JC1 H3 Mathematics Tutor Bukit Timah | Proof by Cases and Construction. Did you know? An apparently difficult proof may become simple when you divide the possible inputs into a few complete categories—or when you build exactly the object that the question asks you to show exists.
For students learning JC1 H3 Mathematics preparation in Bukit Timah, Sixth Avenue and Beauty World, proof by cases and proof by construction strengthen the connection between H2 algebra and H3-style reasoning. This guide gives quick answers, original examples, practice questions and checks that parents or tutors can use.
The Singapore–Cambridge H3 Mathematics 9820 syllabus explicitly includes these reasoning methods and assumes H2 Mathematics. A JC1 student can develop the skills ahead of possible later H3 study; this is not a claim that the college offers an H3 class in JC1 or that selection is guaranteed.
The quick difference between cases and construction
A proof by cases covers all possible inputs by dividing the domain into categories and proving the claim in every category. A proof by construction supplies an object that satisfies the conditions of an existence question, then verifies every requested property.
- Cases must be exhaustive: for integers, even or odd covers all inputs.
- For real x, x<0, x=0 or x>0 covers all inputs; omitting zero would leave a gap.
- A construction may depend on the parameter given in the problem.
- A counterexample is a construction used to disprove a universal claim.
- A construction proves existence but not automatically uniqueness.
Worked example 1: the product of consecutive integers is even
Let n be any integer. We want to prove n(n+1) is even. The integer n is either even or odd, and those cases cover every possible n.
If n=2k for some integer k, the product is 2k(n+1), which is even. If n=2k+1, then n+1=2(k+1) is even, and the product is again even. Thus the statement holds for every integer n.
This argument is much stronger than testing n=1,2,3 and finding even answers. It also covers negative integers and zero. The logical reason is that among two consecutive integers, one is necessarily even.
Worked example 2: squares modulo four
Every integer is even or odd. An even integer n=2k has square n²=4k², divisible by four. An odd integer n=2k+1 has square 4k(k+1)+1, leaving remainder one on division by four.
Therefore an integer square can have only remainder zero or one modulo four. It can never have remainder two or three. For example, 2027 leaves remainder three when divided by four, so it cannot be a perfect square.
Did you know? This proof rules out infinitely many possibilities with only two cases. Remainder classes are a powerful way to make the hidden structure behind an integer statement visible.
Worked example 3: three consecutive integers contain a multiple of three
Take n,n+1,n+2. Depending on n, its remainder modulo three is zero, one or two. If it is zero, n itself is divisible by three. If it is one, n+2 is divisible by three. If it is two, n+1 is divisible by three.
These are all possible remainders, so the argument proves that one of the three consecutive integers is a multiple of three. In fact exactly one is, because the three numbers belong to different remainder classes.
Combining this fact with the presence of an even factor shows that n(n+1)(n+2) is always divisible by six. Factorisation and case reasoning work together without requiring a cubic expansion.
Worked example 4: absolute values are naturally piecewise
For any real x, |x|=x if x≥0 and |x|=−x if x<0. If x≥0, then |x| is nonnegative directly. If x<0, then −x is positive. Hence |x|≥0 for all real x.
The boundary zero belongs to the first branch. A rushed explanation that says “absolute values are positive” is technically wrong, because |0|=0. Precise language matters when an inequality is strict.
The same discipline appears in piecewise functions, geometry and optimisation: define the branches, solve within each and check what happens at the switching point.
Worked example 5: solve an absolute-value equation
Solve |x−2|=3. For x≥2, we have x−2=3, giving x=5; this satisfies x≥2. For x<2, we have 2−x=3, giving x=−1; this satisfies x<2.
The solution set is {−1,5}. Always check that each candidate actually belongs to the case that produced it. Solving an algebraic branch can otherwise create invalid candidates.
Try |x−4|≤2 as an extension. The branches give 4≤x≤6 and 2≤x<4; combining them gives 2≤x≤6, including the endpoints where the inequality is satisfied.
