Can we sketch the graph of $g \colon x \to x - [x]$?

Can we sketch the graph of $g \colon x \to x - [x]$?

Question

For any real number

$x$

,

$[x]$

denotes the greatest integer not exceeding

$x$

; e.g.

$[3.6] = 3$

,

$[2] = 2$

,

$[-1.4] = -2$

, etc. Functions

$f$

and

$g$

are defined on the domain of all real numbers as follows:

$$\begin{equation*} f \colon x \to [x]; \quad g \colon x \to x - [x]. \end{equation*}$$

Find the ranges of $f$ and $g$, and sketch the graph of $g$.

Determine the solution sets of the equations

1. $f(x) = g(x),\qquad$ (ii) $fg(x) = g\,f(x)$.

Solution

The function $[x]$ defined here is known as the floor function.

From the definition of the floor function, the output value of $f$ will always be an integer. Since the input $x$ can be any real number the range of $f$ is all the integers, $\mathbb{Z}$.

The function $g(x)=x-[x]$ which means it subtracts the whole number part, leaving only the fractional part of the input value $x$.

For integer values of $x$, $[x]=x$ which means that $g(x)=0$. So the graph of the function looks like this.

The range is $0\leq g(x)<1$.

Determine the solution sets of the equations

1. $f(x) = g(x)$

We want to find for which values of

$x$

,

$$\begin{equation*} f(x) = g(x) \iff [x] = x - [x]. \end{equation*}$$

The left-hand side is always an integer, while the only integer value that the right-hand side can take is zero, which occurs whenever $x$ is an integer.

The solution set is therefore $x=0$.

Since the range of

$g$

is

$0\leq g(x)<1$

we have that

$$\begin{equation*} fg(x) = [g(x)] = 0. \end{equation*}$$

We also have that

$$\begin{equation*} g\,f(x) = g([x]) = [x] - [[x]] = 0. \end{equation*}$$

The equation is therefore satisfied regardless of the value of $x$. The solution set is the real numbers, $\mathbb{R}$.