JC2 H3 Mathematics Tutor Bukit Timah

Did you know? Removing two opposite corners from a chessboard makes it impossible to cover the remaining squares with 31 ordinary dominoes. Even though there are exactly 62 squares left—just enough area for 31 two-square pieces—the colour pattern makes the task impossible. A simple symmetry and counting argument reveals what measuring area alone cannot.

For JC2 H3 Mathematics students in Bukit Timah, this is the appeal of proof by cases and symmetry. Rather than calculate every possibility, we identify a small number of exhaustive situations or a transformation that preserves the problem. Both are named reasoning principles in the Singapore–Cambridge H3 Mathematics (9820) syllabus.

Quick answer: Proof by cases works when you divide all permitted situations into categories and prove the claim in every category. A symmetry argument works when swapping, reflecting or rotating the objects preserves the assumptions and the claim. You may use ‘without loss of generality’ only after justifying that the omitted situation follows by symmetry.

What makes proof by cases a proof?

Imagine a statement about every integer n. One natural division is that n is even or n is odd. Every integer lies in at least one of these two categories, and no integer belongs to both. Establish the claim in each category, and the universal statement follows.

The coverage is essential. A proof that checks only positive and negative integers may accidentally omit zero. Likewise, if the claim concerns all real numbers, checking only integers and fractions is not enough because irrational real numbers remain.

Cases need not always be mutually exclusive, but they must collectively cover every permitted input. Disjoint cases are often easier to present because they avoid repeated work and make completeness visible.

The most common H3 error is writing ‘Case 1’ and ‘Case 2’ without explaining why there cannot be a third case. The first sentence should establish the categories and the domain before any calculation.

For a constructive proof, the same discipline matters. If an algorithm handles different circumstances differently, identify each circumstance, prove the chosen action is valid, and show that one of the cases must occur.

Worked example: every integer square is 0 or 1 modulo four

Claim: For every integer n, n² is congruent to either 0 or 1 modulo four.

Case 1 — n is even. Write n=2k for an integer k. Then n²=4k², which is divisible by four. Therefore n²≡0 (mod 4).

Case 2 — n is odd. Write n=2k+1. Then n²=4k²+4k+1=4k(k+1)+1, so n²≡1 (mod 4).

Conclusion: Every integer is either even or odd, and both cases have been established. Thus every integer square is congruent to zero or one modulo four.

A useful consequence is that the equation n²=4m+2 has no integer solution. If such an integer existed, its square would have remainder two modulo four, contradicting the two exhaustive cases.

This example connects proof by cases with modular arithmetic: the possible remainders provide a systematic way to choose the cases.

An absolute-value case proof without guesswork

The absolute value of a real number x is defined piecewise: |x|=x when x≥0, and |x|=−x when x<0. That definition suggests the natural cases for a problem involving |x|.

Claim: For every real x, |x|≥x. If x≥0, then |x|=x and equality holds. If x<0, then |x|=−x>x because −x−x=−2x>0. The two cases cover all real x, so the claim is proved.

Now consider |x|+|x−2| for real x. The expressions change form at x=0 and x=2, so the natural cases are x<0, 0≤x≤2, and x>2.

For x<0 the sum equals −x+(2−x)=2−2x>2. For 0≤x≤2 it equals x+(2−x)=2. For x>2 it equals x+(x−2)=2x−2>2. Therefore the global minimum is 2 for every x in [0,2].

The key was choosing case boundaries where the formulas change. That is a general tactic: first locate sign changes, then perform each calculation under the correct assumptions.

Symmetry: when ‘without loss of generality’ is justified

A problem is symmetric under a transformation when exchanging, reflecting or rotating certain objects leaves its assumptions and requested conclusion unchanged. If two cases are interchangeable in this way, proving one may automatically prove the other.

Suppose a and b are real numbers. The inequality min(a,b)≤(a+b)/2≤max(a,b) is symmetric when a and b are swapped. We may assume a≤b without loss of generality because the swapped case has exactly the same claim.

