Question

Prove that `6-4sqrt5` is an irrational number, given that `sqrt5` is an irrational number.

Answer 1

Let us assume that `6-4sqrt5` be a rational number

Let `6-4sqrt5` = `a/b`     ...[b ≠ 0; a and b are integers]

∴ `sqrt5 = ([6 - a/b])/4`

We know that,

`([6 - a/b])/4` is a rational number.

But this contradicts the fact that `sqrt5` is an irrational number.

So, our assumption is wrong.

Therefore, `6-4sqrt5` is an irrational number.