Question

Prove that √3 is an irrational number
(Easy method)

Answer 1

Answer:

$do \: not \: take \: it \: hardly \: see \: and \: understand \: this \: its \: easy \\ \\ t.p = \sqrt{3} is \: an \: irrational \: number \\ it \: can \: be \: solved \: by \: fundamental \: theorem \: of \: arithmetic(f.t.a) \: means \: if \: a \: divides \: {b}^{2} then \: a \: divides \: b \\ proof = \\ let \: \sqrt{3} \: is \: a \: rational \: number \\ so \: \sqrt{3} = \frac{p}{q} (p \: and \: q \: are \: real \: no. \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: q \: is \: not \: equal \: to \: 0) \\ squaring \: both \: sides \\ 3 = \frac{ {p}^{2} }{ {q}^{2} } (p \: and \: q \: are \: co \: primes) \\ {p}^{2} = 3 {q}^{2} \\ {p}^{2} is \: a \: multiple \: of \: 3 \\ if \: \frac{3}{ {p}^{2} } \\ then \: \frac{3}{p} (by \: f.t.a) \\ 3 \: is \: a \: factor \: of \: p \\ p = 3m \\ {(3m)}^{2} = 3 {q}^{2} \\ 9 {m}^{2} = 3 {q}^{2} \\ {q}^{2} = 3 {m}^{2} \\ if \: \frac{3}{ {q}^{2} } \\ then \: \frac{3}{q} (by \: f.t.a) \\ 3 \: is \: a \: factor \: of \: q \\ therefore \: 3 \: is \: common \: factor \: of \: p \: and \: q \\ so \: p \: and \: q \: are \: not \: co \: primes \\ this \: contradicts \: our \: assumption \\ therefore \: \sqrt{3} \: is \: an \: irrational \: number. \\ \\ \\ hope \: it \: helps............ \\ plz \: mark \: as \: brainliest................$