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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....

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

Explanation

Solution

Hint: To prove that the given function is one-one, assume two elements x1 and x2{{x}_{1}}\text{ and }{{x}_{2}} in the set of the domain of the given function and show that, if f(x1)=f(x2)f({{x}_{1}})=f({{x}_{2}}) then, x1=x2{{x}_{1}}={{x}_{2}}. To prove that the given function is into, show that the set of f(x)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 x1 and x2{{x}_{1}}\text{ and }{{x}_{2}} in the set of the domain of the given function. Therefore,
f(x1)=f(x2)f({{x}_{1}})=f({{x}_{2}})
Substituting, x1 and x2{{x}_{1}}\text{ and }{{x}_{2}} in the function, we get,

& 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{aligned}$$ 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.