Question

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

Answer 1

Step 1: Assume the opposite (proof by contradiction):
To prove that $ \sqrt{3} $ is irrational, we will assume the opposite, i.e., assume that $ \sqrt{3} $ is a rational number.
This means that $ \sqrt{3} $ can be expressed as a fraction of two integers $ p $ and $ q $, where $ p $ and $ q $ are coprime (i.e., they have no common factors other than 1) and $ q \neq 0 $.
So, we assume that:
\[ \sqrt{3} = \frac{p}{q} \] where $ p $ and $ q $ are integers with $ \gcd(p, q) = 1$.

Step 2: Square both sides:
Next, square both sides of the equation to eliminate the square root:
\[ 3 = \frac{p^2}{q^2} \] Multiplying both sides by $ q^2 $ gives:
\[ 3q^2 = p^2 \] This equation shows that $ p^2 $ is divisible by 3, which means that $ p $ must also be divisible by 3 (since if a square of a number is divisible by 3, the number itself must be divisible by 3).
So, let $ p = 3k $ for some integer $ k $.

Step 3: Substitute $ p = 3k $ into the equation:
Substitute $ p = 3k $ into the equation $ 3q^2 = p^2 $:
\[ 3q^2 = (3k)^2 = 9k^2 \] Now, divide both sides by 3:
\[ q^2 = 3k^2 \] This shows that $ q^2 $ is divisible by 3, and hence $ q $ must also be divisible by 3.

Step 4: Contradiction:
We have now shown that both $ p $ and $ q $ are divisible by 3. But this contradicts our original assumption that $ p $ and $ q $ are coprime (i.e., that they have no common factors other than 1).
Therefore, our assumption that $ \sqrt{3} $ is a rational number must be false.

Conclusion:
Since assuming that $ \sqrt{3} $ is rational leads to a contradiction, we conclude that $ \sqrt{3} $ is irrational.