Step-by-step explanation:
[tex]\dfrac{\sqrt3}{\sqrt8}=\dfrac{\sqrt3\cdot\sqrt2}{\sqrt8\cdot\sqrt2}=\dfrac{\sqrt{3\cdot2}}{\sqrt{8\cdot2}}=\dfrac{\sqrt6}{\sqrt{16}}=\dfrac{\sqrt6}{4}[/tex]
Used:
[tex]\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\\sqrt{a}=b\iff b^2=a\ \text{for}\ a\geq0\ \wedge\ b\geq0\\\\\sqrt{16}=4\ \text{because}\ 4^2=16[/tex]
