Discrete mathematics

The definitive reference. Compiled 15 September 2026 for a study site aimed at a study score of 45+. Every substantive claim carries a source tag. Where the archive is silent, that is said, not filled in.

This document is the companion to research/sm/01-study-design.md and does not contradict it. The single most important fact it inherits from that document: proof sat in Specialist Mathematics Units 1 and 2 throughout the 2016–2022 accreditation period and arrived in Units 3 and 4 only with the 2023–2027 design. There is therefore no legacy proof practice in the pre-2023 archive — but there is a great deal of legacy proof-writing practice, in the form of "show that" questions, and the reports on those are the best available evidence about how VCAA marks a written argument.

Contents

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0. Sources and evidence tags

0.1 Corpus limits that constrain this document

These are real and they bound what can honestly be claimed here.

• questions.json contains no NHT Examination 1 at all, and nothing from 2026. It holds November papers 2006–2025 plus NHT Examination 2 Section A only for 2024 and 2025. So the three live NHT induction proofs — 2024 NHT Exam 1 Q8, 2025 NHT Exam 1 Q3, 2026 NHT Exam 1 Q2 — exist in [PAPERS] and [GUIDE] but carry no percentage, because VCAA does not publish NHT mark distributions.
• The 2024 November papers have no text layer. Documents_exams_mathematics_2024_2024specmaths1-w.pdf and ...2024specmaths2-w.pdf are image-only; the corresponding .txt files are empty headers. The literal wording of 2024 Exam 1 Q2 (the direct-proof question) and 2024 Exam 2 Section A Q1 (the contrapositive item) cannot be quoted from the archive. Everything said about them below comes from [RPT24 E1], [RPT24 E2] and [GUIDE], and is labelled as such.
• The 2025 NHT papers are likewise image-only. 2025 NHT Exam 2 Section A Q1 is tagged Discrete mathematics in [QJSON] with correct answer D, but its text cannot be recovered. It is counted in the archive tally and excluded from the catalogue.
• The 2011 papers are unreadable (CID-shifted font encoding). [QJSON] carries 13 duplicate ref values, twelve of them from 2011, where part labels collided during reconstruction. Any 2011 percentage quoted below is flagged.
• The topic field is null for a substantial block of 2009–2010 entries, so topic-based filtering understates the archive. This document therefore searches the paper and report text as well as the topic tag.

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1. What the study design puts in this area

1.1 The area of study, verbatim

[SD] overview:

"In this area of study students cover the development of mathematical argument and proof. This includes conjectures, connectives, quantifiers, examples and counter-examples, and proof techniques including mathematical induction. Proofs will involve concepts from topics such as: divisibility, inequalities, graph theory, combinatorics, sequences and series including partial sums and partial products and related notations, complex numbers, matrices, vectors and calculus. The concepts, skills and processes from this area of study are to be applied in the other areas of study."

[SD] content:

This area of study includes:
- conjecture – making a statement to be proved or disproved
- implications, equivalences and if and only if statements (necessary and sufficient conditions)
- natural deduction and proof techniques: direct proofs using a sequence of direct implications, proof by cases, proof by contradiction, and proof by contrapositive
- quantifiers 'for all' and 'there exists', examples and counter-examples
- proof by mathematical induction.

Five dot points. That is the whole content list. It is the smallest area of study in the design and, on the evidence of 2023–2026, it generates between 4 and 6 marks per year across both papers.

1.2 The key knowledge and key skills that carry proof

[SD] attaches key knowledge and key skills to the three outcomes, not to the areas of study (see [01SD] §1.1). Six items carry this area:

Outcome 1 key knowledge, verbatim:

• principles of proof and deduction techniques
• the proof scheme and method for mathematical induction

Outcome 1 key skills, verbatim:

• apply deductive reasoning and language, including mathematical induction, to mathematical arguments and proofs involving concepts and contexts from the areas of study

Outcome 2 key knowledge, verbatim:

• the role of examples, counter-examples and general cases in working mathematically
• the role of proof in establishing a general result

Outcome 2 key skills, verbatim:

• develop mathematical formulations of specific and general cases used to derive results for analysis within a given context and establish proofs for general case results

Outcome 3 key skills, verbatim:

• produce results, using a technology, which identify examples or counter-examples for propositions

Four consequences that a 45+ candidate should treat as operational:

1. "the proof scheme and method for mathematical induction" — the words are scheme and method. VCAA is examining the ritual, not only the algebra. This is why [GUIDE] awards a mark for the base case and a separate mark for stating the assumption, before any algebra happens (§3.1).
2. "deductive reasoning and language" — language. Sentences count, and [RPT25 E1] lists "Misstating the assumption" as a named error.
3. Outcome 2's "establish proofs for general case results" is the sentence that licenses a "show that" or "prove that" instruction inside a Calculus, Vectors or Complex Numbers question. [SPEC] puts the emphasis of Examination 2 on Outcome 2, which is why the proof-flavoured load on Examination 2 falls mostly on Section B "show that" parts rather than on a standalone proof question.
4. Outcome 3's counter-example skill is technology-active, so a counter-example question is natural on Examination 2 and has in fact only ever appeared there (2025 Exam 2 Section A Q2).

1.3 Exactly when each part entered the course

The decisive negative finding, reproduced from [01SD] §5.4 and re-verified here: the words induction, contradiction and contrapositive appear in no Specialist Mathematics examination paper before 2023. A case-insensitive search across all 2006–2022 paper files returns zero hits for all three. The 2023 NHT papers are also clean — [01SD] §5.6 establishes that the 2023 NHT sitting was still a 2016–2022-design paper, and the corpus confirms it: neither 2023 NHT paper contains any of the three words.

