Question

Give an example of a function
(i) which is one-one but not onto
(ii) which is not one-one but onto
(iii) which is neither one-one nor onto

Answer 1

(i) which is one-one but not onto.

f: Z → Z given by f(x)=3x+2

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
f(x)=f(y)
$\Rightarrow$3x + 2 =3y + 2
$\Rightarrow$3x = 3y
$\Rightarrow$x = y
$\Rightarrow$f(x) = f(y) $\Rightarrow$x = y
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$\Rightarrow$3x + 2 = y
$\Rightarrow$3x = y - 2

$$\begin{gathered} \Rightarrow x=\frac{y-2}{3}.\text{It may not be in the domain (}\mathit{\text{Z}}\text{)} \\ \text{because if we take y = 3,} \\ x=\frac{y-2}{3}=\frac{3-2}{3}=\frac{1}{3}\text{∉ domain}\mathit{\text{Z}}\text{.} \end{gathered}$$
So, for every element in the co domain there need not be any element in the domain such that f(x) = y.
Thus, f is not onto.

(ii) which is not one-one but onto.
f: Z → N $\cup$ {0} given by f(x) = |x|

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
$\Rightarrow$|x| = |y|
$\Rightarrow$x= $\pm$ y
So, different elements of domain f may give the same image.
So, f is not one-one.

Surjectivity:
Let y be any element in the co domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$\Rightarrow$|x| = y
$\Rightarrow$x = $\pm$ y, which is an element in Z (domain).
So, for every element in the co-domain, there exists a pre-image in the domain.
Thus, f is onto.

(iii) which is neither one-one nor onto.

f: Z → Z given by f(x) = 2x² + 1

Injectivity:
Let x and y be any two elements in the domain (Z), such that f(x) = f(y).
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow 2x^2+1=2y^2+1 \\ \Rightarrow 2x^2=2y^2 \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=\pm y \end{gathered}$$
So, different elements of domain f may give the same image.
Thus, f is not one-one.

Surjectivity:
Let y be any element in the co-domain (Z), such that f(x) = y for some element x in Z (domain).
f(x) = y
$$\begin{gathered} \Rightarrow 2x^2+1=y \\ \Rightarrow 2x^2=y-1 \\ \Rightarrow x^2=\frac{y-1}{2} \\ \Rightarrow x=\pm \sqrt{\frac{y-1}{2}},\text{∉}\mathit{\text{Z}}\text{always.} \\ \text{For example, if we take,}\mathit{\text{y}}\text{= 4,} \\ x=\pm \sqrt{\frac{y-1}{2}}=\pm \sqrt{\frac{4-1}{2}}=\pm \sqrt{\frac{3}{2}},\text{∉ Z} \\ \text{So,}\mathit{\text{x}}\text{may not be in}\mathit{\text{Z}}\text{(domain).} \end{gathered}$$

Thus, f is not onto.