Question

Prove that
$\sqrt{3}$
is an irrational number.

​

Answer 1

Answer:

$let \sqrt{3}$  is rational, then we express it as  $\frac{a}{b}$ where, a and b are co primes

$\sqrt{3} =\frac{a}{b}$

$=> 3= \frac{a^{2} }{b^{2} }$

$=> 3b^{2} = a^{2}$

$=> b^{2}$ is divisible by 3

$=> a^{2}$ is divisible by 3

=> a is also divisible by 3

Now,

            let $a = 2c$

            $=> 3=\frac{3c^{2}}{b^{2} } \\=> 3b^{2} = 9c^{2} \\=> b^{2} = 3.c^{2}$

            => therefore   $3c^{2}$ is divisible by 3

            => c is also divisible by 3

             => b is also divisible by 3

here a and b are not coprime as they have common factors other than 1

∴ √3 is a irrational no.

Step-by-step explanation: Hope it will help you.

Thank you.