Deduction reaches a conclusion that must follow from stated information. A mathematical statement is checked against its precise meaning, its permitted objects and the conditions under which it is claimed. A convincing example can suggest a rule, but it does not by itself establish an unrestricted rule. A single valid counterexample can defeat an “always” claim.
Consider the statement “Every multiple of six is even.” A collection arranged in groups of six can also be arranged in pairs, so the statement holds for whole-number multiples. Now reverse it: “Every even number is a multiple of six.” Eight disproves that claim. Reversing the sentence has changed the relationship, even though the same words and number appear.
This guide is Primary Mathematics enrichment, with optional bridges to later formal reasoning. Begin with sorting numbers, matching clues and explaining counterexamples. The tables involving “if” and “only if” are for learners ready to discuss them, not a requirement that every Primary child learn formal logic notation. Refer to the MOE Primary curriculum page and the learner’s school programme for required scope.
The purpose is not to train children to sound argumentative. It is to help them state what the evidence establishes, notice missing conditions and explain why an answer follows. The fictional clue puzzles concern precisely defined objects; a false puzzle statement is not a basis for judging a real person’s character or motives.
Statements and domains · And, or and not · If and only if · Examples and proof · Clue puzzles · Incomplete information · 24 questions · Worked answers · Teaching
1. Decide what is being claimed and which objects are allowed
“Eighteen is divisible by three” is a statement whose truth can be checked: eighteen equals three times six. “Find a number divisible by three” is an instruction, not a true-or-false statement. “This is a nice number” expresses a preference unless “nice” has been mathematically defined. These sentences do different jobs.
A variable statement needs either a specified value or a description of the values it ranges over. “n is even” cannot be assigned one truth value when n is unspecified. “For every positive whole number n, twice n is even” is a universal claim about a named domain. The domain is the collection of objects the statement is about.
Why the domain matters
The claim “multiplying a positive whole number by two makes it larger” holds because two copies exceed one copy when the original amount is positive. Allow zero and the statement fails at zero: doubling zero leaves zero. Allow negative numbers and “larger” requires further care. A boundary case can expose a missing condition without making the original restricted statement wrong.
Primary examples can stay within non-negative whole numbers, positive whole numbers, or familiar fractions. State which set is being used. Do not silently switch from positive numbers to all numbers midway through an explanation, or mark an excluded example as a counterexample.
A condition is not merely a descriptive extra
“Two different positive whole numbers” excludes equality and zero. “At most twelve” includes twelve; “less than twelve” does not. “Exactly two blue counters” differs from “at least two.” Each phrase changes the permitted cases and therefore the conclusion that may follow.
A useful first step is to rewrite a long question as a few short statements without discarding those qualifiers. This connects to the existing Simultaneous Conditions guide: every condition remains active until it has been satisfied or shown redundant.
Definitions supply a starting agreement
In this guide, a rectangle is a quadrilateral with four right angles, and a square is a rectangle with four equal sides. Under these definitions, every square is a rectangle. The statement is not settled by whether a square is drawn tilted or whether a worksheet uses separate labelled boxes for the two shape names.
Different classroom conventions sometimes package categories differently. When a naming issue matters, state the definition rather than arguing from appearance alone. Mathematical reasoning needs an agreed meaning before it can establish a consequence.
2. “And,” “or” and “not” change the set of accepted cases
Let P mean “the number is even” and Q mean “the number is a multiple of three.” The statement P and Q accepts only numbers satisfying both. Fourteen satisfies P but fails Q, so it fails the combined statement. Twelve satisfies both and is accepted.
For P or Q, this guide uses the mathematical inclusive or: P, Q or both may hold. Fourteen is accepted because it is even; nine is accepted because it is a multiple of three; twelve is accepted because it satisfies both. Five is rejected because neither property holds.
| P | Q | P and Q | P or Q, including both |
|---|---|---|---|
| True | True | True | True |
| True | False | False | True |
| False | True | False | True |
| False | False | False | False |
This four-case table is complete because each of two true-or-false statements has two possible truth values. It does not say those cases are equally likely. Logic classifies acceptance; probability needs an additional model before it assigns chances.
