Question

prove that 4 - root3 is irrational

Answer 1

Step-by-step explanation:

$4 - \sqrt{3} = \frac{a}{b} \\ - \sqrt{3} = \frac{a}{b} - 4 \\ - \sqrt{3} = \frac{a - 4b}{b} \\ \sqrt{3} = - ( \frac{a - 4b}{b}) \\ \sqrt{3} = \frac{4b - a}{b} \\ \sqrt{3} \: is \: an \: irrational \: number \: while \: a \: and \: b \: are \: rational \: numbers \: \\ this \: cannot \: be \: possible \\ hence \: 4 - \sqrt{3} \: is \:an \: irrational \: number$

Answer 2

We have to first assume wrong
Then we have to do contradiction
The note the aumilaraties