(i) Let f: Z → Z given by f(x) = 3x + 2

We have to check one-one condition on f(x) = 3x + 2

Injectivity:

Let x and y be any two elements in domain (Z), such that f(x) = f(y).

 f(x) = f(y)

⇒ 3x + 2 = 3y + 2

⇒ 3x = 3y

⇒ x = y

⇒ f(x) = f(y) 

⇒ x = y

Therefore, 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).

Let f(x) = y

⇒ 3x + 2 = y

⇒ 3x = y – 2

⇒ x = (y – 2)/3. It may not be in the domain (Z)

Because if we take y = 3,

x = (y – 2)/3 = (3 - 2)/3 = 1/3 ∉ domain Z.

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) Example for function which is not one-one but onto

Let f: Z → N ∪ {0} given by f(x) = |x|

Injectivity:

Let x and y be any two elements in domain (Z),

Such that f(x) = f(y).

⇒ |x| = |y|

⇒ x = ± y

So, different elements of domain f may give the same image.

Therefore, f is not one-one.

Surjectivity:

Let y be any element in co domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒ |x| = y

⇒ x = ± 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) Example for function which is neither one-one nor onto.

Let 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).

f(x) = f(y)

⇒ 2x² +1 = 2y² +1

⇒ 2x² = 2y²

⇒ x² = y²

⇒ x = ± y

So, different elements of domain f may give the same image.

Thus, f is not one-one.

Surjectivity:

Let y be any element in co-domain (Z), such that f(x) = y for some element x in Z (domain).

f(x) = y

⇒ 2x² + 1 = y

⇒ 2x² = y − 1

⇒ x² = (y - 1)/2

⇒ x = √((y - 1)/2) ∉ Z always.

For example, if we take, y = 4,

x = ± √((y - 1)/2)

= ± √((4 - 1)/2)

= ± √(3/2) ∉ Z

Therefore, x may not be in Z (domain).

Hence, f is not onto.