# Problem 1 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 1 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 1.2 Problem 1 Solution
1. Show that the function f : R* → R* defined by {f(x) = \dfrac{1}{x}} is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R*?
To show that f is one-one
Consider two elements x_1, x_2 such that {f(x_1) = f(x_2)}
{⇒ \dfrac{1}{x_1} = \dfrac{1}{x_2}}
{x_1 = x_2}
f is one-one.
To show that f is onto
As R* is defined as {f(x) = \dfrac{1}{x}},
And also R* is defined as the set of non-zero real numbers.
For any element y ∈ R*, there exists a corresponding element \dfrac{1}{y}, in the domain R*
f is onto.
To check whether f is one-one when f is defined as f : N → R*
Consider two elements x_1, x_2N such that
f(x_1) = f(x_2)
{\Rightarrow \dfrac{1}{x_1} = \dfrac{1}{x_2}}
{x_1 = x_2}
f is one-one.
To check whether f is onto when f is defined as f : N → R*
To be onto, every element yR*, should have a corresponding mapping element \dfrac{1}{y} in the domain N
As the numbers in the form of \dfrac{1}{y} can not all be natural numbers, there exists some elements in the co-domain R* which are not mapped to any element in the domain N. In otherwords, some elements like 5, 11 in the co-domain R* will not be the images of any elements in the domain N, as {\dfrac{1}{5}, \dfrac{1}{11} ∉ \bf{N}}. In otherwords, all the elements in the co-domain R* are not mapped.
f is not onto.