1.4 What is explicitly not in this area

• No number theory as content. "Euclidean", "modular", "prime" and "greatest common divisor" appear nowhere in the Units 3 and 4 section of [SD], and a corpus search for Euclidean algorithm|gcd|highest common factor across all 97 files returns zero hits. Every modul hit in the corpus is "modulus". Divisibility appears only as a context for proof ([SAMPLE] Exam 1 Q3), leaning on Specialist Units 1 and 2 as assumed knowledge. A question cannot examine modular arithmetic for its own sake.
• No truth tables, no formal logic notation. ∧, ∨, ¬, ⇒ as symbols to be manipulated are absent from [SD] and from every paper. The contrapositive items are all set in English sentences, not symbols (§2.11).
• No set-theoretic proof apparatus.
• No strong induction named — though "proof by mathematical induction" is unqualified, so a strong-induction step is not excluded by the wording. It has never been required.
• No proof of irrationality. A search of all 97 corpus files for irrational returns zero hits. The √2 irrationality proof is the archetype every textbook teaches and VCAA has never set it in Units 3 and 4. It is nonetheless the canonical illustration of contradiction and remains the most likely unexamined contradiction question; treat it as drill, not as a prediction.
• No graph theory or combinatorics proofs, despite both being named in the overview. They are Units 1 and 2 content used as proof contexts; neither has appeared.

1.5 Which parts of the archive are out of scope, and which older questions still drill useful skills

Out of scope, do not drill:

In scope and still the best drill available:

The honest summary: the pre-2023 archive contains zero induction, contradiction, contrapositive or counter-example questions, and roughly ninety "show that" question parts whose marking commentary is directly transferable. Drill the first group from [SAMPLE] and 2023–2026; drill the second group from everywhere.

1.6 How much this area is worth

The pattern is stable and worth memorising. Every Examination 1 since 2023 has carried exactly one proof question, worth 3 or 4 marks, and in five of the six sittings it has been induction. Every Examination 2 since 2023 has opened Section A with a logic item at Question 1 — a contrapositive four times out of five. That is roughly 5%–10% of Examination 1 and 1.25–2.5% of Examination 2, but it is a slice that is fully learnable in an afternoon, which is why a 45+ candidate should treat it as non-negotiable.

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2. The complete catalogue of question types

Twenty types. For each: the name, the literal VCAA wording template with its source clearly labelled, what it tests, the required structure of the model answer, every archive instance with ref and pct, the typical mark allocation, and the traps the reports name.

Throughout, bold percentages are separators (pct ≤ 50). NHT questions carry no percentage — see §0.1.

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2.1 Induction on a finite sum (series identity)

Wording template — real paper. 2025 Exam 1 Q7 (4 marks) [PAPERS]:

"Use mathematical induction to prove that Σ_{i=1}^{n} (i+1)² = (1/6)·n(2n² + 9n + 13) for n ∈ N, where Σ_{i=1}^{n}(i+1)² = 2² + 3² + 4² + … + (n+1)²."

2024 NHT Exam 1 Q8 (4 marks) [PAPERS]:

"Prove by mathematical induction that 1×7 + 2×15 + 3×23 + … + n(8n − 1) = (1/6)·n(n + 1)(16n + 5) for all n ∈ N."

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q1 (4 marks), scaffolded into three parts:

"Consider the statement 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 − 1/2ⁿ, where n ∈ N.
a. Show that if n = 1, the statement is true. (1 mark)
b. Assume that the statement is true for n = k. Write down the assumption in terms of k. (1 mark)
c. Hence, prove by mathematical induction that 1/2 + 1/4 + 1/8 + … + 1/2ⁿ = 1 − 1/2ⁿ, where n ∈ N. (2 marks)"

What it tests. The proof scheme, plus one piece of algebra: adding the (k+1)th term to the assumed sum and factorising the result into the target form with n replaced by k+1.

Required structure of the model answer (this is [GUIDE] for 2025 Exam 1 Q7, reproduced in VCAA's own order):

1. Name the proposition. "Let P(n) be the proposition that …, for all n ∈ N." ([RPT25 E1] opens its sample answer this way.)
2. Base case, computed on both sides separately. "Show that it is true for n = 1: LHS = …, RHS = …" — [GUIDE]: M1.
3. Assumption, stated as an equation in k. "Assume true for n = k: …" — [GUIDE]: M1.
4. Statement of what is required. "Show that it is true for n = k + 1. Required to prove: …"
5. The inductive step, starting from the LHS at n = k+1 and substituting the assumption — [GUIDE]: A1 — "must see use of assumption".
6. Conclusion. "By the principle of mathematical induction, P(n) is true for n ∈ N." — [GUIDE]: A1* — "working required".

Archive instances.

Typical marks: 4. One for the base case, one for the assumption, one for using the assumption, one for the conclusion with complete working.

Traps the reports name. [RPT25 E1] on Q7, verbatim and in full:

"This question was not answered well. Some common errors included:
- Not properly verifying the base case n = 1.
- Misstating the assumption. For example, 'Suppose the proposition is true for n = k. Then …'
- Assuming equality at the beginning of the inductive step."

The third bullet is the killer and is discussed at length in §5.2: writing LHS(k+1) = RHS(k+1) on the first line of the inductive step and then "simplifying both sides" is assuming what is to be proved, and it costs the A1*.

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2.2 Induction on divisibility

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q3 (4 marks):

"Prove by mathematical induction that the number 9ⁿ − 5ⁿ is divisible by 4 for all n ∈ N."

Wording template — real paper, parity variant. 2026 NHT Exam 1 Q2 (3 marks) [PAPERS]:

"Prove by mathematical induction that the number given by n² + 5n is even for all n ∈ N."

What it tests. The same scheme, with one extra move: the inductive step must produce an explicit multiple of the divisor. The candidate must write the n = k+1 expression as d × (integer) where d is the divisor, using the assumption 9ᵏ − 5ᵏ = 4m for some m ∈ Z.

Required structure.