Exactly one is a different condition
“P or Q, but not both” accepts the middle two rows only. In the number example, fourteen and nine pass, while twelve fails because both properties hold. Say “exactly one” when that is intended. Everyday uses of “or” can be ambiguous, so a mathematical question should make the convention clear where it affects the answer.
Suppose a child must choose a red card or a triangle card. Does a red triangle qualify? Under inclusive or, yes. Under “exactly one of the properties,” no. The picture has not changed; the condition has.
Negation must cover every way the original can fail
The negation of “n is greater than ten” is “n is at most ten.” It includes ten. “n is less than ten” would omit the boundary and is therefore too narrow. The negation of “exactly one counter is red” allows zero red counters and also two or more.
For a finite non-empty row of counters, “not every counter is blue” means at least one is not blue. It does not mean every counter is non-blue. One red counter among many blue counters is enough to make “all blue” false.
Negating a combined condition
If a candidate must be even and greater than ten, it fails when it is not even or is not greater than ten. One failed part is enough. It need not fail both. Thus twelve passes, fourteen passes, nine fails and eight fails, but they fail for different reasons.
Conversely, to fail “even or greater than ten,” a number must fail both parts: it must be odd and at most ten. These relationships can be checked by the four truth cases above. The names of formal laws are optional; the complete case comparison supplies the explanation.
3. An if-then rule has a direction
Suppose a fictional collection obeys the rule: if a card has a red sticker, then it is a triangle card. A red sticker is enough to conclude triangle within that collection. The rule excludes a red-sticker card that is not a triangle.
It does not say that every triangle has a red sticker. Some triangles might have blue stickers or no sticker. Seeing a triangle therefore does not, by itself, permit the conclusion that the sticker is red. This mistaken reversal is common because the sentence contains a genuine connection but only in one direction.
Four cases for a rule
Let P mean red sticker and Q mean triangle. The rule P implies Q is contradicted only by a card for which P is true and Q is false. A triangle without a red sticker does not break the rule because the rule did not promise that triangles must be red.
| Card case | Does it contradict “if red, then triangle”? |
|---|---|
| Red triangle | No; it follows the rule |
| Red non-triangle | Yes; it is the forbidden case |
| Non-red triangle | No; the rule allows it |
| Non-red non-triangle | No; the rule allows it |
This is a rule-checking table, not a claim that a non-red card causes anything. Logical implication in these puzzles describes which combinations are permitted. It should not be confused with a scientific claim about cause and effect.
A justified negative conclusion
If the rule holds and a card is not a triangle, it cannot have a red sticker. A red sticker would force triangle and contradict the observed non-triangle condition. This reverse-negative reasoning is valid.
In contrast, a card without a red sticker may still be a triangle. Absence of the sufficient condition does not show absence of the consequence. Test the proposed deduction against a blue triangle: it satisfies the original rule and defeats the unjustified conclusion.
“Only if” identifies a necessary condition
“A card may enter the tray only if it is red” means entering requires redness. Every admitted card must be red. It does not promise that every red card is admitted, because another rule might also be required. For example, a red card might additionally need a printed star.
“If a card is red, it may enter” gives the opposite direction: redness is sufficient for permission under that stated rule. Read the two sentences slowly and test them with a red card that is not admitted. The first sentence can permit that situation; the second cannot if permission and admission are identified in the puzzle.
To avoid mixing actual action with eligibility, the practice questions use “eligible” as a defined status. Being allowed to enter is not the same real-world fact as actually entering. An eligible person might simply choose not to participate.
“If and only if” supplies both directions
A box is marked ready if and only if it contains exactly four counters. Ready therefore guarantees four counters, and four counters guarantees ready. Neither a ready box with five nor an unready box with exactly four fits the rule.
The phrase is optional enrichment vocabulary. The underlying idea can be expressed in two plain sentences. Introducing the notation before the learner understands the two directions can add difficulty without adding insight.
4. Examples suggest; explanations establish; counterexamples refute
Suppose several tested multiples of six are even: six, twelve, eighteen and twenty-four. These support a pattern but do not, as a list alone, cover every multiple. A general explanation says that each group of six contains three pairs. Any whole number of such groups can therefore be paired completely, proving evenness.
