Sec 3 G3 Maths Bukit Timah | Mathematical Reasoning

Secondary 3 G3 Mathematics Tutor Bukit Timah | Mathematical Reasoning: Counterexamples and Justifying Answers — Did you know? Finding ten examples that support a mathematical statement does not always prove it. But one valid counterexample can disprove a statement that claims to work for every case. This is the difference between noticing a pattern and explaining why the pattern must hold. Students can build that reasoning with familiar G3 Mathematics—odd and even numbers, algebraic identities, triangle properties and inequalities.

The quick answer: how to justify or disprove a claim

First read the full statement, including words like every, some, always, if, only if, integer, positive and real.
A counterexample is one permitted case for which a universal claim fails. One is enough to disprove an ‘all’ or ‘every’ statement.
Checking a few examples can reveal a pattern, but examples alone usually cannot establish a universal mathematical rule.
For a true algebraic claim, write a general variable and use lawful transformations to show why it holds for every permitted input.
To prove a claim about consecutive integers, write n, n+1, n+2 rather than relying only on three chosen numbers.
To prove a claim about even or odd integers, write an even number as 2n and an odd one as 2n+1, where n is an integer.
In geometry, state the conditions that justify an angle, side or congruence rule. A not-to-scale sketch alone is not proof.
A valid answer must respect the domain: an integer argument is not automatically an argument about every real number.
Use a clear conclusion stating whether the claim is true, false, conditional or unsupported by the given data.

The 2027 Singapore–Cambridge SEC G3 Mathematics syllabus K310 assesses reasoning and communication under AO3, including justifying statements, explaining conclusions and writing mathematical arguments. The examples in this article illustrate transferable reasoning using syllabus Mathematics; they are not a claim that ‘counterexample proofs’ constitute a separately named content chapter in every school timetable.

Sections: Understand claims · 40 worked examples · Reasoning mistakes · Six-week plan · FAQs.

When examples are useful and when they are not enough

Universal claims need universal reasons

Consider ‘every even number is a multiple of four’. It might seem believable after testing 4, 8 and 12, but the even number 2 is not divisible by four. Therefore the universal statement is false. The counterexample meets the stated domain—an even integer—and breaks the conclusion.

A counterexample is useful because it pinpoints the first place where a statement fails. The learner should explicitly show both facts: the example satisfies the condition, and the claim’s alleged consequence is false. A random unrelated number is not a valid counterexample.

One counterexample does not prove an opposite universal claim

Showing that not every even number is divisible by four does not mean no even number is divisible by four. The number four itself is a counterexample to that new exaggerated claim. Logical language matters: ‘not all’ means at least one failure, not necessarily none succeed.

Teach students to rewrite a claim carefully before reacting. Some, every, none and not every represent different statements. A good short answer avoids changing a modest conclusion into an unjustified extreme.

Consecutive integers have an algebraic structure

If n is an integer, three consecutive integers are n, n+1 and n+2. Their sum is 3n+3=3(n+1), a multiple of three. This is a general explanation: whatever integer n is chosen, the factor three remains.

A small numerical check such as 5+6+7=18 is useful for testing the algebra, but it is not the proof of the universal result. The factorised expression explains why every permitted starting integer works.

Even and odd can be represented symbolically

An even integer has form 2n, and an odd integer has form 2n+1, where n is any integer. The sum of two odd integers 2m+1 and 2n+1 equals 2(m+n+1), so it is even. The algebra shows a factor of two regardless of the chosen odd numbers.

One odd pair such as 3+5=8 illustrates the result, but a general representation demonstrates it. This is useful whenever a problem asks why a parity pattern must hold rather than requesting only a particular numerical total.

A conditional claim has two parts

The statement ‘if a number is divisible by six, it is divisible by three’ has a condition and a consequence. Write the starting number as 6n for integer n; then 6n=3(2n), proving divisibility by three. The argument is about every number meeting the initial condition.

The reverse statement may require separate reasoning. ‘If divisible by three, then divisible by six’ is false, with three itself as a counterexample. Reversing an implication without checking is a common logical mistake.

