Answer
False
Work Step by Step
**Proof by Contradiction:**
Assume, for the sake of contradiction, that \(\frac{\sqrt{2}}{6}\) is rational. Then there exists a rational number \(q\) such that
\[
\frac{\sqrt{2}}{6} = q.
\]
Multiplying both sides by 6 (which is rational and nonzero), we obtain
\[
\sqrt{2} = 6q.
\]
Since the product of two rational numbers is rational, \(6q\) is rational. Therefore, \(\sqrt{2}\) would be rational. This contradicts the well-known fact that \(\sqrt{2}\) is irrational.
Hence, our assumption is false, and \(\frac{\sqrt{2}}{6}\) is irrational.
