Question

Consider the given function f:N\to z:f\left( n \right)\left\\{ \begin{matrix} \dfrac{n}{2},n-even \\\ -\left( \dfrac{n-1}{2} \right),n-odd \\\ \end{matrix} \right. , what type of function is this?
a.Not one-one but onto
b.One-one but onto
c.One-one and onto
d.Not one-one and onto.

Answer 1

Hint: Consider the case of odd by putting n = 2m+1. For the case of even take n = 2m. Substitute these values for n-even and n-odd functions by considering function $f\left( {{m}_{1}} \right)$and $f\left( {{m}_{2}} \right)$.

Complete Step-by-step answer:
Given a function f\left( n \right)=\left\\{ \begin{matrix} \dfrac{n}{2},n-even \\\ -\left( \dfrac{n-1}{2} \right),n-odd \\\ \end{matrix} \right.
Now, let us consider the case where n is odd.
Let, $n=2{{m}_{1}}+1$, is odd.
$\therefore f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)$can be taken as one-one so, ${{m}_{1}}\ne {{m}_{2}}$.
Now checking the condition of $f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)$, substituting the value of n in the function for n-odd.
$\dfrac{-\left( 2{{m}_{1}}+1 \right)-1}{2}=\dfrac{-\left( 2{{m}_{2}}+1 \right)-1}{2}$
Cancelling out the like terms, we get
${{m}_{1}}={{m}_{2}}$
Which means that $f\left( {{m}_{1}} \right)$will only be equal to $f\left( {{m}_{2}} \right)$when ${{m}_{1}}={{m}_{2}}$.
$\therefore f\left( {{m}_{1}} \right)=f\left( {{m}_{2}} \right)$
As it is odd, the function is one-one.
$\because$One-one function is a function that comprises individually that never discrete elements of its co-domain (z).
Now let us consider the case where n is even.
$\therefore$Let $n=2{{m}_{1}}$
$f\left( 2{{m}_{1}} \right)=f\left( 2{{m}_{2}} \right)$
Now, checking this condition by putting the value of n in the function for n-even

& \Rightarrow \dfrac{-2{{m}_{1}}}{2}=\dfrac{-2{{m}_{2}}}{2} \\\ & \Rightarrow {{m}_{1}}={{m}_{2}} \\\ \end{aligned}$$ $$\therefore $$The function is one-one. Now let us consider giving values for $${{m}_{1}}$$and $${{m}_{2}}$$. $$\therefore $$Let $${{m}_{1}}$$be even which becomes $$2{{m}_{1}}$$. $$\therefore $$$$f\left( n \right)=\dfrac{n}{2}=\dfrac{2{{m}_{1}}}{2}={{m}_{1}}$$ If taking odd $$\left( 2{{m}_{1}}+1 \right)$$. $$\begin{aligned} & \Rightarrow f\left( n \right)=\dfrac{-\left( n-1 \right)}{2}=\dfrac{-\left( 2{{m}_{1}}+1-1 \right)}{2} \\\ & f\left( n \right)=-{{m}_{1}} \\\ \end{aligned}$$ Now, if we are considering natural numbers, N = 1, 2, 3, …… If we are putting them in the function, we get, For N = 1, we get (1, -1). N = 2, we get (2, -2) etc. $$\therefore $$The range of the function becomes$$\Rightarrow \left\\{ +1,+2.....-1,-2.... \right\\}\in $$integers. In the question, z is given as a co-domain. $$\therefore $$Range becomes equal to co-domain, so it becomes onto. $$\therefore $$The function is one-one onto. Hence, the correct option is (c). Note: Check for one-one. $$f\left( 1 \right)=\dfrac{-\left( n-1 \right)}{2}=\dfrac{-\left( 1-1 \right)}{2}=0$$(since 1 is odd) $$f\left( 2 \right)=\dfrac{n}{2}=\dfrac{2}{2}=1$$ (since 2 is even) $$f\left( 1 \right)=0$$and $$f\left( 2 \right)=1$$but $$1\ne 2$$. $$\therefore $$f is one-one, as both $$f\left( 1 \right)$$and $$f\left( 2 \right)$$don't have the same image. Check for onto $$f\left( n \right)=\left\\{ \begin{matrix} \dfrac{n}{2},n-even \\\ -\left( \dfrac{n-1}{2} \right),n-odd \\\ \end{matrix} \right.$$ Let $$f\left( x \right)=y$$, such that $$y\in N$$. When n is odd, $$\begin{aligned} & y=\dfrac{-\left( n-1 \right)}{2}\Rightarrow 2y=-n+1 \\\ & \therefore n=1-2y \\\ \end{aligned}$$ Hence, for y is integer number, n = 1 – 2y is an integer. When n is even, $$y=\dfrac{n}{2}\Rightarrow 2y=n$$ Hence, for y is a natural number, n = 2y is also a natural number. Thus, for every $$y\in N$$, there exists $$x\in N$$such that, $$f\left( n \right)=y$$ $$\therefore $$f is onto. $$\therefore $$Function $$f\left( n \right)$$is both one-one and onto.