Algebraic identities can be checked structurally

To assess (a+b)²=a²+b² for all real a,b, test a=2, b=3: the left is 25 and the right 13, so the universal claim is false. The correct identity (a+b)²=a²+2ab+b² follows from expanding (a+b)(a+b).

Numerical substitution can disprove a false universal identity, but a valid general expansion is what proves the corrected identity. This is why reverse checks and expansion are important in the algebraic expressions strand.

Triangle drawings need valid conditions

A triangle’s interior angles total 180°. That conclusion can be supported by a geometrical argument using a line drawn through a vertex parallel to the opposite side, transferring the other two angles onto a straight angle.

A diagram that seems to contain a right angle does not establish that it is 90° unless the angle is marked or logically deduced. Geometry proofs require precisely stated givens and the appropriate parallel, circle, triangle or congruence facts.

Proving two triangles congruent takes enough evidence

If corresponding sides of two triangles are all equal, SSS establishes congruence. But equal corresponding angles alone prove similarity, not necessarily congruence: one triangle could be twice as large as the other.

This gives a counterexample to the claim that AAA always proves congruence. To make the counterexample clear, specify two triangles with equal angles and different corresponding side lengths. Then explain why their size difference prevents congruence.

The domain can change whether a statement is true

The claim x²≥x holds for nonnegative integers x because x(x−1) is nonnegative at integers x=0,1,2,… . But the same claim is false for all real numbers: at x=1/2, x²=1/4 is less than 1/2.

This is a powerful illustration of reading domain restrictions. A pupil should not automatically transfer an integer argument to the real-number line. A counterexample must belong to the domain the statement actually names.

Checking a solution is not the same as proving uniqueness

Suppose a learner finds one pair satisfying two equations. Substitution confirms that pair works, but it does not automatically prove there are no other solutions. The structure of the simultaneous equations, factorisation or graph may be needed to establish completeness.

In a quadratic, two different roots can satisfy the same equation. Checking only one positive square-root candidate will not prove the solution set is complete. The question may ask for all solutions, and the working should address that word explicitly.

An explanation should identify the rule and its condition

A convincing short reasoning answer has three parts: state the given facts, name the relevant relationship and show why the conclusion follows. For example, a radius is perpendicular to a tangent at the point of contact—not at every arbitrary point on the circle.

The details of the rule prevent overgeneralisation. Writing ‘because circle theorem’ is too vague to show which angle is being used or whether the configuration is legitimate. Accurate concise language often saves work rather than adding it.

Counterexamples are a learning tool, not a trick

An efficient tutor can show one true claim and one near-identical false claim, then ask which condition changed. That helps students learn a theorem’s boundaries as well as its content. The aim is not to find a loophole in every question but to read what is actually asserted.

A student who can produce a valid counterexample or explain a general factorisation is beginning to reason independently. Those habits support the K310 objectives in contextual problem-solving and mathematical communication across multiple syllabus strands.

40 practice examples: justify, disprove and check

These are original teaching examples rather than official examination questions. Ten types are repeated with four numerical variations. Some ask for one counterexample; others show a general identity or argument and use the chosen number only as an illustration. Decide first whether the claim concerns every possible case or only one specified case.

Counterexample to an overgeneralised divisor claim

This cluster practises counterexample to an overgeneralised divisor claim. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 1. Question: Claim: Every even integer is divisible by four. Use the even integer 2 to test the claim. Reasoning: The integer 2=2(1) is even, but dividing it by four leaves remainder 2, so it is not a multiple of four. Therefore the universal claim is false. Check: The counterexample satisfies ‘even’ and fails ‘divisible by four’, exactly the two required conditions. Common trap: Finding several multiples of four that are even does not prove every even number is a multiple of four.

