Question

2 under root3 prove that this no is irrational

Answer 1

Let √3-√2 be a rational number.

A rational number can be written in the form of p/q.

√3-√2 = p/q

Squaring on both sides,

(√3-√2)² = (p/q)²

√3²+√2²-2(√3)(√2) = p²/q²

3+2-2√6 = p²/q²

5-2√6 = p²/q²

2√6 = 5-p²/q²

2√6 = (5q²-p²)/q²

√6 = (5q²-p²)/2q²

p,q are integers then (5q²-p²)/q² is a rational number.

Then √6 is also a rational number.

But this contradicts the fact that √6 is an irrational number.

So,our supposition is false.

Therefore,√3-√2 is an irrational number.

Hence proved