Question

Prove that √3 is irrational.

Answer 1

Step-by-step explanation:

let us assume that

$\sqrt{3} \ \: is \: rational \\ therefore \: \: it \: is \: in \: the \: form : \frac{a}{b} \: \: where \: a \: and \: b \: are \: coprime.$

$so \: \sqrt{3} b \: = \: a \\ squaring \: both \: sides \: we \: get$

3b² = a²

Therefore, 3 divides a² , and by theorem 1.3 (of real numbers chapter)

3 divides a .

so, we can write a = 3c , for some integer c.

substituting for a , we get 2b² = 4c², or b² =2c²

Therefore 3 divides b² , and also 3 divides b.

Therefore a and b have atleast 3 as a common factor.

But this contradicts the fact that a and b are coprime.

Therefore, our assumption is wrong and,

$\sqrt{3 } \: is \: a \: irrational$

hope it helps you , mate