Singapore School Mathematics Operating Manual · Chapter 6
Many school problems are easiest when one rule works everywhere. But some mathematical situations change behaviour when a sign changes, a threshold is crossed, an object takes a different type, or a path branches. In those problems, insisting on one universal line of working can hide part of the solution.
Case splitting is the disciplined decision to divide a problem into mutually exclusive situations, solve each situation under its own conditions, and then combine the valid results. It is not a last resort. It is often the clearest representation of the structure already present.
This chapter develops case reasoning from whole-number parity and absolute value through geometry, piecewise functions, probability and optimisation. The emphasis is always the same: every case must have a reason for existing, the cases must cover the allowed possibilities, and the final answer must reunite them without duplication or omission.
Why cases appear · Sign cases · Geometry · Probability branches · Practice · Worked answers
1. One formula is not always the whole model
Consider |x|. If x is non-negative, |x| = x. If x is negative, |x| = −x. The symbol is compact, but underneath it sits a two-case definition.
Likewise, the cost of a service may change after a usage threshold. A geometry problem may have one configuration when a point lies inside a segment and another when it lies outside. A probability tree branches because different first outcomes change the second stage. A greatest-integer or whole-number decision can jump when a continuous input crosses an integer boundary.
In all these examples, the mathematics itself is piecewise. The learner is not creating complexity by splitting cases. The learner is revealing the complexity that was already present.
A good case split has three properties. The cases do not overlap unless overlap is deliberately handled. Together, they cover every admissible situation. Each case carries the conditions that justify the formula or method used inside it.
2. Parity: the simplest structural split
Every integer is either even or odd. This makes parity a natural two-case system.
Suppose we want to prove that n(n + 1) is even for every integer n. One route is to note that consecutive integers have opposite parity. If n is even, the product has an even factor. If n is odd, then n + 1 is even, so again the product has an even factor.
The two cases are exhaustive because every integer is even or odd. They are mutually exclusive because no integer is both. The conclusion is the same in both cases, so the proof covers all integers.
An alternative proof can avoid explicit case splitting by observing directly that one of two consecutive integers must be even. The lesson is not that case splitting is compulsory. It is that a case structure can be compressed when a more global observation captures the same logic.
3. Absolute value equations are branch equations
Solve |2x − 5| = 7. The inside expression can equal 7 or −7.
Case 1: 2x − 5 = 7, so x = 6. Case 2: 2x − 5 = −7, so x = −1. Both satisfy the original equation.
The important point is not the memorised phrase “plus or minus”. The reason is that the absolute value function maps two opposite inputs to the same positive output.
Now solve |2x − 5| < 7. The inequality asks for inside values whose distance from zero is less than 7. This becomes −7 < 2x − 5 < 7, hence −1 < x < 6.
For |2x − 5| > 7, the solution lies outside the central interval: 2x − 5 < −7 or 2x − 5 > 7, giving x < −1 or x > 6. The branch connector changes from “and” to “or”. Case reasoning must preserve the correct logical relationship.
4. Sign-dependent algebra can force separate methods
Consider the expression x/|x| for x ≠ 0. If x > 0, then |x| = x and the expression equals 1. If x < 0, then |x| = −x and the expression equals −1.
The expression therefore defines a sign function: it returns 1 on the positive side and −1 on the negative side. There is no value at zero because the original denominator would be zero.
A student who replaces |x| with x everywhere would obtain 1 for negative inputs and lose the structure. A student who replaces it with −x everywhere would fail on positive inputs. The correct simplification depends on the case.
This pattern appears frequently with square roots too. Since √(x²) = |x|, simplifying √(x²) may require a sign condition on x before it becomes x or −x.
5. Piecewise functions make the cases explicit
Define f(x) = x + 2 for x < 1, and f(x) = 3x − 1 for x ≥ 1. To evaluate f(−4), use the first formula because −4 < 1. The result is −2. To evaluate f(5), use the second formula, giving 14.
At x = 1, only the second formula applies because the first case is strict. Thus f(1) = 2.
A graph of the function should represent the endpoint conditions. The first branch approaches the point (1, 3) but does not include it; the second branch includes (1, 2). The open and closed endpoint markers encode the same inequality information as the algebraic definition.
When solving f(x) = 5, solve inside each branch and then check the branch condition. First branch: x + 2 = 5 gives x = 3, but x = 3 does not satisfy x < 1, so reject it. Second branch: 3x − 1 = 5 gives x = 2, and x = 2 satisfies x ≥ 1. Therefore the solution is x = 2.
6. A formula can change at a threshold
In an invented pricing model, the first 100 units cost $0.20 each and each additional unit costs $0.12. Let q be the non-negative whole-number usage count.