Worked example 6: construct a larger even integer
Given any integer n, we want to show that an even integer m exists with m>n. Choose m=2(|n|+1). The expression is an even positive integer because |n|+1 is a positive integer.
Also m=2|n|+2>|n|≥n, proving m>n. The same formula works for negative numbers, positive numbers and zero, so the construction covers the complete specified domain.
The number m is allowed to depend on n. We have not claimed that there is a single largest even integer; no such integer exists.
Worked example 7: construct a larger odd integer
For an arbitrary integer n, choose m=2(|n|+1)+1. This is odd and greater than n, because m>|n|≥n.
Alternatively, consider the two cases of n’s parity and choose whichever of n+1 and n+2 is odd. The second method is both a construction and a proof by cases. Either is valid when all requirements are checked.
The object has to satisfy the claimed properties simultaneously. Writing a plausible formula without demonstrating its parity and inequality conditions is an incomplete proof.
Worked example 8: construct a rational between two rationals
Suppose a and b are rational numbers with a<b. Define c=(a+b)/2. The number c is rational because rational numbers are closed under addition and division by nonzero integers.
Since a<b, we have 2a<a+b<2b. Divide by positive two to obtain a<c<b. Thus a rational always exists strictly between two distinct rationals.
The formula works for every eligible pair. Checking only one interval, such as from 1/3 to 1/2, would be an example rather than proof of the general statement.
Worked example 9: construct two distinct interior real numbers
Given reals a<b, define c=(2a+b)/3 and d=(a+2b)/3. Direct subtraction gives c−a=(b−a)/3, d−c=(b−a)/3 and b−d=(b−a)/3.
All three expressions are positive because a<b. Therefore a<c<d<b: the constructed numbers are strictly interior and distinct.
This approach generalises to any finite number of points by dividing the interval into equal parts. A good construction makes its required properties follow transparently from its definition.
Worked example 10: infinitely many positive multiples
Let d be a positive integer. Define mₖ=dk for each positive integer k. Every mₖ is positive and divisible by d.
If k₁ and k₂ differ, then dk₁ and dk₂ differ because d is positive. Hence the construction provides infinitely many distinct positive multiples of d.
A short list d,2d,3d is suggestive, but a general rule with a distinctness argument proves the result for infinitely many objects.
Worked example 11: three different integers with a required sum
To show that three distinct positive integers can sum to 18, choose 3,6,9 and verify the total. But suppose the claim is stronger: for every integer N≥6, construct three distinct positive integers with sum N.
The triple 1,2,N−3 works. Its sum is N and, because N≥6, its third entry is at least three. All three numbers are positive and distinct.
The smallest allowed input matters. If N=5, the same formula gives 1,2,2 and fails distinctness. This is why a construction must be checked against the precise parameter range.
Worked example 12: construct a nonsquare rectangle
For any positive real perimeter P, choose the rectangle’s side lengths to be P/6 and P/3. Both lengths are positive and unequal, so the rectangle is not a square.
Its perimeter is 2(P/6+P/3)=P. We have verified the positivity, required perimeter and unequal sides, establishing existence for every P>0.
One example such as a 4-by-6 rectangle only solves the special case P=20. A rule expressed using the given parameter is needed for the general assertion.
Worked example 13: construct points with no three collinear
For an arbitrary positive integer N, choose N distinct points (k,k²), with k=1,2,…,N. They lie on the parabola y=x² and have distinct first coordinates.
A nonvertical line y=mx+c meets the parabola where x²−mx−c=0, an equation with at most two distinct real roots. A vertical line meets the graph only once.
Consequently, no line contains three of the chosen points. The construction turns an existence problem in geometry into a clear algebraic property.
Worked example 14: construct a counterexample by weakening assumptions
When a and b are positive, (a+b)²>a²+b² because the difference equals 2ab>0. Does this remain true for any real a and b? No.
