Question

show that 3 root 6 is an irrational number​

Answer 1

Step-by-step explanation:

$to \: prove \: that \: 3 \sqrt{6} \: is \: an \: irrational \: number \: \\ 3 \sqrt{6} = 3 \sqrt{2 \times 3} \\ = 3 \times \sqrt{2 } \times \sqrt{3} \\ we \: know \: that \: \sqrt{2} \: and \: \sqrt{3} \: are \: \\ irrationals \: so \: their \: product \: \sqrt{6} \\ \: is \: irrational \: now \: 3 \: is \: rational \: \\ and \: \sqrt{6 } \: irrationl \\ product \: of \: a \: rational \: and \: an \\ \: irrational \: is \: also \: irrational \\ thus \: 3 \sqrt{6} \: is \: an \: irrational \: no. \\ proved$