You have seen the proof that \sqrt{2} is irrational. So its decimal 1.41421356\dots goes on forever without a repeating pattern, and no fraction p/q will ever equal it. That raises an uncomfortable question — if you cannot write \sqrt{2} down exactly as a decimal and cannot write it as a fraction, how do you locate it on the number line? Which point is it?

The answer is older than algebra. You locate \sqrt{2} not by computing digits but by building it, with a ruler and a compass, out of a shape you already understand: a square of side 1.

The two-step idea

The plan has two moves. First, build a line segment of length exactly \sqrt{2} — not 1.414, not 1.41421, but the real thing. Second, swing that segment down onto the number line so its endpoint marks the point you want.

The first move uses Pythagoras. Why: the hypotenuse of a right triangle with legs a and b has length \sqrt{a^2 + b^2}, so a right triangle with both legs equal to 1 gives a hypotenuse of \sqrt{1 + 1} = \sqrt{2}. The irrational number is forced into existence by ordinary integer geometry.

The second move uses a compass. A compass keeps a fixed distance between its two legs as you swing it. So if you put the needle at 0 and the pencil at a point that is \sqrt{2} away, and then sweep the pencil down until it hits the number line, the landing point is exactly \sqrt{2} units from 0.

Step-by-step construction

Step 1. Draw the number line and mark the integers 0, 1, 2.

Step 2. Above the segment from 0 to 1, build a unit square. Its four corners are at (0, 0), (1, 0), (1, 1), and (0, 1).

Step 3. Draw the diagonal from the bottom-left corner (0, 0) to the top-right corner (1, 1). This diagonal has length

d = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2}.

Why: you now have a physical line segment whose length is exactly \sqrt{2}. Holding it against a ruler would show something a little longer than 1.4 and a little shorter than 1.5, but the length itself is exact — it is whatever \sqrt{2} happens to be.

Step 4. Put a compass needle at (0, 0) and open it until the pencil touches (1, 1). The compass is now holding a distance of \sqrt{2}.

Step 5. Keeping the needle at (0, 0), swing the pencil downward in an arc until it meets the number line. Mark that point.

That point sits at a distance of \sqrt{2} from 0, along the number line, to the right. By construction, it is the point labelled \sqrt{2}.

See it happen

Drag the slider below to animate the construction. At t = 0 you see just the number line and the unit square with its diagonal. As t increases, the compass sweeps the diagonal around the origin. At t = 1 the arc has completed its swing and the landing point is exact.

Sweep the diagonal of the unit square down to the number line A horizontal number line near the bottom of the frame with marks at zero, one and two. Above the segment from zero to one sits a unit square with its diagonal drawn from the bottom-left corner to the top-right corner. A slider controls a parameter t from zero to one. At t equal to zero, a radius line lies along the diagonal pointing to the top-right corner. As t increases toward one, the radius rotates clockwise about the bottom-left corner, tracing a circular arc. When t reaches one, the radius points horizontally to the right along the number line, and its tip sits at the point square root of two, a little past one point four. Readouts show the current sweep angle in degrees and the x-coordinate of the compass tip. 0 1 2 diagonal = √2 √2 ↔ drag the red slider
Drag the red slider at the bottom from the diagonal position ($t = 0$, sweep angle $0°$) toward the number line ($t = 1$, sweep angle $45°$). The "pencil x" readout shows the horizontal position of the compass tip. At $t = 1$ the tip sits at $\sqrt{2} \approx 1.4142$ — the compass has swung the diagonal's length down onto the number line.

The arc drawing in the SVG above is simplified — the real visual intuition is captured by the readout. As t goes from 0 to 1, the sweep angle grows from 0° to 45° and the pencil drops from the square's corner down onto the line. At t=1, pencil x reads 1.4142\dots — the true landing point.

A static version makes the geometry clearer.

Finished construction of root two on the number line A horizontal number line with marks at zero, one and two. A unit square sits on the segment from zero to one, with its diagonal drawn. A circular arc, centred at the origin and of radius equal to the diagonal, sweeps from the top-right corner of the square down to the number line. It lands at the point marked root two, just past one point four. 0 1 2 √2 √2 ≈ 1.4142…
The compass is pinned at $0$. Its opening is the length of the diagonal, which is $\sqrt{2}$. Swinging the pencil from the square's top-right corner down to the number line places the pencil at the point $\sqrt{2}$. The entire construction uses only ruler and compass — no decimals, no calculators, no approximations.

Why the landing point is exact

The compass preserves distance. That is its defining property. Why: a circle is, by definition, the set of all points that lie at a fixed distance from a centre. Every point the pencil visits while you swing the compass is exactly that distance from the needle.

Before you swung, the pencil was at the corner (1, 1), at a distance of \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{2} from the needle at (0,0). After you swing it down onto the number line, the pencil is still at distance \sqrt{2} from the needle. Since it is now on the x-axis, at some positive coordinate a, its distance from (0,0) equals |a|. So a = \sqrt{2}, with no rounding.

The construction produces \sqrt{2} as a position, not as a decimal. The point is what it is. You never had to approximate.

What you can build this way

The same two-step trick — build a length using Pythagoras, then sweep it onto the number line — places any \sqrt{n} for a positive integer n.

Chain them and you get the spiral of Theodorus, a sequence of right triangles whose hypotenuses are \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots — each one built off the previous, each one landing a new irrational on the number line.

Build √3 on the number line

Step 1. First, construct \sqrt{2} on the number line by the method above. Call this point A.

Step 2. At A, erect a perpendicular to the number line of length 1. The top of this perpendicular is the point (\sqrt{2}, 1).

Step 3. Join this top point back to the origin. By the distance formula,

\text{length} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}.

Step 4. With the compass pinned at the origin and opened to this new length, swing the pencil down onto the number line. The landing point is \sqrt{3}.

Result: \sqrt{3} sits on the number line at the exact foot of the arc swung from (\sqrt{2}, 1) with radius \sqrt{3} — no decimals used.

The construction is beautiful precisely because it side-steps the "but I cannot write it down" problem. You do not need to write an irrational number as a decimal to know where it is. You just need a ruler, a compass, and a right triangle with the right legs.

This satellite sits inside Number Systems.