Answer to Question #237988 in Discrete Mathematics for lavanya

Question

Define a bijective function. Explain with reasons whether the following functions are bijective or not. Find also the inverse of each of the functions.

             i.    f(x) = 4x+2, A=set of real numbers

             ii.   f(x) = 3+ 1/x, A=set of non zero real numbers

             iii.  f(x) = (2x+3) mod7, A=N7

Answer 1

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.

i.   f(x) = 4x+2, A=set of real numbers

One-to-one:

$f(x_1)=f(x_2) \\\Rightarrow 4x_1+2=4x_2+2 \\\Rightarrow 4x_1=4x_2 \\\Rightarrow x_1=x_2$

Thus, it is one-to-one.

Onto:

$f(y)=x \\\Rightarrow 4y+2=x \\\Rightarrow y=\dfrac{x-2}4$

Now, for all y in R, there exists $\dfrac{x-2}4$ in R such that

$f(\dfrac{x-2}4)=x$

Thus, it is onto.

Hence, it is bijective.

$f^{-1}(y)=\dfrac{y-2}4 \forall y\in A$

          ii.  f(x) = 3+ 1/x, A=set of non zero real numbers

One-to-one:

$f(x_1)=f(x_2) \\\Rightarrow 3+\dfrac1{x_1}=3+\dfrac1{x_2} \\\Rightarrow \dfrac1{x_1}=\dfrac1{x_2} \\\Rightarrow x_1=x_2$

Thus, it is one-to-one.

Onto:

f is not onto, because $3 \in A$ (codomain) has no preimage in A (domain).

Hence, it is not onto. Thus, not bijective.

So, we cannot find $f^{-1}$

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

Then f is one-one and onto.

$\Rightarrow$ f is bijection.

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