Question

prove that 2 irrational number (3+ root 2) and (3- root 2) are rational number​

Answer 1

Answer:

let Assume 3 + √2 is a rational number

rational number will be in the form of p/q , a, b, E, Z, b ≠ 0 .

$3 + \sqrt{2 \: } = \frac{a}{b}$

$= \sqrt{2} = \frac{a}{b} - 3$

$= \sqrt{2} = \frac{a}{b} - \frac{3}{1}$

$= \sqrt{2} = \frac{a}{b} - \frac{3}{1} \times \frac{b}{b}$

$= \sqrt{2} = \frac{a}{b} - \frac{3b}{b}$

$= \sqrt{2} = \frac{a - 3b}{b}$

$\frac{a - 3b}{b} \: is \: of \: the \: form \: of \: \frac{a}{b}$

$hence \: it \: is \: a \: rational \: number \: but \: \sqrt{2} is \: a \: irrational \: number$

$the \: assume \: is \: false$

$3 + \sqrt{2} \: is \: an \: irrational \: number \:$

$hence \: proved$

HOPE THIS HELPS U

#Ravalika Rajula

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