Question

Prove that root 2 is an irrational number hence show that 2/5 root 3 is an irrational number

Answer 1

Answer:

That is 3 is a common factor of a and b contradicting the fact that a and b are co-primes . This contradiction arises due to the wrong assumption that √3 is rational. Hence √3 is irrational.

Answer 2

$\sqrt{2} is \: rational \: number \: . \\ in \: the \: form \: of \: \frac{a}{b} \\ as \: squaring \: both \: sides \: .as \\ \sqrt{2 { \: }^{2} } = \: \binom{a}{b}^{2} \\ 2 = \frac{a^{2} }{b^{2} } \\ 2 {b}^{2} = {a}^{2} \: \\ suppose \: a \: is \: 2c \: . \\ such \: that \: if \: {a}^{2} is \: divisible \: by \: 2 {b }^{2} \\ so \: a \: is \: also \: divide \: with \: 2{b}^{2} \\ ...................i \\ next \: it \: is \: 2 {b}^{2} = 4 {c }^{2} \\ {b}^{2} = 2 {c}^{2} \\ suppose \: that \: {b}^{2} = 2a \: \\ as \: 4 {a}^{2} = 2 {c}^{2} \\ 2 {a}^{2} = {c}^{2} .....................ii \\ so \: by \: one \: and \: two \: we \: get \: that \: \sqrt{2} \\ is \: an \: irrational \: number \: as \: well \: as \: our \: assumption \: is \: wrong \: so \: \sqrt{2} \\ is \: rational \: number \ \: similarly \: \\ \frac{2}{5 \sqrt{3} } \: is \: an \: irrational \: number \: because \: \sqrt{3} is \: an \: irrational \: number \:$

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