Take a=1 and b=−1. Then (a+b)²=0 while a²+b²=2. This constructed pair disproves the unrestricted statement.
A counterexample is not just any numerical example. It must satisfy the claim’s hypothesis and violate its conclusion. Here the positivity condition is exactly what protected the original inequality.
A checklist for complete cases and valid constructions
- State the domain and decide whether you need a universal proof or a witness.
- Choose exhaustive cases, including zero and any relevant boundary.
- Prove the same required conclusion in every case.
- For construction, give an explicit object or rule and verify each condition.
- Check positivity, integrality, distinctness, nonzero denominators and parameter bounds.
- Do not overclaim uniqueness simply because one object has been constructed.
Eight quick classroom questions
- 1. Prove n(n+1) is even for all integers.
- 2. Show that square numbers have only remainders zero or one modulo four.
- 3. Show that among n,n+1,n+2 one is divisible by three.
- 4. Solve |x+1|=2 by splitting into cases.
- 5. Construct an odd m>n for any given integer n.
- 6. Construct four distinct integers that are each congruent to one modulo three.
- 7. Construct two rational numbers strictly between 1/5 and 1/2.
- 8. Explain what goes wrong if a proof only checks positive n but claims every integer.
Answer key with reasoning
For 1, one consecutive factor is even. For 2, use even and odd cases. For 3, the three consecutive values cover all remainder classes modulo three. For 4, x=1 or x=−3, and both check in the original equation.
For 5, use m=2(|n|+1)+1. For 6, choose 1,4,7,10. For 7, choose 3/10 and 2/5. For 8, zero and negative integers are part of the original domain and remain unproved.
Independent transfer problems
- A. Show n²+n is always even.
- B. List possible remainders of n² modulo three.
- C. Construct a positive multiple of seven greater than any real T.
- D. Solve |2x−1|=5 with sign checks.
- E. Construct four different positive integers whose sum is 22.
- F. Prove every real x satisfies x≥0 or −x>0.
- G. Construct three distinct numbers inside any real interval (a,b).
- H. Disprove (a+b)²≥a²+b² for all real a,b.
Transfer answers and corrections
A: factor n(n+1). B: residues 0,1,2 square to 0,1,1 modulo three. C: choose a positive integer k>max(T/7,0) and take 7k. D: x=3 and x=−2 both satisfy the original.
E: one solution is 2,4,6,10. F: split into x≥0 and x<0. G: divide the interval into four equal pieces and take its three internal division points. H: a=1,b=−1 gives zero on the left and two on the right.
A practical JC1 tutoring and revision routine
Begin with parity and absolute-value cases, then move into parameterised constructions. Let learners write a short explanation of why their cases cover the whole domain and why each construction meets the question’s conditions.
After several days, give altered problems without indicating the method. A learner who can choose between cases, direct factorisation, counterexample and construction independently has developed a more transferable skill than one who simply recognises yesterday’s worksheet.
Parents can ask one changed question: does the result still work when n is negative, when the parameter is smallest, or when a boundary equals zero? The quality of the explanation matters more than how quickly the answer is spoken.
Advanced application 1: prove a square’s parity using both directions
Try to prove: for every integer n, n is even if and only if n² is even. A two-way claim requires two implications. For the forward direction, let n=2k. Then n²=4k², which is even.
For the reverse direction, suppose n² is even. Split n into the exhaustive cases even and odd. If n is odd, write n=2k+1 and calculate n²=4k(k+1)+1, which is odd, contradicting the assumption. Therefore n cannot be odd and must be even.
This example combines cases and logical equivalence. It also shows why a theorem’s reverse direction sometimes benefits from a different style of proof than the forward direction.
Advanced application 2: every integer has a nearby multiple of five
Given an integer n, construct an integer k chosen from {0,1,2,3,4} such that n+k is divisible by five. Instead of guessing k, divide into the five possible remainder cases for n.
