Question

Prove that the greatest integer function f : R → R, given by f(x) = [x], is neither one-one nor onto

Answer 1

Given: \( f : \mathbb{R} \to \mathbb{R} \) defined by \( f(x) = [x] \), where \( [x] \) is the greatest integer less than or equal to \( x \).
To prove: \( f(x) = [x] \) is neither one-one nor onto. 
 
1. Not One-One (Injective):
A function is one-one if different inputs give different outputs.
Take two real numbers: \( x = 1.2 \) and \( y = 1.9 \)
\[ f(1.2) = [1.2] = 1, f(1.9) = [1.9] = 1 \] But \( 1.2 \neq 1.9 \) and \( f(1.2) = f(1.9) \)
Therefore, \( f \) is not one-one. 
2. Not Onto (Surjective):
A function is onto if every real number has a pre-image in the domain.
The function \( f(x) = [x] \) always gives an integer as output.
For example, there is no real number \( x \in \mathbb{R} \) such that \( f(x) = \pi \) or \( f(x) = 2.5 \), since the range of the function is only integers.
Therefore, all real numbers are not covered in the co-domain \( \mathbb{R} \).
Hence, \( f \) is not onto. 
Conclusion: The greatest integer function \( f(x) = [x] \) is neither one-one nor onto.