The proof refers to the structure shared by every permitted case. It does not depend on having checked a very large final number. A drawing of repeated six-counter groups can express the same argument before letters such as 6k are introduced.
A counterexample has two jobs
To refute “every even positive whole number is divisible by six,” choose eight. It meets the premise because it is even, but fails the conclusion because division by six leaves remainder two. Both parts are necessary. Choosing nine would not be a valid counterexample because nine is not in the stated even-number domain.
A counterexample does not show that the claim is false in every case. Twelve is even and divisible by six. It shows only that the word “every” is too strong. The statement becomes “some even positive whole numbers are divisible by six, and some are not.”
Always, sometimes and never
A statement about a named domain is always true when every permitted case satisfies it, never true when none do, and sometimes true when at least one does and at least one does not. To establish “sometimes,” provide one working case and one failing case. To establish “always,” explain why no permitted case can fail.
NRICH’s Always, Sometimes or Never? Number provides further teacher examples of this kind of reasoning. Our statements below are independently constructed and use explicit domains so that zero, equality and boundary cases are not silently excluded.
Example A: Two neighbouring positive whole numbers
The product of two consecutive positive whole numbers is even. One of the neighbours must be even: an odd number is followed by an even number, and an even number is followed by an odd. Multiplying by an even factor gives a product that can be paired. This covers every permitted pair.
The consecutive condition matters. Two arbitrary odd numbers can have an odd product, such as three times five. The correct generalisation retains the condition that supports the argument.
Example B: Sum compared with product
For two positive whole numbers, “their sum equals their product” is sometimes true. Two and two give four both ways. One and three give a sum of four but a product of three, so the statement is not always true.
The two examples are enough for the requested sometimes classification. They do not list every pair giving equality. If the task asked for all such pairs, further reasoning would be needed. Match the strength of the answer to the question’s actual demand.
Example C: Repair a claim rather than discard it
For a non-negative whole number n, the statement n squared is greater than n fails at zero and one. It holds for every integer n at least two: n copies of a positive amount n are more than one copy when there are at least two copies. The repaired condition is n at least two.
A useful correction identifies precisely where the original rule works. “The rule is wrong” is less informative than naming the excluded boundary cases and supplying the valid restricted version.
5. Solve a clue puzzle by testing complete possibilities
A star is in exactly one of boxes A, B or C. Three printed statements read: Statement 1, “The star is in A”; Statement 2, “The star is in A”; Statement 3, “The star is in B.” Exactly one statement is true. Where is the star?
Test the three possible locations. In A, the first two statements are true and the third false, giving two true statements. In B, only the third is true. In C, all three are false. Only location B satisfies the condition of exactly one true statement.
| Star location | Statement 1 | Statement 2 | Statement 3 | True count |
|---|---|---|---|---|
| A | True | True | False | 2 |
| B | False | False | True | 1 |
| C | False | False | False | 0 |
The repeated first two statements are deliberate. Two separately printed statements with identical content have the same truth value, but they still contribute two to a count of true printed statements. The question must define whether it counts statements, different claims or speakers; those are not always the same.
Why a truth table is a complete argument here
The premise says the star is in exactly one of the three named boxes. There are therefore exactly three location cases to test. The table evaluates every statement in each case and shows that one case alone satisfies the final condition. No guess about a person’s intention is needed.
A table is especially helpful when a learner tries to make every clue true despite a rule stating otherwise. The task is to find a location that makes the specified truth pattern hold, not to satisfy the content of all printed clues simultaneously.
A simpler assignment puzzle
Three pupils, Kai, Lina and Noor, each receive one different badge: circle, star or square. Kai has the star. Lina does not have the square. Because each badge is used once, Lina cannot have the star either and must have the circle. Noor then has the square.
Every exclusion has a reason. If Kai’s badge were not supplied, Lina’s restriction alone would not identify the complete assignment uniquely. A familiar story does not license filling gaps with a likely or aesthetically pleasing arrangement.
Contradictory clues need not identify a liar
In a fictional number puzzle, a hidden whole number is required to be both odd and even. No number satisfies those conditions. The mathematical conclusion is inconsistency. It does not establish who wrote an error, whether it was intentional or what a real person’s motives were.
