Section 1.3: Functions
Example Exercise
Example 10
Show that the function \(f : \mathbb{N} \to
\mathbb{N}\) given by \(f(1) = f(2) =
1\) and \(f(x) = x - 1\), for
every \(x > 2\), is onto but not
one-one.
Example 10
Show that the function \(f : \mathbb{N} \to
\mathbb{N}\) given by \(f(1) = f(2) =
1\) and \(f(x) = x - 1\), for
every \(x > 2\), is onto but not
one-one.
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
\(f(1) = 1\).