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Primary Mathematics: Explaining, Convincing and Proving | Worked Learning Guide

A correct answer tells us what happened in one problem. A convincing mathematical argument tells us why the result had to happen.

Primary Mathematics already contains the beginnings of proof. A learner explains why two odd numbers add to an even number, why a list contains every valid possibility, why a proposed route is impossible, why a square gives the smallest perimeter among listed rectangles, why the same operation on both sides of an equality preserves the balance, or why a counterexample defeats the word always.

None of these requires a formal theorem-proof format. What they require is responsibility for the reasoning. The conclusion must follow from stated facts, accepted definitions and steps that another person can inspect. “It looks right” is weaker than “these are all the cases”. “I tried several examples” is weaker than “the same structure applies to any allowed number”.

This guide develops a ladder from explanation to convincing argument to proof. It uses Primary-accessible methods—pairing, decomposition, exhaustive cases, invariants, lower bounds, constructions and contradiction by an impossible consequence—while keeping the language proportionate to the learner’s stage.

Explain · Convince · Prove · Primary proof methods · Limits and counterexamples · 24 questions · Worked solutions

1. Explanation connects steps; it does not merely repeat the answer

Question: Why is 37 + 48 equal to 35 + 50?

Weak answer: “Because both equal 85.”

Stronger explanation: two units were moved from the first addend to the second. One addend decreased by two and the other increased by two, so the total was unchanged.

The second response identifies the preserved relationship. It would still make sense if the numbers changed.

Worked example: equivalent fractions

Why is 3/4 equal to 6/8?

Split each of the four original quarter-parts into two equal smaller parts. The whole now contains eight eighths, and the three selected quarters become six selected eighths. The amount of the whole represented has not changed.

This is more informative than cross-multiplying 3×8 and4×6. The numerical check confirms equality; the partition argument explains the equivalence.

2. A convincing argument anticipates the obvious doubt

Suppose a learner counts 18 rectangles in a 2×3 grid. “I counted carefully” gives no way to know whether one rectangle was missed or counted twice.

A convincing argument groups rectangles by height and width, or chooses top/bottom and left/right boundary lines. It explains why every valid rectangle appears in exactly one category.

Convincing mathematics is therefore not necessarily longer. It answers the reader’s strongest question: How do you know there are no missing cases or hidden exceptions?

Worked example: all possibilities

Using digits 1,2,3 without repetition, how many three-digit numbers can be formed?

Choose the hundreds digit in three ways. For each choice, the tens digit has two remaining choices. The final digit is then forced. Total 3×2×1=6.

This is convincing because every valid number has exactly one first choice, one second choice and one final choice. The multiplication structure covers the complete search.

3. Proof answers an “all cases” question

A proof is a reasoning chain showing that a statement must be true under its stated conditions. At Primary level, many proofs can be visual or verbal.

Worked example: odd plus odd is even

Represent each odd number as complete pairs plus one leftover object. When two odd collections are combined, the two leftover objects form another pair. Everything is now paired, so the sum is even.

The argument does not depend on the specific sizes of the odd numbers. That generality is what turns the explanation into proof.

Worked example: three consecutive numbers

Let the middle whole number be n. The three consecutive numbers are n−1, n and n+1. Their total is 3n. Therefore the sum is divisible by 3.

The variable compresses the same structural fact for every possible middle number.

4. Method One: prove by structure

Structural proof explains how the mathematical object is built.

Joined squares

A row of n joined unit squares has perimeter 2n+2. The top contributes n unit edges, the bottom n, and the two outer ends contribute one each. Therefore perimeter=n+n+1+1=2n+2.

This proves the rule for every positive whole-number n in that construction. Listing 4,6,8,10 is evidence of the pattern, but the boundary decomposition explains why it continues.

Triangular numbers

Two identical staircases containing 1+2+…+n unit counters can be fitted together to form an n by n+1 rectangle. Therefore twice the staircase total is n(n+1), so one staircase contains n(n+1)/2 counters.

The picture can carry the proof if the pairing construction works for any n.