1. Base case, evaluated: 9¹ − 5¹ = 4, which is divisible by 4.
2. Assumption with an explicit integer witness. "Assume 9ᵏ − 5ᵏ = 4m for some m ∈ Z." The witness variable is what makes the step work; omitting it leaves nothing to substitute.
3. Inductive step: rewrite 9^{k+1} − 5^{k+1} so the assumption can be substituted — e.g. 9^{k+1} − 5^{k+1} = 9·9ᵏ − 5·5ᵏ = 9(9ᵏ − 5ᵏ) + 4·5ᵏ = 9(4m) + 4·5ᵏ = 4(9m + 5ᵏ).
4. State the conclusion of the step in words: "which is divisible by 4, since 9m + 5ᵏ ∈ Z."
5. Concluding sentence invoking the principle of mathematical induction.

Archive instances. [SAMPLE] Exam 1 Q3 (sample question, no percentage). 2026 NHT Exam 1 Q2 (3 marks, no percentage — NHT).

[GUIDE] for 2026 NHT Exam 1 Q2, verbatim structure:

"Prove true for n = 1: L.S. = … A1
Assume true for n = k: … M1
Prove for n = k + 1: … A1* given, must have working"

Note the three-mark split: base case, assumption, and a single combined mark for the step plus conclusion. When the proof is short, VCAA compresses; when it is long (four marks), it separates "use of assumption" from "working required". The ritual does not change; only the mark count does.

Typical marks: 3 or 4.

Traps. No report exists for [SAMPLE] or for any NHT sitting, so there is no VCAA commentary on this sub-type. The transferable warning is [RPT24 E1] on the direct-proof question (§2.8): "Occasional arithmetic or algebraic errors were seen." Divisibility proofs fail on algebra, not on scheme.

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2.3 Induction on an inequality, with a threshold

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q2 (4 marks), in two parts:

"a. Consider the inequality 2ⁿ > n² for n ≥ n₀, where n ∈ N. Show that n₀ = 5. (1 mark)
b. Prove by mathematical induction that 2ⁿ > n² for n ≥ 5, where n ∈ N. (3 marks)"

What it tests. Three things at once: (i) locating the threshold by testing small cases — the one place in this area of study where checking examples is the intended method; (ii) an induction whose base case is n = 5, not n = 1; (iii) an inductive step that chains inequalities rather than equalities, which requires each step to be justified in the correct direction.

Required structure.

1. Part a: tabulate n = 1, 2, 3, 4, 5 showing 2ⁿ ≤ n² for n = 2, 3, 4 and 2⁵ = 32 > 25 = 5², and state that the inequality then holds from n = 5. Note carefully: part a asks you to show n₀ = 5, which is a finite check, and is legitimate. Part b is what proves it continues.
2. Part b base case: n = 5, both sides evaluated.
3. Assumption: "Assume 2ᵏ > k² for some k ≥ 5, k ∈ N." The k ≥ 5 must be carried — it is used in the step.
4. Step: 2^{k+1} = 2·2ᵏ > 2k² (by the assumption), then show 2k² ≥ (k+1)² for k ≥ 5 — i.e. k² − 2k − 1 ≥ 0, true for k ≥ 3. Chain: 2^{k+1} > 2k² ≥ (k+1)².
5. Conclusion.

Archive instances. [SAMPLE] Exam 1 Q2 only. Never set in a live paper. The inequality context is named in the [SD] overview ("Proofs will involve concepts from topics such as: divisibility, inequalities, …") and VCAA demonstrated it in the sample, so it is squarely in scope and is the most conspicuous gap in the live record.

Typical marks: 4 (1 + 3 in the sample's own split).

Traps. None documented. The structural trap is obvious from the mathematics: the inductive step needs a second, subsidiary inequality (2k² ≥ (k+1)²) which itself needs the k ≥ 5 condition. Students who write 2^{k+1} > 2k² > (k+1)² without justifying the second > have a gap.

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2.4 Induction on a recursively generated object (the nth derivative)

Wording template — real paper. 2023 Exam 1 Q8 (4 marks) [PAPERS], verbatim:

"A function f has the rule f(x) = x·e^{2x}.
Use mathematical induction to prove that f^{(n)}(x) = (2ⁿ x + n·2^{n−1})·e^{2x} for n ∈ Z⁺, where f^{(n)}(x) represents the nth derivative of f(x). That is, f(x) has been differentiated n times."

What it tests. That the candidate understands the inductive step as applying the defining operation once more. Here the operation is differentiation: f^{(k+1)}(x) = d/dx [f^{(k)}(x)], and the assumption is substituted inside the derivative. This is the hardest induction sub-type in the live archive and it produced the lowest full-mark rate of any proof question ever set.

Required structure.

1. Base case n = 1: differentiate f(x) = x e^{2x} by the product rule to get f'(x) = e^{2x} + 2x e^{2x} = (2x + 1)e^{2x}, and check this against the formula at n = 1: (2¹x + 1·2⁰)e^{2x} = (2x + 1)e^{2x}. ✓
2. Assumption: "Assume f^{(k)}(x) = (2ᵏx + k·2^{k−1})e^{2x} for some k ∈ Z⁺."
3. Step: "Then f^{(k+1)}(x) = d/dx[(2ᵏx + k·2^{k−1})e^{2x}]" — differentiate, do not apply index laws.
4. Product rule, simplify, and match the target (2^{k+1}x + (k+1)2ᵏ)e^{2x}.
5. Conclusion.

Archive instances.

Read that distribution: 39% of the state scored exactly 2 of 4 — base case and assumption, then nothing. The scheme is learnable and was learned; the step was not.

Typical marks: 4.

Traps the reports name. [RPT23 E1], verbatim:

"Many students were able to begin the proof by showing the base step and making an assumption for the n = k case. Students were then required to differentiate f^{(k)}(x) with respect to x to show that the n = k + 1 case followed. A number of students either did not differentiate the function or differentiated incorrectly. Many students appeared to be thinking of index laws and assumed that f^{(k+1)}(x) was equal to f^{(k)}(x) × f(x)."

