Constructing an irrational length and marking it on the number line

What to assume they know

• Proof by contradiction: why √2 cannot be a ratio of integers — that √2 admits no fractional writing, so subdividing cannot find it
• Placing a rational number on the line, and distance as |a − b| — locating a rational by cutting the unit interval
• The Baudhāyana–Pythagoras Theorem, used forwards to compute a hypotenuse
• Constructing a perpendicular at a point on a line with ruler and compass
• Using a compass to transfer a length, and the idea that all points of a circle are equidistant from its centre

What they should be able to do

• Explain why the subdividing method of §3.4.1 cannot locate √2
• Carry out the printed three-step construction and state what each step contributes
• Compute the constructed hypotenuse from the theorem and confirm it is √2
• Explain why swinging a compass arc from the origin transfers the length faithfully
• Identify the constructed point's position between two integers, and say which two
• Extend the construction to √3 by re-using the previous hypotenuse as a leg
• Generalise to any √n for a positive whole number n, and state how many triangles that takes
• Read a number line carrying rationals and irrationals together and identify which marks are which
• Describe the square root spiral as an iteration of the same construction

Where it usually goes wrong

• "If you can draw it, you can write it as a fraction." The construction and the proof together are the chapter's demonstration that this is false. Drawable and writable are different properties, and holding them apart is the point of the module.
• "The arc is just an eyeballed transfer." A compass held at a fixed opening traces points all exactly the same distance from its centre. That is why the transfer is exact rather than approximate, and it is worth one sentence saying so.
• "The perpendicular has to be at the origin." It is raised at A, one unit along, and the arc is centred at O. Students who swap the two get a length of 1 rather than √2. Fig. 3.11 makes the distinction visible.
• "P is at 1.414." P is at √2. 1.414 is a fraction, and by the previous topic no fraction is at P. Any decimal must be labelled as an approximation.
• "You need a new construction for each new root." You need one more triangle. The recursion is the elegant part and it is what the spiral figure exists to show.
• "The spiral's triangles all have a hypotenuse of 1." Every triangle has one leg of 1; the hypotenuses all differ, which is why the exercise asks for them. The right-angle marks tell you which side is which.
• "The construction can reach any irrational number." It reaches roots of whole numbers. The chapter explicitly says the remaining irrationals are taken on trust, and π in Fig. 3.12 is marked without being constructed.

Questions to check understanding

• Construct √2 on the number line and describe each step
• Construct √3 and √5, stating which previous length is re-used as a leg
• Given the spiral figure, find the hypotenuses of all the triangles — the chapter's own end-of-chapter Q16
• State how many triangles are needed to reach a given √n
• Explain why a compass arc transfers a length exactly
• Identify, on a supplied mixed number line, which marks are rational and which are not
• Explain why the construction method does not reach every irrational number

Examples worth working on the board

Values marked verified are worked out here on the chapter's stated inputs. The chapter prints no answers.

