Question

prove that root3 is irrational number​

Answer 1

let root 3 is rational no.

therefore root 3 =a/b ( a and b is a coprime no.)

now,squaring on both sides

we get,

(root3)^2=(a/b)^2

3=a^2/b^2

3b^2=a^2.....(i)

b^2=a^2/3

Since a^2 is divisible by 3

therefore a is also divisible by 3

a=3c

put a =3c in eq (i)...

3b^2=(3c)^2

3b^2=9c^2

3b^2/9=c^2

b^2/3=c^a

Since b^2 is divisible by 3

therefore b is also divisible by 3

3 is a common no. in a and b ,which is contradiction.

therefore root 3 is an irrational no.

-------hence proved-----------

Answer 2

Answer:

Step-by-step explanation: assume that √3 is a rational number

So√3 = a/b

Squaring on both the sides

(√3)²=(a/b)²

Square and root be cancelled

Then 3=a²/b²

3b²=a². ------------- 1

b²=a²/3

So 3 divides a

3divides a²

Let we take a²= 3c for a integer

By equation 1

3b²=a²

3b²=(3c)²

3b²=9c²

b²=9c²/3

b²=3c²

c²(b²)=3

c²=b²/3

So 3 divides b

And 3 divides b²

Our assumption is wrong √3 is a irrational number