Question

prove
$\sqrt{3}$
is an irrational no​

Answer 1

Let $\sqrt{3}$ is rational

$\therefore \sqrt{3} = \dfrac{a}{b}$

Here, $a$ and $b$ are integers and $b \neq 0$

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

and $a^2 = 3b^2$

$\implies a^2$ is divisible by 3 and so $a$ is also divisible by 3.

Let $a = 3c$

$\therefore \quad a^2 = 3b^2$

$\implies 9c^2 = 3b^2$

and, $\ \ \ \ \ b^2 = 3c^2$

$\implies b^2$ is divisible by 3 and so $b$ is divisible by 3.

$\because a$ is divisible by 3 and $b$ is also divisible by 3

$\implies \sf \dfrac{a}{b} \: is \: not \: rational$

$\implies \sf \sqrt{3} \: is \; not \: rational$

and, so $\bold{ \sqrt{3} \: is \: irrational.}$

Remember,

In a rational number, numerator and denominator can't have a common factor except 1

Answer 2

$\tt\huge\blue{\underbrace{ Question : }}$

Prove that √3 is an irrational number.

$\tt\huge\blue{\underbrace{ Solution : }}$

Let us assume that √3 is a rational number.

$\bf\:\implies \sqrt{3} = \frac{a}{b}$

• [ a & b are co - primes. ]

$\bf\:\implies b\sqrt{3} = a$

• [ Squaring on both sides. ]

$\bf\:\implies (b\sqrt{3})^{2} = (a)^{2}$

$\bf\:\implies 3b^{2} = a^{2}$

$\bf\:\implies b^{2} = \frac{a^{2}}{3}$

★ 3 divides a² and 3 divides a.

★ So, we can write " a = 3c " .

• Substitute the value of a.

$\bf\:\implies b^{2} = \frac{(3c)^{2}}{3}$

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

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

$\bf\:\implies \frac{b^{2}}{3} = c^{2}$

★ Hence, 3 divides b² and 3 divides b.

↪ Therefore, both a & b have 3 as a common factor.

↪ But this contradicts the fact that, a & b are co - primes.

↪ This contradictions has arisen because of our assumption that √3 is a rational number.

↪ Thus our assumption is false. So, I concluded that...

$\underline{\boxed{\bf{\purple{ \therefore \sqrt{3} \: is\: an\:irrational\:number.}}}}\:\orange{\bigstar}$