Question

Prove that $\sqrt{3}$ is an irrational number.

Hint:

Use the method of contradiction to prove irrationality — assume rationality and reach a contradiction.

Answer 1

Step 1: Assume the opposite. 
Let $\sqrt{3}$ be rational. Then it can be expressed as: \[ \sqrt{3} = \dfrac{p}{q}, \quad \text{where } p, q \text{ are integers and } \gcd(p, q) = 1 \] Step 2: Square both sides. 
\[ 3 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 3q^2 \] Step 3: Analyze divisibility. 
Since $p^2$ is divisible by 3, $p$ must also be divisible by 3. Let $p = 3k$. 
Step 4: Substitute back. 
\[ p^2 = (3k)^2 = 9k^2 \Rightarrow 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2 \] Thus, $q^2$ is also divisible by 3, so $q$ is divisible by 3. 
Step 5: Contradiction. 
This contradicts the assumption that $p$ and $q$ have no common factor other than 1. 
Step 6: Conclusion. 
Therefore, $\sqrt{3}$ is irrational.