Question

prove that √3 is an irrational number.

Answer 1

let the √3 is rational so it can be written in the form of p/q where p and q are co prime

√3=a/b
squaring on both sides we get

3=a²/b²
3b²=a²


hence a²divide 3b² so it also divide 3b

now le a =3c


3b²=(3c)²

3b²= 9c²

hence 3b² divide 9c² and hence 9c


thus a and b are not co prime



hence √3 is irrational

Answer 2

Let √3 be rational number

Hence, √3 can be written in the form a/b where a and b are coprime.

√3 = a/b

Squaring both sides

3= a²/b²
3b² = a²

Hence 3 divides a² 
⇒ 3 divides a

Now let a= 2c for some integer c

3b² = 9c²
b²= 3c²
Hence 3 divides b²
⇒ 3 divides b

But a and b were co prime which means they were supposed to have no common factor other than 1.

Therefore our assumption was wrong

√3 is irrational.