Section 1.3: Functions

Solution
\(f\) is not one-one, as \(f(1) = f(2) = 1\).
But \(f\) is onto, as given any \(y \in \mathbb{N}, y \ne 1\), we can choose \(x\) as \(y + 1\) such that
\[f(y + 1) = y + 1 - 1 = y\]
Also for \(1 \in \mathbb{N}\), we have \(f(1) = 1\).