Question

If the function $f:\mathbb{N}\to \mathbb{N}$ is defined as $f=x-{{\left( -1 \right)}^{x}}$then the function $f$ is
(a) One – one and into
(b) Many – one and into
(c) One – one and onto
(d) Many – one and onto

Answer 1

For solving this problem by dividing the function into two parts one for even numbers and other for odd numbers.
We solve this problem by using the definition of one – one, into, many – one and onto function. The checking theorems of each type of functions are
(1) For one – one
If we can prove that ${{x}_{1}}={{x}_{2}}$ by assuming $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ such that ${{x}_{1}},{{x}_{2}}$ belongs to domain of function $f$ then we can say that the function is one – one.
(2) For many – one
If a function is not one – one then the function will be many = one.
(3) For onto
If the range of a function is same as the domain then the function is onto
(4) For into
If the function is not onto then the function is into.

Complete step by step answer:
We are given that the function as$f:\mathbb{N}\to \mathbb{N}$
$f=x-{{\left( -1 \right)}^{x}}$
Now, let us divide the given function into two parts one for even numbers and other for odd numbers then we get

& x+1,x\text{ is odd} \\\ & x-1,x\text{ is even} \\\ \end{aligned} \right.$$ Let us check whether this function is one – one or not. (1) We know that if we can prove that $${{x}_{1}}={{x}_{2}}$$ by assuming $$f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$$ such that $${{x}_{1}},{{x}_{2}}$$ belongs to domain of function $$f$$ then we can say that the function is one – one. Now, let us check for the odd numbers. Let us assume that $${{x}_{1}},{{x}_{2}}$$ are two odd numbers and $$\Rightarrow f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$$ Then by using the given function to above equation we get $$\begin{aligned} & \Rightarrow {{x}_{1}}+1={{x}_{2}}+1 \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$$ Here, we can say that the given function is one – one for all odd numbers. Similarly, let us check for the even numbers. Let us assume that $${{x}_{1}},{{x}_{2}}$$ are two even numbers and $$\Rightarrow f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$$ Then by using the given function to above equation we get $$\begin{aligned} & \Rightarrow {{x}_{1}}-1={{x}_{2}}-1 \\\ & \Rightarrow {{x}_{1}}={{x}_{2}} \\\ \end{aligned}$$ Here, we can say that the given function is one – one for all even numbers. Therefore, we can conclude that the given function is one – one. (2) We know that if a function is not one – one then the function will be many - one. Here, we can see that the given function is one – one so that it is not many – one. (3) Let us check for onto We know that if the range of a function is same as the domain then the function is onto We are given that the mapping of functions is $$\mathbb{N}\to \mathbb{N}$$ which means that the domain and range of given function both are same and equal to natural numbers $$\mathbb{N}$$ So, we can conclude that the given function is onto. (4) We know that if the function is not onto then the function is into. Here, we can see that the function is onto which means that this function is not into. Therefore the given function $$f:\mathbb{N}\to \mathbb{N}$$ is defined as $$f=x-{{\left( -1 \right)}^{x}}$$ is one – one and onto. **So, the correct answer is “Option c”.** **Note:** Students may do this problem in a lengthy manner. The checking of one – one functions has another method that is by definition. If for every element in the domain of function there exist only one value in the range of function, then the function is one – one. Based on this definition we should check for every element in the domain that is natural numbers $$\mathbb{N}$$ whether there exists one or more values in range that is natural numbers $$\mathbb{N}$$ This process is limitless because we have infinite numbers in the domain which is impossible to check for every element. So, we need to use the shortcut method that we used in the solution.