Singapore School Mathematics Operating Manual · Chapter 20
Different mathematical claims require different kinds of evidence.
To prove that something can happen, one valid construction may be enough. To prove that a proposed answer works, verification may be enough for that candidate. To prove that an answer is the only one, verification alone is not enough. To prove that something is impossible, testing many failed examples is not enough unless the search is genuinely exhaustive or a structural obstruction rules out every case.
This chapter develops a practical evidence ladder: construct → verify → establish coverage → conclude. It also develops the negative side: counterexample → contradiction → invariant obstruction → impossibility.
The goal is to help students match the strength of the evidence to the strength of the claim.
Existence · Verification · Uniqueness · Impossibility · Counterexamples · Geometric construction · Practice · Worked answers
1. One valid witness proves existence
Suppose the claim is: “There exists an even prime number.”
The number 2 is even and prime.
That one witness proves existence.
No classification of all prime numbers is required for the existence claim.
Existential statements are therefore mathematically economical: one admissible example can be decisive.
2. A witness must satisfy every stated condition
Suppose the claim is: “There exists a positive integer x satisfying x² = 9.”
x = −3 satisfies the equation but fails positivity.
x = 3 satisfies both conditions and is therefore a valid witness.
A construction that satisfies only part of the contract proves nothing about the full claim.
3. A numerical example can prove possibility but not a universal rule
To prove that two different positive integers can have the same square remainder modulo 5, one example pair may be enough.
1² ≡ 4² ≡ 1 (mod 5).
This proves possibility.
It does not prove that every pair has the same remainder, nor classify all such pairs.
The logical target determines what the witness establishes.
4. Construction can be algebraic
Suppose we are asked to find two consecutive integers whose sum is 41.
Let the smaller integer be n. Then n + (n + 1) = 41, so 2n + 1 = 41 and n = 20.
The pair (20, 21) is a constructed solution.
Substitution verifies the sum.
Because the equation is linear and has one solution for n, the construction is also unique.
5. Verification answers “does this candidate work?”
Suppose a student claims x = 4 solves 3x − 5 = 7.
Substitute: 3(4) − 5 = 12 − 5 = 7.
The candidate works.
Verification is direct and independent of the route used to obtain the candidate.
This makes substitution one of the strongest school-level checking methods for equations.
6. Verification does not explain completeness
If x = 4 solves x² = 16, substitution confirms it.
But x = −4 also works.
Verification certifies membership in the solution set. It does not prove the set contains no other members.
This is the distinction developed in Exhaustive Solutions, Completeness and Coverage.
7. Independent verification is stronger than repeating the same calculation
If a quadratic root is found by factorisation, substitute it into the original equation.
If an area is found by decomposition, compare with an alternative decomposition when practical.
If a probability is found by branch addition, check that complementary outcomes produce a total of 1.
A check is strongest when it uses a different relationship from the main solution route.
8. Units can verify the type of a result
Suppose a volume calculation ends in cm².
The numerical arithmetic may be correct, but the output type is wrong.
Unit checking is a verification certificate for dimensional consistency.
It cannot prove the entire formula correct, but it can reject type-invalid answers immediately.
9. Boundary checks verify qualitative behaviour
Suppose a formula claims to give the number of handshakes among n people: n(n − 1)/2.
At n = 1, it gives 0. At n = 2, it gives 1.
These small cases support consistency.
They do not prove the formula globally, but they can expose obvious off-by-one errors.
10. Graphical verification can support algebra
Suppose an equation’s algebra gives two roots.
A graph can show whether two intersections are plausible and approximately where they lie.
The graph is useful as a check, especially for sign and number of roots.
But an approximate graph should not replace exact algebra when exact roots or proof are required.
11. Uniqueness requires ruling out every second candidate
To prove “there exists exactly one x satisfying…” we need two parts:
Existence: at least one solution works.
Uniqueness: no two distinct solutions can both work.
These are separate obligations.
A candidate check proves existence for that candidate. It does not prove uniqueness.
12. Linear equations often give uniqueness automatically
For ax + b = c with a ≠ 0, rearrangement gives x = (c − b)/a.
Because division by non-zero a produces one real number, there is exactly one solution.
The algebraic structure establishes both existence and uniqueness.
If a = 0, the classification changes: there may be no solution or infinitely many.
13. Strict monotonicity can prove uniqueness
If a function is strictly increasing, it cannot take the same output value at two distinct inputs.
Therefore f(x) = c has at most one solution.
If we separately find one solution, it is unique.
This is a powerful way to upgrade a constructed witness into a unique solution.
14. Geometry may require construction plus rigidity
Suppose two points A and B are fixed and we seek points equidistant from both.
The perpendicular bisector of AB is the full locus.
Constructing one point on the perpendicular bisector proves one equidistant point exists.
