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

\[\sqrt{2} =\frac{p}{q} where,\; p\; and\; q,\; are \;co-prime\; i.e. \;their\; HCF \;is \;1\]

\[ OR \; (\sqrt{2})^2 =\frac{p^2}{q^2} \]

\[OR\; 2 =\frac{p^2}{q^2} \]

\[OR\; 2 \times q^2 = p^2 \]

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.