The Sierpinski carpet and gasket: what repeated removal leaves behind

What to assume they know

• Self-similarity: a shape that contains copies of itself — self-similarity, and the idea that a fractal is approached by a sequence of steps
• Powers with whole-number exponents, and reading 8ⁿ and 3ⁿ as repeated multiplication
• Area of a square and of a triangle, and the idea of a fraction of an area
• Multiplying fractions, enough to iterate 8/9 and 3/4
• Midpoint of a segment; the midsegment fact is derived in this topic, not assumed
• Angles of an equilateral triangle, and the congruence criteria (SAS is enough)
• Reading a subscripted symbol such as Rₙ as "the value of R at step n"

What they should be able to do

• Prove that joining the midpoints of an equilateral triangle cuts it into four identical equilateral triangles, using the chapter's own hint about the corner triangles
• State the carpet's construction rule and the gasket's construction rule, and say what plays the part of "one piece" in each
• Justify Rₙ₊₁ = 8Rₙ by an argument about a single surviving square
• Justify Hₙ₊₁ = Hₙ + Rₙ by two separate observations, one about new holes and one about old ones
• Turn the recursion Rₙ₊₁ = 8Rₙ into the closed form Rₙ = 8ⁿ and say why the step from one to the other is legitimate
• Compute the corresponding counts for the gasket and compare them with the carpet's
• Compute the area remaining at step n for both fractals, taking the starting figure as 1 square unit
• Explain how the piece count can grow without limit while the area shrinks towards zero

Where it usually goes wrong

• "Rₙ₊₁ = 8Rₙ is just an observed pattern." It is not — it is proved by one sentence about one square. Each surviving square is subdivided into 9 and loses 1, so it contributes exactly 8, and this happens to every survivor independently. Show the argument on a single square, then multiply.
• "Hₙ₊₁ = Hₙ + Rₙ because a hole appears in each hole." No. New holes appear in the surviving squares, one apiece — a hole is gone and does nothing further. The Rₙ in that formula is the survivor count, not the hole count, and students routinely substitute the wrong one.
• "Rₙ = 8ⁿ because the pattern 1, 8, 64 looks like powers of 8." The reason is that multiplying by 8 exactly n times starting from 1 gives 8ⁿ. The chapter writes R₁ = 8 × 1 and R₂ = 8 × 8 precisely so the repeated multiplication is visible before the exponent is claimed.
• "Joining midpoints obviously gives four identical triangles." It is not obvious and the chapter asks for a proof. The claim that needs work is that the middle triangle is equilateral and the same size as the corners; the hint about isosceles corner triangles is the route in.
• "Removing pieces forever must leave nothing at all." The area does tend to zero, but the set of leftover points is not empty — every corner of every surviving piece survives every step. The chapter does not raise this and an explanation should not resolve it, but it should not assert emptiness either.
• "More pieces means more area." The single sharpest error this topic can correct. Eight pieces at one-ninth each is less than one piece at full size. Put 8 × (1/9) beside 1.
• "The gasket is a different kind of thing from the carpet." Same machine, different starting shape and different keep-fraction. Every formula in the gasket half is the carpet formula with 8 → 3 and 1/9 → 1/4.

Questions to check understanding

• Given a construction described in words, write the recursion for the number of surviving pieces and justify it in one sentence
• Given a recursion of the form aₙ₊₁ = k·aₙ with a₀ given, write the closed form
• Compute the number of holes at a stated step from the chapter's own recursion
• Prove that joining the midpoints of an equilateral triangle produces four congruent equilateral triangles
• Find the fraction of the original area remaining after a stated number of steps
• Decide, with a reason, whether a given count can grow while the corresponding area shrinks
• Adapt every result to a variant rule — for example, a square cut into 25 pieces with the middle one removed — which is the standard competency-based extension

Examples worth working on the board

Values marked not in the book are worked out here or an added proof on the chapter's stated inputs; the chapter prints no answers.