Independent extension: Read every domain word and identify what qualifies as a valid example before deciding if a claim holds. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 2. Question: Claim: Every even integer is divisible by four. Use the even integer 6 to test the claim. Reasoning: The integer 6=2(3) is even, but dividing it by four leaves remainder 2, so it is not a multiple of four. Therefore the universal claim is false. Check: The counterexample satisfies ‘even’ and fails ‘divisible by four’, exactly the two required conditions. Common trap: Finding several multiples of four that are even does not prove every even number is a multiple of four.

Independent extension: To disprove a universal claim, show both that the selected case meets the condition and that the proposed conclusion fails. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 3. Question: Claim: Every even integer is divisible by four. Use the even integer 10 to test the claim. Reasoning: The integer 10=2(5) is even, but dividing it by four leaves remainder 2, so it is not a multiple of four. Therefore the universal claim is false. Check: The counterexample satisfies ‘even’ and fails ‘divisible by four’, exactly the two required conditions. Common trap: Finding several multiples of four that are even does not prove every even number is a multiple of four.

Independent extension: To prove an integer pattern, use n as a general integer rather than relying on a particular sample of numbers. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 4. Question: Claim: Every even integer is divisible by four. Use the even integer 14 to test the claim. Reasoning: The integer 14=2(7) is even, but dividing it by four leaves remainder 2, so it is not a multiple of four. Therefore the universal claim is false. Check: The counterexample satisfies ‘even’ and fails ‘divisible by four’, exactly the two required conditions. Common trap: Finding several multiples of four that are even does not prove every even number is a multiple of four.

Independent extension: Distinguish a numerical check that illustrates a formula from an argument that justifies it for every permitted value. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

General sum of three consecutive integers

This cluster practises general sum of three consecutive integers. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 5. Question: Explain why the sum of 3, 4 and 5 is divisible by three, then justify the pattern generally. Reasoning: For general integer m, m+(m+1)+(m+2)=3m+3=3(m+1). Thus the sum is always divisible by three. In this example the total is 12. Check: Dividing 12 by three gives 4, and the general proof retains factor three for every integer m. Common trap: The numerical example alone is not the general proof.

Independent extension: Look for an exact factor such as two or three when claiming divisibility or parity. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 6. Question: Explain why the sum of 4, 5 and 6 is divisible by three, then justify the pattern generally. Reasoning: For general integer m, m+(m+1)+(m+2)=3m+3=3(m+1). Thus the sum is always divisible by three. In this example the total is 15. Check: Dividing 15 by three gives 5, and the general proof retains factor three for every integer m. Common trap: The numerical example alone is not the general proof.

Independent extension: Do not reverse an ‘if…then’ statement without separately checking whether its converse is true. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 7. Question: Explain why the sum of 5, 6 and 7 is divisible by three, then justify the pattern generally. Reasoning: For general integer m, m+(m+1)+(m+2)=3m+3=3(m+1). Thus the sum is always divisible by three. In this example the total is 18. Check: Dividing 18 by three gives 6, and the general proof retains factor three for every integer m. Common trap: The numerical example alone is not the general proof.

Independent extension: After drawing a geometry diagram, identify which marked fact justifies each equality or angle calculation. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 8. Question: Explain why the sum of 6, 7 and 8 is divisible by three, then justify the pattern generally. Reasoning: For general integer m, m+(m+1)+(m+2)=3m+3=3(m+1). Thus the sum is always divisible by three. In this example the total is 21. Check: Dividing 21 by three gives 7, and the general proof retains factor three for every integer m. Common trap: The numerical example alone is not the general proof.

Independent extension: State whether an example establishes a counterexample, a check, or a general proof, and do not mix these purposes. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Parity of the sum of two odd numbers

This cluster practises parity of the sum of two odd numbers. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 9. Question: Numbers 1 and 3 are odd. Explain generally why the sum of any two odd integers is even. Reasoning: Write odd integers as 2m+1 and 2n+1. Their sum is 2(m+n+1), divisible by two. For this case 1+3=4 is even. Check: Both the chosen example and the general symbolic expression have factor two. Common trap: A proof should not depend on this one pair of odd numbers.

