Exercise 1
This proof is by contradiction.
Suppose the claim is false: \(\sqrt{4}\) is rational.
We can write \(\sqrt{4}=\frac{n}{d}\), where \(n\) and \(d\) are integers with no common factors.
Squaring both sides and moving \(d^2\) to the left-hand side, we get \(4d^2 = n^2\).
This implies that \(n^2\) is a multiple of 4, and thus 4 is a factor of \(n\).
Therefore, \(n^2\) is a multiple of 16.
But since \(4d^2 = n^2\), \(d^2\) must also be a multiple of 4, and thus 4 is a factor of \(d\).
The numerator and denominator have a common factor of 4, which is a contradiction. Hence, \(\sqrt{4}\) must be irrational. \(\blacksquare\)
