The Real Number Line and Intervals

Intervals and inequalities

Every point on the number line is a real number, and every real number is a point on the line. A set of real numbers, such as "all numbers from $-1$ up to, but not including, $4$," can be written as an inequality or as an interval. Both say exactly the same thing.

A closed dot (or square bracket) means the endpoint is included. An open dot (or parenthesis) means it is not included.

Absolute value and distance

The absolute value $|x|$ is the distance from $x$ to $0$ on the number line. More generally, the distance between two points $A$ and $B$ is $|A - B|$.

Some problems call absolute value the modulus. It is the same thing.

Writing $|x - a|$ means "the distance from $x$ to $a$." That turns equations and inequalities with absolute value into questions about distance:

Neighborhoods: $E(a, r)$

"A value $a$, give or take $r$" is such a common idea that it has its own name. The neighborhood of center $a$ and radius $r$ is the open interval of all points closer to $a$ than $r$.

To go from an interval $(p, q)$ back to a neighborhood, find the midpoint and the half-length:

Tolerances and measurement error

Every measurement comes with a margin of error. A statement like "$L = 8 \pm 0.05$ in." means the true length is within $0.05$ of $8$, so $|L - 8| \le 0.05$ and $L \in [7.95, 8.05]$.

Quality control often states the rejection condition instead of the acceptance condition. The accepted values are then the complement: flip the inequality.

Union, intersection, and complement

Because intervals are sets, we can combine them.

• Union $A \cup B$: numbers in $A$ or $B$ (or both).

• Intersection $A \cap B$: numbers in $A$ and $B$.

• Complement $\mathbb{R} \setminus I$: all real numbers not in $I$.

The best method is to draw each set on its own number line, one above the other, with the endpoints lined up.

The intersection is only where both lines are thick. The union is where at least one is thick.

Intervals can also have an empty intersection. For example, $(-\infty, 1) \cap [3, \infty) = \varnothing$. And two intervals that touch can merge: $[1, 3] \cup (3, 5) = [1, 5)$.

Absolute value inequalities

To solve an absolute value inequality, isolate the absolute value, then split it into one of two forms:

• $|u| \le c$ (with $c \ge 0$) becomes $-c \le u \le c$ (one chain).

• $|u| \ge c$ becomes $u \le -c$ or $u \ge c$ (two cases).

Here $u$ can be any expression, such as $3x + 1$.

Quadratic inequalities and sums of absolute values

For a quadratic inequality, move everything to one side, find the zeros, and check the sign on each piece of the number line. If $a > 0$, the parabola opens upward, so it is negative between its zeros and positive outside them.

Absolute values can also be added. Split the number line at the points where each absolute value changes behavior, and solve each case separately.

Irrational numbers on the line

Rational numbers are fractions, and their decimals stop or repeat. Irrational numbers, such as $\sqrt{2}$, $\pi$, and $\varphi$, cannot be written as a fraction. Both kinds fill the real line, with no gaps.

Plotting a square root exactly

The Pythagorean Theorem lets you place $\sqrt{n}$ exactly. Write $n$ as a sum of two perfect squares, $n = p^2 + q^2$. Draw a right triangle with legs $p$ and $q$ starting at $0$ on the number line. Its hypotenuse has length $\sqrt{p^2 + q^2} = \sqrt{n}$. Swing a compass with center $0$ and that radius until it hits the line.

The golden ratio

The golden ratio $\varphi = \dfrac{1 + \sqrt{5}}{2}$ also has a compass-and-straightedge construction on the line. Take a unit square on $[0, 1]$. Let $M$ be the midpoint of its base, at $\tfrac{1}{2}$. Draw the segment from $M$ to the top-right corner. Swing an arc centered at $M$ onto the number line.

Why this lands exactly at $\varphi$: the right triangle has legs $1$ and $\tfrac{1}{2}$, so the hypotenuse is $\sqrt{1 + \tfrac{1}{4}} = \tfrac{\sqrt{5}}{2}$. Starting at $M = \tfrac{1}{2}$ and moving that far to the right reaches $\tfrac{1}{2} + \tfrac{\sqrt{5}}{2} = \tfrac{1 + \sqrt{5}}{2} = \varphi$. You can also check that $\varphi$ is a zero of $x^2 - x - 1$: $\varphi^2 = \tfrac{6 + 2\sqrt{5}}{4} = \tfrac{3 + \sqrt{5}}{2} = \varphi + 1$.

Irrational numbers between rationals

Between any two real numbers there are both rational and irrational numbers. To find an irrational number between two rationals $p < q$:

1. Find the gap $q - p$.

2. Choose a whole number $N$ so large that $\sqrt{2}/N$ is smaller than the gap.

3. Then $p + \sqrt{2}/N$ is between $p$ and $q$.

It is irrational because a rational number plus an irrational number is irrational: if $p + \sqrt{2}/N$ were a rational $s$, then $\sqrt{2}/N = s - p$ would be rational, which is false.

Back to the start: a can is accepted when it holds between $11.8$ and $12.2$ oz. As an interval that is $[11.8, 12.2]$. As a distance condition it is $|c - 12| \le 0.2$. And as a closed interval it has center $12$ and radius $0.2$, which is exactly what $|c - 12| \le 0.2$ says. Three ways to say the same thing.

Summary