Under a≤b, 2a≤a+b≤2b. Dividing by two gives the inequality. If b≤a instead, swap their names and the proof is identical. The symmetry justifies the reduction.

However, an expression involving only a, such as a²+3b, is not generally unchanged when a and b are swapped. Claiming ‘without loss of generality a≤b’ can be invalid unless the entire condition and conclusion are truly symmetric.

One clean way to audit a symmetry step is to describe its inverse. If swapping the labels transforms any omitted case into a covered one and back again while preserving the statement, no case has been lost.

The H3 content outline lists the symmetry principle under mathematical proof and reasoning. Symmetry can turn a long enumeration into a short argument, but the preservation of the problem must be shown.

The chessboard puzzle: equal area, impossible tiling

Question: Remove two diagonally opposite corner squares from an 8×8 chessboard. Can the remaining 62 squares be tiled by 31 dominoes, each covering two edge-adjacent squares?

Observation: In the standard alternating colouring, diagonally opposite corners have the same colour. The original board contains 32 black and 32 white squares. Removing two corners of the same colour leaves 30 squares of that colour and 32 of the other.

Invariant: Every two-square domino placed over edge-adjacent squares covers one black and one white square. Therefore any tiling with 31 dominoes would cover exactly 31 black squares and 31 white squares.

Contradiction: The modified board has a 30–32 split, not a 31–31 split. Hence no such tiling exists.

Notice what did not solve the puzzle. The area is 62 squares and the total domino area is also 62, so the area equation alone is satisfied. The colouring argument identifies an additional constraint preserved by every placement.

This is a lesson in choosing the right representation. A problem that appears to be about geometry and physical tiling becomes a short proof about two colours and a conserved difference.

A shortest-path proof using reflection

Problem: Let A=(0,3) and B=(4,1), both above the x-axis. Choose a point P on the x-axis to minimise the broken path AP+PB.

Step 1 — reflect one endpoint. Reflect B across the x-axis to B′=(4,−1). For every point P on the x-axis, PB=PB′ because reflection preserves distances from points on the mirror line.

Step 2 — apply the triangle inequality. AP+PB=AP+PB′≥AB′. The distance AB′=√[(4−0)²+(−1−3)²]=√32=4√2.

Step 3 — check equality. Equality occurs when A, P and B′ lie on a straight segment. The segment from (0,3) to (4,−1) crosses the x-axis at P=(3,0).

Answer: The minimum broken-path distance is 4√2, attained at P=(3,0). Indeed AP=3√2 and PB=√2, summing to 4√2.

Reflection changed a constrained optimisation problem into a straight-line distance calculation. The proof relies on a preserved distance, a general triangle inequality and an explicit equality condition—not on visually guessing where the path should touch the axis.

A symmetry-based counting proof: odd and even subsets

Take a nonempty finite set with n elements. Its subsets come in two categories: those containing an even number of elements and those containing an odd number.

Choose and fix one distinguished element a. Pair each subset S with the subset obtained by toggling membership of a: if a belongs to S, remove it; otherwise add it.

This toggling is reversible and changes subset size by exactly one. Thus every even-sized subset is paired with exactly one odd-sized subset, and every odd-sized subset with exactly one even-sized subset.

There are 2ⁿ subsets in total. The pairing splits them into equally numerous groups, so for n≥1 there are 2ⁿ⁻¹ even-sized subsets and 2ⁿ⁻¹ odd-sized subsets.

This is a bijection proof with a symmetry-like involution: applying the toggle twice restores the original subset. It provides a concrete bridge to the bijection principle, where exact correspondence replaces complicated direct counting.

Twelve H3-style proof clinics

1. Even or odd

For every integer n, n(n+1) is even: if n is even the first factor is even; otherwise n+1 is even.

2. Squares modulo four

An integer square is 0 or 1 modulo four, using the exhaustive even and odd cases.

3. A three-way sign split

For real x, the cases x<0, x=0 and x>0 are disjoint and exhaustive.

