Review question

# Can we sketch the graph of $g \colon x \to x - [x]$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5764

## Solution

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 $g$.

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 $x=0$.

1. $fg(x) = g\,f(x)$
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, $\mathbb{R}$.