Question

Prove that $\sqrt{2}$ is an irrational number.

Hint:

To prove irrationality, always start by assuming the opposite and reach a contradiction using the properties of even and odd numbers.

Answer 1

Step 1: Assume the contrary.
Let us assume, to the contrary, that $\sqrt{2}$ is a rational number. Then it can be expressed as \[ \sqrt{2} = \frac{p}{q} \] where \( p \) and \( q \) are co-prime integers (having no common factors other than 1), and \( q \neq 0 \).

Step 2: Square both sides.
\[ 2 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2 \]
Step 3: Analyze divisibility.
Since \( p^2 \) is even, \( p \) must also be even. Let \( p = 2r \).
Step 4: Substitute and simplify.
\[ (2r)^2 = 2q^2 \Rightarrow 4r^2 = 2q^2 \Rightarrow q^2 = 2r^2 \] Thus, \( q^2 \) is also even, and therefore \( q \) is even.
Step 5: Contradiction.
If both \( p \) and \( q \) are even, they have a common factor 2, contradicting the assumption that \( p \) and \( q \) are co-prime.
Step 6: Conclusion.
Hence, our assumption is false, and therefore \( \sqrt{2} \) is an irrational number.