Intervals on the number line

At a water park, a sign stands by the highest slide: you must be at least 1.20 m tall. If you are exactly 1.20 m, you may go. If you are 1.19 m, you may not. Call your height h. The sign then says:

h ≥ 1.20

A condition like this is an inequality. All the numbers that satisfy it lie side by side on the number line. Such a piece of the number line is an interval. In this post you learn to draw intervals and to work with them.

A piece of the number line

Take all the numbers between −2 and 3. Whole numbers are not the only ones. 0.5 and √2 lie in between too. At the ends you still have a choice to make.

• A filled dot means the end belongs to the interval. The inequality uses ≤ or ≥. The interval uses a square bracket: [ or ].
• An open circle means the end does not belong to it. The inequality uses < or >. The interval uses a round bracket: ( or ).

Three examples with the same ends:

−2 < x < 3 is (−2, 3)
−2 ≤ x ≤ 3 is [−2, 3]
−2 < x ≤ 3 is (−2, 3]

Drag the ends below. The buttons make an end open or closed.

Drag the ends

−2 < x ≤ 3(−2, 3]

Left end: openRight end: closed

Inequality

−2 < x ≤ 3

Interval

(−2, 3]

Whole numbers inside

5: −1, 0, 1, 2, 3

The smaller number always comes first, just as on the number line. You read (−2, 3] as: all numbers greater than −2 and less than or equal to 3.

On to infinity

The sign at the slide has only one end. Every height from 1.20 m up is fine, however tall you are. An interval like this goes on forever on one side. It is called a half-line.

On that side there is no number. You write ∞ for infinity:

x ≥ 4 is [4, ∞)
x < 1 is (−∞, 1)

The bracket next to ∞ is always round. Infinity is not a number. You never reach it, so it never belongs to the interval.

Pick a sign below and drag the end.

A half-line

x <x ≤x >x ≥

x ≥ 4[4, ∞)

Inequality

x ≥ 4

Interval

[4, ∞)

End included

yes

Going back works the same way. (−1, 5] becomes −1 < x ≤ 5. And [0, ∞) becomes x ≥ 0.

Solving an inequality

You solve an inequality almost the same way as an equation. You may add or subtract the same number on both sides. You may multiply or divide both sides by the same positive number. A negative number works differently.

Take 3 − 2x < 7.

3 − 2x < 7
−2x < 4 (subtract 3 from both sides)
x > −2 (divide both sides by −2, the sign flips)

The solution is x > −2, so (−2, ∞).

The flip comes from a mirror. Times −1 mirrors the number line in 0. What lay on the left ends up on the right. Drag a and b below and watch the order.

Times −1 mirrors the number line

• a
• b

Before times −1

2 < 5

After times −1

−2 > −5

Show the numbers

point	number	times −1
a	2	−2
b	5	−5

2 is less than 5. But −2 is greater than −5. Dividing by −2 is the same as times −1 and then dividing by 2. That is why the sign flips.

Check your answer with a number. Take x = 0. It lies in (−2, ∞). You get 3 − 0 = 3, and 3 < 7 is true. Take x = −3. It lies outside. You get 3 + 6 = 9, and 9 < 7 is false.

Two signs at once

Sometimes x sits between two bounds in one inequality. Then you do each step to all three parts.

−1 ≤ 2x + 3 < 9
−4 ≤ 2x < 6 (subtract 3 from all parts)
−2 ≤ x < 3 (divide all parts by 2)

The solution is [−2, 3).

Two conditions: and, or

Sometimes there are two conditions. The word between them matters a lot.

• And: a number must satisfy both. You keep only the part where the two intervals overlap.
• Or: a number must satisfy at least one. You take both pieces together. You write this with the sign ∪.

Two examples:

x > −1 and x ≤ 3 gives (−1, 3]
x < −2 or x ≥ 1 gives (−∞, −2) ∪ [1, ∞)

Drag the ends of A and B below. Then pick and or or.

A and B, A or B

A and BA or B

• A
• B
• Solution

A

[−4, 2]

B

[−1, 5]

A and B

[−1, 2]

Solution

−1 ≤ x ≤ 2

If A and B lie apart, and leaves no number at all. With or you then keep two separate pieces.

Quadratic inequalities

This part is mathematics B, MULO 4.

Take x² − 7x + 12 < 0. Look at the graph of y = x² − 7x + 12. You want the x-values where the graph lies below the x-axis.

First find the roots. Factor:

x² − 7x + 12 = (x − 3)(x − 4) = 0
x = 3 or x = 4

The number in front of x² is 1. It is positive, so the graph is a parabola that opens upward. It lies below the x-axis only between the roots. The ends do not belong, because y = 0 there. The solution is (3, 4).

Check with x = 3.5: 12.25 − 24.5 + 12 = −0.25. That is less than 0, so it fits.

Move c with the slider below and pick a sign. The thick piece on the x-axis is the solution.

Below or above the x-axis

c = 12

< 0≤ 0> 0≥ 0

x² − 7x + 12 < 0(3, 4)

• y = x² − 7x + c
• Solution

Roots

x = 3 and x = 4

Solution

(3, 4)

Show the numbers

x	y	sign of y
0	12	+
1	6	+
2	2	+
3	0	0
4	0	0
5	2	+
6	6	+
7	12	+

At c = 12.25 the parabola touches the x-axis at one point. At c = 14 it no longer dips below the x-axis. Then x² − 7x + c < 0 has no solution.

The 2017 MULO exam in mathematics B asked this: −x² + 7x − 12 < 0. First multiply by −1. The sign flips:

x² − 7x + 12 > 0

Now you want where the parabola lies above the x-axis: left of 3 and right of 4. The answer is (−∞, 3) ∪ (4, ∞). The exam wrote it as ⟨←, 3⟩ ∪ ⟨4, →⟩.

Try it yourself