Answer
True
Work Step by Step
**Proof by Contradiction:**
Let \(a\) be a positive irrational number. We wish to prove that \(\sqrt{a}\) is irrational.
Suppose, for the sake of contradiction, that \(\sqrt{a}\) is rational. Then there exist integers \(p\) and \(q\) (with \(q \neq 0\) and in lowest terms) such that
\[
\sqrt{a} = \frac{p}{q}.
\]
Squaring both sides gives
\[
a = \left(\frac{p}{q}\right)^2 = \frac{p^2}{q^2}.
\]
Since \(p^2\) and \(q^2\) are integers and \(q^2 \neq 0\), it follows that \(\frac{p^2}{q^2}\) is a rational number. Thus, \(a\) is rational. This contradicts our assumption that \(a\) is irrational.
Hence, our assumption must be false, and \(\sqrt{a}\) is irrational.
This completes the proof.
