Final answer:
By assuming \(\sqrt{2} + \sqrt{3}\) is rational, we derive that \(\sqrt{6}\) is rational, which contradicts the theorem stating that the square root of a non-perfect square is irrational. Hence, \(\sqrt{2} + \sqrt{3}\) must be irrational.
Step-by-step explanation:
To prove that \(\sqrt{2} + \sqrt{3}\) is irrational, we start by assuming, for the sake of contradiction, that it is rational. Let us suppose that \(\sqrt{2} + \sqrt{3}\) = p/q, where p and q are integers with no common factors, and q is not zero. Squaring both sides of the equation, we get \(2 + 2\sqrt{6} + 3 = p^2/q^2\), which leads to \(2\sqrt{6} = (p^2 - 5q^2)/q^2\). Since the left side involves a square root and the right side is a rational number, it implies that \(\sqrt{6}\) is rational. However, this contradicts the given theorem because 6 is not a perfect square, and therefore \(\sqrt{6}\) must be irrational.
Thus, our assumption that \(\sqrt{2} + \sqrt{3}\) is rational leads to a contradiction, and it must be irrational.