If n has remainder zero, choose k=0. If its remainder is one, choose k=4. For remainder two choose k=3; remainder three choose k=2; remainder four choose k=1. In every case, n+k leaves remainder zero modulo five.
The construction is exhaustive because there are exactly five remainder classes. It also controls the size of k. Simply choosing k=5n would not generally make n+k a multiple of five, and choosing an arbitrary large k might violate the specified set.
Advanced application 3: build a Pythagorean triple for every integer m≥2
A constructive proof can generate infinitely many geometric examples. Let m≥2 be an integer and define a=m²−1, b=2m and c=m²+1. These are all positive integers: m²−1≥3, 2m≥4, and m²+1≥5.
Now verify the Pythagorean property. The sum a²+b² equals (m²−1)²+4m² = m⁴−2m²+1+4m² = m⁴+2m²+1 = (m²+1)² = c². Thus (a,b,c) forms a right-triangle integer triple for every m≥2.
Taking m=2 yields (3,4,5); m=3 yields (8,6,10). The family does not claim that every possible Pythagorean triple has this form. It establishes an infinite construction, not a classification of all solutions.
Advanced application 4: choosing between cases and direct algebra
Suppose a question asks you to show that n³−n is divisible by six for every integer n. One possible route splits n into six congruence classes modulo six and checks them all. That method works, but it creates six separate branches.
A clearer route factors n³−n=(n−1)n(n+1). The three factors are consecutive integers. Exactly one is divisible by three, and at least one is even, so the product is divisible by six.
The lesson is not to reject proof by cases. It is to compare candidate methods for clarity and completeness. A good mathematician asks what structure makes the theorem true and chooses the shortest transparent demonstration.
Advanced application 5: an inequality that needs a split at zero
Show for every real x that |x|≥x and |x|≥−x. If x≥0 then |x|=x, so the first inequality is equality, and the second becomes x≥−x, which follows from 2x≥0.
If x<0 then |x|=−x, so the second inequality is equality, and the first becomes −x≥x, which follows from −2x>0. Both inequalities hold in both cases, completing the proof.
This argument is small but useful: the expression |x| is the larger of x and −x. That observation often provides a quick bound in future inequality or optimisation problems.
Advanced application 6: a construction may require two separate checks
Suppose for every positive integer N≥4 you must construct two different positive integers a and b with a+b=N. One simple choice is a=1 and b=N−1. Both are positive, but are they always distinct? At N≥4, b≥3 and differs from a.
So this construction works for every N≥4. It would fail the distinctness condition at N=2, where it gives (1,1). This is why a parameter lower bound appears in a rigorous proof, not as an arbitrary technical decoration.
For a harder extension, suppose the two numbers must both be even. That is impossible if N is odd, because the sum of two evens is even. A case distinction on N’s parity first tells us when a construction is even possible.
A 20-minute proof workshop for tutor and student
Begin with five minutes of reading. Present three claims: a universal integer statement, an existence statement and an existence statement depending on an arbitrary input. Ask the student to mark the quantifier, the domain and all restrictions without solving anything.
Use the next seven minutes to attempt an integer remainder proof, such as the multiple-of-five construction. The student should choose and label each residue case, produce k and verify n+k≡0 (mod 5). If one case is forgotten, do not provide its answer immediately; ask the learner to list all possible remainders.
Use five more minutes for a Pythagorean-triple construction. The student should check positivity and the identity a²+b²=c² independently. Encourage algebraic expansion rather than accepting a few numerical triples as universal proof.
Finish with a three-minute reflection: which method established existence, which established a universal statement and which required a parameter restriction? Ask the learner to write a one-sentence summary of the first wrong inference they made and how to correct it next time.
Study notes: what a fully checked construction looks like
- A clear opening defines every input and specifies its allowed set.
- The proposed object is written explicitly as a function of the inputs.
- Every required property receives a separate verification, using the original definitions.
- A boundary case is checked if a parameter is restricted by a lower or upper bound.
