Numbers on the number line

The number line is a straight line on which every point corresponds to exactly one real number, and every real number corresponds to exactly one point. Integers sit at equal unit spacing in both directions from $0$; between any two integers lie infinitely many rationals and irrationals, packed so densely that between any two numbers, however close, another number always exists.

Locating rational numbers. A rational number $\dfrac{p}{q}$ is placed by dividing the unit segment into $q$ equal parts and counting off $p$ of them from $0$. To mark $\dfrac{3}{4}$, divide the segment from $0$ to $1$ into $4$ equal parts and take the $3$rd division mark. A negative rational is marked the same way on the left side of $0$. This only needs a ruler and careful equal spacing — no special construction.

Locating irrational surds. Numbers such as $\sqrt2, \sqrt3, \sqrt5$ cannot be written as an exact fraction, so equal subdivision cannot place them — you would only ever get a decimal approximation, which ICSE does not accept as a valid representation. Instead, geometry supplies an exact answer through the Pythagoras theorem,

where: $a$ and $b$ are the two perpendicular legs of a right triangle and $c$ is the hypotenuse. If you choose whole-number legs $a$ and $b$ so that $a^2 + b^2$ equals the number under the root, the hypotenuse is exactly the surd you need — and a compass can carry that exact length onto the number line. To mark $\sqrt2$: at the point representing $1$, draw a perpendicular of length $1$ unit. The hypotenuse of this right triangle, with both legs equal to $1$, has length

Open a compass to this hypotenuse length, place the pointer at $0$, and draw an arc cutting the number line. The intersection point is exactly $\sqrt2$ — not an approximation, because the compass has physically transferred the true length $OB$. This idea extends step by step: once $\sqrt2$ is marked, a new perpendicular of length $1$ at that point gives a hypotenuse $\sqrt{(\sqrt2)^2+1^2} = \sqrt3$ from $0$, and repeating the process gives $\sqrt4 = 2$, $\sqrt5$, and so on — a "square-root spiral" of right triangles. There is also a quicker single-triangle route whenever the surd can be written directly as $\sqrt{a^2+b^2}$ for whole numbers $a, b$: for $\sqrt5 = \sqrt{2^2+1^2}$, legs of $2$ and $1$ give the hypotenuse in one triangle, skipping the intermediate $\sqrt2, \sqrt3, \sqrt4$ steps. Both methods are equally valid geometrically; the spiral is convenient when several surds are wanted on one diagram, and the direct-leg method is faster for a single required surd.

Represent $\sqrt2$ on the number line using ruler and compasses.

Step 1. Draw a number line and mark $O$ at $0$ and $A$ at $1$, so $OA = 1$ unit.

Step 2. At $A$, draw $AB \perp OA$ with $AB = 1$ unit, and join $OB$ to form right triangle $OAB$.

Step 3. Apply the Pythagoras theorem in triangle $OAB$: $OB^2 = OA^2+AB^2 = 1^2+1^2 = 2$, so $OB = \sqrt2$.

Step 4. With centre $O$ and radius $OB$, draw an arc cutting the number line at point $P$ on the positive side. Since the compass keeps radius $OB$ constant, $OP = OB = \sqrt2$.

Answer: $P$ represents $\sqrt2$ exactly, because it lies at distance $\sqrt2$ from $O$ on the number line — an exact geometric result, not an estimate.

Competitive stretch

Suppose the question instead asked for $\sqrt{13}$. Since $13 = 3^2+2^2$, choose legs $a=3$ and $b=2$: mark $O$ at $0$, $A$ at $3$, draw $AB \perp OA$ with $AB=2$, and $OB^2 = 9+4=13$ gives $OB=\sqrt{13}$ directly in a single triangle — no need to build up through $\sqrt2, \sqrt3, \ldots$ first, since $13$ already splits neatly into two perfect squares.

• Trying to "guess-mark" $\sqrt2$ or $\sqrt3$ by eye using a decimal approximation like $1.414\ldots$ instead of the exact compass-and-Pythagoras construction — eye-marking is not accepted as a valid ICSE representation.
• Forgetting that the perpendicular leg must be a clean whole-number length matching the unit already used for spacing on the line, not an arbitrary or unmeasured segment.
• Confusing which length the compass actually transfers — it is always the hypotenuse $OB$, swung from centre $O$, that lands on the required point $P$, never one of the legs.
• Mixing up the leg values — for $\sqrt5$, legs $2$ and $1$ give $4+1=5$ correctly, but legs $1$ and $1$ would only give $\sqrt2$, not $\sqrt5$.
• Leaving the construction points $O$, $A$, $B$, $P$ unlabelled in the diagram — ICSE answers expect each stage named and the Pythagoras step shown explicitly, not just a final arc drawn on the line.

Ask the tutor: If choosing the right leg lengths for a construction like $\sqrt7$ or $\sqrt{10}$ feels tricky, ask your AI tutor to walk through picking $a$ and $b$ step by step.