Question

Let a function $f:N\to N;f(x)=2x$ for all $x\in N$. Show that f is one-one and into function.

Answer 1

Hint: To prove that the given function is one-one, assume two elements ${{x}_{1}}\text{ and }{{x}_{2}}$ in the set of the domain of the given function and show that, if $f({{x}_{1}})=f({{x}_{2}})$ then, ${{x}_{1}}={{x}_{2}}$. To prove that the given function is into, show that the set of $f(x)$, that is co-domain, contains such elements which do not have a pre-image in the set of ‘x’ or domain.

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 prove that 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,
\[f({{x}_{1}})=f({{x}_{2}})\]
Substituting, ${{x}_{1}}\text{ and }{{x}_{2}}$ in the function, we get,
\[\begin{align}
  & 2{{x}_{1}}=2{{x}_{2}} \\
 & \Rightarrow 2{{x}_{1}}-2{{x}_{2}}=0 \\
 & \Rightarrow 2({{x}_{1}}-{{x}_{2}})=0 \\
 & \Rightarrow {{x}_{1}}-{{x}_{2}}=0 \\
 & \Rightarrow {{x}_{1}}={{x}_{2}} \\
\end{align}\]
Hence, it is proved that if $f({{x}_{1}})=f({{x}_{2}})$ then, ${{x}_{1}}={{x}_{2}}$. Therefore, $f(x)$ is a one-one function.
Now, let us prove that the given function is into.
Clearly, we can see that the co-domain of the function contains the set of natural numbers. But every natural number in the set of co-domain does not have a pre-image in the domain. For example: ‘1’ is a natural number and hence an element in the set of co-domain but for no value of ‘x’, the value of $f(x)$ is 1. Hence, the element 1 in the co-domain does not have a pre-image in domain.
Therefore, it is proved that $f(x)$ is into function.

Note: One may note that if we will get any relation between ${{x}_{1}}\text{ and }{{x}_{2}}$ other than ${{x}_{1}}={{x}_{2}}$, then the function will be not one-one. It will be many-one. Now, if every element in the co-domain has a preimage in domain, then the function will be onto and not into. So, remember all the definitions of the types of functions, so that you may not get confused.