Singapore School Mathematics Operating Manual · Chapter 18
Finding a valid answer is not always the same as solving the problem completely.
If x² = 25, x = 5 is a valid answer—but it is not the complete real solution set because x = −5 also works. If sin θ = 1/2 on 0° ≤ θ ≤ 360°, 30° is valid—but 150° is missing. If two counters are drawn and the target event is “one red and one blue”, the branch red-then-blue is valid—but blue-then-red must also be counted.
This chapter develops solution completeness: the discipline of showing that every admissible case has been found, every listed case is valid, and no case has been counted twice.
The key question is not only “does this answer work?” but also “why are there no other answers?”
Existence versus completeness · Branches · Periodic solutions · Counting without duplication · Proof coverage · Practice · Worked answers
1. Existence proves at least one; completeness finds them all
Suppose x² = 9.
Substituting x = 3 verifies that one solution exists.
But the equation factors as (x − 3)(x + 3) = 0, so x = 3 or x = −3.
A complete solution must include both.
Existence asks: “is there at least one?” Completeness asks: “what is the entire admissible set?”
2. Verification and exhaustion answer different questions
Checking x = 3 in x² = 9 shows x = 3 works.
Checking x = −3 shows that value works too.
But even verifying several candidates does not by itself prove that no others exist.
The algebraic structure x² − 9 = (x − 3)(x + 3) provides the exhaustion argument because a product is zero only when at least one factor is zero.
A complete solution therefore combines validation with a reason that the search space has been fully covered.
3. A complete answer may be a set, interval, family or classification
For x² = 9, the answer is the finite set {−3, 3}.
For x > 2, the solution is an interval.
For sin θ = 0 over all real θ, the complete family is θ = nπ for integer n, if radians are used.
For a parameter question, the answer may be a classification such as k < 9, k = 9, k > 9.
Completeness depends on representing the whole set in a form suited to its structure.
4. Many-to-one operations create branch obligations
When |x| = 7, the absolute value has hidden sign branches.
The complete solution is x = 7 or x = −7.
When x² = 49, the square hides the same two-sign structure.
When sin θ = 1/2, periodic and symmetry branches appear.
A solver who uses an inverse operation must ask whether that inverse returns one representative or every original input.
5. Taking a square root is a common completeness trap
From x² = 16, writing x = √16 = 4 gives only the principal square root.
The equation asks for every x whose square is 16, so x = ±4.
The symbol √16 names one non-negative number. It is not shorthand for both equation solutions.
This distinction prevents systematic missing branches.
6. Dividing by a variable expression can delete a case
Solve x(x − 5) = 0.
If we divide both sides by x, we obtain x − 5 = 0 and lose x = 0.
The correct zero-product split is x = 0 or x = 5.
Whenever cancellation assumes a factor is non-zero, the zero case must be handled separately before division.
The companion chapter on Reversible Steps and Extraneous Solutions explains why this information can disappear.
7. Squaring can create extra cases while roots can delete cases
These two errors point in opposite directions.
Squaring may enlarge the candidate set by forgetting signs.
Taking only a principal root may shrink the candidate set by ignoring a negative branch.
A complete solver therefore watches both sides of the problem: missing candidates and extraneous candidates.
The final solution set should equal the original problem’s admissible set exactly—not merely contain it or sit inside it.
8. Absolute-value inequalities require correct branch connectors
|x| < 5 means −5 < x < 5.
|x| > 5 means x < −5 or x > 5.
The first solution is a connected interval. The second is two disjoint branches.
A common incompleteness error is to solve only the positive side.
Another is to use “and” where “or” is required, accidentally demanding a number be less than −5 and greater than 5 simultaneously.
9. Case splits must be exhaustive
For integers, “even or odd” is exhaustive.
For the sign of a real number, the exhaustive split is negative, zero or positive.
Using only “negative or positive” omits zero.
For a discriminant, the exhaustive classification is positive, zero or negative.
Completeness begins with partitioning the universe into cases that cover every possibility.
10. Cases should also be disjoint when they will be added
Suppose we count integers from 1 to 30 that are divisible by 2 or 3.
Counting multiples of 2 and multiples of 3 separately and adding gives duplicates for multiples of 6.
There are 15 multiples of 2 and 10 multiples of 3, but 5 multiples of 6 belong to both groups.
The correct count is 15 + 10 − 5 = 20.
Coverage without duplicate control can overcount just as missing branches undercount.
11. Inverse trigonometric functions return principal values, not complete periodic solutions
sin θ = 1/2.
A calculator may return θ = 30°.
On 0° ≤ θ ≤ 360°, another solution is 150°.
Over all real angles, the complete family repeats every 360°.
The inverse key helps locate a reference angle. The domain and periodicity determine the complete solution set.
12. Domain limits tell us when the search can stop
If the problem asks for 0° ≤ θ ≤ 360°, there is no need to list 390° because it lies outside the domain.
If the domain is all real θ, stopping after two angles is incomplete.
Completeness is always relative to the stated universe.
A finite interval can turn an infinite periodic family into a finite list.