5. Method Two: prove by exhaustive cases

Some finite problems can be proved by showing that every allowed case has been considered.

Worked example: two-digit numbers from 1,2,3

Without repetition, the numbers are 12,13,21,23,31,32. Organise by tens digit:

• 1_: 12,13
• 2_: 21,23
• 3_: 31,32

Every valid number has one of the three possible tens digits, and each row lists both remaining unit choices. Therefore the six-number list is complete.

Worked example: whole-number rectangles of area 24

Factor pairs are 1×24,2×12,3×8 and4×6. These are all positive whole-number pairs because factors have been listed in increasing order until the pair would reverse. Comparing perimeters 50,28,22,20 shows 4×6 has the minimum perimeter among the allowed rectangles.

The conclusion is conditional on positive whole-number side lengths. State the domain that made the exhaustive list complete.

6. Method Three: use parity

Parity classifies whole numbers as odd or even. Many impossibility arguments become short once parity is tracked.

Worked example: consecutive sum

Two consecutive whole numbers have opposite parity. Their sum is therefore odd.

Worked example: impossible total

Can three odd whole numbers add to 20?

Odd+odd=even; even+odd=odd. Therefore the sum of three odd numbers is odd. Since 20 is even, the requested total is impossible.

No trial list is required because parity rules out every possible triple at once.

7. Method Four: use an invariant

An invariant is a quantity or relationship that does not change during allowed operations.

Worked example: internal transfer

Two boxes contain a total of 80 counters. Move any number of counters from one box to the other. The individual amounts change, but the total remains 80 because no counter enters or leaves the two-box system.

Therefore any proposed final pair whose total is not 80 is impossible under transfer-only moves.

Worked example: equal additions

If A is 12 greater than B, adding the same 7 to both preserves the difference. The difference is invariant under equal addition.

This method links directly to Invariants, Before-and-After and Unchanged Quantities.

8. Method Five: prove a minimum with a lower bound and a construction

To prove that a solution is optimal, showing one good construction is only half the job. We also need to prove that nothing better can work.

Worked example: grid path

Move from (0,0) to (8,5) using one-unit horizontal or vertical steps. Every route must change the horizontal coordinate by eight and the vertical by five, so at least 13 moves are unavoidable. A route using exactly eight right and five up moves attains 13. Therefore 13 is the minimum.

The lower bound says “no fewer than 13”. The construction says “13 is achievable”. Together they prove optimality.

Worked example: rectangle perimeter

If the task is restricted to positive whole-number rectangles of area 24, exhaustive factor pairs can act as the lower-bound proof: no other permitted rectangle shape exists. The 4×6 candidate has the smallest perimeter among all listed possibilities, so it is optimal within the stated domain.

See Optimisation, Minimum Moves and Efficient Constructions for a full collection of this proof pattern.

9. Method Six: prove impossibility by contradiction

A contradiction argument assumes the requested object exists and shows that this would force an impossible condition.

Worked example: equal groups

Can 25 counters be split into six equal whole-number groups with none left over?

If six equal groups existed, 25 would be divisible by 6. But multiples of 6 around 25 are 24 and30; 25 is not divisible by 6. Therefore the exact equal grouping is impossible.

Worked example: exact post spacing

Can a 25 m straight route have posts at both ends with every gap exactly 6 m?

If it could, the total length would be a whole number of 6 m gaps, hence a multiple of 6. But 25 is not a multiple of 6. Therefore no such exact-spacing arrangement exists.

10. Method Seven: prove by a reversible transformation

Sometimes two problems are equivalent because a transformation can be undone.

Worked example: compensation

37+48 can be changed to 35+50 by moving two from the first addend to the second. The operation is reversible: move two back. Because the total amount distributed between the two addends is unchanged, the sums are equal.

Worked example: scaling a ratio

3:5 and12:20 are equivalent because multiplying both terms by4 changes the common unit size but not the relative number of units. Dividing both terms of12:20 by4 reverses the transformation.

11. Method Eight: use a counterexample to disprove

A proof establishes a universal statement. A counterexample can destroy one.