Proving the entire locus requires showing both directions: every point on the perpendicular bisector is equidistant, and every equidistant point lies on it.
Construction and classification are different evidence levels.
15. Failed attempts do not usually prove impossibility
Suppose a student tries five pairs of positive integers and cannot find a pair whose sum is 10 and product is 30.
Five failures do not prove impossibility unless all feasible pairs have been tested.
A structural argument is better.
Let x + y = 10 and xy = 30. Then x and y would be roots of t² − 10t + 30 = 0.
The discriminant is 100 − 120 = −20, so there are no real roots. Therefore no real pair, and hence no positive-integer pair, satisfies both.
16. Contradiction can prove impossibility globally
Suppose an integer is assumed to be both even and odd.
Then n = 2a and n = 2b + 1 for integers a and b.
Equating gives 2a = 2b + 1, so 2(a − b) = 1.
The left side is even while the right side is odd, impossible.
The contradiction rules out the assumed state completely.
17. Invariants can prove impossibility without enumerating moves
Suppose a token begins on a number congruent to 1 modulo 4 and every legal move adds a multiple of 4.
The remainder modulo 4 stays 1.
A target congruent to 2 modulo 4 is unreachable.
No search through move sequences is required.
The invariant separates the entire reachable set from the target.
18. Parity can be an impossibility certificate
Suppose three odd integers are claimed to sum to an even number.
Odd + odd = even, and even + odd = odd.
Therefore the sum of three odd integers is always odd.
The claimed even total is impossible.
Parity gives a global obstruction.
19. Bounds can prove impossibility
Suppose x and y are both positive and x + y = 6.
The maximum possible product occurs at x = y = 3, giving xy = 9.
Therefore a requirement xy = 12 is impossible over positive reals under the sum constraint.
A bound can rule out an entire region without checking every pair.
20. Domain restrictions can prove impossibility instantly
log₂(x − 1) is defined over the reals only when x > 1.
If another condition requires x ≤ 0, the feasible set is empty.
No logarithmic calculation is needed.
Conflicting domain constraints provide an impossibility certificate.
21. The discriminant is an impossibility certificate for real quadratic roots
For x² + 4x + 8 = 0, the discriminant is 16 − 32 = −16.
Therefore no real roots exist.
The discriminant does more than fail to find roots. It proves the entire real solution set is empty.
22. One counterexample disproves a universal claim
Claim: every even number is divisible by 4.
Counterexample: 6 is even but not divisible by 4.
The universal claim is false.
No second counterexample is needed.
A counterexample is a witness for the existence of failure.
23. The counterexample must satisfy the premise
Claim: every prime greater than 2 is odd.
Using 9 as a supposed counterexample fails because 9 is not prime.
A valid counterexample must enter the statement through its premise and then break the conclusion.
This is why information-status checking matters before refutation.
24. Counterexamples disprove, but examples do not usually prove universals
Testing 3, 5, 7 and 11 supports the claim that primes greater than 2 are odd.
But the proof comes from the fact that every even integer greater than 2 has factor 2 and is therefore composite.
The logical asymmetry is important:
One counterexample kills a universal claim.
Many confirming examples do not automatically prove it.
25. A contradiction may expose a bad assumption rather than a bad theorem
Suppose a long derivation reaches 0 = 1.
Something earlier must be invalid under ordinary arithmetic.
Perhaps there was division by zero, an invalid square-root step, a false assumption or an inconsistent set of givens.
The contradiction is evidence that the current premise chain cannot all be true together.
It does not identify the exact bad line automatically.
26. Construction should preserve all constraints
Suppose a problem asks for a rectangle with perimeter 20 and integer side lengths.
The pair 3 and 7 is a valid construction because 2(3 + 7) = 20 and both sides are positive integers.
The pair 2.5 and 7.5 satisfies the perimeter but fails the integer-side condition.
A witness is only valid if it lies in the complete feasible set.
27. Geometric construction is an existence method
Using ruler-and-compass style reasoning, drawing the perpendicular bisector of segment AB constructs every point equidistant from A and B.
Drawing a circle of radius r centred at O constructs the locus of points at distance r from O.
Intersections of loci can construct points satisfying multiple conditions simultaneously.
The geometry converts conditions into objects whose intersection acts as a witness set.
28. Two loci can show zero, one or multiple constructions
Two circles can intersect in two points, be tangent at one point, or fail to intersect.
These three cases correspond to multiple constructions, a unique construction or impossibility.
Distance between centres and radii determine which case occurs.
Construction geometry therefore naturally contains existence and uniqueness questions.
29. A “show that” task needs a derivation, not only a witness
If a question says “show that the area is 20x − x²”, substituting one value of x and obtaining the matching area does not prove the formula.
The task is universal over the allowed variable range.