Independent extension: Change the domain from integers to reals only after reconsidering whether the original argument still works. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 10. Question: Numbers 3 and 5 are odd. Explain generally why the sum of any two odd integers is even. Reasoning: Write odd integers as 2m+1 and 2n+1. Their sum is 2(m+n+1), divisible by two. For this case 3+5=8 is even. Check: Both the chosen example and the general symbolic expression have factor two. Common trap: A proof should not depend on this one pair of odd numbers.

Independent extension: Return after two days to a new mathematical statement and explain which first reasoning step you choose. Now alter one condition, such as integer to real, included to excluded angle, or a universal claim to a particular case. Explain why that change does or does not invalidate the original argument. Keep the student’s first reasoning attempt and compare it with a delayed new example.

Worked example 11. Question: Numbers 5 and 7 are odd. Explain generally why the sum of any two odd integers is even. Reasoning: Write odd integers as 2m+1 and 2n+1. Their sum is 2(m+n+1), divisible by two. For this case 5+7=12 is even. Check: Both the chosen example and the general symbolic expression have factor two. Common trap: A proof should not depend on this one pair of odd numbers.

Worked example 12. Question: Numbers 7 and 9 are odd. Explain generally why the sum of any two odd integers is even. Reasoning: Write odd integers as 2m+1 and 2n+1. Their sum is 2(m+n+1), divisible by two. For this case 7+9=16 is even. Check: Both the chosen example and the general symbolic expression have factor two. Common trap: A proof should not depend on this one pair of odd numbers.

Parity of the product of an even integer

This cluster practises parity of the product of an even integer. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 13. Question: Explain why 4×3 is even, and show why an even integer times any integer is even. Reasoning: An even integer is 2m; multiplying by another integer n gives (2m)n=2(mn), still a multiple of two. The specific product 4×3=12 has that property. Check: The factor two is visible before choosing numerical values and remains after multiplication. Common trap: Do not reason from the digits of a product alone when a general factorisation is available.

Worked example 14. Question: Explain why 6×4 is even, and show why an even integer times any integer is even. Reasoning: An even integer is 2m; multiplying by another integer n gives (2m)n=2(mn), still a multiple of two. The specific product 6×4=24 has that property. Check: The factor two is visible before choosing numerical values and remains after multiplication. Common trap: Do not reason from the digits of a product alone when a general factorisation is available.

Worked example 15. Question: Explain why 8×5 is even, and show why an even integer times any integer is even. Reasoning: An even integer is 2m; multiplying by another integer n gives (2m)n=2(mn), still a multiple of two. The specific product 8×5=40 has that property. Check: The factor two is visible before choosing numerical values and remains after multiplication. Common trap: Do not reason from the digits of a product alone when a general factorisation is available.

Worked example 16. Question: Explain why 10×6 is even, and show why an even integer times any integer is even. Reasoning: An even integer is 2m; multiplying by another integer n gives (2m)n=2(mn), still a multiple of two. The specific product 10×6=60 has that property. Check: The factor two is visible before choosing numerical values and remains after multiplication. Common trap: Do not reason from the digits of a product alone when a general factorisation is available.

Conditional divisibility by three

This cluster practises conditional divisibility by three. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 17. Question: If n is a multiple of six, explain why n+3 is divisible by three. Check with n=6. Reasoning: Write n=6m. Then n+3=6m+3=3(2m+1), so it is divisible by three. Here 6+3=9=3(3). Check: The symbolic expression proves the claim for any integer m; the chosen value gives a separate check. Common trap: Do not conclude n+3 must be divisible by six; the expression 6m+3 has remainder three on division by six.

Worked example 18. Question: If n is a multiple of six, explain why n+3 is divisible by three. Check with n=12. Reasoning: Write n=6m. Then n+3=6m+3=3(2m+1), so it is divisible by three. Here 12+3=15=3(5). Check: The symbolic expression proves the claim for any integer m; the chosen value gives a separate check. Common trap: Do not conclude n+3 must be divisible by six; the expression 6m+3 has remainder three on division by six.

