# Check the injectivity and surjectivity of the following functions:

Question:

(i) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{2}$

(ii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{2}$

(iii) $f: \mathbf{R} \rightarrow \mathbf{R}$ given by $f(x)=x^{2}$

(iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by $f(x)=x^{3}$

(v) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ given by $f(x)=x^{3}$

Solution:

(i) $f: \mathbf{N} \rightarrow \mathbf{N}$ is given by,

$f(x)=x^{2}$

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^{2}=y^{2} \Rightarrow x=y$

$\therefore f$ is injective.

Now, $2 \in \mathbf{N}$. But, there does not exist any $x$ in $\mathbf{N}$ such that $f(x)=x^{2}=2$.

$\therefore f$ is not surjective.

Hence, function f is injective but not surjective.

(ii) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ is given by,

$f(x)=x^{2}$

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.

$\therefore f$ is not injective.

Now, $-2 \in \mathbf{Z}$. But, there does not exist any element $x \in \mathbf{Z}$ such that $f(x)=x^{2}=-2$.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iii) $f: \mathbf{R} \rightarrow \mathbf{R}$ is given by,

$f(x)=x^{2}$

It is seen that $f(-1)=f(1)=1$, but $-1 \neq 1$.

∴ is not injective.

Now, $-2 \in \mathbf{R}$. But, there does not exist any element $x \in \mathbf{R}$ such that $f(x)=x^{2}=-2$.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

(iv) $f: \mathbf{N} \rightarrow \mathbf{N}$ given by,

$f(x)=x^{3}$

It is seen that for $x, y \in \mathbf{N}, f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$.

f is injective.

Now, $2 \in \mathbf{N} .$ But, there does not exist any element $x$ in domain $\mathbf{N}$ such that $f(x)=x^{3}=2$.

Hence, function f is injective but not surjective.

(v) $f: \mathbf{Z} \rightarrow \mathbf{Z}$ is given by,

$f(x)=x^{3}$

It is seen that for $x, y \in \mathbf{Z}, f(x)=f(y) \Rightarrow x^{3}=y^{3} \Rightarrow x=y$.

∴ f is injective.

Now, $2 \in \mathbf{Z}$. But, there does not exist any element $x$ in domain $\mathbf{Z}$ such that $f(x)=x^{3}=2$.

∴ f is not surjective.

Hence, function f is injective but not surjective.