Question

prove that 2-√3is irrational ,given that √3 is irrational​

Answer 1

Answer:

Step-by-step explanation:

AS GIVEN IN THE QUESTION ROOT 3 IS GIVEN IRRATIONAL AND 2 IS A RATIONAL NUMBER. SO RATIONAL NUMBER - IRRATIONAL NUMBER WILL ALWAYS BE AN IRRATIONAL NUMBER .

Answer 2

Answer:

$let \: 2 - \sqrt{3 } \: be \: a \: rational \: number \: where \: 2 - \sqrt{3 } = \frac{a}{b} where \: b \: is \: not \: equal \: to \: zero \: and \: hcf \\ of \: a \: and \: b \: = 1 \\ 2 \: = \: \frac{a}{b} + \sqrt{3 } \\ \sqrt{3 } = 2 - \frac{a}{b} \\ 2 - \frac{a}{b} = \: rational \: number \: as \: we \: \: know \: that \\ \: rational \: number - \frac{rational \: number}{rational \: number} = rational \: number \\ therefore \sqrt{3 } = rational \: number \\ but \: it \: is \: given \: that \sqrt{3} = irrational \: number \\ but \: it \: contradict \: our \: assumption \: that \: \sqrt{3} \: is \: a \: rational \: number \: therefore \\ 2 - \sqrt{3 } \: is \: a \: irrational \: number$