Show that the function $$f:R \cdot \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$ is one-one and onto, where ‘$$R \cdot $$’ is the set of all non-zero real numbers. Is the result true, if the domain ‘$$R \cdot $$’ is replaced by $$N$$ with a co-domain being the same as ‘$$R \cdot $$’?

Question

Show that the function $$f:R \cdot  \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$ is one-one and onto, where ‘$$R \cdot $$’ is the set of all non-zero real numbers. Is the result true, if the domain ‘$$R \cdot $$’ is replaced by $$N$$ with a co-domain being the same as ‘$$R \cdot $$’?

Vedantu Content Team · Accepted Answer

Hint: To find whether the given function is one-one or not, we will take two elements $${x_1}$$ and $${x_2}$$ in the set of the domain of the given function and the we will substitute $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$. If $${x_1} = {x_2}$$, then $$f(x)$$ is a one-one function. To check that the given function is onto or not, we will check if the range of $$f(x)$$ is equal to the co-domain or not.Complete step-by-step solution:It is given that the function is defined for all non-zero real numbers and over all non-zero real numbers. Therefore, both domain and co-domain of the given function consist of the set of all non-zero real numbers.Assume two elements $${x_1}$$ and $${x_2}$$ in the set of the domain of the given function. Therefore,$$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$Substituting $${x_1}$$ and $${x_2}$$ in the function, we get$$ \Rightarrow \dfrac{1}{{{x_1}}} = \dfrac{1}{{{x_2}}}$$On cross multiplying, we get$$ \Rightarrow {x_2} = {x_1}$$On rewriting we get$$ \Rightarrow {x_1} = {x_2}$$Hence, we can see that, when $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$ then $${x_1} = {x_2}$$. Therefore, $$f$$ is one-one.Now we will check whether $$f$$ is onto or not.Given, $$f:R \cdot  \to R \cdot $$ where ‘$$R \cdot $$’ is the set of all non-zero real numbers defined by $$f(x) = \dfrac{1}{x}$$.Let $$y = f(x)$$, such that $$y \in R \cdot $$. Therefore, we can write$$ \Rightarrow y = \dfrac{1}{x}$$On cross multiplying we get$$ \Rightarrow x = \dfrac{1}{y}$$Since, the denominator can’t be zero. $$\therefore y \ne 0$$Therefore, we can say that if $$y \in R - \left\{ 0 \right\}$$, then $$x \in R - \left\{ 0 \right\}$$ also.Now, we will check for $$y = f(x)$$.Putting the value of $$x$$ in $$f(x)$$, we get$$ \Rightarrow f(x) = f\left( {\dfrac{1}{y}} \right)$$Substituting $$\dfrac{1}{y}$$ in the function, we get$$ \Rightarrow f(x) = \dfrac{1}{{\dfrac{1}{y}}}$$On simplification,$$ \Rightarrow f(x) = y$$Thus, for every $$y \in R \cdot $$, there exists $$x \in R \cdot $$ such that $$f(x) = y$$.Hence, $$f$$ is onto.Now, when the domain $$R \cdot $$ is replaced by  $$N$$ with co-domain being same as ‘$$R \cdot $$’So, we get $$f:N \to R \cdot $$Again, we will check if $$f:N \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$ is one-one or not.Assume two elements $${x_1}$$ and $${x_2}$$ in the set of the domain of the given function. Therefore,$$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$Substituting $${x_1}$$ and $${x_2}$$ in the function, we get$$ \Rightarrow \dfrac{1}{{{x_1}}} = \dfrac{1}{{{x_2}}}$$On cross multiplying, we get$$ \Rightarrow {x_2} = {x_1}$$On rewriting we get$$ \Rightarrow {x_1} = {x_2}$$Hence, we can see that, when $$f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$$ then $${x_1} = {x_2}$$. Therefore, $$f$$ is one-one.Now, we will check that $$f$$ is onto or not.We have, $$f:N \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$.Let $$y = f(x)$$, such that $$y \in R \cdot $$. Therefore, we can write$$ \Rightarrow y = \dfrac{1}{x}$$On cross multiplying we get$$ \Rightarrow x = \dfrac{1}{y}$$Here, $$x$$ cannot always be a natural number because $$y$$ is a real number except zero.For example, let $$y = 3$$ then $$x = \dfrac{1}{3}$$, which is not a natural number. Hence, $$f$$ is not onto. Note: We can also show that $$f:R \cdot  \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$ is onto by a graphical method.Consider the graph of $$f(x) = \dfrac{1}{x}$$ as shown below,\n    \n    \n    \n    \n  Clearly, we can see that the range of the function $$f:R \cdot  \to R \cdot $$ defined by $$f(x) = \dfrac{1}{x}$$  is $$R - \left\{ 0 \right\}$$.The range of the given function is all real numbers except zero. Hence, the range and codomain of the function are equal. Therefore, $$f$$ is not onto.