Claim: “Taking the largest coin first always uses the fewest coins.” With coin values1,3,4 and target6, greedy gives4+1+1 using three coins; 3+3 uses two. The claim is false.

We do not need to test every target. One valid counterexample is logically enough because the original statement said always.

12. Necessary is not the same as sufficient

Being even is necessary for a whole number to be divisible by 6: every multiple of 6 is even. But being even is not sufficient; 8 is even and not divisible by 6.

Divisibility by both 2 and3 is sufficient for divisibility by6. At Primary level, this language can be introduced through questions such as “What must be true?” versus “What is enough to guarantee it?”

Worked example: circuit of route edges

In the route-network enrichment model, having an even number of odd-degree vertices is necessary for any ordinary network because odd vertices occur in pairs. But that condition alone is not sufficient for a one-stroke Euler trail: four odd vertices is even in count, yet a one-stroke trail requires specifically zero or two odd vertices, together with connectivity.

13. Verification is not always proof of the original method

Suppose a learner solves x+17=45 and obtains x=28. Substituting 28 gives 28+17=45, confirming that 28 satisfies the equation.

This verifies the candidate. It does not necessarily justify every step used to obtain it. A learner could have made an invalid step and accidentally landed on the correct answer. When method matters, explain why each transformation preserves equality.

Worked example: arithmetic checks

A remainder-nine digit-sum check can reject many wrong calculations, but a candidate that passes is not thereby proved correct. The Digit Sums and Arithmetic Error Checks guide develops this distinction.

14. A diagram can be part of a proof when its structure is general

A picture of 5+7=12 does not prove every odd+odd sum is even. But a diagram showing each odd number as pairs plus one, then combining the two leftover ones into a pair, is a reusable construction for any odd inputs.

The difference is not “pictures versus algebra”. It is one example versus a representation whose logic covers every permitted case.

15. Proof should match the claim

Claim type	Useful evidence
This one calculation is correct	exact inverse or substitution check
Every case in a finite list has been counted	exhaustive organised cases
A rule holds for any whole number	general structural or algebraic argument
No solution exists	contradiction, parity, divisibility or invariant
This is the minimum	lower bound plus attaining construction
An always-claim is false	one valid counterexample

Using more mathematics than the claim needs can hide rather than strengthen the argument.

16. Model limits and unstated assumptions

A proof belongs to its conditions. A factor-pair proof for minimum perimeter among whole-number rectangles does not automatically cover rectangles with arbitrary real side lengths. A counting proof on a complete grid may fail when line segments are missing. A ratio-scaling argument requires the same non-zero scale factor on both terms.

The final sentence of a proof should therefore be no broader than the assumptions that support it.

17. Common proof errors

• Giving several examples for an “all cases” claim and calling that proof.
• Listing cases without explaining why the list is complete.
• Showing a construction but not proving it is minimal.
• Using a counterexample that violates the conditions.
• Reversing a true implication automatically.
• Using a diagram that assumes an unstated property such as equal sides.
• Checking a final answer but ignoring invalid intermediate transformations.
• Proving a stronger statement than the given assumptions justify.

18. Practice: 24 original questions

Questions 1–8: Explain and convince

1. Explain why 37+48=35+50 without calculating both sums independently.
2. Explain visually why 3/4=6/8.
3. Using digits1,2,3 without repetition, prove that exactly six three-digit numbers can be formed.
4. Prove that the sum of two odd whole numbers is even.
5. Prove that the sum of three consecutive whole numbers is divisible by3.
6. Explain why a row of n joined unit squares has perimeter2n+2.
7. Prove that n(n+1) is even for every whole number n.
8. A learner counts rectangles in a small grid. What must the explanation include before the count is convincing?