• Fig. 3.11 (p. 56), read from a close-up as well as the printed page, because this figure carries more lettering than any other in the module. What it shows: a horizontal number line with arrowheads, ticked and labelled from −3 to 3. The point O sits at 0 and A at 1, so OA is one unit. A vertical segment rises from A to a point B labelled at the top right, and the segment AB carries the label 1. A fourth point C is labelled at the top left, directly above O, and the four points are drawn as a closed unit square. The square's diagonal from O to B is drawn and labelled √2. A dashed arc leaves B, curves down to the right and meets the number line at a point P between 1 and 2, and a short arrow points at that meeting labelling it √2. The caption describes the figure as building irrational lengths and then finding their places on the line.
• The printed steps (§3.5.2, pp. 55–56):
  • On the number line, mark OA one unit, and raise a perpendicular to OA at A.
  • On that perpendicular, mark B with AB one unit, and join O to B; then OB is √2 units.
  • Centre the compass at O, open it to OB, and draw an arc meeting the line at P; then OP is √2 units, so P is where the irrational number sits.
• The arithmetic behind Step 2 (working added here, from the theorem the chapter already used at §3.5, p. 53). Verified: with both legs one unit, 1² + 1² = 2, so the hypotenuse is √2. Show it, because the chapter states OB is √2 without recomputing it here.
• Where P falls (the explanation's estimate, for the figure only). Verified: √2 ≈ 1.41421, so P sits a little under halfway from 1 to 2. Fig. 3.11 draws it consistently with this. Do not show the decimal as though it were the length — it is an approximation, which is the whole point of the module.
• The extension prompt (Think and Reflect, p. 56). Asks the student to construct √3 and √5 with ruler and compass, and then to generalise the method to any √n for a positive whole number n. Verified for √3: take OB, which is √2, as one leg; raise a unit perpendicular at B; the new hypotenuse squares to 2 + 1 = 3. For √5, one further round gives 3 + 1 = 4 and then 4 + 1 = 5, so √5 is reached at the fourth triangle counting from the first. The chapter works none of these.
• Fig. 3.14, the square root spiral (p. 66, inside the end-of-chapter exercises), read from the figure and a second, tighter close-up of its centre. A fan of coloured right triangles all sharing one apex, each with a small square right-angle mark and each with one edge labelled 1. From the tighter close-up: the smallest triangle in the fan has both of its legs marked 1, which is the first triangle of the construction; every later triangle takes the previous hypotenuse as a leg and adds a fresh unit edge. I counted ten right-angle marks — so ten triangles — and eleven edges labelled 1, the extra one being the second leg of that starting triangle. On the construction rule the hypotenuses run √2, √3, √4, √5 and onwards in order, so ten triangles carry hypotenuses from √2 to √11. Exercise 16 on the same page asks for the lengths of all the hypotenuses in this figure.
• Fig. 3.12 (p. 57), read from the printed page. A number line from −5 to 5 with labels split above and below the axis. Above: −22/5, −√10, −12/5, 0, √2, √5, π, 7/2, 9/2. Below: −5, −4, −3, −2, −3/2, −1, −1/3, 5/6, 1, 3/2, 2, 3, 4, 5. This is the figure that shows rationals and irrationals sharing one line, and it is the natural closing image for this topic. Verified: −22/5 = −4.4, −√10 ≈ −3.162, −12/5 = −2.4, √2 ≈ 1.414, √5 ≈ 2.236, π ≈ 3.142, 7/2 = 3.5, 9/2 = 4.5 — all consistent with the drawn order.
• What the chapter concedes (§3.5.3 closing, p. 57). After the constructions, the chapter says only some irrationals have been marked, and that the rest are simply taken to lie on the line. That is an honest limitation.

Figures to have open

• Fig. 3.11 (p. 56) redrawn as a build sequence rather than one static image. The chapter's own figure. Keep the point C and the closed square: the printed figure draws them even though the printed steps mention only O, A and B, and a redrawn version that omits C will not match the page a teacher is holding.
• Fig. 3.14 (p. 66) redrawn as a fan of right triangles with a common apex, each outer edge marked 1, each right angle marked, and the hypotenuses labelled in sequence. The chapter's own figure, printed as flat colour with no hypotenuse labels; adding the labels is what turns it from decoration into the exercise it belongs to.
• Fig. 3.12 (p. 57) redrawn for the closing beat, with labels split above and below the axis as printed, and the constructed points visually distinguished from the assumed ones.
• A compass-transfer inset showing the equidistance of arc points from the centre. Standard schematic; the chapter prints nothing for section 7.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Manjari, Class 9 Mathematics, printed Chapter 3 on the world of numbers. Section §3.5.2, printed heading "Construction of Length" followed by a surd symbol, opens on p. 55 with a Think and Reflect box and Step 1; Steps 2 and 3 and Fig. 3.11 sit on p. 56.
• Think and Reflect, p. 56, setting √3, √5 and the general case.
• Fig. 3.12 and the concession about unmarked irrationals, p. 57, at the end of §3.5.3.
• Fig. 3.14 and end-of-chapter exercise Q16, p. 66.
• The theorem used at Step 2 is introduced at §3.5, p. 53.