13. Endpoints can be solutions too
Suppose cos θ = 1 on 0° ≤ θ ≤ 360°.
Both 0° and 360° satisfy the equation.
Because the interval includes both endpoints, both belong to the listed solution set even though they represent coterminal directions.
If the interval were 0° ≤ θ < 360°, only 0° would be listed.
Endpoint completeness depends on the exact interval notation.
14. Quadratic equations have at most two real roots—but may have fewer
A non-zero quadratic ax² + bx + c = 0 has at most two real solutions.
The discriminant identifies whether the complete real solution set contains two distinct roots, one repeated root or no real roots.
Thus the discriminant is not merely a calculation tool. It gives a completeness classification for the entire real root set.
15. A repeated root should not be double-counted as two distinct solutions
(x − 3)² = 0 has the root x = 3 with multiplicity 2.
As a set of real solutions, the answer is simply {3}.
Multiplicity matters in polynomial structure, but the same numerical solution should not be listed twice as though it were two different x-values.
Completeness includes duplicate control.
16. Rational equations require both coverage and exclusions
Solve (x + 1)/(x − 2) = 0.
A rational expression is zero when its numerator is zero and denominator is non-zero.
Thus x + 1 = 0 gives x = −1, and x = −1 is allowed because the denominator is −3.
The excluded value x = 2 is not a solution and does not need to be “added back”.
A complete answer accounts for numerator roots and denominator exclusions.
17. Logarithmic equations can generate polynomial roots that need filtering
Suppose log₂x + log₂(x − 2) = 3.
The domain requires x > 2.
Combining gives x(x − 2) = 8, hence x² − 2x − 8 = 0.
The polynomial roots are x = 4 and x = −2.
The complete logarithmic solution set is only {4} because −2 is outside the original domain.
Completeness means every original solution is kept and every transformed-only candidate is removed.
18. Probability events often have multiple successful routes
A bag contains 3 red and 2 blue counters. Two are drawn without replacement. Find the probability of one red and one blue.
The successful branches are red-then-blue and blue-then-red.
P(RB) = (3/5)(2/4) = 3/10.
P(BR) = (2/5)(3/4) = 3/10.
Total = 3/5.
Counting only one order gives a valid branch probability but an incomplete event probability.
19. Counting problems require a coverage plan before arithmetic
When counting arrangements, ask:
What constitutes one distinct outcome? What choices generate every outcome? Are the branches disjoint? Does any outcome appear in more than one branch? Are symmetries making apparently different constructions identical?
A counting formula is trustworthy only if its generation scheme has correct coverage and multiplicity.
20. Permutations can overcount indistinguishable objects
The letters in LEVEL contain five positions but repeated letters E and L.
5! counts all label-distinct arrangements as though the two Es and two Ls were distinguishable.
Each visible arrangement is counted 2!2! times.
The number of distinct arrangements is 5!/(2!2!) = 30.
Completeness in counting includes not only finding all outcomes but counting each exactly once.
21. Symmetry can create duplicate geometric configurations
Suppose a construction produces a triangle and its mirror image.
If the problem treats mirror images as distinct placements, count both.
If the problem asks only for shape up to reflection, they may represent one equivalence class.
The counting rule depends on the definition of “distinct”.
A complete answer must match the problem’s identity convention.
22. Sequence rules may require every branch of a recurrence
If a recurrence defines different rules for even and odd n, checking only one parity branch is incomplete.
Every admissible index type must be covered.
Similarly, a piecewise function solution must solve inside every branch and then check each branch condition.
The local rules combine into one global solution set.
23. Piecewise equations need branch validation
Suppose f(x) = x + 2 for x < 0 and x² for x ≥ 0. Solve f(x) = 4.
First branch: x + 2 = 4 gives x = 2, but x = 2 violates x < 0, so reject it.
Second branch: x² = 4 gives x = ±2, but only x = 2 satisfies x ≥ 0.
The complete solution set is {2}.
Solving each formula without checking its branch domain would create false coverage.
24. Geometry can have ambiguous configurations
Given side-angle information in a triangle, more than one geometric configuration can sometimes satisfy the data.
A calculator may produce one angle, while the supplementary angle may also be feasible.
The remaining side and angle constraints determine which configurations survive.
A complete geometric answer should account for every configuration permitted by the conditions, not merely the one matching the sketch.
25. Diagrams can hide missing configurations
A point drawn inside a segment may suggest one case, but the wording may allow the point outside.
A line drawn crossing a circle twice may suggest two intersections, but a parameter could produce tangency or no intersection.
The diagram helps visualisation but should not silently reduce the case space.
The actual conditions own the geometry.
26. A proof by cases must cover the whole universe
To prove a statement for all integers using parity, even and odd cases are sufficient because every integer belongs to exactly one.
To prove a statement for all real numbers using sign, negative and non-negative cases cover the reals, or negative, zero and positive cases can be used if zero behaves differently.
A proof is incomplete if one admissible class never enters the argument.
27. Proof by contradiction must exclude every alternative under the assumption
Suppose we assume the negation of a target statement and derive a contradiction.
The contradiction must follow from the assumption together with accepted premises.