- The conclusion states exactly what exists, without claiming uniqueness or completeness of the classification unless separately proved.
At home, a helpful follow-up is to hide the model construction and ask the student to make one of their own. Comparing different valid constructions reveals that a proof often has more than one possible route. The aim is to understand why the route works, not to reproduce an answer word for word.
A proof editor’s clinic: repair four tempting but incomplete arguments
Draft A: “I tested n=1,2,3,4 and n(n+1) is always even, so the theorem is proved.” The calculation is correct for those inputs, but it does not establish a universal result. Repair it by splitting an arbitrary integer n into even and odd cases and proving each one.
Draft B: “Take sides 1 and N−1; the pair is distinct for any positive N.” The assertion fails for N=2, when the sides are both one, and can fail positivity for N=1. Repair it by specifying an adequate lower bound on N, such as N≥4, and checking the properties.
Draft C: “For x>0, |x|=x, and for x<0, |x|=−x, so |x|≥0 for every real x.” Although the conclusion is true, the proof omitted x=0. Repair it by using x≥0 in the first case, or by adding a separate zero case.
Draft D: “I constructed one triple satisfying a²+b²=c²; therefore there are infinitely many such triples.” One successful example does not establish infinitude. Repair it by giving the parameterised construction (m²−1,2m,m²+1) for integers m≥2 and explaining why the generated triples are distinct as m increases.
How to critique a proof without merely saying it is wrong
A constructive critique names the exact gap, explains why it matters, and supplies a valid replacement. For instance: “The proof only handles x>0 and x<0. Since the stated domain includes zero, we must also check x=0; the absolute-value claim then follows.” That sentence is far more informative than “you forgot something”.
Sometimes the gap is not a missing case but a theorem used outside its assumptions. A construction requiring division by d is not automatically valid when d=0. A formula describing positive values must be checked when a parameter approaches its smallest permitted value.
In a JC Mathematics group, have learners swap anonymous proof drafts and identify whether the issue concerns coverage, existence, uniqueness, domain, reversibility or an invalid inference. Asking for a corrected proof helps students develop the mathematical communication expected in more advanced work.
An independent mastery check: choose the method before calculating
- Can you tell the difference between a universal claim, an existential claim and a uniqueness claim?
- Can you state why your cases cover the entire given domain?
- Can you distinguish a numerical example from a general argument?
- Can you produce a parameterised construction and verify every required property?
- Can you handle zero, negative inputs and the smallest allowed parameter?
- Can you identify exactly where an otherwise plausible proof becomes invalid?
If a student can answer these questions and solve a changed problem after a few days, they have developed a more durable foundation than memorising an impressive-looking proof. Clear reasoning, careful definitions and honest checking make subsequent H3 preparation more manageable.
Frequently asked questions
Are proof by cases and construction explicitly included in H3 Mathematics?
Yes. The official 2027 H3 Mathematics 9820 syllabus lists both under mathematical proofs and reasoning principles, with H2 Mathematics content assumed.
Are a few examples the same as proof by cases?
No. A valid proof by cases uses categories that cover every allowed object. Checking a handful of integers does not prove a statement about every integer.
Does one constructed example prove uniqueness?
No. It establishes existence. A separate uniqueness argument must rule out two different objects satisfying the conditions.
Does JC1 proof practice guarantee H3 enrolment?
No. The student’s school controls H3 offerings, eligibility and timing. These exercises strengthen H2 mathematical thinking and possible later H3 readiness.
Where to go next
Continue with quantifiers and negation, existence and uniqueness and induction. For the broader H2 foundation, see JC Mathematics tuition in Bukit Timah.
The immutable eduKateSG small-group Mathematics reference remains unchanged. For JC subject support near Sixth Avenue MRT, enquire with eduKate Singapore and confirm relevant tutor expertise, actual class availability, fees and times. Additional resources are at the Bukit Timah Tuition Hub.