That last sentence is the whole lesson: f^{(k+1)} means "differentiate once more", not "multiply by f". Whatever the recursive object is — a derivative, a matrix power, a term of a recurrence — the step is apply the defining operation to the assumed form.

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2.5 Induction with the scheme scaffolded into separate parts

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q1, quoted in full at §2.1: part a asks for the base case (1 mark), part b asks the candidate to "Write down the assumption in terms of k" (1 mark), part c asks for the rest (2 marks).

What it tests. Nothing the unscaffolded version does not — but it is diagnostic evidence about how VCAA values the scheme. VCAA was willing to pay a full mark for writing down the assumption. That is the clearest possible signal that in an unscaffolded question, the assumption line is worth a mark on its own.

Archive instances. [SAMPLE] Exam 1 Q1 only. No live paper has scaffolded an induction. Expect the unscaffolded form, and supply the scaffolding yourself.

Typical marks: 4 (1 + 1 + 2).

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2.6 Induction on De Moivre's theorem for integral powers

Wording template — no exam instance exists. [SD] Area of Study 3 content, verbatim:

"De Moivre's theorem, proof for integral powers, powers and roots of complex numbers in polar form, and their geometric representation and interpretation"

The plausible wording, constructed from the study design and the house style of §2.1 — this is a reconstruction, not a VCAA question: "Use mathematical induction to prove that (cis θ)ⁿ = cis(nθ) for all n ∈ Z⁺."

What it tests. The induction scheme applied to the compound-angle identities. The step is cis(kθ)·cis(θ) = cis((k+1)θ), which is expanded with cos(A+B) and sin(A+B).

Required structure. Base case n = 1 is trivial and must still be written. Assumption (cis θ)ᵏ = cis(kθ). Step: (cis θ)^{k+1} = (cis θ)ᵏ · cis θ = cis(kθ)·cis θ = (cos kθ + i sin kθ)(cos θ + i sin θ), expand, collect real and imaginary parts, apply the compound-angle formulae (both are on the formula sheet), conclude = cis((k+1)θ). Concluding sentence.

Archive instances. None. The nearest live relative is 2023 Exam 2 Section B Q2fii, which uses De Moivre inside a "show that" (see §2.17) and scored 7%.

Why it belongs in the catalogue. [SD] names the proof explicitly and [SD] also says "The concepts, skills and processes from this area of study are to be applied in the other areas of study." An induction proof of De Moivre is the single most obvious unexamined question in the design.

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2.7 Induction on a recursive sequence or partial product

Wording template — no exam instance exists. [SD] overview, verbatim, names the context:

"Proofs will involve concepts from topics such as: … sequences and series including partial sums and partial products and related notations, complex numbers, matrices, vectors and calculus."

What it tests. Given t₁ = a and t_{n+1} = f(t_n), prove a closed form for t_n. The step substitutes the assumed closed form into the recurrence. The matrix variant proves a closed form for Aⁿ.

Required structure. Identical to §2.1, with one addition: the step must begin t_{k+1} = f(t_k) — quote the recurrence — and only then substitute the assumption.

Archive instances. None under the current design. Note that "partial products" and "matrices" are named in [SD] but are Units 1 and 2 machinery; a Units 3 and 4 question would have to supply the recurrence or the matrix in the stem.

Typical marks: 3–4 by analogy.

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2.8 Direct proof by integer parametrisation (odd / even)

Wording template — real paper, reconstructed from the report because the 2024 papers are image-only. 2024 Exam 1 Q2 (3 marks). [RPT24 E1] sample answer, verbatim:

"As n is an odd integer, we may let n = 2k + 1 where k ∈ Z.
Alternatively, it may be observed directly that if n is odd then n² is even… n is odd, and … is odd, so that … is the sum of one even and two odd integers, hence even."

(The report's mathematical expressions are MathType images and do not extract; the exact expression proved is not recoverable from the corpus. [GUIDE] confirms the structure — see below.)

What it tests. That "let n be an odd integer" is turned into an algebraic object: n = 2k + 1, k ∈ Z. Everything after that is expansion.

Required structure, and this is [GUIDE] for 2024 Exam 1 Q2 verbatim:

"Let n = 2k + 1 where k ∈ Z. M1 (specify k ∈ Z not required)
… M1 substitution.
… which is even A1* answer given"

Three moves, three marks: parametrise, substitute, conclude in words. Note VCAA's own concession — "specify k ∈ Z not required" — which means the parametrisation earns the mark even without the quantifier. Write it anyway; it costs three characters and the concession may not be repeated.

Archive instances.

[RPT24 E1] lists "direct proof (Question 2)" among the areas of strength for the paper — the only time a proof question has appeared on a strengths list.

Typical marks: 3.

Traps the reports name. [RPT24 E1]: "This question was answered well by students. Substituting 2k + 1 (or 2k − 1) for n in the expression and obtaining … hence a multiple of 2 and so even, was a reasonable approach. Occasional arithmetic or algebraic errors were seen." The failure mode here is expansion, not logic.

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2.9 Proof by contradiction — surd inequality

Wording template — real paper. The proof is given and the candidate identifies the technique; see §2.13 for the item type. The proof itself, 2024 NHT Exam 2 Section A Q1 [PAPERS], verbatim:

"Prove that √15 + √7 > √19.
Assume √15 + √7 ≤ √19.
Then (√15 + √7)² ≤ 19
15 + 2√105 + 7 ≤ 19
2√105 ≤ −3
hence √15 + √7 > √19."

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q5 (3 marks):

"Use proof by contradiction to prove that √3 + √5 > √11."

What it tests. (i) Correctly negating the statement — the negation of > is ≤, not <; (ii) squaring both sides of an inequality between positive quantities, which preserves the direction; (iii) recognising the absurdity (2√105 ≤ −3 is impossible because the left side is positive) and saying so.

