Real Numbers on the Number Line

A number that you cannot write as a fraction can still live as a point on the number line. This lesson shows how to draw $\sqrt{2}, \sqrt{3}, \sqrt{5}, \ldots$ exactly using only a ruler and compass, and how to "zoom in" on any decimal so precisely that its location is determined to as many digits as you wish. After this, the words "real number" really do mean every point of the line.

Definitions

The real number line is a horizontal line on which $0$ is marked, $1$ unit to the right is $1$ , and every other real number has its own point, with negatives to the left of $0$ . This correspondence between numbers and points is a one-to-one match: every real number is some point, and every point is some real number.

A number is constructible (in the geometric sense of this chapter) if it can be drawn on the line using only a straight-edge and compass, starting from the markings $0$ and $1$ . We will see that $\sqrt{n}$ is constructible for every natural number $n$ .

Concept and construction

Step-by-step construction of $\sqrt{2}$ . Mark $O = 0$ and $A = 1$ on the line. At $A$ , draw a perpendicular of length $1$ to reach a point $B$ . By the Pythagoras theorem, $OB = \sqrt{OA^2 + AB^2} = \sqrt{1+1} = \sqrt{2}.$ With $O$ as centre and radius $OB$ , swing an arc to cut the number line at $P$ . Then $OP = \sqrt{2}$ , so $P$ represents $\sqrt{2}$ .

Building $\sqrt{3}, \sqrt{4}, \sqrt{5}, \ldots$ from $\sqrt{2}$ . At $B$ , draw a new perpendicular of length $1$ to reach $C$ . Then $OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}.$ Continue: $\sqrt{OC^2 + 1^2} = \sqrt{4} = 2$ , then $\sqrt{5}, \sqrt{6}, \ldots$ . The resulting spiral is called the Theodorus spiral or "spiral of $\sqrt{n}$ ". Each new hypotenuse jumps to the next square root.

General principle. If $\sqrt{n}$ has been drawn, then $\sqrt{n+1}$ is the hypotenuse of a right triangle whose legs are $\sqrt{n}$ and $1$ . So every $\sqrt{n}$ for $n \ge 1$ can be placed on the line in finitely many steps.

Successive magnification (zooming in). Locating a decimal like $2.665$ uses a different idea: not Pythagoras but repeated subdivision. First locate the integer part $2 < 2.665 < 3$ on a stretched portion of the line. Divide that segment into $10$ equal parts and find $2.6 < 2.665 < 2.7$ . Divide that subsegment into $10$ equal parts to find $2.66 < 2.665 < 2.67$ . Divide once more to land exactly on $2.665$ . With one zoom per digit you can locate any decimal , terminating or not , to any precision.

Why this matters. For a non-terminating non-recurring decimal like $0.1010010001\ldots$ , no finite zoom lands you on it exactly, but each zoom traps it more tightly. This is the geometric meaning of "the real line has no holes" , irrationals are exactly the points the rationals are zooming in on.

A cleaner construction for $\sqrt{n}$ for any specific $n$ . Mark $0, n$ and $n+1$ on the line. On the segment from $0$ to $n+1$ , draw a semicircle. At the point $n$ , erect a perpendicular meeting the semicircle at $Q$ . Then the length of that perpendicular equals $\sqrt{n}$ . (This uses the fact that, inside a semicircle, the perpendicular from a point on the diameter to the arc has length equal to the geometric mean of the two pieces of the diameter.) So with $n+1$ marked, one perpendicular yields $\sqrt{n}$ in a single step.

Worked examples

Example 1. Construct $\sqrt{5}$ on the number line.

Mark $O = 0$ , $A = 2$ on the line. Erect a perpendicular of length $1$ at $A$ to reach $B$ . Then $OB = \sqrt{2^2 + 1^2} = \sqrt{5}$ . With centre $O$ and radius $OB$ , cut the line at $P$ . Then $P$ represents $\sqrt{5}$ .

Example 2. Locate $3.234$ on the number line by successive magnification.

Zoom 1: enlarge $[3,4]$ . Zoom 2: enlarge $[3.2, 3.3]$ and mark $3.23$ . Zoom 3: enlarge $[3.23, 3.24]$ and mark $3.234$ on the fourth division.

Example 3. Locate $\sqrt{7}$ on the number line.

Use the semicircle method. Mark $0$ , $7$ and $8$ on the line. Draw a semicircle on the segment of length $8$ . The perpendicular at $7$ to the arc has length $\sqrt{7 \cdot 1} = \sqrt{7}$ . Transfer this length onto the number line with a compass.

Example 4. Why does the spiral construction give exactly $\sqrt{n}$ ?

Because each new triangle is right-angled with legs $\sqrt{n-1}$ and $1$ , the hypotenuse is $\sqrt{(\sqrt{n-1})^2 + 1^2} = \sqrt{n}$ . The construction is just the Pythagoras theorem applied repeatedly.

Example 5. Show that $0.\overline{6}$ corresponds to the same point as $\dfrac{2}{3}$ .

Let $x = 0.\overline{6}$ . Then $10x = 6.\overline{6} = 6 + x$ , giving $9x = 6$ , $x = \dfrac{2}{3}$ . So the repeating decimal and the fraction live at the same point.

Try it yourself

Construct $\sqrt{6}$ on the number line.
Construct $\sqrt{2}, \sqrt{3}, \sqrt{4}$ in a single Theodorus-style spiral.
Use successive magnification to locate $4.265$ on the line.
Locate $\sqrt{10.5}$ using the semicircle construction.
Where does the decimal $1.\overline{3}$ sit on the line? Convert and mark.
Is the point at $\sqrt{2}$ closer to $1.4$ or $1.5$ ? Justify with a single inequality.
Locate $\sqrt{2} + 1$ on the number line, given $\sqrt{2}$ already constructed.
Use the semicircle method to construct $\sqrt{9.3}$ given $1$ unit length.
Mark all of $-\sqrt{2}, -1, 0, 1, \sqrt{2}, \sqrt{3}$ on the same line and order them.
Explain in one sentence why every point on the line corresponds to a unique real number.

Pitfalls / Insight

Constructions are exact, not approximate. $OP = \sqrt{2}$ by the Pythagoras theorem , not "about $1.414$ ".
Don't confuse "rational" with "constructible". $\sqrt{2}$ is irrational and constructible; not all irrationals (like $\pi$ ) are constructible.
Zoom needs only one new digit per stage. Trying to zoom by, say, $100$ -ths is harder to draw and harder to read.

Insight. Pythagoras is the secret weapon of the chapter. Whenever you need to draw $\sqrt{n}$ , ask yourself, "can I write $n$ as $a^2 + b^2$ ?" If yes, a right triangle with legs $a$ and $b$ does the job.