Let $f(x) = x + \left[ x \right]$. What is ${f^{ - 1}}(x)$?

Question

Let $f:\Bbb{R}\to\bigcup_{n\in \Bbb Z}[2n,2n+1)$ and $f(x) = x + \left[ x \right]$. What is ${f^{ - 1}}(x)$?

Answer 1

The inverse will exist iff $f(x)$ is bijective.

$f(x)=x+[x]$ is bijective if it is defined as $$f:\Bbb{R}\to\bigcup_{n\in \Bbb Z}[2n,2n+1)$$

To find $f^{-1}$:

$y=x+[x]=2[x]+\{x\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $Note: $x=[x]+\{x\}$

$\Rightarrow y-\{x\}=2[x]$

The above equation implies $y-\{x\}$ is an integer, therefore it can be written as $y-\{y\}$.Hence, it implies $\{x\}=\{y\}.$

$\Rightarrow y-\{y\}=2[x]$

$\Rightarrow [x]=\frac{[y]}{2}$

Adding {x} both sides we get,

$\Rightarrow [x]+\{x\}=\frac{[y]}{2}+\{x\}$

$\Rightarrow x=\frac{[y]}{2}+\{y\}=y-\frac{[y]}{2}$

Hence,

$$f^{-1}(x)=x-\frac{[x]}{2}$$

Answer 2

The inverse is $g(y) = [y]/2 + \{y\} = y - [y]/2.$