Keep that boundary explicit when using truth puzzles with children. Logic can analyse the statements supplied. Social interpretation often needs evidence that a puzzle does not contain.
6. Unknown is not the same as false
Suppose a hidden whole number is greater than eight and less than twelve. Its possibilities are nine, ten and eleven. Is it even? Ten is even, but nine and eleven are not. The given information does not determine the answer. “Not established” is different from “false.”
To demonstrate insufficient information, give two complete possibilities satisfying the premises but producing different answers to the queried conclusion. Nine and ten do that here. A single example of ten proves evenness is possible, not that it is necessary.
Unique, multiple and impossible cases
A hidden number from one to twelve that is odd, greater than eight and divisible by three must be nine. The three conditions narrow the candidate set to one. Remove the divisibility condition and both nine and eleven remain. Add an evenness condition while retaining oddness and no candidate remains.
These three outcomes should be distinguished in the answer: one forced value, several permitted values, or inconsistent conditions. The process is not a failure merely because it ends with more than one value or with none.
Some information can be redundant
If n is divisible by eight, it is already even. Adding the clue “n is even” does not further narrow the candidates. A condition can be true without adding a new restriction. Recognising redundancy can simplify a proof, but first explain why one condition implies the other.
A repeated clue does not become independent evidence just because it is written twice. That is different from the earlier puzzle counting true printed statements, where two repeated sentences intentionally counted as two statements. Always identify what the task is measuring.
Preserve the strength of the conclusion
“Some rectangles are squares” is true under the definitions used here. “All rectangles are squares” is false. “No rectangles are squares” is also false. Quantifiers such as some, all, none and exactly one determine how much evidence the claim needs.
In a finite classroom set, checking every listed object can prove a statement about that set. It does not automatically prove the same statement about all objects in the wider mathematical category. Four checked rectangles are not the whole universe of rectangles.
7. Practice: 24 questions
Keep the answers covered. Use the stated domain for every question. “Or” is inclusive unless “exactly one” is written. Under this guide’s definitions, squares are rectangles with four equal sides. Give a reason, counterexample or complete case record, not only true or false.
Questions 1–8: Read statements precisely
1. Is “12 is a multiple of three” true? Supply a multiplication check.
2. For n = 14, is “n is even and a multiple of three” true?
3. For n = 14, is “n is even or a multiple of three” true under inclusive or?
4. For n = 6, does “even or a multiple of three” hold? Does “exactly one of these properties” hold?
5. A row contains four blue counters and one red counter. Is “not every counter is blue” true? Does it mean every counter is non-blue?
6. State the negation of “n is greater than ten.”
7. Under the definitions above, is every square a rectangle? Explain.
8. Is every rectangle a square? Give a counterexample if not.
Questions 9–16: Direction, necessity and negation
9. Explain why every whole-number multiple of six is even without checking a long list.
10. Does being even force a positive whole number to be a multiple of six? Test the inference.
11. In a card collection, every red-sticker card is a triangle. A particular card is not a triangle. What follows about its sticker?
12. Under the same rule, a card does not have a red sticker. Must it be a non-triangle?
13. The rule says “A card is eligible only if it is red.” An eligible card is observed. Must it be red? Must every red card be eligible?
14. A box is ready if and only if it contains exactly four counters. A box contains four counters. What follows? What follows from a box being ready?
15. Negate “at least one counter in this row is red.”
16. Negate “exactly one counter in this row is red.” Describe all allowed counts under the negation.
Questions 17–24: Deduction, examples and information
17. A star is in exactly one of boxes A, B, C. Three statements say “in A,” “in A,” and “in B.” Exactly one statement is true. Find the star and check all three locations.
18. A hidden whole number from one to twelve is odd, greater than eight and divisible by three. Find it.
19. Every number in the finite set {4, 8, 12, 16} is checked and found divisible by four. Does this establish the claim for that set? Does it establish that every positive whole number is divisible by four?
20. For two positive whole numbers, classify “the sum equals the product” as always, sometimes or never true. Give suitable evidence.
