Prove that \(\sqrt{p}+\sqrt{q}\) is irrational, where \(p, q\) are primes.
Irrational.
Recall. For a prime \(p\), the number \(\sqrt{p}\) is irrational. If \(\sqrt{p}\) were rational, say \(\sqrt{p}=\dfrac{m}{n}\) in lowest terms, then squaring would give \(p=\dfrac{m^2}{n^2}\). Hence \(pn^2=m^2\), so \(p\) divides \(m^2\), and therefore divides \(m\), contradicting lowest terms unless \(n=1\) and \(p\) were a perfect square. But a prime is not a perfect square. So \(\sqrt{p}\) is irrational.
Goal. Show that the sum \(\sqrt{p}+\sqrt{q}\) cannot be rational.
Step 1. Assume, for contradiction, that the sum is rational. Let \(s\in\mathbb{Q}\) with
\(\sqrt{p}+\sqrt{q}=s.\)
Step 2. Isolate one square root and square. From
\(\sqrt{q}=s-\sqrt{p}\)
square both sides to get
\(q=s^2+p-2s\,\sqrt{p}.\)
Step 3. Solve for \(\sqrt{p}\). Rearranging gives
\(2s\,\sqrt{p}=s^2+p-q.\)
If \(s\neq 0\) (the sum of two positive square roots cannot be \(0\)), then
\(\sqrt{p}=\dfrac{s^2+p-q}{2s}.\)
Step 4. Reach the contradiction. The right-hand side is a quotient of rationals, hence rational. Thus \(\sqrt{p}\) is rational, which contradicts the recall above.
Special case \(p=q\). Then \(\sqrt{p}+\sqrt{q}=2\sqrt{p}.\) Since \(\sqrt{p}\) is irrational and \(2\) is a nonzero rational, the product \(2\sqrt{p}\) is irrational.
Conclusion. In all cases, the assumption that \(\sqrt{p}+\sqrt{q}\) is rational leads to a contradiction. Therefore \(\sqrt{p}+\sqrt{q}\) is irrational.