Question

Show that the function $f:N\to N$, defined by f(x)=\left\\{ \begin{aligned} & x+1,\text{ if }x\text{ is odd} \\\ & x-1,\text{ if }x\text{ is even} \\\ \end{aligned} \right\\} is one-one and onto.

Answer 1

Hint: To check whether the given function is one-one or not, assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function and substitute $f({{x}_{1}})=f({{x}_{2}})$. If ${{x}_{1}}={{x}_{2}}$, then $f(x)$ is one-one and if there is any more relation between ${{x}_{1}}\text{ and }{{x}_{2}}$ other than ${{x}_{1}}={{x}_{2}}$ then $f(x)$ is not one-one. To check whether the given function is onto or not, check if the range of $f(x)$ is equal to co-domain or not. Here, co-domain is the set of all values of $f(x)$ in which the range of the function is constrained.

Complete step-by-step solution -
It is given that function is defined for all natural numbers and over all natural numbers. Therefore, both domain and co-domain of the given function consists of the set of all natural numbers.
First let us show the function is one-one.
Assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function. Therefore,
$f({{x}_{1}})=f({{x}_{2}})$
(i) Considering the case that ‘x’ is odd.
Substituting, ${{x}_{1}}\text{ and }{{x}_{2}}$ in the function, we get,
$\begin{aligned} & {{x}_{1}}+1={{x}_{2}}+1 \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$
(ii) Considering the case that ‘x’ is even.
Substituting, ${{x}_{1}}\text{ and }{{x}_{2}}$ in the function, we get,
$\begin{aligned} & {{x}_{1}}-1={{x}_{2}}-1 \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$
Clearly, we can see that, in both the cases when $f({{x}_{1}})=f({{x}_{2}})$ then, ${{x}_{1}}={{x}_{2}}$. Therefore, $f(x)$ is one-one.
Now, let us show that the given function is onto.
We know that the given function is, f(x)=\left\\{ \begin{aligned} & x+1,\text{ if }x\text{ is odd} \\\ & x-1,\text{ if }x\text{ is even} \\\ \end{aligned} \right\\}. We can see that if the value of ‘x’ is chosen as an odd natural number then the value of $f(x)$ will be an even natural number. Also, if the value of ‘x’ is chosen as an even natural number then the value of $f(x)$ will be an odd natural number.
Therefore, the range of $f(x)$ is all the natural numbers. Hence, range and codomain of the function are equal.
Therefore, it is proved that $f(x)$ is onto.

Note: One may note that we have to show that the function is one-one for two different cases just as we did above. If in any case the rule fails then the function will not be one-one. Also, we can show that the function is onto by graphical method but that will be somewhat lengthier. Therefore, we have applied a theoretical approach.