Required structure.

1. "Assume, for contradiction, that √3 + √5 ≤ √11."
2. "Both sides are positive, so squaring preserves the inequality: 8 + 2√15 ≤ 11."
3. "Hence 2√15 ≤ 3, i.e. 4 × 15 ≤ 9, i.e. 60 ≤ 9." — push to a numerical absurdity.
4. "This is a contradiction. Therefore √3 + √5 > √11."

Note the model proof VCAA prints in 2024 NHT Exam 2 Section A Q1 stops at 2√105 ≤ −3 and writes "hence" — VCAA is content with an absurdity that is visibly impossible without a further line. But that proof was written by VCAA to be recognised, not marked.

Archive instances. [SAMPLE] Exam 1 Q5 (sample, 3 marks). 2024 NHT Exam 2 Section A Q1 (1 mark, as a recognition item, no percentage). No live written contradiction proof exists.

Typical marks: 3.

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2.10 Proof by contradiction — parity / number property

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 1 Q4 (3 marks):

"Use proof by contradiction to prove that if n is odd, where n ∈ N, then n³ + 1 is even."

What it tests. That the candidate negates the conclusion while keeping the hypothesis — this is the single most common logical error in the whole area. The assumption is "n is odd and n³ + 1 is odd", not "n is even".

Required structure.

1. "Suppose, for contradiction, that there exists an odd n ∈ N such that n³ + 1 is odd."
2. "n odd ⟹ n = 2k + 1, k ∈ Z."
3. Expand n³ + 1 = 8k³ + 12k² + 6k + 2 = 2(4k³ + 6k² + 3k + 1), which is even.
4. "So n³ + 1 is both odd and even — a contradiction. Therefore if n is odd, n³ + 1 is even."

A remark worth internalising. This particular statement is more naturally proved directly (§2.8) or by contrapositive. VCAA asked for contradiction anyway. The command word chooses the technique; the mathematics does not. A correct direct proof of a question that says "Use proof by contradiction" is not answering the question asked.

Archive instances. [SAMPLE] Exam 1 Q4 only.

Typical marks: 3.

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2.11 Contrapositive identification (multiple choice)

Wording template — real papers. Four live instances, all at Section A Question 1, all in English sentences.

2023 Exam 2 Section A Q1 [PAPERS], verbatim:

"Consider the following statement. 'If my football team plays badly, then they are not training enough.' Which one of the following statements is the contrapositive of the statement above?
A. If they are not training enough, then my football team plays badly.
B. If my football team plays badly, then they need more training.
C. If they are training enough, then my football team does not play badly.
D. If my football team doesn't play badly, then they are training enough.
E. If they are training enough, then my football team will most likely win."

2025 Exam 2 Section A Q1 [PAPERS], verbatim:

"A tiger is a type of cat. Consider the following statement. 'If I have a tiger, then I have a cat.' The contrapositive of this statement is
A. if I do not have a tiger, then I do not have a cat.
B. if I have a cat, then I have a tiger.
C. if I do not have a cat, then I do not have a tiger.
D. if I do not have a tiger, then I have a different type of cat."

2026 NHT Exam 2 Section A Q1 [PAPERS], verbatim:

"Consider the following statement. 'If the temperature is 35 °C or more, then the students do not play sport.' The contrapositive of this statement is
A. if the temperature is 36 °C, then the students do not play sport.
B. if the students do not play sport, then the temperature is 35 °C or more.
C. if the temperature is less than 35 °C, then the students play sport.
D. if the students play sport, then the temperature is less than 35 °C."

2024 Exam 2 Section A Q1 is a contrapositive item with correct answer C; its text is unrecoverable (§0.1).

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 2 Section A Q1, which is the quantified variant and the hardest of the five:

"Consider the following statement. 'For all integers n, if n² is even, then n is even.' Which one of the following is the contrapositive of this statement?
A. For all integers n, if n² is odd, then n is odd.
B. There exists an integer n such that n² is even and n is odd.
C. There exists an integer n such that n is even and n² is odd.
D. For all integers n, if n is odd, then n² is odd.
E. For all integers n, if n is even, then n² is even."

What it tests. The transformation P ⟹ Q becomes ¬Q ⟹ ¬P. Both halves swap and both are negated. The distractors are always: the converse (Q ⟹ P), the inverse (¬P ⟹ ¬Q), and one or two irrelevant sentences. In the sample's quantified version the distractors also test whether "for all" survives contraposition (it does) and whether a negated implication is an existence statement (it is — options B and C are the negation of the whole statement, not its contrapositive).

Required answer. One letter.

Archive instances.

Note the option count: 2023 had five options, 2024 November onwards has four (confirmed by the report tables and by the Multiple-Choice Answer Sheet in corpus/sm/text/2025-10_MCAS_SpecialistMaths2.txt, which prints A–D only). The 2024 NHT paper still used five.

Typical marks: 1.

Traps the reports name. [RPT24 E2], verbatim: "Asked for contrapositive — therefore, switch the hypothesis and the conclusion and negate both." [RPT25 E2], verbatim: "The question asked for contrapositive, which occurs when switching the hypothesis and the conclusion and negating both."

That both reports felt the need to state the rule is the tell. And look at 2024: 20% of the state chose option A, which was the converse or the inverse. Those 20% did one of the two operations, not both. The whole item is: swap, then negate. Both. Every time.

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2.12 Disproof by counter-example (multiple choice)

Wording template — real paper. 2025 Exam 2 Section A Q2 [PAPERS], verbatim:

"Consider the following statement. 'If f''(0) = 0, then the graph of f necessarily has a point of inflection at x = 0.' A counter-example that disproves this statement is when
A. f(x) = sin⁻¹(x)
B. f(x) = 2x/(x² + 1)
C. f(x) = x^{1/3}
D. f(x) = x⁴ + x"

What it tests. Two things simultaneously: (i) that a single counter-example disproves a universal claim; (ii) that f''(a) = 0 is necessary but not sufficient for a point of inflection at a — the second derivative must also change sign. f(x) = x⁴ + x has f''(x) = 12x², which is zero at x = 0 but does not change sign there.

