Answer
True
Work Step by Step
Suppose, for the sake of contradiction, that
\[
6 - 7\sqrt{2}
\]
is rational. Then there exists a rational number \( q \) such that
\[
6 - 7\sqrt{2} = q.
\]
Rearrange the equation to isolate \(\sqrt{2}\):
\[
-7\sqrt{2} = q - 6 \quad \Longrightarrow \quad \sqrt{2} = \frac{6 - q}{7}.
\]
Since \(q\) and $6$ are rational, their difference \(6 - q\) is rational, and dividing a rational number by 7 (a nonzero rational) yields a rational number. Thus, \(\sqrt{2}\) would be rational.
However, it is a well-known fact that \(\sqrt{2}\) is irrational. This contradiction shows that our assumption is false.
Therefore, \(6 - 7\sqrt{2}\) is irrational.