A derivation from geometric relationships is required.
One numerical witness proves only one case.
30. A “find an example” task needs much less
If the task asks “give one pair of integers with product 12 and sum 7”, the pair (3, 4) is enough.
There is no need to classify every factor pair unless the question says “find all”.
Efficient Mathematics matches effort to proof obligation.
31. A certificate can be shorter than the search that found it
A student may discover a factorisation after trial and error.
The final proof can simply present the factorisation and conclusion.
A computer may search many possibilities and return one witness that is easy to verify.
Discovery cost and verification cost can be very different.
This distinction is useful in both school problem solving and later computational Mathematics.
32. Some certificates prove non-existence efficiently
A negative discriminant certifies no real quadratic roots.
A parity mismatch certifies an impossible integer condition.
An invariant mismatch certifies an unreachable state.
A violated triangle inequality certifies an impossible non-degenerate triangle.
These are compact reasons that replace exhaustive failed search.
33. Approximate witnesses need tolerance-aware verification
Suppose a numerical solver returns x ≈ 1.41421356 for x² = 2.
Exact substitution cannot hold in decimal form because the approximation is finite.
Instead, check the residual |x² − 2| and whether it meets the requested tolerance.
Numerical verification must match approximate output type.
The Iteration and Convergence chapter develops this further.
34. Verification should return to the original, not only the transformed problem
If a square-root equation was squared, verify candidates in the original equation.
If denominators were cleared, check excluded values.
If a trigonometric equation was transformed using an identity, check the stated interval.
If a model was simplified, return to the original context and units.
The original problem is the final authority.
35. A complete mathematical conclusion should state the evidence level
Useful final forms include:
“Therefore x = 5 is a solution.”
“Therefore the only real solution is x = 5.”
“Therefore at least one such triangle exists.”
“Therefore no positive-integer solution exists.”
“Therefore the claim is false; 6 is a counterexample.”
The wording should not claim more than the evidence supports.
36. A practical evidence audit
Ask:
Is the task asking for one example, all examples, uniqueness, proof or impossibility? If I constructed a candidate, have I verified every condition? If I verified one, have I shown there are no others? If I claim impossibility, what global obstruction rules out every case? If I use a counterexample, does it satisfy the premise? If I use approximation, what tolerance certifies it?
This audit links proof, solving, modelling and checking into one operating discipline.
37. Independent practice
1. Give one witness proving that an even prime exists.
2. Verify that x = 5 solves 2x + 3 = 13.
3. Explain why that verification is enough to prove the solution is unique for this linear equation.
4. Give a counterexample to “every multiple of 3 is odd”.
5. Prove there is no real solution to x² + 2x + 5 = 0.
6. Prove three odd integers cannot sum to an even integer.
7. Construct positive integers whose sum is 9 and product is 20.
8. A token starts at a number congruent to 2 modulo 5 and every move adds 10. Prove it cannot reach a multiple of 5.
9. Explain why five failed guesses do not normally prove impossibility.
10. A numerical root estimate satisfies |f(x)| < 10⁻⁶. What has been verified?
11. Why should candidates from a squared equation be checked in the original equation?
12. What two obligations are required to prove “exactly one solution exists”?
38. Worked answers
1. 2. It is even and prime.
2. 2(5) + 3 = 13, so x = 5 is a solution.
3. Rearranging 2x + 3 = 13 gives 2x = 10 and x = 5. Division by non-zero 2 is reversible, so there can be no second real solution.
4. 6 is a multiple of 3 and is even, not odd.
5. Discriminant = 2² − 4(1)(5) = 4 − 20 = −16 < 0, so no real roots exist.
6. Odd + odd = even, and even + odd = odd. Therefore the sum of three odd integers is always odd.
7. 4 and 5: 4 + 5 = 9 and 4 × 5 = 20.
8. Adding 10 changes the number by 0 modulo 5, so remainder 2 is invariant. A multiple of 5 has remainder 0, so it is unreachable.
9. The untested search space may still contain a valid case. Impossibility needs exhaustive coverage or a structural obstruction.
10. The candidate has a small residual in the original equation. Whether that guarantees small input error depends on the problem’s sensitivity and requested tolerance.
11. Squaring can create extraneous candidates by losing sign information. Original substitution restores the correct problem conditions.
12. Existence of at least one solution and uniqueness—proof that no distinct second solution can exist.
39. Batch 05 as one control system
Mathematical Type Checking asks whether the operations and outputs are legal for the objects involved.
Exhaustive Solutions and Completeness asks whether every admissible case has been covered exactly once.
Iteration, Convergence and Stopping asks whether an approximate process has become accurate enough to trust.
This chapter asks what evidence is sufficient to move from candidate to conclusion—or from failed possibility to impossibility.
Return to the BTT Mathematics Hub for Batch 05.