Question

Using (√7+√3)(√7-√3) = 4 and the fact that (√7+√3) is irrational prove that √7-√3 is irrational.

Answer 1

Given, (√7 + √3)(√7 - √3) = 4 and we know, (√7 + √3) is an irrational number.

now, $\bf{(\sqrt{7}-\sqrt{3})=(\sqrt{7}-\sqrt{3})\times\frac{(\sqrt{7}+\sqrt{3})}{(\sqrt{7}+\sqrt{3})}}$
[multiplying numerator and denominator with (√7 + √3) ]
use (a - b)(a + b) = a² - b²
$(\sqrt{7}-\sqrt{3})= \frac{(\sqrt{7}^2-\sqrt{3}^2)}{(\sqrt{7}+\sqrt{3})}=\frac{4}{(\sqrt{7}+\sqrt{3})}$

here it is given that (√7 + √3) is an irrational number and we know 4 is a rational number.
therefore, (√7 - √3) = 4/(√7 + √3) = rational/irrational

we know, rational divided by irrational is always irrational number.
so, (√7 - √3) is an irrational number.