Required answer. One letter. In the written form (which has never been set but is licensed by [SD]'s "examples and counter-examples"), the answer would be: state the example, evaluate the relevant quantity, and state which part of the claim fails.

Archive instances.

40% chose C, f(x) = x^{1/3}. That is a function with a point of inflection at x = 0 where f''(0) does not exist — i.e. a counter-example to the converse, not to the stated claim. This is the single most instructive distractor in the whole area of study.

Typical marks: 1.

Traps the reports name. [RPT25 E2], verbatim:

"To show a point of inflection exists at x = 0, the second derivative must equal zero at x = 0 and there must be a change of sign of the second derivative either side of x = 0. CAS can be used to determine this in the algebra menu, or students could use the graphing menu to see the shape of the graph."

The archive has been warning about this since 2006. 2006 Exam 2 Section B Q4cii (8%) [RPT06 E2]: "A large number of students included y = 0 and y = 1 as the y-coordinates of the points of inflection. Only a few attempted to verify the point of inflection. … Most thought that it was sufficient to show that the second derivative needed to be zero in order to establish a point of inflection." And 2026 NHT Exam 1 Q3 [GUIDE] marks a sign table and the sentence "f'' does not change sign at x = …. So not a point of inflection." Nineteen years apart, same idea, still separating.

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2.13 Identify the proof technique (multiple choice)

Wording template — real paper. 2024 NHT Exam 2 Section A Q1 [PAPERS], verbatim:

"Consider the following proof.
Prove that √15 + √7 > √19.
Assume √15 + √7 ≤ √19.
Then (√15 + √7)² ≤ 19
15 + 2√105 + 7 ≤ 19
2√105 ≤ −3
hence √15 + √7 > √19.
This proof can be best described as a
A. direct proof.
B. proof by contrapositive.
C. proof by contradiction.
D. proof by counter-example.
E. proof by mathematical induction."

What it tests. That the candidate can read a proof and name its logical shape. The tell is the second line: the proof assumes the negation of the conclusion. A contrapositive proof would instead assume the negation of the conclusion and derive the negation of the hypothesis — there is no hypothesis here, so it cannot be contrapositive.

Required answer. One letter. Note the option list is itself a checklist of [SD]'s named techniques, minus "proof by cases" — the five options are, verbatim, the five things the study design names.

Archive instances.

Typical marks: 1.

Why this type matters out of proportion to its one mark. It is the only item type that tests the taxonomy directly. Anyone who can tell contradiction from contrapositive on sight will also write the right opening line in a 3-mark written proof.

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2.14 Proof by cases

Wording template — none exists. [SD] content, verbatim: "natural deduction and proof techniques: direct proofs using a sequence of direct implications, proof by cases, proof by contradiction, and proof by contrapositive".

What it would test. Partitioning the domain exhaustively (n even / n odd; x ≥ 0 / x < 0; n = 3m, 3m+1, 3m+2) and proving the claim in each case, then stating that the cases are exhaustive.

Required structure. State the partition and justify that it is exhaustive; prove each case; conclude. The exhaustiveness sentence is the mark that is lost.

Archive instances. None. Not in [SAMPLE], not in any live paper. It is the only named technique in [SD] with no exemplar anywhere in the corpus. Treat it as a live risk for a 3-mark Examination 1 question.

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2.15 Vector proof of a geometric result — perpendicularity

Wording template — real papers.

2015 Exam 1 Q1b (2 marks) [PAPERS], verbatim — this is [SD]'s own canonical example:

"Consider the rhombus OABC shown below, where OA = a·i and OC = i + j + k, and a is a positive real constant.
a. Find a. (1 mark)
b. Show that the diagonals of the rhombus OABC are perpendicular. (2 marks)"

2022 Exam 1 Q6bii (3 marks) [PAPERS], verbatim — [SD]'s "angle subtended by a diameter in a circle is a right angle":

"OPQ is a semicircle of radius a with equation y = √(a² − (x − a)²). P(x, y) is a point on the semicircle OPQ, as shown below.
i. Express the vectors OP and QP in terms of a, x, y, i and j … (1 mark)
ii. Hence, using the vector scalar (dot) product, determine whether OP is perpendicular to QP. (3 marks)"

2013 Exam 1 Q3b (2 marks) [PAPERS], verbatim:

"The points A, B and C are the vertices of a triangle. Prove that the triangle has a right angle at A."

2010 Exam 2 Section B Q1dii (3 marks) [PAPERS], verbatim: "Use a vector method to show that OQ is perpendicular to AB."

2006 Exam 2 Section B Q2a (2 marks) [PAPERS]: "Show that AC and BD are perpendicular."

What it tests. That perpendicularity is proved by a dot product being zero, in the general case, with the two vectors written in terms of the given symbols and not in terms of numbers.

Required structure.

1. Write both vectors in terms of the stated symbols, using the triangle rule (AC = OC − OA).
2. Compute the dot product symbolically.
3. Use whatever relation the stem gives (|a| = |b| for a rhombus; the circle equation for the semicircle) to show the dot product is zero.
4. A concluding sentence: "Since OP · QP = 0 and neither vector is the zero vector, OP is perpendicular to QP."

Archive instances.

Typical marks: 2–3.

Traps the reports name. [RPT13 E1] on Q3b: "Students were required to show that there was a right angle at A using the dot product or Pythagoras's theorem. Most students used one of the correct approaches. Many students made a mistake in finding AC (or CA). In a small number of cases the dot product was evaluated as a vector. A significant number of students used a·b/(|a||b|) = cos A, wasting time finding the modulus values in the denominator."

