Define a bijective function. Explain with reasons whether the following functions are bijective or not. Find also the inverse f(x) = (2x+3) mod7, A=N7

Solution:

A bijective function has no unpaired elements and satisfies both injective (one-to-one) and surjective (onto) mapping of a set A to a set B. Thus, bijective functions satisfy injective as well as surjective function properties and have both conditions to be true. In mathematical terms, let f: A → B is a function; then, f will be bijective if every element ‘b’ in the co-domain B, has exactly one element ‘a’ in the domain A, such that f (a) =b.

Given

$f: \mathrm{A} \rightarrow A \\ defined\ by\ f(x)=(2 x+3)(\bmod 7) \\ where \quad \mathrm{A}=\mathrm{N}_{7}=\{0,1,2,3,4,5,6\}$

We have $\quad f(0)=3, f(1)=5, f(2)=0, f(3)=2, f(4)=4, f(5)=6, f(6)=1$ ...(i)

Then f is one-one and onto.

$\Rightarrow$ f is bijection.

Then, from (i) $0=f^{-1}(3),1=f^{-1}(5), 2=f^{-1}(0),3=f^{-1}(2),4=f^{-1}(4),5=f^{-1}(6),6=f^{-1}(1),$

$\therefore f^{-1}(x)=\{0,1,2,3,4,5,6\}$