Worked example 19. Question: If n is a multiple of six, explain why n+3 is divisible by three. Check with n=18. Reasoning: Write n=6m. Then n+3=6m+3=3(2m+1), so it is divisible by three. Here 18+3=21=3(7). Check: The symbolic expression proves the claim for any integer m; the chosen value gives a separate check. Common trap: Do not conclude n+3 must be divisible by six; the expression 6m+3 has remainder three on division by six.

Worked example 20. Question: If n is a multiple of six, explain why n+3 is divisible by three. Check with n=24. Reasoning: Write n=6m. Then n+3=6m+3=3(2m+1), so it is divisible by three. Here 24+3=27=3(9). Check: The symbolic expression proves the claim for any integer m; the chosen value gives a separate check. Common trap: Do not conclude n+3 must be divisible by six; the expression 6m+3 has remainder three on division by six.

Counterexample to a false binomial identity

This cluster practises counterexample to a false binomial identity. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 21. Question: Claim: For all real a,b, (a+b)²=a²+b². Test a=2, b=3. Reasoning: Left=(a+b)²=25; right=a²+b²=13. Since 25≠13, this permitted input disproves the universal claim. The correct expansion includes 2ab. Check: The discrepancy 12 equals 2ab=12, the missing cross-product. Common trap: Testing a=0 or b=0 would not disprove the claim, because the missing cross-product would vanish there.

Worked example 22. Question: Claim: For all real a,b, (a+b)²=a²+b². Test a=3, b=4. Reasoning: Left=(a+b)²=49; right=a²+b²=25. Since 49≠25, this permitted input disproves the universal claim. The correct expansion includes 2ab. Check: The discrepancy 24 equals 2ab=24, the missing cross-product. Common trap: Testing a=0 or b=0 would not disprove the claim, because the missing cross-product would vanish there.

Worked example 23. Question: Claim: For all real a,b, (a+b)²=a²+b². Test a=4, b=5. Reasoning: Left=(a+b)²=81; right=a²+b²=41. Since 81≠41, this permitted input disproves the universal claim. The correct expansion includes 2ab. Check: The discrepancy 40 equals 2ab=40, the missing cross-product. Common trap: Testing a=0 or b=0 would not disprove the claim, because the missing cross-product would vanish there.

Worked example 24. Question: Claim: For all real a,b, (a+b)²=a²+b². Test a=5, b=6. Reasoning: Left=(a+b)²=121; right=a²+b²=61. Since 121≠61, this permitted input disproves the universal claim. The correct expansion includes 2ab. Check: The discrepancy 60 equals 2ab=60, the missing cross-product. Common trap: Testing a=0 or b=0 would not disprove the claim, because the missing cross-product would vanish there.

A geometric claim about every quadrilateral

This cluster practises a geometric claim about every quadrilateral. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 25. Question: Claim: Every quadrilateral has four right angles. Give a parallelogram with opposite angles 55° and 125° as a counterexample. Reasoning: A parallelogram can have alternating interior angles 55°, 125°, 55°, 125°. Opposite angles are equal, adjacent angles add to 180°, and none is 90°. Thus the universal claim is false. Check: The angle sum is 2×55+2×125=360°, consistent with a valid parallelogram. Common trap: The fact that rectangles are quadrilaterals with right angles does not make every quadrilateral a rectangle.

Worked example 26. Question: Claim: Every quadrilateral has four right angles. Give a parallelogram with opposite angles 60° and 120° as a counterexample. Reasoning: A parallelogram can have alternating interior angles 60°, 120°, 60°, 120°. Opposite angles are equal, adjacent angles add to 180°, and none is 90°. Thus the universal claim is false. Check: The angle sum is 2×60+2×120=360°, consistent with a valid parallelogram. Common trap: The fact that rectangles are quadrilaterals with right angles does not make every quadrilateral a rectangle.

