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.

(N/A) To check if $f$ is one-one:
$A$ function $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$.
Here,$f(1) = 1$ and $f(2) = 1$.
Since $f(1) = f(2)$ but $1 \neq 2$,the function $f$ is not one-one.
To check if $f$ is onto:
$A$ function $f: N \rightarrow N$ is onto if for every $y \in N$,there exists an $x \in N$ such that $f(x) = y$.
Case $1$: If $y = 1$,we have $f(1) = 1$. Thus,$1$ has a pre-image.
Case $2$: If $y > 1$,then $y+1 > 2$. We can choose $x = y+1$. Then $f(x) = f(y+1) = (y+1) - 1 = y$.
Since every $y \in N$ has a pre-image in $N$,the function $f$ is onto.