4. Zero must be covered

Positive and negative numbers are not exhaustive cases for all real numbers, because zero belongs to neither.

5. Absolute value

The inequality |x|≥x holds by the cases x≥0 and x<0.

6. An interval of minimisers

|x|+|x−2| has minimum 2 throughout 0≤x≤2; the breakpoints are 0 and 2.

7. Symmetric average

For real a,b, the arithmetic mean lies between min(a,b) and max(a,b); swapping a and b preserves the claim.

8. Chessboard colouring

Removing opposite corners of an even chessboard removes two same-coloured squares; edge-adjacent dominoes still cover one of each colour.

9. Reflection path

For A=(0,3), B=(4,1) and P on y=0, the minimum AP+PB is 4√2 at P=(3,0).

10. Even and odd subsets

For n≥1, toggling one fixed element pairs even and odd subsets, giving 2ⁿ⁻¹ of each.

11. Invalid WLOG

You cannot assume a≤b without justification if the problem treats a and b differently, such as a fixed coefficient on a alone.

12. Symmetry of complex roots

Multiplying all nth roots of unity by the same primitive root rotates their polygon and permutes the root set.

The four most common logic gaps

Missing case: The chosen situations do not cover the full domain. For real numbers, positive and negative omit zero. Changing the assumption: A student’s working inside one case quietly uses an inequality from a different case.

Unjustified symmetry: Swapping labels changes the original problem, yet the proof announces ‘without loss of generality’. Unproved equality: A geometric bound is established, but the student never checks whether an allowed configuration actually attains it.

Before writing the final answer, make a small checklist: domain, case coverage, argument inside each case and final conclusion. For symmetry, add the exact transformation and the reason it preserves the relevant conditions.

A short proof is not necessarily a weak proof. In fact, once the correct invariant or symmetry has been found, a valid proof can become only a few lines. Precision matters more than length.

A practical JC2 learning plan

Session 1: Write three parity proofs with explicit, exhaustive cases. Session 2: Solve absolute-value inequalities, identifying every sign-change breakpoint. Session 3: Rewrite a long two-variable proof using a justified symmetry transformation.

Session 4: Explain the removed-corners chessboard contradiction from first principles. Session 5: Derive the reflected-path minimum, including the equality condition. Session 6: Build the even/odd subset pairing and explain why it is reversible.

An effective advanced mathematics tutor should ask, ‘Why do those cases cover everything?’ and ‘What does your symmetry preserve?’ The aim is not to train a phrase like ‘WLOG’, but to teach when it is logically warranted.

Near Bukit Timah, the eduKateSG three-student Mathematics reference near Sixth Avenue MRT describes a Secondary 1 tutorial model. It should not be interpreted as confirmation of a JC2 H3 class. Ask eduKateSG about current JC2 H3 scope and availability.

Frequently asked questions

Are proof by cases and symmetry examined in H3 Mathematics?

Both are named in the Singapore–Cambridge H3 Mathematics (9820) syllabus under mathematical proofs and reasoning principles.

Must proof cases be mutually exclusive?

No, but together they must cover every permitted input. Disjoint cases are often clearer and prevent repeated reasoning.

When may I write ‘without loss of generality’?

Only when a symmetry or equivalence shows the excluded situations follow from those retained, with all assumptions and the target statement preserved.

Why is the missing-corners chessboard impossible to tile?

Removing opposite corners disrupts the equality between black and white squares. Each domino covers one of each, so a full tiling is impossible.

How do I know a symmetry proof is complete?

State the transformation, show it preserves the relevant conditions, identify the covered case and explain how the transformation handles the other cases.

Official syllabus and related proofs

The 2026 official H3 Mathematics 9820 syllabus and 2027 syllabus both name proof by cases and the symmetry principle, with H2 content assumed.

Continue with induction and the pigeonhole principle, triangle inequalities and contradiction and counterexamples. These combine naturally when a problem requires a strategic change of viewpoint.

JC2 H3 Mathematics Tutor Bukit Timah | Proof by Cases and the Symmetry Principle