Worked example 27. Question: Claim: Every quadrilateral has four right angles. Give a parallelogram with opposite angles 65° and 115° as a counterexample. Reasoning: A parallelogram can have alternating interior angles 65°, 115°, 65°, 115°. Opposite angles are equal, adjacent angles add to 180°, and none is 90°. Thus the universal claim is false. Check: The angle sum is 2×65+2×115=360°, consistent with a valid parallelogram. Common trap: The fact that rectangles are quadrilaterals with right angles does not make every quadrilateral a rectangle.

Worked example 28. Question: Claim: Every quadrilateral has four right angles. Give a parallelogram with opposite angles 70° and 110° as a counterexample. Reasoning: A parallelogram can have alternating interior angles 70°, 110°, 70°, 110°. Opposite angles are equal, adjacent angles add to 180°, and none is 90°. Thus the universal claim is false. Check: The angle sum is 2×70+2×110=360°, consistent with a valid parallelogram. Common trap: The fact that rectangles are quadrilaterals with right angles does not make every quadrilateral a rectangle.

AAA is insufficient to prove congruence

This cluster practises aaa is insufficient to prove congruence. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 29. Question: A triangle has sides 3, 4, 5 cm. Another has sides 6, 8, 10 cm. Explain why matching angle sets do not prove congruence. Reasoning: The second triangle is a scaled copy with length factor 2, so it has the same angles and is similar to the first. Its side lengths differ, so it is not congruent. This refutes the claim that AAA alone proves congruence. Check: The corresponding side ratio is 2, not 1; a rigid movement could not change side length. Common trap: Similarity is not the same as congruence when the scale factor differs from one.

Worked example 30. Question: A triangle has sides 6, 8, 10 cm. Another has sides 12, 16, 20 cm. Explain why matching angle sets do not prove congruence. Reasoning: The second triangle is a scaled copy with length factor 2, so it has the same angles and is similar to the first. Its side lengths differ, so it is not congruent. This refutes the claim that AAA alone proves congruence. Check: The corresponding side ratio is 2, not 1; a rigid movement could not change side length. Common trap: Similarity is not the same as congruence when the scale factor differs from one.

Worked example 31. Question: A triangle has sides 9, 12, 15 cm. Another has sides 18, 24, 30 cm. Explain why matching angle sets do not prove congruence. Reasoning: The second triangle is a scaled copy with length factor 2, so it has the same angles and is similar to the first. Its side lengths differ, so it is not congruent. This refutes the claim that AAA alone proves congruence. Check: The corresponding side ratio is 2, not 1; a rigid movement could not change side length. Common trap: Similarity is not the same as congruence when the scale factor differs from one.

Worked example 32. Question: A triangle has sides 12, 16, 20 cm. Another has sides 24, 32, 40 cm. Explain why matching angle sets do not prove congruence. Reasoning: The second triangle is a scaled copy with length factor 2, so it has the same angles and is similar to the first. Its side lengths differ, so it is not congruent. This refutes the claim that AAA alone proves congruence. Check: The corresponding side ratio is 2, not 1; a rigid movement could not change side length. Common trap: Similarity is not the same as congruence when the scale factor differs from one.

Domain-dependent inequality counterexample

This cluster practises domain-dependent inequality counterexample. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 33. Question: Claim: For every real x, x²≥x. Test x=1/2. Reasoning: The real number x=1/2 lies strictly between zero and one. Then x²=1/4, which is smaller than x=1/2. This disproves the claim over all real numbers. Check: Multiplying both sides of 1/4<1/2 by the positive 4 gives 1<2, true. Common trap: A claim that works for nonnegative integers need not work for every real number.

Worked example 34. Question: Claim: For every real x, x²≥x. Test x=1/3. Reasoning: The real number x=1/3 lies strictly between zero and one. Then x²=1/9, which is smaller than x=1/3. This disproves the claim over all real numbers. Check: Multiplying both sides of 1/9<1/3 by the positive 9 gives 1<3, true. Common trap: A claim that works for nonnegative integers need not work for every real number.

