Question

Assuming that √7 is an irrational number, prove that 4-3√7 is an irrational number

Answer 1

It is given that √7 is an irrational number.

Now, let us assume that 4 - 3√7 is a rational number.

∴ 4 - 3√7 = p/q 
as any rational number can be expressed in the form p/q where p and q are integers and q ≠ 0.
Also p and q are co-prime.

∵ $4 - 3 \sqrt{7} = \frac{p}{q}$
∴ $3 \sqrt{7} = 4 - \frac{p}{q}$
∴ $3 \sqrt{7} = \frac{4q - p}{q}$
∴ $\sqrt{7}= \frac{4q - p}{3q}$

Now, we know that √7 is an irrational number.
But, 4q - p / 3q is a rational number since p and q are integers.
∴ An irrational number cannot be equated to a rational number.
∴ Our assumption that 4 - 3√7 is a rational number is incorrect.
∴ 4-3√7 is an irrational number.