• The carpet construction (Part II p.70). Start with a square. Break it into 9 smaller squares. Remove the central one. Repeat on each of the 8 that remain. Step 0, Step 1 and Step 2 are printed, followed by an ellipsis.
• The two properties the chapter lists for every step (Part II p.71): the surviving squares at a given step are all the same size, and that common size shrinks as the step number rises; and the holes are square, each having been made by deleting a square piece. These two are the hypotheses the counting arguments use.
• The carpet recursions, exactly as printed (Part II p.71). Rₙ = the number of surviving squares at step n, Hₙ = the number of holes at step n. Printed: Rₙ₊₁ = 8Rₙ, with the reason that each surviving square at step n yields 8 survivors at step n + 1. Printed: Hₙ₊₁ = Hₙ + Rₙ, with two reasons — each survivor at step n creates one hole at step n + 1, and every hole already present persists.
• The printed values (Part II p.71). R₀ = 1; R₁ = 8 × 1 = 8; R₂ = 8 × 8 = 8²; and in general Rₙ = 8ⁿ. Then the paired column: R₀ = 1 with H₀ = 0; R₁ = 8 with H₁ = 1; R₂ = 8² with H₂ = 1 + 8; R₃ = 8³ with H₃ = 1 + 8 + 8²; followed by vertical dots. The chapter stops there and never gives a closed form for Hₙ — checked against the page image.
• Not in the book, carpet: Hₙ = 1 + 8 + 8² + … + 8ⁿ⁻¹ for n ≥ 1, which equals (8ⁿ − 1) / 7. Check it against the printed rows: H₁ = 1, H₂ = 9, H₃ = 73, and (8³ − 1)/7 = 511/7 = 73. Do not present the closed form as the book's; present it as the pattern the book's own dots invite.
• The gasket construction (Part II p.72). Take an equilateral triangle. Join the three midpoints of its sides; the chapter states that this splits it into four congruent equilateral pieces. Delete the middle piece. Repeat on each of the 3 that survive. Step 0, Step 1 and Step 2 are printed, plus a near-final picture captioned Sierpinski Triangle.
• The proof the chapter demands (Part II p.72). It asks the reader to show that joining the midpoints divides an equilateral triangle into 4 identical equilateral triangles, and gives the Hint that the corner triangles are isosceles. An added proof, in the order the hint suggests: let the big triangle have side a and let the three midpoints be joined. Each corner triangle has two sides of length a/2 — the two half-sides of the parent — with the parent's 60° angle between them. Two equal sides and a 60° included angle force the base angles to be equal and to sum to 120°, so each is 60° and the corner triangle is equiangular, hence equilateral with side a/2. That makes every joining segment a side of a corner triangle, so all three joining segments have length a/2, and the central triangle has three sides of length a/2 as well. Four equilateral triangles of side a/2, all congruent.
• The gasket's exercise set (Part II p.72, Figure it Out, three items). Item 1: draw the first steps of the gasket sequence, at least to Step 2. Item 2: find the number of holes and the number of surviving triangles at each step. Item 3: for both Sierpinski fractals, work out how much area is left at step n, with the starting square or triangle taken to have an area of 1 sq. unit.
• Not in the book, gasket counts: Tₙ₊₁ = 3Tₙ with T₀ = 1, so Tₙ = 3ⁿ surviving triangles. Holes: Hₙ₊₁ = Hₙ + Tₙ with H₀ = 0, so H₁ = 1, H₂ = 1 + 3 = 4, H₃ = 1 + 3 + 9 = 13, and in general Hₙ = (3ⁿ − 1)/2. Every one of these follows from the same two sentences that justified the carpet's recursions, with 8 replaced by 3 — which is the point of putting the two fractals in one video.
• Not in the book, areas (this is item 3, and it is the payoff). Carpet: each step keeps 8 of the 9 equal pieces of every surviving square, so the surviving area is multiplied by 8/9 each time. Starting from 1 sq. unit, the area at step n is (8/9)ⁿ. Numerically 1, then 0.888…, 0.790…, 0.702…, and after ten steps about 0.308. Gasket: each step keeps 3 of the 4 equal pieces, so the multiplier is 3/4 and the area at step n is (3/4)ⁿ: 1, 0.75, 0.5625, 0.421875, and after ten steps about 0.056. Cross-check that ties the two halves of the explanation together: Rₙ = 8ⁿ pieces each of area (1/9)ⁿ gives (8/9)ⁿ, and Tₙ = 3ⁿ pieces each of area (1/4)ⁿ gives (3/4)ⁿ. The area formula is forced by the count formula; they are not two results.
• The number. Not in the book: at step 3 the carpet has 512 surviving squares and 73 holes, and covers about 70% of the original area. At step 10 it has over a billion surviving squares and covers about 31%. A count in the billions and an area still visibly present is the concrete form of the thesis.

Figures to have open

• The carpet step panels, Step 0 to Step 2, with the 3 × 3 subdivision lines drawn on Step 1. The chapter's own figures (Part II p.70) do not show the subdivision lines; the explanation needs them for the counting argument, so redraw.
• The gasket step panels, Step 0 to Step 2 (Part II p.72). Redraw as a schematic. Step 1 must show the three joining segments, since they carry the proof.
• A single equilateral triangle with midpoints marked, the three joining segments drawn, one corner triangle shaded, and its two half-sides and included angle labelled. This is the proof figure and the chapter prints nothing like it. Standard schematic, must be drawn.
• A "one square becomes eight" inset: one square, the 3 × 3 grid, the centre removed, the eight survivors shaded. Standard schematic.
• A twin-axis plot of piece count against step and remaining area against step, for both fractals. Standard schematic; the chapter plots nothing.
• The near-final gasket picture (Part II p.72, captioned Sierpinski Triangle) is useful once, to show what many steps look like. It is a step, not the fractal.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 4, §4.1 "Fractals", printed subheadings "Sierpinski Carpet" (Part II pp.70–71) and "Sierpinski Gasket" (Part II p.72).
• The two recursions and the paired R–H columns are printed on Part II p.71.
• The midpoint claim, its Hint, and the alternative name for the fractal are on Part II p.72, immediately above the Figure it Out set.
• The Figure it Out set for this material is Part II p.72, items 1–3. Item 3 covers both Sierpinski fractals, which is why one brief owns both.
• Part II p.102, SUMMARY, bullet 2 names the Sierpinski Carpet and Sierpinski Gasket among the chapter's mathematical fractals.