The greatest integer function

The great­est inte­ger func­tion rounds every real num­ber down to an inte­ger. It sounds sim­ple, but it behaves in a sur­pris­ing way for neg­a­tive num­bers, and its graph is a stair­case rather than a smooth curve. It appears in many exam­i­na­tion ques­tions on domains, ranges and inequal­i­ties. This les­son defines it, works through many val­ues, draws the graph, lists its prop­er­ties and solves inequal­i­ties with it.

The step of a real num­ber#

Def­i­n­i­tion. If $x \in \mathbb{R}$, $n \in \mathbb{Z}$ and

$$n \leq x < n + 1$$

then $[x] = n$. We read $[x]$ as "step $x$" or "great­est inte­ger of $x$". It is the great­est inte­ger that is less than or equal to $x$.

So $[x]$ is con­stant on each inter­val from one inte­ger up to (but not includ­ing) the next:

Work­ing out val­ues#

For non-neg­a­tive num­bers, $[x]$ sim­ply drops the dec­i­mal part:

$$[0] = 0, \quad [0.1] = 0, \quad [0.5] = 0, \quad [0.9999] = 0$$

$$[1] = 1, \quad [1.1] = 1, \quad [1.9999] = 1, \quad [2] = 2, \quad [2.5] = 2, \quad [2.999] = 2$$

For neg­a­tive num­bers, think of the num­ber line. $[x]$ is the inte­ger you reach by mov­ing left from $x$ (or stay­ing put if $x$ is already an inte­ger). For exam­ple, $-0.1$ lies between $-1$ and $0$; mov­ing left from $-0.1$ the first inte­ger you meet is $-1$. So $[-0.1] = -1$, not $0$.

$$[-0.1] = -1, \quad [-0.2] = -1, \quad [-0.9] = -1, \quad [-1] = -1$$

$$[-1.2] = -2, \quad [-1.7] = -2, \quad [-1.999] = -2, \quad [-2] = -2$$

Com­pare pos­i­tive and neg­a­tive val­ues side by side:

For inte­gers there is noth­ing to round: $[0] = 0$, $[1] = 1$, $[2] = 2$, $[-1] = -1$, $[-2] = -2$.

The great­est inte­ger func­tion#

The func­tion $y = f(x) = [x]$ is called the step func­tion, the great­est inte­ger func­tion or the floor func­tion.

$$\text{Domain} = \mathbb{R}, \qquad \text{Range} = \mathbb{Z}$$

Every real num­ber can be rounded down, and the result is always an inte­ger; every inte­ger $n$ is the value of $[n]$.

The graph of $y = [x]$#

The graph is a stair­case of hor­i­zon­tal seg­ments, one unit long:

• $y = -2$ on $[-2, -1)$,
• $y = -1$ on $[-1, 0)$,
• $y = 0$ on $[0, 1)$,
• $y = 1$ on $[1, 2)$,
• $y = 2$ on $[2, 3)$, and so on in both direc­tions.

Each seg­ment has a filled dot at its left end (the value is reached there) and an open cir­cle at its right end (the value jumps up at the next inte­ger). At each inte­ger the graph jumps up by $1$.

Prop­er­ties#

1. $[x] \leq x$, for all $x \in \mathbb{R}$.
2. $[x] = x$, for all $x \in \mathbb{Z}$.
3. $[x] < x$, for all $x \in \mathbb{R} - \mathbb{Z}$.
4. $[x] > x$ for no real $x$.
5. $[x + k] = [x] + k$, for all $x \in \mathbb{R}$ and $k \in \mathbb{Z}$. For exam­ple, $[x + 1] = [x] + 1$. But this works only for an inte­ger $k$: $\displaystyle \left[x + \frac{1}{2}\right] \neq [x] + \frac{1}{2}$ in gen­eral (the right side is not even an inte­ger).
6. $[-x] = -[x]$, for all $x \in \mathbb{Z}$.
7. $[-x] = -[x] - 1$, for all $x \in \mathbb{R} - \mathbb{Z}$.
8. $[x] + [-x] = 0$ for $x \in \mathbb{Z}$, and $[x] + [-x] = -1$ for $x \in \mathbb{R} - \mathbb{Z}$.
9. $[[x]] = [x]$, for all $x \in \mathbb{R}$.
10. $x - 1 < [x] \leq x$, for all $x \in \mathbb{R}$.

