Question

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

Answer 1

Step 1: Assume the opposite (proof by contradiction):
To demonstrate that $ \sqrt{3} $ is irrational, we commence by assuming its opposite: that $ \sqrt{3} $ is a rational number.
This implies $ \sqrt{3} $ can be represented as the ratio of two integers, $ p $ and $ q $, where $ p $ and $ q $ are coprime (sharing no common factors other than 1) and $ q eq 0 $.
Our initial assumption is therefore:
\[ \sqrt{3} = \frac{p}{q} \] with $ p, q \in \mathbb{Z} $ and $ \gcd(p, q) = 1$.

Step 2: Square both sides:
Squaring both sides of the equation eliminates the radical:
\[ 3 = \frac{p^2}{q^2} \] Multiplying by $ q^2 $ yields:
\[ 3q^2 = p^2 \] This equation signifies that $ p^2 $ is a multiple of 3, which consequently means $ p $ must also be a multiple of 3 (as if the square of an integer is divisible by 3, the integer itself must be).
Thus, we can express $ p $ as $ p = 3k $ for some integer $ k $.

Step 3: Substitute $ p = 3k $ into the equation:
Substituting $ p = 3k $ into $ 3q^2 = p^2 $ gives:
\[ 3q^2 = (3k)^2 = 9k^2 \] Dividing both sides by 3 results in:
\[ q^2 = 3k^2 \] This shows that $ q^2 $ is divisible by 3, implying that $ q $ is also divisible by 3.

Step 4: Contradiction:
We have now established that both $ p $ and $ q $ are divisible by 3. This directly contradicts our initial premise that $ p $ and $ q $ are coprime (having no common factors other than 1).
Therefore, our assumption that $ \sqrt{3} $ is rational must be incorrect.

Conclusion:
The derivation of a contradiction from the assumption that $ \sqrt{3} $ is rational leads us to conclude that $ \sqrt{3} $ is indeed irrational.