[RPT22 E1] on Q6bii: "In this question students were required to make use of the scalar (dot) product. Some algebraic errors were made and incorrect conclusions drawn."

[RPT10 E2] on Q1dii: "numerous instances of values being adjusted within the scalar product calculation to this end. For students working with the general case, it was common to see |b| = |a| = 1 used for their scalar product to give zero." — i.e. students silently assumed the vectors were unit vectors, which is assuming what is not given.

Three separate reports naming three separate versions of the same failure: substituting a convenient special case for the general one.

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2.16 Vector proof of a quadrilateral property

Wording template — real papers.

2010 Exam 2 Section B Q1b (3 marks) [PAPERS], verbatim:

"Let N be the midpoint of the line segment OB. Use a vector method to prove that the quadrilateral MNQA is a parallelogram."

2008 Exam 1 Q8c (1 mark) [PAPERS], verbatim:

"The coordinates of three points are A(1, 0, 5), B(−1, 2, 4) and C(3, 5, 2).
a. Express the vector AB in the form xi + yj + zk. (1 mark)
b. Find the coordinates of the point D such that ABCD is a parallelogram. (2 marks)
c. Prove that ABCD is a rectangle. (1 mark)"

What it tests. That the candidate knows the minimal sufficient condition for the named figure, and proves that one condition and no more. This is the purest test in the archive of "necessary and sufficient conditions", which [SD] names in its second content dot point.

• Parallelogram: one pair of opposite sides equal and parallel, i.e. MN = AQ as vectors. That is it.
• Rectangle, given it is already a parallelogram: one adjacent pair of sides perpendicular. That is it.

Required structure. Write the two vectors, show they are equal (or that the dot product is zero), and write the sentence that names the property established.

Archive instances.

Typical marks: 1–3.

Traps the reports name. [RPT08 E1] on Q8c, verbatim and in full — the best single paragraph in the archive on over-proving:

"This was a straightforward question but many students did not recognise what additional property would lead to a parallelogram being a rectangle. Many wasted time showing that the opposite pairs of sides had equal length or were parallel. Some students tried to show that all four angles were right angles, or that all four angles were right angles and opposite pairs of sides had equal length. A few students correctly showed an adjacent pair of sides were at right angles, but then wasted time showing that adjacent sides were unequal in length, not realising that a square is a type of rectangle. Students could also have shown that diagonals have equal length."

One mark. Sixty-four per cent of the state got nothing for it.

[RPT10 E2] on Q1b, verbatim:

"The most popular approach to this question was to show opposite sides to be parallel. A number of students attempted to show the diagonals intersected at right angles. Others attempted to show that opposite sides were equal using simplifying assumptions such as |a| = |b| = 1/2, believing that a and b were orthogonal unit vectors. It was evident that some students did not read the question carefully enough and as a result these students worked with the wrong quadrilateral. Not all students understood clearly what they needed to show to prove that a given quadrilateral is a parallelogram."

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2.17 Proof and "show that" on complex numbers

Wording template — real papers.

2023 Exam 2 Section B Q2fii (2 marks) [PAPERS], verbatim:

"Given that w = cis(2π/7) satisfies (z − 1)(z⁶ + z⁵ + z⁴ + z³ + z² + z + 1) = 0, use De Moivre's theorem to show that cos(2π/7) + cos(4π/7) + cos(6π/7) = −1/2."

2023 Exam 2 Section B Q2e (1 mark) [PAPERS], verbatim: "Verify that the equation z⁷ − 1 = 0 can be expressed in the form (z − 1)(z⁶ + z⁵ + z⁴ + z³ + z² + z + 1) = 0."

2022 Exam 2 Section B Q2ai (2 marks) [PAPERS], verbatim: "Given that u·v̄ = 2 − √6 + (2 + √6)i, show that a² + (1 − √3)a − √3 = 0." (coefficients as extracted; the text layer is imperfect)

2007 Exam 2 Section B Q1e and Q1f [PAPERS], verbatim: "Show that the cartesian equation for the relation |z| = |z − z₁| is given by y = √3 x + 2." / "Show that z₁ satisfies the relation |z| = |z − z₁|."

2017 Exam 2 Section B Q3b, Q3d [PAPERS]: "Show that the roots of z² + 4z + 16 = 0 are z = −2 − 2√3 i and z = −2 + 2√3 i." / "Show that the cartesian form of the relation |z| = |z − (2 − 2√3 i)| is x − √3 y − 4 = 0."

What it tests. That the candidate manipulates from one side to the other in a stated direction, rather than starting from the answer.

Required structure. Start from the given object, transform by stated steps, arrive at the target. Never start from the target.

Archive instances.

Typical marks: 1–3.

Traps the reports name. [RPT23 E2] on Q2fii, verbatim:

"This question was not well done. Many students were able to express the given equation in terms of powers of w but most students did not 'show that' the required result arose through a series of logical steps."

[RPT22 E2] on Q2ai, verbatim: "A number of students apparently used a CAS to solve the given equation and then substituted their answers, again using CAS to verify the given result. … In a 'show that' question, students are required to clearly and logically show the steps that lead to the given result."

[RPT07 E2] on Q1f: "This question was not very well done, with many students simply reproducing their efforts for Question 1e. Very few students worked separately on the right hand side to show that it became the left hand side."

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2.18 "Show that" as a general-case derivation inside a modelling question

Wording template — real papers. This is the workhorse of Examination 2 Section B and it is where Outcome 2's "establish proofs for general case results" lives.

2016 Exam 2 Section B Q3c (2 marks) [PAPERS]: "Show that the differential equation relating y and t is dy/dt + y/(10 + t) = 1/3."

