Give an example of two rational numbers, whose quotient is a rational number.
A) $\sqrt {\text{5}} {\text{,}}\sqrt {\text{2}} $
B) $\sqrt {\text{8}} {\text{,}}\sqrt {\text{2}} $
C) $\sqrt {\text{3}} {\text{,}}\sqrt {\text{2}} $
D) $\sqrt 7 {\text{,}}\sqrt {\text{2}} $

Hint: In this question, we will approach our answer from the options given to us.
If two rational number on dividing has a rational number as the quotient one must a rational multiple of another. For example, a and b are two irrational numbers and on dividing has a rational number as the quotient then,
${\text{a = rb}}$ where, ${\text{r}} \in {\text{Q}}$

Complete step by step solution: From option(A)
On dividing $\sqrt {\text{5}} {\text{ and }}\sqrt {\text{2}} $, we get either $\dfrac{{\sqrt {\text{5}} }}{{\sqrt {\text{2}} }}{\text{ or }}\dfrac{{\sqrt {\text{2}} }}{{\sqrt {\text{5}} }}$ and neither is a rational number.
From option(B)
On dividing $\sqrt {\text{8}} {\text{ and }}\sqrt {\text{2}} $, we get either $\dfrac{{\sqrt {\text{8}} }}{{\sqrt {\text{2}} }}{\text{ = 2 or }}\dfrac{{\sqrt {\text{2}} }}{{\sqrt {\text{8}} }}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$ and both are rational number.
From option(C)
On dividing $\sqrt {\text{3}} {\text{ and }}\sqrt {\text{2}} $, we get either $\dfrac{{\sqrt {\text{3}} }}{{\sqrt {\text{2}} }}{\text{ or }}\dfrac{{\sqrt {\text{2}} }}{{\sqrt {\text{3}} }}$ and neither is a rational number.
From option(C)
On dividing $\sqrt {\text{7}} {\text{ and }}\sqrt {\text{2}} $, we get either $\dfrac{{\sqrt {\text{7}} }}{{\sqrt {\text{2}} }}{\text{ or }}\dfrac{{\sqrt {\text{2}} }}{{\sqrt {\text{7}} }}$ and neither is a rational number.

Therefore, (B) is correct,

Note: We can also check the options as $\sqrt {\text{2}} $ is there in every option, so we have to find an option having a multiple of $\sqrt {\text{2}} $, i.e. option(B) $\sqrt {\text{8}} {\text{,}}\sqrt {\text{2}} $, which we can also write as${\text{2}}\sqrt {\text{2}} {\text{,}}\sqrt {\text{2}} $, where the quotient is either ${\text{2 or }}\dfrac{{\text{1}}}{{\text{2}}}$.