Question

Prove that √6-√3 is an irrational number

Answer 1

Answer:

Let 3√6 be rational. an equal to a/b. where a,b≠0 are rational.. then a/b must be rational. but this contradicts that 3√6 is rational. hence 3√6 is irrational.

Answer 2

$\sqrt{6} - \sqrt{3}$

let us assume that it is rational so

root 6 - root 3=a/b

root6 = a/b + root3

root6 = a+ root 3 / b

here, a+root3 / b is rational but root 6 is irrational so this contradict our fact that it is irrational.