Question

If 'n' is a prime number, then \(\sqrt{n}\)  is

• A. Prime number
• B. Composite number
• C. Rational number
• D. Irrational number

Answer 1

The Correct Option is D

Step 1: Understand what a prime number is.

A prime number is a number that has exactly two distinct positive divisors: 1 and itself. Examples include 2, 3, 5, 7, 11, etc.

Step 2: Consider the square root of a prime number.

Let’s take examples:

• \(\sqrt{2} \approx 1.4142...\) — not a rational number
• \(\sqrt{3} \approx 1.7320...\) — not a rational number
• \(\sqrt{5} \approx 2.2360...\) — not a rational number

Step 3: Can the square root of a prime number be rational?

Suppose \( \sqrt{n} \) is rational, where \( n \) is a prime.

Then it can be expressed as: \[ \sqrt{n} = \frac{p}{q} \] where \( p \) and \( q \) are integers, \( \gcd(p, q) = 1 \), and \( q \ne 0 \).

Squaring both sides: \[ n = \frac{p^2}{q^2} \Rightarrow n q^2 = p^2 \] This implies that \( p^2 \) is divisible by \( n \), hence \( p \) is divisible by \( n \). So \( p = nk \) for some integer \( k \).

Substituting back: \[ n q^2 = (nk)^2 = n^2 k^2 \Rightarrow q^2 = n k^2 \] Now \( q^2 \) is divisible by \( n \), which implies \( q \) is divisible by \( n \). But that contradicts our assumption that \( p \) and \( q \) have no common factor.

Step 4: So, \( \sqrt{n} \) is irrational when \( n \) is a prime number.

The correct option is (D): Irrational number