Question

Give an example of two irrational numbers, whose product is a rational number.

• A. $\sqrt{8},\sqrt{2}$
• B. $\sqrt{5},\sqrt{2}$
• C. $2+\sqrt{8},\sqrt{2}$
• D. $\sqrt{8},2+\sqrt{2}$

Answer 1

The correct option is A $\sqrt{8},\sqrt{2}$

Consider option $A$.

The number are $\sqrt{8}$ and $\sqrt{2}$.
Their product $=\sqrt{8}\times \sqrt{2}=2\sqrt{2}\times \sqrt{2}=2\times 2=4$, which is a rational number.

Consider option $B$.

The number are $\sqrt{5}$ and $\sqrt{2}$.
Their product $=\sqrt{5}\times \sqrt{2}=\sqrt{10}$, which is an irrational number.

Consider option $C$.

The number are $2+\sqrt{8}$ and $\sqrt{2}$.
Their product $=(2+\sqrt{8})\times \sqrt{2}=2\sqrt{2}+(\sqrt{2}\times \sqrt{8})=2\sqrt{2}+4$, which is an irrational number.

Consider option $D$.

The number are $\sqrt{8}$ and $2+\sqrt{2}$

Their product $=\sqrt{8}\times (2+\sqrt{2})=2\sqrt{8}+(\sqrt{8}\times \sqrt{2})=4\sqrt{2}+4$, which is an irrational number.

Therefore, option $A$ is correct.