Question

prove that 3 + root 2 is an irrational number​

Answer 1

Hope it helps

As we know✓2 is irrational

3+✓2=p/q

✓2=(p/q)-3

✓2=p-3q/q

Irrational no. =Rational no.

So 3+✓2=irrational

Answer 2

Answer:

$let \: be \: assume \: that \: 3 + \sqrt{2} \\ \: is \: a \: rational \: number. \\ therefore \: it \: can \: be \: written \\ \: in \: the \: form \: of \: \frac{a}{b} \\ \\ 3 + \sqrt{2} = \frac{a}{b} \\ \\ \sqrt{2} = \frac{a}{b} - 3 \\ \\ \sqrt{2} = \frac{a - 3b}{b} \\ \\ \frac{a - 3b}{b} \: is \: a \: rational \\ number \: therefore \: \sqrt{2} \\ is \: also \: a \: rational \: number \\ but \: it \: is \: contradict \\ \: that \: \sqrt{2} \: is \: irrational \\ therefore \: our \: assumption \\ is \: wrong \: and \: 3 + \sqrt{2} \: is \: \\ an \: irrational \: number.$

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