Question

Show that the function $f:N\to N,$ given by
$f\left(1\right)=f\left(2\right)=1$ and $f\left(x\right)=x-1$ for every $x>2,$ is onto but not one-one.

Answer 1

Solve for one-one:
Here $f:N\to N$
Given, $f\left(1\right)=f\left(2\right)=1$ and $f\left(x\right)=x-1,$ for every $x>2$
Since, different elements $1,2$ having same image $1$,
$\therefore f$ is not one-one.

Solve for onto:
$f:N\to N$
Given, $f\left(1\right)=f\left(2\right)=1$ and $f\left(x\right)=x-1$, for every $x>2$
Let $f\left(x\right)=y$ such that $y\in N$
Here,$y$ is a natural number $\&$ for every $y$, there is a value of $x$ which is a natural number.
$\therefore f$ is onto.
Hence, $f$ is onto but not one-one.