Question

prove √3 is irrational​

Answer 1

REQUIRED ANSWER:-

Let $\sqrt{3}$ is a rational number.

Hence,

$\sqrt{3} = \frac{a}{b} \:where\: b\neq o \: and \: a \: ,\: b \: \: are \:coprimes.$

$a=\sqrt{3}b$

Squaring both sides

$a^{2} = 3b^{2}$ -----> (1)

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

∴ 3 is a factor of a.

a = 3c (where c is a constant)

Substitute,

a=3c in $eq^{n}$ (1)

(3c)² = 3b²

9c²=3a²

c²= $\frac{a^{2} }{3}$

∴ 3 is also a factor of b.

But, this is a contradiction.

∴ 3 is an irrational number.

hope it helps :)

Answer 2

Answer:

√3 = 1.7320508

which is non terminating and non recurring

Therefore √3 is an irrational number