Question

EXERCI
1. Prove that V5 is irrational.
man in the cont​

Answer 1

Answer:

because root 5 =2.236...... as this decimal expansion is non terminating non recurring, it is irrational.

Answer 2

$let \: \sqrt{5 \: is \: rational} \: \\ we \: take \: two \: integers \: \:a \: and \: b \: \\ \sqrt{5 = a \div b \: \\ b \sqrt{5 = a \: on \: squaring \: both \: sides \: \\ \sqrt{5b {?}^{2} = \: a {?}^{2} } \: \\ a ^{2} is \: divisible \: by \: 5 \: so \: a \: is \: also \: divisible \: by \: 5 \: \\ then \: a \: = 5c \: \\ for \: some \: integer \: \: c \: \\ 5b ^{2} \: = \: 25c \: \\ b ^{2} \: = \: 5c \: \\ then \: b ^{2} \: is \: divisible \: by \: 5 \: so \: b ^{2} \: is \: also \: divisible \: by \: 5 \: \\ so \: our \: assumption \: that \: the \: \sqrt{5 \: is \: rational \: is \: false. \\ \sqrt{5 \: is \: irrational.$