Question

prove that root 2 +3/root 2 is irrational number​

Answer 1

$let \sqrt{2} + 3 \div \sqrt{2} \: be \: a \: rational \: number \\ \sqrt{2} + 3 \div \sqrt{2 } = p \div q \\ 2 + 3 \div \sqrt{2} = p \div q \\ 6 \div \sqrt{2} = p \div q \\ rtionalise \: that \: equatoion \\ 6 \div \sqrt{2} \times \sqrt{2} \div \sqrt{2} = p \div q\\ 6 \sqrt{2} \div 2 = p \div q \\ 6 \sqrt{2} = 2p \div q \\ \sqrt{2} = 2p \div 6q \\ \sqrt{2} = p \div 3q$

p/3q is a rational number

$\sqrt{2} is \: a \: rationl \: number \\ but \: this \: is \: a \: contradiction \\ hence \: \sqrt{2} + 3 \div \sqrt{2} is \: a \: irrational \: number$