Question

Fill in the blanks to complete the proof of $1+7\sqrt{3}$ as an irrational number, provided $\sqrt{3}$ is an irrational number.

Let’s assume $1+7\sqrt{3}$ be a rational number
$\Rightarrow 1+7\sqrt{3}=\frac{a}{b},b\ne 0$
$\Rightarrow \sqrt{3}=$ (i)___________ which is a (ii)_________ number.

$\Rightarrow \sqrt{3}$ is a rational number.
But $\sqrt{3}$ is a/an irrational number.
Hence, we have arrived at a contradiction.
$\because 1+7\sqrt{3}$ is an irrational number.

• A. (i) $\left(\frac{a-b}{7b}\right)$ , (ii) irrational
• B. (i) $\left(\frac{a-1}{7b}\right)$ , (ii) rational
• C. (i) $\left(\frac{a-b}{7b}\right)$ , (ii) rational
• D. (i) $\left(\frac{a-1}{7b}\right)$ , (ii) irrational

Answer 1

The correct option is C (i) $\left(\frac{a-b}{7b}\right)$ , (ii) rational