Question

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

Answer 1

Step 1: Assume $\sqrt{3}$ is rational. If $\sqrt{3}$ is rational, it can be expressed as:
$\sqrt{3} = \frac{p}{q}, \quad \text{where } p \text{ and } q \text{ are coprime integers, and } q \neq 0.$
Step 2: Square both sides:
$3 = \frac{p^2}{q^2} \implies p^2 = 3q^2.$
This implies $p^2$ is divisible by 3, so $p$ is also divisible by 3 (property of primes).
Step 3: Express $p$ as $p = 3k$. Substitute $p = 3k$ into $p^2 = 3q^2$:
$(3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies q^2 = 3k^2.$
This implies $q^2$ is divisible by 3, so $q$ is also divisible by 3.
Step 4: Contradiction. If both $p$ and $q$ are divisible by 3, they are not coprime, which contradicts our initial assumption. Thus, $\sqrt{3}$ is irrational.
Correct Answer: Proved

Answer 2

Step 1: Assume $\sqrt{3}$ is rational. If $\sqrt{3}$ is rational, it can be expressed as:
$\sqrt{3} = \frac{p}{q}, \quad \text{where } p \text{ and } q \text{ are coprime integers, and } q \neq 0.$
Step 2: Square both sides:
$3 = \frac{p^2}{q^2} \implies p^2 = 3q^2.$
This implies $p^2$ is divisible by 3, so $p$ is also divisible by 3 (property of primes).
Step 3: Express $p$ as $p = 3k$. Substitute $p = 3k$ into $p^2 = 3q^2$:
$(3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies q^2 = 3k^2.$
This implies $q^2$ is divisible by 3, so $q$ is also divisible by 3.
Step 4: Contradiction. If both $p$ and $q$ are divisible by 3, they are not coprime, which contradicts our initial assumption. Thus, $\sqrt{3}$ is irrational.
Correct Answer: Proved