Questions 9–16: Cases, invariants and optimality

9. List all positive whole-number rectangles of area24 and prove which has minimum perimeter.
10. Three odd whole numbers are claimed to sum to20. Prove this is impossible.
11. Two boxes contain80 counters. Only internal transfers are allowed. Prove that a final state of31 and50 counters is impossible.
12. A is12 greater than B. Add7 to both. Prove the difference remains12.
13. From (0,0) to(8,5), use unit horizontal/vertical moves. Prove the shortest path length is13.
14. Can25 counters be divided into six equal whole-number groups with none left? Prove your answer.
15. Can a25 m path have endpoint posts with every gap exactly6 m? Prove your answer.
16. Coin values1,3,4 are available. Disprove “largest coin first always gives the fewest coins.”

Questions 17–24: Logic and proof control

17. If a whole number is divisible by6, it is even. Is the reverse true? Prove or disprove.
18. If a quadrilateral is a square, it is a rectangle. Is the reverse true? Prove or disprove.
19. A learner solves x+17=45 and writes x=28. Give an exact verification and explain what it does and does not prove.
20. Show that consecutive square numbers differ by an odd number.
21. Explain why multiplying both terms of a ratio by the same positive factor preserves the ratio.
22. A complete2×3 grid has18 rectangles. Describe a proof method that shows no rectangle was omitted or duplicated.
23. Give one example of a claim for which a lower bound plus a construction is the right proof structure.
24. Create one false universal Primary Mathematics statement and disprove it with one valid counterexample.

19. Worked solutions

Solutions 1–8

1. Moving two units from48 to37 gives35 and50 while preserving the total. One addend decreases by2 and the other increases by2, so the changes cancel.

2. Divide each quarter into two equal eighths. Three selected quarters become six selected eighths while the represented portion of the whole remains unchanged.

3. Choose the hundreds digit in3 ways, tens in2 remaining ways and units in1 way. Every permitted number has exactly one such choice sequence, giving3×2×1=6.

4. Each odd number is complete pairs plus one leftover. Combining two odds pairs the two leftovers, leaving only complete pairs. Therefore the sum is even.

5. Let the middle number be n. The three numbers are n−1,n,n+1. Sum=3n, hence divisible by3.

6. Top boundary has n unit edges, bottom n, and the two ends one each. Total perimeter=n+n+1+1=2n+2.

7. Consecutive whole numbers have opposite parity, so one of n and n+1 is even. Their product therefore contains a factor2 and is even.

8. It must organise all allowed rectangles into complete non-overlapping categories—for example by height and width or by boundary-line pairs—and explain why every rectangle belongs to exactly one category.

Solutions 9–16

9. Factor pairs:1×24,2×12,3×8,4×6. Perimeters:50,28,22,20. These are all positive whole-number factor pairs, so4×6 has minimum perimeter20 within the stated domain.

10. Odd+odd=even; even+odd=odd. Therefore any sum of three odd whole numbers is odd. Since20 is even, such a triple cannot total20.

11. Internal transfers preserve total80. Proposed31+50=81, so the state cannot be reached without adding a counter from outside or changing the rules.

12. New difference=(A+7)−(B+7)=A−B=12. Equal additions cancel in the difference.

13. Any route must make eight horizontal and five vertical units of net progress, requiring at least13 moves. Eight right plus five up moves achieve13, so13 is minimal.

14. Exact equal groups would require25 to be divisible by6. It is not:24 and30 are neighbouring multiples of6. Therefore the requested grouping is impossible.

15. Exact6 m gaps between endpoint posts would make25 a whole multiple of6. It is not, so no such arrangement exists.

16. Target6: greedy gives4+1+1 using3 coins;3+3 uses2. One valid counterexample disproves the universal claim.

Solutions 17–24

17. Reverse false. Example8 is even but not divisible by6. Therefore evenness is necessary for divisibility by6 but not sufficient.

18. Reverse false. A3×5 rectangle is a rectangle but not a square. Every square is a rectangle, but not every rectangle is a square.

19. Substitute:28+17=45, so28 satisfies the original equation. This verifies the candidate but does not by itself justify every step the learner may have used to obtain it.

20. (n+1)²−n²=n²+2n+1−n²=2n+1, which is odd. Visually, growing an n×n square adds an L-shaped border of2n+1 cells.