If the argument reaches only “this one attempted construction fails”, the proof may show that one route is impossible without proving that every alternative route is impossible.
Global impossibility requires exhaustive logical coverage.
28. Proving existence is easier than proving uniqueness
To prove an equation has at least one solution, exhibit or establish one.
To prove it has exactly one, also show that no second distinct solution is possible.
For a strictly increasing function, for example, the equation f(x) = c can have at most one solution because the function never repeats an output.
If existence is separately shown, uniqueness follows.
This is one clean route from local construction to global completeness.
29. Data sufficiency and completeness are related but distinct
Data sufficiency asks whether the information determines the target.
Completeness asks whether the final answer includes every admissible target value.
A problem can be fully specified but still solved incompletely by the student.
Conversely, a student may correctly report “multiple answers are possible” when the data are insufficient for uniqueness.
The Data Sufficiency chapter handles the information side of that distinction.
30. A completeness certificate should explain why the search is finished
Good final statements often contain a small completeness reason:
“These are the two roots of the quadratic.”
“These are all angles in the stated interval.”
“The two parity cases exhaust the integers.”
“The successful probability branches are mutually exclusive and exhaustive for the target event.”
“The only stationary points and endpoints have been checked.”
This need not be verbose. It tells the reader why no hidden case remains.
31. Search limits should be mathematical, not psychological
A student should not stop because “I found two answers and usually there are two”.
Stop because the structure proves there can be no more.
Examples: degree bounds for polynomials, interval bounds for trigonometry, case partitions, factorisation, monotonicity, exhaustive tree branches or finite enumerations.
Completeness should come from the mathematics.
32. Exhaustive search can be valid when the feasible set is genuinely finite
Suppose x and y are positive integers with x + y = 8 and x < y.
The feasible pairs are (1,7), (2,6), (3,5). The pair (4,4) fails x < y, and larger x values would reverse the order.
A finite enumeration can be a complete method if the bounds are justified.
“Guess-and-check” becomes rigorous when the search space is finite, coverage is systematic and every candidate is tested.
33. Duplicate answers can arise from different methods
Two branches may lead to the same numerical result.
For a solution set, list the value once.
For a probability calculation, if the branches are genuinely different events, their probabilities may still need to be added even if they end at the same numerical outcome label.
The distinction is between duplicate values and distinct paths.
34. A practical completeness audit
Before finalising, ask:
What is the full domain? Which branches exist? Did any transformation delete a case? Did any transformation create extra candidates? Are all interval endpoints handled? Are periodic solutions exhausted? Are probability routes mutually exclusive and complete? Did I double-count equivalent outcomes? Why can there be no additional solution?
This audit is especially valuable whenever the question says “find all”, “list all”, “hence solve”, “number of ways”, “all possible values” or “show that the only”.
35. Independent practice
1. Solve x² = 49 over the reals.
2. Solve |2x − 1| = 5.
3. Solve x(x − 4) = 0 without dividing away a case.
4. Solve sin θ = 1/2 for 0° ≤ θ ≤ 360°.
5. Solve cos θ = 1 for 0° ≤ θ < 360°.
6. A bag contains 4 red and 3 blue counters. Two are drawn without replacement. Find the probability of exactly one red.
7. Count integers from 1 to 24 divisible by 2 or 3.
8. How many distinct arrangements does the word MAMA have?
9. Solve f(x) = 4 if f(x) = x + 1 for x < 0 and f(x) = x² for x ≥ 0.
10. Explain why verifying x = 2 in x² = 4 does not prove the solution set is complete.
11. A proof for all integers covers only the even case. What is missing?
12. State one reason a finite exhaustive search can be rigorous.
36. Worked answers
1. x = 7 or x = −7.
2. 2x − 1 = 5 or 2x − 1 = −5, giving x = 3 or x = −2.
3. x = 0 or x − 4 = 0, so x = 0 or 4.
4. θ = 30° or 150°.
5. θ = 0°. The endpoint 360° is excluded.
6. Successful branches are RB and BR. P(RB) = (4/7)(3/6) = 2/7. P(BR) = (3/7)(4/6) = 2/7. Total = 4/7.
7. Multiples of 2: 12. Multiples of 3: 8. Multiples of 6 counted twice: 4. Total = 12 + 8 − 4 = 16.
8. 4!/(2!2!) = 6.
9. First branch: x + 1 = 4 gives x = 3, but x < 0 fails. Second branch: x² = 4 gives x = ±2, but x ≥ 0 keeps only x = 2. Complete solution: x = 2.
10. It proves one value works but does not rule out x = −2.
11. The odd-integer case. Even and odd together exhaust all integers.
12. If justified bounds create a finite candidate set and every candidate is systematically tested exactly once, the search is exhaustive.
37. Continue through the School Mathematics manual
Use Mathematical Type Checking to confirm candidates have the right object type, Iteration, Convergence and Stopping when the candidate set is reached numerically rather than exactly, and Construction, Verification and Impossibility for the evidence needed to prove existence or rule it out.
Return to the BTT Mathematics Hub for Batch 05.