CBSE Class 10 Mathematics Standard 2024: question 26(a)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Assume, on the contrary, that $\sqrt{3}$ is rational.

   Then $\sqrt{3}=\frac{p}{q}$, where $p,q$ are integers, $q\ne 0$, and $p,q$ are coprime.

   Squaring: $3q^2=p^2$.

   ½ mark

2. So $p^2$ is divisible by 3. Since 3 is prime, 3 divides $p^2$ means 3 divides $p$. ... (1)

   1 mark

3. Let $p=3a$ for some integer $a$. Then $9a^2=3q^2$, so $q^2=3a^2$.

   So $q^2$ is divisible by 3, and hence $q$ is divisible by 3. ... (2)

   1 mark

4. From (1) and (2), 3 divides both $p$ and $q$. This contradicts the fact that $p$ and $q$ are coprime.

   So our assumption is wrong. Hence $\sqrt{3}$ is irrational.

   ½ mark

Answer: Assuming $\sqrt{3}=p/q$ (coprime) leads to 3 dividing both p and q, a contradiction; hence $\sqrt{3}$ is irrational

Revise the chapter

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