Question

Prove that root 3 is irrational

Answer 1

hello.....

Firstly let's assume √3 to be a rational number..

so therefore,

√3 = a/b ,

▶a and b both r integers and b = not zero ..

▶ 'a ' and 'b ' have no common factor than 1

▶therefore root 3 = a/b ....

b √ 3 = a

3b2 = a2 

(squaring both sides)............(1)

b2 = a2/3

Here...

3 divides a2 ...

so 3 divides a too!

so....

a = 3c..........
'c ' is an integer...........(2)

a2 = (3c)2 .....

(squaring both sides)

therefore,3b2 = 9c2 
substuting (2) in (1)

therefore, b2 = 3 c2

c2 = b2 / 3

3 divides b2 , means 3 divides b

so therefore...

▶'a ' and 'b ' have at least 3 as common factor...

▶ but a and b had no common factors other than 1.

▶this contradicts our assumption...

▶therefore, root 3 is irritional number...

thank you...

Answer 2

Let us assume that √3 is a rational number.

That is, we can find integers a and b (≠ 0) such that √3 = (a/b)

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

√3b = a

⇒ 3b2=a2 (Squaring on both sides) → (1)

Therefore, a2 is divisible by 3

Hence ‘a’ is also divisible by 3.

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

Equation (1) becomes,

3b2 =(3c)2

⇒ 3b2 = 9c2

∴ b2 = 3c2

This means that b2 is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are co prime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.

HOPE IT HELPS YOU!!!