CBSE Class 10 Mathematics Standard 2023: question 27

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Step-by-step solution

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1. Assume, to the contrary, that $\sqrt{5}$ is rational.

   Then $\sqrt{5}=\frac{p}{q}$, where $p,q$ are integers, $q\ne 0$, and $p,q$ are co-prime.

   ½ mark

2. Squaring gives $5q^2=p^2$. So $p^2$ is divisible by 5.

   Since 5 is prime, $p$ is divisible by 5 [if a prime divides $p^2$, it divides $p$].

   So $p=5a$ for some integer $a$.

   1 mark

3. Substituting: $25a^2=5q^2$, so $q^2=5a^2$.

   Thus $q^2$ is divisible by 5, and hence $q$ is divisible by 5.

   ½ mark

4. So 5 is a common factor of $p$ and $q$. This contradicts the assumption that $p$ and $q$ are co-prime.

   The assumption is wrong. Hence $\sqrt{5}$ is irrational.

   1 mark

Answer: $\sqrt{5}$ is irrational (proved by contradiction).

Revise the chapter

• Chapter: Real Numbers (notes, formulas, where marks are lost and every past question)
• Topic: Proving Numbers Irrational