Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.
Show that the function $f: N \rightarrow N ,$ given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2,$ is onto but not one-one.
$f$ is not one-one, as $f(1)=f(2)=1 .$ But $f$ is onto, as given any $y \in N ,\, y \neq 1$ we can choose $x$ as $y+1$ such that $f(y+1)$ $=y+1-1=y .$ Also for $1 \in N$, we have $f(1)=1$.
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