Question

Prove that \( \sqrt{5} \) is an irrational number.

Hint:

If a prime \( p \) divides \( a^2 \), then \( p \) must divide \( a \). This is the core logic of irrationality proofs.

Answer 1

Step 1: Understanding the Concept:
We use the method of contradiction by assuming \( \sqrt{5} \) is rational.
Step 2: Detailed Explanation:
Suppose \( \sqrt{5} \) is rational. Then \( \sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Squaring both sides: \( 5 = \frac{a^2}{b^2} \implies a^2 = 5b^2 \).
This implies \( a^2 \) is divisible by 5, which means \( a \) is also divisible by 5.
Let \( a = 5c \) for some integer \( c \).
Substitute: \( (5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2 \).
This implies \( b^2 \) is divisible by 5, which means \( b \) is also divisible by 5.
Thus, \( a \) and \( b \) have a common factor 5, contradicting that they are co-prime.
Our assumption is wrong; therefore, \( \sqrt{5} \) is irrational.
Step 3: Final Answer:
\( \sqrt{5} \) is irrational. Hence Proved.