21. If A:B=a:b, multiplying both actual quantities by positive k produces ka:kb. Dividing both by k returns a:b, so the multiplicative comparison is unchanged.

22. Three horizontal boundary lines have3 pairs; four vertical lines have6 pairs. Each pair-of-pairs creates exactly one rectangle, and every rectangle has exactly one top/bottom and left/right pair. Total=18.

23. Example: shortest unit grid path. Prove at least13 moves are unavoidable from coordinate displacement, then construct a13-move route.

24. Example false claim: “every even whole number is divisible by4.” Counterexample6. The single valid example defeats the word every.

20. Proof laboratory: one result, several levels of justification

Claim: the sum of the first n odd positive whole numbers is n².

Level 1 — examples: 1=1,1+3=4,1+3+5=9,1+3+5+7=16.

Level 2 — visual structure: begin with one square cell. Add an L-border of3 cells to create2×2. Add5 to create3×3. Add7 to create4×4. At stage n, the new border contains2n−1 cells and turns an (n−1)×(n−1) square into an n×n square.

Level 3 — algebraic check: if the first n−1 odd numbers total (n−1)², adding the nth odd number2n−1 gives (n−1)²+2n−1=n².

The levels are not competitors. The examples help notice, the visual construction convinces geometrically, and the algebra compresses the repeating structure. A learner does not need the most formal route available; the proof needs to be general enough for the claim.

21. Parent and tutor guide

When a child says “because that’s the formula,” ask what the formula is counting or preserving. When the child lists examples, ask whether the question concerns those examples or every allowed case. When the child says “I checked twice,” ask whether the two checks are genuinely different or repetitions of the same method.

Use sentence starters that point to reasoning rather than performance:

• “This must be true because…”
• “These are all the cases because…”
• “No smaller answer can work because…”
• “This construction works because…”
• “The statement is false because this valid case…”
• “The quantity that does not change is…”

Do not require formal proof vocabulary before the mathematical relationship is understood. A precise drawing with a general explanation can be stronger Primary mathematics than copied symbolic notation.

The first weak link

Proof failure can begin at different points: the learner may not understand the claim, may use examples instead of general reasoning, may omit cases, may have no lower bound for an optimisation, may use an invalid counterexample, or may prove something broader than the conditions allow. Repair the earliest point.

Return to Conjecturing and Generalising if the claim itself is unstable. Return to Visualising and Representing when the learner needs a structure that can carry the explanation.

22. Delayed return

Three days later, give four claims with no method labels: one parity statement, one exhaustive-count problem, one impossible arrangement and one minimum problem. Ask the learner to choose an appropriate proof style before calculating.

The transfer goal is method selection: knowing whether the job needs a counterexample, complete cases, invariant, contradiction, structural explanation or lower bound plus construction.

23. Mastery receipt

• I distinguish a correct answer from an explanation.
• I distinguish supporting examples from proof of an all-cases claim.
• I make finite lists convincing by showing completeness and no duplication.
• I use parity and invariants for impossibility arguments.
• I prove minima with a lower bound and an attaining construction.
• I use counterexamples to disprove universal statements.
• I understand that verification of a candidate is not always proof of the method.
• I keep conclusions within the assumptions that support them.

Sources and scope

University of Cambridge NRICH — Thinking Mathematically, Primary Students identifies explaining, convincing and proving as a central strand of primary mathematical thinking. NRICH — Developing Mathematical Thinking, Primary Teachers provides a related teacher-facing route across proof and reasoning tasks.

For official Singapore subject documents, use the MOE Primary curriculum and syllabus page alongside the learner’s current school programme. This BTT guide is enrichment and reasoning support, not an official marking scheme or examination specification.

Continue the Working Mathematically collection

• Exploring, Noticing and Finding Structure
• Visualising, Representing and Seeing the Same Mathematics
• Conjecturing, Generalising and Testing a Rule
• BTT Primary Mathematics Learning Hub

The Quiet Return

The mature question is not only “What is the answer?” It is “Why should another careful reader believe it?” Primary Mathematics becomes deeper when the learner can carry that responsibility without losing the simplicity of the idea.