21. Explain why the product of two consecutive positive whole numbers is always even.
22. For non-negative whole numbers n, test “n squared is greater than n.” Identify the failures and give a valid restricted version.
23. All square cards are rectangular cards. Some rectangular cards are red. Does it follow that some square cards are red? Construct a permitted collection to test the claim.
24. Compare two hidden-number questions: A requires a whole number from one to nine that is both even and odd. B requires a whole number from one to nine and asks whether it is even, with no other clue. Explain the different information problems.
8. Worked answers and justifications
Answers 1–8
1. True. Twelve equals three times four, so division by three is exact. The multiplier four is a whole number, matching the definition of multiple used here. This checks the particular statement about twelve; it is not intended as a proof about all other numbers.
2. False. Fourteen is even, but it is not divisible by three: four groups of three make twelve with two left. “And” requires both conditions. Satisfying the evenness part alone is insufficient to make the combined statement true.
3. True. Inclusive or accepts a case when either condition or both hold. Fourteen satisfies evenness, so it passes even though it is not a multiple of three. This answer differs from Question 2 because the connective changed, not because the arithmetic changed.
4. Inclusive or: true; exactly one: false. Six is both even and a multiple of three. Inclusive or permits both. The exactly-one condition excludes a case with two true properties. Stating the convention prevents everyday ambiguity about the word “or.”
5. True; no. The single red counter is enough to refute “every counter is blue.” However, four counters are still blue, so it would be wrong to conclude that all are non-blue. The negation of all supplies at least one exception, not necessarily an entirely opposite collection.
6. “n is at most ten,” or n ≤ 10. The boundary ten must be included because ten is not greater than ten. Saying “n is less than ten” would omit a case where the original statement is false, so it is not the complete negation.
7. Yes. A square meets the rectangle requirement of four right angles. Its additional equal-side property makes it a more specific rectangle rather than excluding it from the larger category. A rotated drawing does not change these defining properties.
8. No. A rectangle with adjacent side lengths two and three units has four right angles but does not have all four sides equal. It satisfies the premise of being a rectangle and fails the conclusion of being a square, so it is a valid counterexample.
Answers 9–16
9. Every group of six can be organised as three pairs. Combining any whole number of such groups leaves all objects paired, which is the even-number structure. Equivalently, 6k = 2(3k) for a whole number k. The argument covers all permitted multiples rather than only selected examples.
10. No. Eight is even but not a multiple of six. This tests the reversed implication, not the true rule in Question 9. A correct one-direction rule does not automatically remain correct when its premise and conclusion are exchanged.
11. It cannot have a red sticker. A red sticker would, by the rule, make it a triangle, contradicting the supplied non-triangle fact. The conclusion is restricted to the collection in which the rule holds. No claim about all real red stickers is being made.
12. No. A triangle with a blue sticker is allowed by the rule: it is not a red non-triangle. Thus a non-red card might still be a triangle. The absence of the premise does not establish the absence of the conclusion.
13. The eligible card must be red; not every red card is forced to be eligible. Redness is necessary under the “only if” rule. There may be further eligibility conditions. A red but ineligible card does not contradict the stated one-direction rule.
14. Four counters implies ready, and ready implies exactly four counters. “If and only if” explicitly supplies both directions. A ready box containing five, or an unready box containing exactly four, would contradict the rule. Checking only one direction would use less information than the question gives.
15. “No counter in the row is red.” At least one includes one, two and every larger possible positive red count. Negating it leaves zero red counters. “Not every counter is red” would be too weak, because a mixed row can still contain at least one red.
16. Zero red counters or at least two red counters. Exactly one excludes both of these ranges, so its negation includes both. Saying only “none is red” would miss a row with three reds, while saying only “more than one” would miss a row with none.
Answers 17–24
17. Box B. If the star is in A, the truth pattern is true, true, false: two true statements. In B it is false, false, true: exactly one. In C all three are false. The three locations exhaust the premise, so B is not merely possible but uniquely forced.
18. Nine. Odd numbers from one to twelve that exceed eight are nine and eleven. Nine is divisible by three and eleven is not. Checking all remaining candidates establishes uniqueness. Omitting the divisibility condition would leave two possible answers instead.
