Class X Session 2023-24 Question 22

MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041)

SECTION B - Question 22

Prove that √2 is an irrational number. 

Proof:

Let us assume  √2  is a rational number. Then, there exist positive integers a and b such that

$$\sqrt{2}=\frac{p}{q}where,pandq,areco-primei.e.theirHCFis1$$

$$OR(\sqrt{2}{)}^2=\frac{p^2}{q^2}$$

$$OR2=\frac{p^2}{q^2}$$

$$OR2\times q^2=p^2$$

This implies that 𝑝^2 is even because it is equal to 2𝑞^2. From this, we can conclude that 𝑝 must also be even because the square of an odd number is odd, and the square of an even number is even.

Let 𝑝 = 2k, where k is arbitrary integer , Substituting 𝑝 = 2k into the equation 2q^2 = p^2 we get 

Substituting 𝑝 = 2k into the equation 2q^2 = p^2

$$\Rightarrow 2q^2=(2k{)}^2$$

$$\Rightarrow 2q^2=4(k{)}^2$$

$$\Rightarrow q^2=2(k{)}^2$$

This implies that 𝑞^2  is even. Therefore, 𝑞 must also be even. Now, we have shown that both 𝑝  and  q are even, which contradicts our initial assumption that 𝑝  and 𝑞 have no common factors other than 1. This contradiction arises from our assumption that √2 ​ is rational.

Hence, we can conclude that √2  is an irrational number.