数的假设矛盾， 所以${\displaystyle {\sqrt {2}}}$是有理数的假设不成立。

證：

假設${\displaystyle {\sqrt {2}}+{\sqrt {3}}=p}$是有理數，兩邊平方得到

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

其中因為${\displaystyle p}$是有理數，所以${\displaystyle {\frac {p^{2}-5}{2}}}$也是有理數。

透過證明${\displaystyle {\sqrt {a}}}$為無理數的方法，其中${\displaystyle {a}}$為一非完全平方数

可以證明${\displaystyle {\sqrt {6}}}$是無理數

同樣也推出${\displaystyle {\frac {p^{2}-5}{2}}}$是無理數

但這又和${\displaystyle {\frac {p^{2}-5}{2}}}$是有理數互相矛盾

所以${\displaystyle {\sqrt {2}}+{\sqrt {3}}}$是一無理數

證：

同樣，假設${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p}$是有理數，兩邊平方後得到

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

於是${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}}$是有理數。兩邊再次平方，得：

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

於是${\displaystyle 5{\sqrt {6}}+3{\sqrt {10}}+2{\sqrt {15}}={\frac {{\frac {(p^{2}-10)^{2}}{8}}-31}{2}}}$

由於${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}}$是有理數，所以

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

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

透過證明形如${\displaystyle {\sqrt {a}}+{\sqrt {b}}}$的數是無理數的方法，得出${\displaystyle 3{\sqrt {6}}+{\sqrt {10}}}$也是一無理數

但這結果明顯和${\displaystyle {\frac {{\frac {(p^{2}-10)^{2}}{8}}-31}{2}}}$與${\displaystyle 2({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}})}$皆為有理數出現矛盾，故${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}}$為無理數

另一種證明：

同樣假設${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p}$是有理數，

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

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

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

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

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

透過證明形如${\displaystyle {\sqrt {a}}+{\sqrt {b}}}$的數是無理數的方法，得出${\displaystyle {\sqrt {6}}+p{\sqrt {5}}}$是一無理數

也是矛盾的。

證：

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

${\displaystyle \Rightarrow {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}=p-{\sqrt {7}}}$，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${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}}$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：

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

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

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

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

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

由於${\displaystyle {\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}}$，${\displaystyle p}$皆為有理數

設${\displaystyle {\sqrt {10}}+6{\sqrt {6}}+p(2p^{2}-3){\sqrt {7}}=q=\left(p^{2}-{\frac {3}{2}}\right)^{2}+7p^{2}-4\left({\sqrt {6}}+{\sqrt {10}}+{\sqrt {15}}+p{\sqrt {7}}\right)-31}$，${\displaystyle q}$亦為有理數

透過證明形如${\displaystyle {\sqrt {a}}+{\sqrt {b}}+{\sqrt {c}}}$的數是無理數的方法，可知${\displaystyle {\sqrt {10}}+6{\sqrt {6}}+p(2p^{2}-3){\sqrt {7}}}$為無理數

這和${\displaystyle q}$是有理數衝突

所以得證${\displaystyle {\sqrt {2}}+{\sqrt {3}}+{\sqrt {5}}+{\sqrt {7}}}$為一無理數