Worked example 35. Question: Claim: For every real x, x²≥x. Test x=1/4. Reasoning: The real number x=1/4 lies strictly between zero and one. Then x²=1/16, which is smaller than x=1/4. This disproves the claim over all real numbers. Check: Multiplying both sides of 1/16<1/4 by the positive 16 gives 1<4, true. Common trap: A claim that works for nonnegative integers need not work for every real number.

Worked example 36. Question: Claim: For every real x, x²≥x. Test x=1/5. Reasoning: The real number x=1/5 lies strictly between zero and one. Then x²=1/25, which is smaller than x=1/5. This disproves the claim over all real numbers. Check: Multiplying both sides of 1/25<1/5 by the positive 25 gives 1<5, true. Common trap: A claim that works for nonnegative integers need not work for every real number.

An implication and its false converse

This cluster practises an implication and its false converse. Decide whether the task asks for a counterexample, a general argument or a check of one specific case. Identify the relevant domain and use exact equalities or inequalities so another reader can audit the conclusion.

Worked example 37. Question: Claim: If an integer is divisible by six, it is divisible by three. Justify this generally and test the converse using 3. Reasoning: For any integer m, writing x=6m gives x=3(2m), so every multiple of six is divisible by three. But 3=3(1) is divisible by three and is not divisible by six because it is an odd multiple of three. Hence the converse is false. Check: The separate value 12 is divisible by six and three, illustrating the original true direction; 3 disproves the reverse direction. Common trap: One valid implication does not automatically imply its converse.

Worked example 38. Question: Claim: If an integer is divisible by six, it is divisible by three. Justify this generally and test the converse using 9. Reasoning: For any integer m, writing x=6m gives x=3(2m), so every multiple of six is divisible by three. But 9=3(3) is divisible by three and is not divisible by six because it is an odd multiple of three. Hence the converse is false. Check: The separate value 18 is divisible by six and three, illustrating the original true direction; 9 disproves the reverse direction. Common trap: One valid implication does not automatically imply its converse.

Worked example 39. Question: Claim: If an integer is divisible by six, it is divisible by three. Justify this generally and test the converse using 15. Reasoning: For any integer m, writing x=6m gives x=3(2m), so every multiple of six is divisible by three. But 15=3(5) is divisible by three and is not divisible by six because it is an odd multiple of three. Hence the converse is false. Check: The separate value 24 is divisible by six and three, illustrating the original true direction; 15 disproves the reverse direction. Common trap: One valid implication does not automatically imply its converse.

Worked example 40. Question: Claim: If an integer is divisible by six, it is divisible by three. Justify this generally and test the converse using 21. Reasoning: For any integer m, writing x=6m gives x=3(2m), so every multiple of six is divisible by three. But 21=3(7) is divisible by three and is not divisible by six because it is an odd multiple of three. Hence the converse is false. Check: The separate value 30 is divisible by six and three, illustrating the original true direction; 21 disproves the reverse direction. Common trap: One valid implication does not automatically imply its converse.

Six reasoning errors that deserve a precise correction

Calling several supporting examples a universal proof

A general argument must explain why every permitted case works; a handful of cases cannot normally establish this. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

Using an example outside the stated domain

A counterexample must satisfy the original assumptions before showing a failed conclusion. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

Confusing not all with none

One failure disproves a universal claim but does not prove that every possible case fails. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

Reversing an if–then statement without justification

A valid implication can have a false converse; test the reverse separately. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

Drawing a geometric conclusion from appearance

Use marked angle, side, parallel or congruence conditions, not a not-to-scale sketch alone. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

Checking a proposed solution but not completeness

One working solution may not be the only solution. If the question asks for all, examine whether other candidates remain. The tutor should highlight the first unsupported assertion, replace it with a justified step and ask the learner to test a related unfamiliar claim without a prompt. The aim is valid logic rather than more elaborate wording.

