Question:

Show that the Modulus Function f: R\(\to\)R given by f(x) = IxI, is neither one-one nor onto, where IxI is x, if x is positive or 0 and IxI is −x, if x is negative.

Updated On: Aug 28, 2023
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Solution and Explanation

f: R\(\to\)R is given by,

\[f(x) = |x|=  \begin{cases}     x,       & \quad \text{if } x {\geq0}\\     -x,  & \quad \text{if } n {<0}   \end{cases}\]

It is seen that f(-1)=I-1I=1, f(1)=I1I=1. 
∴f(−1) = f(1), but −1 ≠ 1. 
∴ f is not one-one. 
Now, consider −1 ∈ R. 
It is known that f(x) = is always non-negative. Thus, there does not exist any element x in domain R such that f(x) = = −1. 
∴ f is not onto. 

Hence, the modulus function is neither one-one nor onto.

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Concepts Used:

Types of Functions

Types of Functions

One to One Function

A function is said to be one to one function when f: A → B is One to One if for each element of A there is a distinct element of B. 

Many to One Function

A function which maps two or more elements of A to the same element of set B is said to be many to one function. Two or more elements of A have the same image in B.

Onto Function

If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function. 

One – One and Onto Function

A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function.

Read More: Types of Functions