根式的乘除運算與代數式之推導

乘法運算:

設 $a\ge 0$ 、 $b\ge 0$ ,則: $\sqrt{a}\times \sqrt{b}=\sqrt{ab}$

$\sqrt2\times \sqrt3=\sqrt{2\times 3}=\sqrt6$

說明

$\begin{array} {rl} & (\sqrt2\times \sqrt3)^2 \\ = & (\sqrt2\times \sqrt3)\times (\sqrt2\times \sqrt3) \\ = & (\sqrt2\times \sqrt2)\times (\sqrt3\times \sqrt3) \\ = & (\sqrt2)^2\times (\sqrt3)^2 \\ = & 2\times 3 \\ = & 6\end{array}$

$\Rightarrow \sqrt2\times \sqrt3$ 是 $6$ 的正平方根

$\Rightarrow \sqrt2\times \sqrt3=\sqrt{2\times 3}=\sqrt6$

 

除法運算:

設 $a\ge 0$ 、 $b\ge 0$ ,則: $\sqrt{a}\div \sqrt{b}=\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}=\sqrt{a\div b}$

$\sqrt{2}\div \sqrt{3}=\dfrac{\sqrt{2}}{\sqrt{3}}=\sqrt{\dfrac{2}{3}}=\sqrt{2\div 3}$

說明

$\begin{array} {rl} & (\sqrt2\div \sqrt3)^2 \\ = & (\dfrac{\sqrt2}{\sqrt3})^2 \\ = & (\dfrac{\sqrt2}{\sqrt3})\times (\dfrac{\sqrt2}{\sqrt3}) \\ = & \dfrac{\sqrt2\times \sqrt2}{\sqrt3 \times \sqrt3} \\ = & \dfrac{(\sqrt2)^2}{(\sqrt3)^2} \\ = & \dfrac23 \end{array}$

$\Rightarrow \sqrt2\div \sqrt3$ 是 $\dfrac23$ 的正平方根

$\Rightarrow \sqrt2\div \sqrt3=\dfrac{\sqrt2}{\sqrt3}=\sqrt{\dfrac23}=\sqrt{2\div 3}$

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