Question

Fill in the blanks to complete the proof of $2-\sqrt{5}$ as an irrational number, provided $\sqrt{3}$ is an irrational number.

Let’s assume $2-\sqrt{5}$ be a rational number
$\Rightarrow 2-\sqrt{5}=\frac{a}{b},b\ne 0$
$\Rightarrow \sqrt{5}=$ (i)___________ which is a (ii)_________ number.
$\Rightarrow \sqrt{5}$ is a rational number.
But $\sqrt{5}$ is an irrational number.
Hence, we have arrived at a contradiction.
$\because 2-\sqrt{5}$ is an irrational number.

• A. (i) $\frac{2b-a}{b}$, (ii) irrational
• B. (i) $\frac{2b-a}{b}$, (ii) rational
• C. (i) $\frac{a-2b}{b}$, (ii) rational
• D. (i) $\frac{a-2b}{b}$, (ii) irrational

Answer 1

The correct option is B (i) $\frac{2b-a}{b}$, (ii) rational