For 0 ≤ q ≤ 100, cost C = 0.20q. For q > 100, the first 100 units cost $20, and the remainder q − 100 costs $0.12 each. Thus C = 20 + 0.12(q − 100).
The second formula should not be simplified to 0.12q and applied to all units. The threshold changes the marginal rule, not the historical price of the first 100 units.
If q = 130, the cost is 20 + 0.12(30) = $23.60. The two branches meet consistently at q = 100: both give $20 if the second expression is evaluated there, although the stated model assigns the boundary to the first case.
This is an invented arithmetic model only. Real tariffs may include fixed fees, taxes or other conditions. The teaching point is how a threshold creates a piecewise rule.
7. Geometry changes when a point changes position
Suppose A and B are fixed points on a straight line with AB = 10 cm, and P lies on the same line. Find PA + PB.
If P lies between A and B, then PA + PB = AB = 10.
If P lies outside the segment on the A side, then PB = PA + AB, so PA + PB = 2PA + 10, which is greater than 10.
If P lies outside on the B side, similarly PA = PB + AB and the sum is 2PB + 10.
The original condition “P lies on the same line” is not enough to force one formula for PA + PB. The point’s position determines the case.
Many diagram errors come from assuming the pictured configuration is the only possible one. A robust solution asks whether the conditions themselves force the point to be inside, outside, above, below or on a particular side.
8. Triangle problems can split by ambiguous angle information
Suppose a trigonometric problem gives sin θ = 0.6 for an angle θ inside a triangle-related configuration. The inverse sine gives an acute reference angle of about 36.87°. But sine has the same positive value in the second quadrant, giving 143.13°.
Whether both angles are geometrically possible depends on the remaining triangle information. If another given angle is 60°, the obtuse option would make the angle sum exceed 180° and is impossible. If another angle is 20°, both may remain possible until a side condition distinguishes them.
The calculator gives one principal angle. Geometry decides which branches survive.
This is an example of algebraic or trigonometric case generation followed by geometric filtering. A complete solution respects both layers.
9. Probability trees are structured case splits
A bag contains 3 red and 2 blue counters. Two counters are drawn without replacement. The probability of obtaining one red and one blue can occur in two different orders.
Case 1: red then blue. Probability = (3/5)(2/4) = 3/10.
Case 2: blue then red. Probability = (2/5)(3/4) = 3/10.
The cases are mutually exclusive because the first draw cannot be both red and blue. Together they cover every route producing one of each colour. Therefore the required probability is 3/10 + 3/10 = 3/5.
The tree structure is a case split over time. Multiply probabilities along a branch because every event on that branch must occur. Add probabilities across mutually exclusive successful branches because any one of them is sufficient.
10. Conditional information changes the branch weights
In the same bag, suppose we are told that the first counter drawn was red. The second-draw probability of blue is then 2/4 = 1/2, not 2/5.
The condition has moved us into one branch of the tree. Probabilities must now be calculated inside the reduced sample space for that branch.
Case splitting is therefore not only a presentation device. It changes the information state. Once a branch condition is known, possibilities inconsistent with that branch disappear.
This is the same mathematical idea seen in algebraic domains: entering a case gives extra conditions that govern what operations and values are valid.
11. Integer problems often split by remainder classes
Every integer leaves remainder 0, 1 or 2 when divided by 3. This creates three exhaustive cases.
To show that n³ − n is divisible by 3, one can use algebraic factorisation: n(n − 1)(n + 1), the product of three consecutive integers, so one factor must be divisible by 3.
Alternatively, use remainder cases. If n = 3k, the expression is divisible by 3. If n = 3k + 1, expand or reduce modulo 3 to get 1³ − 1 = 0 modulo 3. If n = 3k + 2, then 2³ − 2 = 6 = 0 modulo 3.
The modular proof is a structured case split over all possible remainder classes. It works because there are no other integer remainders modulo 3.
12. Optimisation can change formula at a capacity boundary
Suppose an invented transport system has vehicles that each hold at most 40 passengers. If n passengers must all travel, the number of vehicles needed is the smallest integer at least n/40.
For 1 ≤ n ≤ 40, one vehicle is needed. For 41 ≤ n ≤ 80, two. For 81 ≤ n ≤ 120, three. The decision function is stepwise rather than smoothly proportional.
A learner who writes n/40 as the final number of vehicles has treated a discrete output as continuous. A learner who always rounds normally can also fail: 61/40 = 1.525 rounds to 2, which works, but 41/40 = 1.025 would round to 1 even though two vehicles are necessary.
The cases arise from capacity thresholds. This connects directly to the Discrete and Continuous Models chapter.
