Question

Consider the function f : R → {0, 1} such that \(f(x)=\left\lbrace \begin{matrix} 1 \ \text{if} \ x \ \text{is rational} \\\ 0 \ \text{if} \ x \ \text{is irrational} \end{matrix} \right.\)

Which one of the following is correct?

1. The function is one-one into
2. The function is many-one into
3. The function is one-one onto
4. The function is many-one onto

Answer 1

Correct Answer - Option 4 : The function is many-one onto

Concept:

Let f(x) be any function.

f (x) is onto if range of f (x ) = Codomain  

The function f is said to be many-one functions if there exist two or more than two different elements in X having the same image in Y.

 

Calculations:

Given function f : R → {0, 1} such that \(f(x)=\left\lbrace \begin{matrix} 1 \ \text{if} \ x \ \text{is rational} \\\ 0 \ \text{if} \ x \ \text{is irrational} \end{matrix} \right.\)

Codomain = {0, 1}

Since, on taking a straight line parallel to the x-axis, the group of given function intersect it at many points. 

⇒ f (x) is many-one. 

Range of function is {0, 1}

As range of f (x ) = Codomain 

⇒ f (x) is onto. 

Hence, f (x) is many-one onto