How a Bukit Timah tutor can make mathematical explanations clearer

A Secondary 3 G3 Mathematics tutor in Bukit Timah can give three statements: ‘all even numbers are multiples of four’, ‘the sum of two odd numbers is even’, and ‘all equal-angle triangles are congruent’. Students classify each and then provide either one valid counterexample or a general reason. This contrasts what it takes to disprove a claim with what it takes to justify a true one.

In a small group, pupils can critique the exact statement rather than each other. The tutor models one symbolic argument, removes the prompt for the next claim and asks students to revise their conclusion when the domain changes. A delayed task shows whether they can recognise an invalid implication or missing condition independently.

The unchanged Clementi Secondary 1 Mathematics small-group guide provides a foundation-level teaching example. The Secondary 3 G3 Mathematics Bukit Timah overview is the relevant year-level context. School chapter order and current class arrangements should be confirmed separately.

A six-week reasoning and communication plan

Week 1 — read claims exactly

Underline every, some, if, then, and domain words. Classify what would be needed to justify or disprove the statement. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Week 2 — practise counterexamples

Find permitted examples that break over-strong divisibility, algebraic and geometric claims. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Week 3 — express integer patterns generally

Use n, n+1 and 2n+1 to explain consecutive-number and parity statements. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Week 4 — explain conditional statements

Compare implications with their converses and identify where an assumption is necessary. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Week 5 — justify geometry conclusions

Name the angle, parallel or congruence condition before stating the resulting equality. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Week 6 — write concise arguments for unseen claims

Give new true and false statements, require a reason or counterexample, and compare explanation quality with Week 1. Keep one dated original claim-analysis attempt and a later unfamiliar one. The key progress measure is whether the student now identifies assumptions and justifies the conclusion without copying a template.

Frequently asked questions

What is a mathematical counterexample?

A permitted case that makes a proposed universal statement false. Try a new statement and require the learner to name the domain and first justified step before concluding.

Can a single counterexample disprove an all-case statement?

Yes, provided the example meets the statement’s assumptions and violates its conclusion. Try a new statement and require the learner to name the domain and first justified step before concluding.

Can ten successful examples prove a universal claim?

Not usually. A general reason is needed unless an explicitly finite domain has been exhaustively checked. Try a new statement and require the learner to name the domain and first justified step before concluding.

How do we represent an even integer?

As 2n where n is an integer. Try a new statement and require the learner to name the domain and first justified step before concluding.

How do we represent an odd integer?

As 2n+1 where n is an integer. Try a new statement and require the learner to name the domain and first justified step before concluding.

Why is the sum of three consecutive integers divisible by three?

It is n+(n+1)+(n+2)=3(n+1), visibly a multiple of three. Try a new statement and require the learner to name the domain and first justified step before concluding.

Is the converse of a true implication always true?

No. Divisible by six implies divisible by three, but the reverse fails for x=3. Try a new statement and require the learner to name the domain and first justified step before concluding.

Can a geometry diagram alone prove a right angle?

No. The right angle must be given or derived from valid geometric conditions. Try a new statement and require the learner to name the domain and first justified step before concluding.

How can a domain change a result?

An assertion valid over integers may be false over real numbers, such as x²≥x for all nonnegative values. Try a new statement and require the learner to name the domain and first justified step before concluding.

Is this reasoning part of 2027 K310?

AO3 explicitly assesses justifying mathematical statements, explaining problems and writing mathematical arguments, though the examples here are instructional rather than official test predictions. Try a new statement and require the learner to name the domain and first justified step before concluding.

Official curriculum reference: 2027 SEC G3 Mathematics K310, Assessment Objectives AO2 and AO3. These examples are original learning material, not a published list of guaranteed exam proofs.

The takeaway

Did you know? A strong mathematical argument does not have to be long. It needs a precise claim, the correct conditions, a valid reason and an appropriate conclusion. One counterexample can disprove a false universal statement; a clear general relationship can justify a true one. Those habits make unfamiliar G3 Mathematics questions easier to evaluate with confidence.

Secondary 3 G3 Mathematics Tutor Bukit Timah | Mathematical Reasoning: Counterexamples and Justifying Answers