Question

prove that root3 is irrational​

Answer 1

$\red\bigstar$ Solution:-

⠀ ⠀⠀

•If possible, let √3 be rational and let its simplest form be a/b.

⠀ ⠀⠀

↬Then , a and b are integer having no common factor other than 1 , and b⇎0

Now,

$\sf \implies\sqrt{3} = \dfrac{a}{b}$

On squaring both sides we get;

$\implies \sf 3 = \dfrac{ {a}^{2} }{ {b}^{2} }$

$\implies \sf \: 3 {b}^{2} = {a}^{2} \: \: \: \: ....(1)$

$\implies \sf3 \: divides \: {a}^{2} \: \: \: | \longrightarrow3 \: divides \: 3{b}^{2} |$

$\implies \sf3 \: divides \: a \: \: \: \: \: | \longrightarrow3 \: is \:prime \: and \: 3 \: divides \: {a}^{2} \implies3 \: divides \: a|$

⠀ ⠀⠀

↬Let a=3c for some integer c

Putting a=3c in (1) we get;

$\implies \sf \: 3 {b}^{2} = 9 {c}^{2}$

$\implies \sf {b}^{2} = 3c$

$\implies \sf \: 3 \: divides \:{ b}^{2} \: \: | \longrightarrow3 \: divides \: 3 {c}^{2} |$

$\implies \sf \: 3 \: divides \: b \: \: | \longrightarrow3 \:is \: prime \: and \: 3 \: divides \: {b}^{2} \implies3 \: divides \: b |$

↬Thus 3 is a common factor of a and b.

↬But, this contradict the fact that a and b have no common factor other than 1

⠀ ⠀⠀

↬The contradict arises by assuming that √3 is rational.

↬Hence, √3 is irrational

⠀ ⠀⠀