Question

Prove that √2 is an irrational no.

Answer 1

Answer :

To prove,

√2 is an irrational no.

Proof :

Let √2 be a rational number in the form of p / q where q is not equal to zero at p and q are co-prime integers.

√2 = p/q

Whole sqauring both sides of this equation :-

2 = p^2/q^2

p^2 = 2q^2 (I)

From (I),

2 divided p^2

So, p divides p. (a)

Now , let p= 2k where k is any integer.

Substituting the values , we get :-

(2k)^2 = 2q^2

4k^2 = 2q^2

q^2 = 2k^2 (ii)

From (ii),

2 divides q^2.

Therefore, 2 divides q also. (b)

From statements (a) and (b) , we can say that :-

p and q have a common factor namely 2.

Hence, our assumption that p and q are co-prime is wrong. Hence , √2 is an irrational no.

Hence proved.

This method is called contradiction method.

Answer 2

Solution:

Let us assume that, √2 is a rational number of simplest form $\frac{a}{b}$, having no common factor other than 1.

√2 = $\frac{a}{b}$

On squaring both sides, we get ;

2 = $\frac{a^{2}}{b^{2}}$

⇒ a² = 2b²

Clearly, a² is divisible by 2.

So, a is also divisible by 2.

Now, let some integer be c.

⇒ a = 2c

Substituting for a, we get ;

⇒ 2b² = 2c

Squaring both sides,

⇒ 2b² = 4c²

⇒ b² = 2c²

This means that, 2 divides b², and so 2 divides b.

Therefore, a and b have at least 2 as a common factor. But this contradicts the fact that a and b have no common factor other than 1.

This contradiction has arises because of our assumption that √2 is rational.

So, we conclude that √2 is irrational.