Question

Check which function is bijective from $Z\to Z$:
(A) $f\left(x\right)=x^3$ (B) $f\left(x\right)=2x+1$

Answer 1

A) For $f\left(x\right)=x^3$, let $f\left(x_1\right)=f\left(x_2\right)$

or, $x_1^3=x_2^3$

or, $x_1=x_2$.[As $x_1,x_2\in Z$]

So for $f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$, so $f\left(x\right)$ is one-one.

The function is not onto as $2$ is point in the co-domain which has no pre-image in the domain.

Let,

$f\left(x\right)=2$.

or, $x^3=2$

or, $x=2^{\frac{1}{3}}\notin Z$.

So $f\left(x\right)$ is not onto.

$\therefore f\left(x\right)$ is one-one but not onto, so not bijective.

(B) For $f\left(x\right)=2x+1$ is an one-one function but not onto.

$f$ is one-one as for $f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$.

But $f$ is not onto.

As for example $2\in$ the co-domain of $f$.

Now for $2=2x+1$

or, $x=\frac{1}{2}\notin Z$.

So there exists one such point in the co-domain of the function which have no pre-image in the domain of the function.

So $f$ is not onto.

So $f$ in this case is not bijective.