Prove that \(\sqrt{3}+\sqrt{5}\) is irrational.
Irrational.
Goal. Show that \(\sqrt{3}+\sqrt{5}\) cannot be a rational number.
Step 1: Assume the opposite.
Let \(s = \sqrt{3} + \sqrt{5}\).
Assume \(s\) is rational.
Step 2: Isolate one radical.
From \(s = \sqrt{3} + \sqrt{5}\), we get
\(\sqrt{5} = s - \sqrt{3}\).
Step 3: Square to remove \(\sqrt{5}\).
Square both sides:
\(5 = (s - \sqrt{3})^2\).
Expand the right-hand side:
\(5 = s^2 + 3 - 2s\sqrt{3}\).
Step 4: Solve for \(\sqrt{3}\).
Rearrange:
\(2s\sqrt{3} = s^2 - 2\).
Therefore,
\(\sqrt{3} = \dfrac{s^2 - 2}{2s}\).
Step 5: Get the contradiction.
Since we assumed \(s\) is rational, both \(s^2\) and \(2s\) are rational.
Hence \(\dfrac{s^2 - 2}{2s}\) is rational.
That means \(\sqrt{3}\) is rational, which is false.
Step 6: Conclude.
Our assumption that \(s\) is rational leads to a contradiction.
Therefore, \(\sqrt{3}+\sqrt{5}\) is irrational.
Note. We divided by \(2s\) in Step 4. This is valid because \(s = \sqrt{3}+\sqrt{5} > 0\).