Question

Show that the function $f:Z\to Z$ defined by $f\left(x\right)=x^2+x\forall x\in Z$, is a many-one function.

Answer 1

Let $x,y\in Z$ such that $f\left(x\right)=f\left(y\right)$

$\Rightarrow x^2+x=y^2+y$

$\Rightarrow x^2-y^2+x-y=0$

$\Rightarrow (x-y)(x+y+1)=0$

$\Rightarrow x=y$ or $x=-y-1$

Since $f\left(x\right)=f\left(y\right)$ does not yield the unique solution $x=y$ but also provides the solution $x=-y-1,$

So it is not a one - one function.

For example, if $y=1,$ then $x=1$ from $x=y$ and also $x=-2$ from $y=-x-1$

This means that $1$ and $-2$ have the same image

Hence, $f$ is a many-one function