Question

Show that \[2 - \sqrt{3}\] is an irrational number.

Answer 1

Let us assume that \[2 - \sqrt{3}\] is rational .Then, there exist positive co primes a and b such that

\[2 - \sqrt{3} = \frac{a}{b}\]
\[\sqrt{3} = 2 - \frac{a}{b}\]

This implies,

\[\sqrt{3}\] is a rational number, which is a contradiction.

Hence, \[\sqrt{3}\] is irrational number.