19. Yes for the finite set; no for all positive whole numbers. Each member of the named four-element set has been checked, so that restricted claim is established. The wider claim is false, for example at five. A complete examination of one set is not a complete examination of a larger domain.
20. Sometimes. Two and two give sum four and product four, so the statement can be true. One and three give sum four and product three, so it can be false. These two examples establish the requested classification but do not constitute a complete classification of every pair satisfying equality.
21. One of two consecutive whole numbers is even. Their parity alternates, so the product has an even factor and can be arranged in pairs. This general argument applies to every consecutive positive pair. Arbitrary non-consecutive pairs would need different analysis because both could be odd.
22. It fails at zero and one; it holds for n ≥ 2. Zero squared is zero, and one squared is one, neither greater. For n at least two, n copies of the positive amount n exceed one copy. The restricted claim states the condition that the original omitted.
23. No. Consider one blue square card and one red non-square rectangle card. Every square card is rectangular and some rectangular card is red, but no square card is red. This complete permitted collection satisfies the premises while refuting the proposed conclusion.
24. A is inconsistent; B is undetermined. No whole number is simultaneously even and odd, so A has no valid candidate. B permits both even values such as two and odd values such as three. Its premises are consistent but do not determine evenness. “No solution” and “not enough information” are different conclusions.
9. Help the learner explain the first justified move
Ask which statement supports the next line. If the learner replies, “It looks right,” return to the named condition or definition. A useful explanation need not be long, but it must connect the conclusion to something established rather than to a familiar-looking picture or preferred answer.
For a child reversing implications, use two overlapping categories with one inside the other. A square belongs inside rectangles, but a non-square rectangle sits outside squares while remaining inside rectangles. This gives a visible counterexample to the reverse direction without requiring formal terminology.
Distinguish checking from proving
Give a finite list and ask whether every member has a property. Then change the question to all positive whole numbers. Discuss why checking the finite list completes one task but not the other. Invite a structural explanation, such as grouping multiples into pairs, when a general claim is made.
Do not insist on algebra before the reasoning is clear. A verbal explanation, a labelled counter arrangement or a complete small table can establish the intended result. Formal notation should preserve the meaning already understood rather than hide a gap behind symbols.
Use one false case constructively
When a child makes an overgeneralisation, ask for a boundary case: zero, one, equality, the smallest allowed number or an endpoint. If the case lies outside the domain, explain why it cannot refute the stated claim. A disciplined counterexample must satisfy the premise before it can challenge the conclusion.
Once a counterexample is found, repair the statement. Change “all” to “some,” add the missing condition or describe exactly which cases remain possible. This keeps mathematical criticism connected to building a more accurate explanation.
A paired-lesson sequence
Start with true or false for specified numbers. Move to two conditions joined by and, then compare inclusive or. Introduce a one-direction card rule and test its four cases. Finish with a small puzzle where the permitted location or number is unique, multiple or impossible. This is a proposed teaching sequence to adapt to the learner’s working, not a claim of measured effectiveness.
Record whether the learner needed the candidate list, the rule direction or the counterexample supplied. A correct conclusion after those decisions are given is useful guided practice, but it is not the same evidence as independently selecting the argument.
Connections to the rest of the collection
A probability answer depends on defining the event correctly; use Probability, Chance and Fairness. A remainder claim depends on a specified divisor and range; use Remainder Cycles, Clock Arithmetic and Modular Patterns. A winning strategy must handle every legal reply; use Mathematical Games, Strategy and Winning Positions. All three are applications of careful conditions rather than separate collections of tricks.
Return to the BTT Primary Mathematics Learning Hub for prerequisite arithmetic, shapes and systematic listing. The central habit is simple: say what is known, say what follows and keep uncertainty where the information does not settle the answer.
Sources and scope
The MOE page is the official reference point for school scope; this guide is an enrichment commission rather than an official syllabus document. The linked NRICH activity supplies further reading on examining mathematical claims. The definitions stated here, original examples, 24 questions, solutions and proposed teaching sequence are self-contained. No school endorsement, student assessment or psychological interpretation is claimed.