Question

Prove that 3+√2 is an irrational number
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Answer 1

Step-by-step explanation:

$let \: 3 + \sqrt{2} \: be \: rational \: number$

$3 + \sqrt{2} = \frac{p}{q}$

$\sqrt{2} = (p - 3q) \div q$

$but \: \sqrt{2} is \: irrational \: no.$

therefore 3+√2 is an irrational no.