13. Case splits should be minimal, not decorative
If one argument works for all allowed values, do not manufacture cases merely to make the solution look sophisticated. Case splitting is useful when a rule genuinely changes or when a many-to-one relationship creates branches.
For example, to prove that the sum of two consecutive integers is odd, one need not split into “first integer even” and “first integer odd”. Let the first integer be n; the next is n + 1. Their sum is 2n + 1, which is odd for every integer n. One algebraic representation handles all cases at once.
On the other hand, solving |x| = 4 does require two branches because the inverse structure itself is two-valued.
The best case split exposes the smallest distinction that changes the mathematics.
14. Avoid overlapping cases unless overlap is intentional
Suppose a piecewise rule says “use formula A for x ≤ 5 and formula B for x ≥ 5”. Both formulas claim x = 5. If they give the same value there, the function can still be well-defined, but the overlap should be deliberate. If they give different values, the definition is contradictory at x = 5.
A cleaner partition often uses x ≤ 5 and x > 5, or x < 5 and x ≥ 5.
Likewise, when counting cases, overlapping categories can cause double counting. “Students who take English” and “students who take Mathematics” overlap if some students take both. Adding the two group sizes without correcting the intersection is not a valid case sum.
Cases used for addition should usually be disjoint, or the overlap must be handled explicitly through inclusion-exclusion or another correction.
15. A complete case solution has a reunion step
Solving the branches is not the end. The learner must combine the valid outputs into one final answer.
For |2x − 5| = 7, the final set is x = −1 or x = 6. For a probability tree, add the successful branch probabilities. For a piecewise graph, draw all branches with correct endpoint markers. For an integer proof, state that the listed cases exhaust all integers.
The reunion step answers a quiet but important question: what has now been established about the original unsplit problem?
Without that step, the reader may see several local calculations without a clear global conclusion.
16. Independent practice
1. Solve |x − 3| = 5.
2. Solve |2x + 1| ≤ 7.
3. Define f(x) = 2x + 1 for x < 0 and x² + 1 for x ≥ 0. Find f(−3), f(0) and f(2).
4. Solve f(x) = 5 for the function in Question 3.
5. A bag contains 4 red and 3 blue counters. Two are drawn without replacement. Find the probability of drawing one of each colour.
6. Prove by cases that n² has the same parity as n for every integer n.
7. Solve x/|x| = −1.
8. A point P lies on the line through A and B, where AB = 12. What can be said about PA + PB if P lies between A and B? What if P lies outside the segment?
9. A pricing rule charges $5 for up to 20 units, then $0.30 for each additional unit. Write a piecewise cost rule for whole-number usage q ≥ 0.
10. Explain why the categories “multiples of 2” and “multiples of 3” are not disjoint cases over the integers.
17. Worked answers
1. x − 3 = 5 or x − 3 = −5, so x = 8 or x = −2.
2. −7 ≤ 2x + 1 ≤ 7. Subtract 1: −8 ≤ 2x ≤ 6. Divide by 2: −4 ≤ x ≤ 3.
3. f(−3) = 2(−3) + 1 = −5. Since 0 belongs to the second branch, f(0) = 1. Also f(2) = 5.
4. First branch: 2x + 1 = 5 gives x = 2, but this violates x < 0, so reject it. Second branch: x² + 1 = 5 gives x = ±2. Only x = 2 satisfies x ≥ 0. Therefore x = 2.
5. Red then blue gives (4/7)(3/6) = 2/7. Blue then red gives (3/7)(4/6) = 2/7. Add the mutually exclusive successful branches: 4/7.
6. If n is even, n = 2k and n² = 4k² = 2(2k²), even. If n is odd, n = 2k + 1 and n² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, odd. The two cases cover every integer.
7. For x > 0, x/|x| = 1. For x < 0, x/|x| = −1. The expression is undefined at x = 0. Therefore the solution is x < 0.
8. If P lies between A and B, PA + PB = 12. If P lies outside the segment, the sum is greater than 12 and depends on how far outside P lies.
9. C(q) = 5 for 0 ≤ q ≤ 20, and C(q) = 5 + 0.30(q − 20) for q > 20, under the stated invented model.
10. Multiples of 6 belong to both categories. For example, 12 is divisible by both 2 and 3. Adding the two case counts would double-count such values unless the overlap is corrected.
18. Continue through the operating manual
Use Reversible Steps, Lost Information and Extraneous Solutions when a transformation can create or remove candidates. Continue to Boundary Cases, Endpoints and Extremes for equality and threshold behaviour, and Discrete and Continuous Models for whole-number and measurement decisions.
Return to the BTT Mathematics Hub for Batch 02.