Question

Show that the function $$f:R\cdot \to R\cdot$$ defined by $$f(x)={\displaystyle \frac{1}{x}}$$ 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 ‘$$R\cdot$$’?

Answer 1

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
$$\Rightarrow {\displaystyle \frac{1}{x_1}}={\displaystyle \frac{1}{x_2}}$$
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 $$f(x)={\displaystyle \frac{1}{x}}$$.
Let $$y=f(x)$$, such that $$y\in R\cdot$$. Therefore, we can write
$$\Rightarrow y={\displaystyle \frac{1}{x}}$$
On cross multiplying we get
$$\Rightarrow x={\displaystyle \frac{1}{y}}$$
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
$$\Rightarrow f(x)=f\left({\displaystyle \frac{1}{y}}\right)$$
Substituting $${\displaystyle \frac{1}{y}}$$ in the function, we get
$$\Rightarrow f(x)={\displaystyle \frac{1}{{\displaystyle \frac{1}{y}}}}$$
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 $$f(x)={\displaystyle \frac{1}{x}}$$ 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
$$\Rightarrow {\displaystyle \frac{1}{x_1}}={\displaystyle \frac{1}{x_2}}$$
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 $$f(x)={\displaystyle \frac{1}{x}}$$.
Let $$y=f(x)$$, such that $$y\in R\cdot$$. Therefore, we can write
$$\Rightarrow y={\displaystyle \frac{1}{x}}$$
On cross multiplying we get
$$\Rightarrow x={\displaystyle \frac{1}{y}}$$
Here, $$x$$ cannot always be a natural number because $$y$$ is a real number except zero.
For example, let $$y=3$$ then $$x={\displaystyle \frac{1}{3}}$$, 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 $$f(x)={\displaystyle \frac{1}{x}}$$ is onto by a graphical method.
Consider the graph of $$f(x)={\displaystyle \frac{1}{x}}$$ as shown below,

Clearly, we can see that the range of the function $$f:R\cdot \to R\cdot$$ defined by $$f(x)={\displaystyle \frac{1}{x}}$$ 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.