Question

Prove that √3 is irrational.​

Answer 1

Answer:

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

Answer 2

Answer:

Let us assume that √3 is irrational number.

So, We can find integers a & b where (b is not equal to zero).

Such that √3 = a/b

a & b have Common Factor other than 1 , then we can divide by the common factor & assume that a & b are co - primes.

=> b√3 = a

Squaring both sides:-

3b² = a²______eq(1)

Here we can see a² is divisible by 3 so, a is also divisible by 3.

Let a = 3k for some integer k.

Putting this value in eq (1)

we get 3b² = 9k²

Here b² = 3k²

That means b² is divisible by 3, so b is also divisible by 3.

It shows that a & b both have atleast 3 as a common factor.

But this contradiction arise that a & b are co - Prime.

Our assumption is wrong.

Therefore√3 is an irrational number.

Hence Proved