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

We have a function $f:N\to N$, defined as

$f\left(1\right)=f\left(2\right)=1$ and $f\left(x\right)=x-1$, for every $x>2$

For one-one :

Since, $f\left(1\right)=f\left(2\right)=1$, therefore 1 and 2 have same image, namely 1.

So, f is not one-one.

For onto :

$y=1$ has two pre-image, namely 1 and 2.

Now, let $y\in N,y\ne 1$ be any arbitrary elements.

Then, $y=f\left(x\right)$

$⟹y=x-1⟹x=y+1>2$ for every $y\in N,y\ne 1$.

Thus, for every $y\in N,y\ne 1$, there exists $x=y+1$ such that

$f\left(x\right)=f(y+1)=y+1-1=y$

Hence, f is onto.