Question

Check the injectivity and surjectivity of the following functions:

(i) $f:N\to N$ given by $f\left(x\right)=x^2$

(ii) $f:Z\to Z$ given by $f\left(x\right)=x^2$

(iii) $f:R\to R$ given by $f\left(x\right)=x^2$

(iv) $f:N\to N$ given by $f\left(x\right)=x^3$

(v) $f:Z\to Z$ given by $f\left(x\right)=x^3$

Answer 1

(i)

$f\left(x\right)=x^2$
It is seen that for $x,y\in N$,

$f\left(x\right)=f\left(y\right)\Rightarrow x^2=y^2\Rightarrow x=y$

$\therefore f$ is injective
Now, $2\in N$, But, there does not exist any $x$

in $N$ such that $f\left(x\right)=x^2=2$.

$\therefore f$ is not surjective.

Hence, function $f$ is injective but not surjective.

(ii)

$f\left(x\right)=x^2$
It is seen that $f(-1)=f\left(1\right)=1$, but $-1\ne 1$
$\therefore f$ is not injective.
Now, $-2\in Z$. But, there does not exist any

element $x\in Z$ such that $f\left(x\right)=x^2=-2$

$\therefore f$ is not surjective.
Hence, function $f$ is neither injective nor surjective.

(iii)

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

It is seen that $f(-1)=f\left(1\right)=1$, but $-1\ne 1$.
$\therefore f$ is not injective.
Now, $-2\in R$. But, there does not exist any element $x\in R$ such that $f\left(x\right)=x^2=-2$
$\therefore f$ is not surjective.
Hence, function $f$ is neither injective nor surjective.

(iv)

$f\left(x\right)=x^3$
It is seen that for $x,y\in N,f\left(x\right)=f\left(y\right)\Rightarrow x^3=y^3\Rightarrow x=y$
$\therefore f$ is injective.
Now, $2\in N$. But, there does not exist any element
$x$ in domain $N$ such that $f\left(x\right)=x^3=2$
$\therefore f$ is not surjective.
Hence, function $f$ is injective but not surjective.

(v)

$f\left(x\right)=x^3$

It is seen that for $x,y\in Z,f\left(x\right)=f\left(y\right)\Rightarrow x^3=y^3\Rightarrow x=y$

$\therefore f$ is injective.

Now, $2\in Z$. But, there does not exist any element

$x$ in domain $Z$ such that $f\left(x\right)=x^3=2$

$\therefore f$ is not surjective.

Hence, function $f$ is injective but not surjective.