根式之積者，兩根式之積也，有式云
a ⋅ b = a b {\displaystyle {\sqrt {a}}\cdot {\sqrt {b}}={\sqrt {ab}}}
同平方可得
[ ( a ) ( b ) ] 2 = ( a ) 2 ( b ) 2 = a b {\displaystyle [({\sqrt {a}})({\sqrt {b}})]^{2}=({\sqrt {a}})^{2}({\sqrt {b}})^{2}=ab}
故證之
由 [ r ] 2 = r , r ∈ [ 0 , + ∞ ) {\displaystyle [{\sqrt {r}}]^{2}=r\quad ,r\in [0,+\infty )}
r = r 1 2 {\displaystyle {\sqrt {r}}=r^{\frac {1}{2}}}
r n = r 1 n = r n − 1 {\displaystyle {\sqrt[{n}]{r}}=r^{\frac {1}{n}}=r^{n^{-1}}}
a r ⋅ b r = ( a b ) r , r ∈ R {\displaystyle a^{r}\cdot b^{r}=(ab)^{r},r\in \mathrm {R} }
![{\displaystyle [({\sqrt {a}})({\sqrt {b}})]^{2}=({\sqrt {a}})^{2}({\sqrt {b}})^{2}=ab}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5cf676d99bad0f7c23b790bda24d1fe9022c3531)
![{\displaystyle [{\sqrt {r}}]^{2}=r\quad ,r\in [0,+\infty )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d502b2889dd5b4690ca48ef609f45b355eed352)
![{\displaystyle {\sqrt[{n}]{r}}=r^{\frac {1}{n}}=r^{n^{-1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbab4150be1055253322103ea4095b1ecf4bfd48)
