Solution: injective function Definition: A function f: A → B is said to be a one - one function or injective mapping if different elements of A have different f images in B. A function f is injective if and only if whenever f(x) = f(y), x = y. Example: f(x) = x + 9 from the set of real number R to R is an injective function. When x = 3,then :f(x) = 12,when f(y) = 8,the value of y can only be 3,so x = y. (ii) surjective function Definition: If the function f:A→B is such that each element in B (co - domain) is the ‘f’ image of at least one element in A , then we say that f is a function of A ‘onto’ B .Thus f: A→B is surjective if, for all b ∈ B, there are some a ∈ A such that f(a) = b. Example: The function f(x) = 2x from the set of natural numbers N to the set of non negative even numbers is a surjective function. (iii) bijective function Definition: A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y. Alternatively, f is bijective if it is a one - to - one correspondence between those sets, in other words, both injective and surjective. Example: If f(x) = x2,from the set of positive real numbers to positive real numbers is both injective and surjective. Thus, it is a bijective function. f:N\rightarrow N \\f(x) = x^2f:N→N f(x)=x 2 \begin{aligned} &\begin{array}{l} x_{1}, x_{2} \in N \\ f\left(x_{1}\right)=f\left(x_{2}\right) \Rightarrow x_{1}^{2}=x_{2}^{2} \end{array} \\ &\Rightarrow x_{1}^{2}-x_{2}^{2}=0 \\ &\Rightarrow\left(x_{1}+x_{2}\right)\left(x_{1}-x_{2}\right)=0 \\ &\Rightarrow x_{1}=x_{2}\left\{\begin{array}{c} x_{1}+x_{2} \neq 0 \\ \text { as } x_{1}, x_{2} \in N \end{array}\right\} \end{aligned} ​ x 1 ​ ,x 2 ​ ∈N f(x 1 ​ )=f(x 2 ​ )⇒x 1 2 ​ =x 2 2 ​ ​ ⇒x 1 2 ​ −x 2 2 ​ =0 ⇒(x 1 ​ +x 2 ​ )(x 1 ​ −x 2 ​ )=0 ⇒x 1 ​ =x 2 ​ { x 1 ​ +x 2 ​  =0 as x 1 ​ ,x 2 ​ ∈N ​ } ​ hence f is injective, for some elements like, 2,3 etc has no preimage in N such that f(x)=2 hence not surjective.