Exam­ple 1#

Ver­ify prop­erty 7 for $x = 1.3$.

Left side: $[-x] = [-1.3] = -2$.

Right side: $-[x] - 1 = -[1.3] - 1 = -1 - 1 = -2$.

The two sides agree.

Inequal­i­ties with $[x]$#

Since $[x]$ only takes inte­ger val­ues, an inequal­ity in $[x]$ is really a state­ment about which steps of the stair­case are allowed. Then trans­late back to $x$ using $[x] = n \iff n \leq x < n + 1$.

Exam­ple 2#

Solve (a) $[x] < 2$ (b) $[x] \leq 2$ (c) $[x] > 2$ (d) $[x] \geq 2$.

(a) $[x] < 2$ means $[x]$ is $1, 0, -1, \ldots$. The high­est allowed step is $[x] = 1$, which cov­ers $1 \leq x < 2$. So $x \in (-\infty, 2)$.

(b) $[x] \leq 2$ allows the step $[x] = 2$ as well, which cov­ers $2 \leq x < 3$. So $x \in (-\infty, 3)$.

(c) $[x] > 2$ means $[x]$ is $3, 4, \ldots$. The low­est allowed step is $[x] = 3$, start­ing at $x = 3$. So $x \in [3, \infty)$.

(d) $[x] \geq 2$ starts with the step $[x] = 2$, at $x = 2$. So $x \in [2, \infty)$.

Exam­ple 3#

Solve (a) $[x] > -3$ (b) $[x] \leq -5$.

(a) The low­est allowed step is $[x] = -2$, which starts at $x = -2$. So $x \in [-2, \infty)$.

(b) The high­est allowed step is $[x] = -5$, which cov­ers $-5 \leq x < -4$. So $x \in (-\infty, -4)$. A com­mon wrong answer is $(-\infty, -5]$: it misses num­bers such as $-4.5$, for which $[-4.5] = -5$.

Domains with $[x]$#

Exam­ple 4#

Find the domain of $\sqrt{x - [x]}$.

We need $x - [x] \geq 0$, that is, $[x] \leq x$, which is true for every real $x$ by prop­erty 1. So the domain is $\mathbb{R}$.

Exam­ple 5#

Find the domain of $\displaystyle \frac{1}{\sqrt{[x] - x}}$.

We need $[x] - x > 0$, that is, $[x] > x$, which is true for no real $x$ by prop­erty 4. So the domain is $\phi$.

Exam­ple 6#

Solve $[x]^2 - 3[x] + 2 = 0$.

Treat $[x]$ as the unknown: $([x] - 1)([x] - 2) = 0$, so $[x] = 1$ or $[x] = 2$. These give $1 \leq x < 2$ or $2 \leq x < 3$. Together, $x \in [1, 3)$.

Prac­tice#

1. Find $[3.7]$, $[-3.7]$, $[-5]$ and $[0.01]$.
2. Find $[\pi]$ and $[-\pi]$, using $\pi \approx 3.14$.
3. Solve $[x] = 4$.
4. Solve $[x] < 7$.
5. Solve $[x] \geq -6$.
6. Solve $[x] > -4$.
7. Solve $[x] \leq 0$.
8. Find $[2.4] + [-2.4]$, and say which prop­erty it shows.
9. Solve $[x + 3] = 5$.
10. Find $[[-1.5]]$.
11. Solve $[x]^2 - 5[x] + 6 = 0$.
12. Find the domain of $\displaystyle \frac{1}{\sqrt{x - [x]}}$.

Answers#