2018 Exam 2 Section B Q3a (2 marks) [PAPERS]: "Show that the volume, V cubic metres, of water in the fountain when it is filled to a depth of h metres is …"

2021 Exam 2 Section B Q4d (2 marks) [PAPERS]: "Show that v in terms of s is given by v = (180s)^{1/3}."

2025 Exam 2 Section B Q4d (3 marks) [PAPERS]: "Show that the speed of the particle, in m s⁻¹, at time t can be expressed as …"

2026 NHT Exam 2 Section B Q6ci [PAPERS]: "Show that the plane Π₃ given by d = 1.5z + x + y, where d ∈ R, is parallel to …"

What it tests. That a general relationship can be derived from a described situation, with every step visible. [SPEC] is explicit: "Students should use command/task words, other instructional information within questions and corresponding mark allocations to guide their responses", and [RPT24 E2] sharpens it: "The number of marks allocated to a question indicates the level of detail required in the response. Answers without supporting work will not earn method marks."

Required structure. Begin from the physical or geometric set-up; write the governing relation; transform to the stated target; write the target as the final line.

Archive instances (separators only). 2016 Exam 2 Section B Q3c 40%; 2018 Exam 2 Section B Q3a 32%; 2021 Exam 2 Section B Q4d 31%; 2025 Exam 2 Section B Q4d 35.6%; 2022 Exam 2 Section B Q2ai 34%; 2024 Exam 1 Q9a 32%.

Typical marks: 1–3.

Traps the reports name. [RPT16 E2] on Q3c, verbatim: "Some students incorrectly started with the given expression with no explanation of its origin. Students frequently did not seem to realise that work done for Question 3b. was useful here." And in the report's own sample answer: "This 'show that' question required students to obtain the expression dy/dt = 1/3 − y/(10 + t) by logical steps."

[RPT18 E2] on Q3a: "Approximately half of the students were able to either set up an appropriate definite integral or find an antiderivative and attempt to evaluate the constant of integration. Of these, many did not explicitly show that the first part of their response yielded the required volume."

[RPT25 E2] on Q4d: "Another 'show that' question which required working that shows the use of trigonometric identities that are given on the formula sheet."

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2.19 Justifying the nature of a stationary point or point of inflection

Wording template — real papers. 2006 Exam 2 Section B Q4ci (1 mark) [PAPERS]: "Show that d²y/dx² = (1 − 2y)·y(1 − y)." followed by Q4cii (2 marks): find the coordinates of the points of inflection. 2026 NHT Exam 1 Q3 (3 marks) [PAPERS]: "Determine the x-coordinate(s) of any point(s) of inflection of the graph of f(x) = x⁵ + x⁴ − x."

What it tests. The necessary-versus-sufficient distinction that [SD] names in its "implications, equivalences and if and only if statements (necessary and sufficient conditions)" dot point, applied to calculus. f''(a) = 0 is necessary, not sufficient.

Required structure. Solve f''(x) = 0; then test the sign of f'' on both sides of each root (a sign table is the compact form); then state which roots give points of inflection and which do not. [GUIDE] for 2026 NHT Exam 1 Q3 prints exactly this — a sign table, then the sentence "f'' does not change sign at x = …. So not a point of inflection."

Archive instances.

Typical marks: 1–3.

Traps the reports name. [RPT06 E2] on Q4cii, verbatim: "Most thought that it was sufficient to show that the second derivative needed to be zero in order to establish a point of inflection." And [RPT25 E2] on the 2025 counter-example item repeats the same content nineteen years later. [RPT24 NHT E1] on its Q3a notes the other direction: "Verification that these points were points of inflection was not required." — so read the command word; VCAA sometimes waives it, and when it does, do not spend the time.

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2.20 Quantifier handling inside a proof

Wording template — real papers, embedded. Every live induction question carries a quantifier in its statement and the domain differs between them:

• 2023 Exam 1 Q8: "for n ∈ Z⁺"
• 2024 NHT Exam 1 Q8: "for all n ∈ N"
• 2025 Exam 1 Q7: "for n ∈ N"
• 2026 NHT Exam 1 Q2: "for all n ∈ N"
• [SAMPLE] Exam 1 Q2: "for n ≥ 5, where n ∈ N"

Wording template — VCAA sample question, not a real paper. [SAMPLE] Exam 2 Section A Q1 uses "For all integers n, …" in the stem and offers "There exists an integer n such that …" as two distractors (options B and C), testing whether the candidate knows that negating a universally quantified implication yields an existential conjunction, not another implication.

What it tests. That the domain is read, carried into the assumption, and reproduced in the concluding sentence. The base case is determined by the quantifier: "n ∈ N" starts at 1, "n ≥ 5" starts at 5, "n ∈ Z⁺" starts at 1.

Required structure. The quantifier appears three times in a complete proof: in the proposition statement, in the assumption (for some k ∈ N, k ≥ 5), and in the conclusion (for all n ∈ N). Note the switch from "for some" to "for all" — that switch is the induction.

Archive instances. No standalone quantifier question exists. It is examined only as part of the induction ritual and as [SAMPLE] Exam 2 Section A Q1 distractors.

Typical marks: 0 as a standalone; it is one of the four marks of an induction.

Traps the reports name. [RPT25 E1] on 2025 Exam 1 Q7: "Misstating the assumption. For example, 'Suppose the proposition is true for n = k. Then …'" The defect VCAA is flagging is the missing quantification on k — "for some k ∈ N" — and, in the version that follows in the report, the assumption being written as something other than the proposition at n = k.

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3. The standard wordings, and what must appear to earn every mark

[SPEC], verbatim: "Students should use command/task words, other instructional information within questions and corresponding mark allocations to guide their responses." This section is that instruction, unpacked, for each wording VCAA actually uses.

3.1 "Prove, by mathematical induction, that …" / "Use mathematical induction to prove that …"

Both forms appear. 2023 Exam 1 Q8 and 2025 Exam 1 Q7 use "Use mathematical induction to prove that"; 2024 NHT Exam 1 Q8 and 2026 NHT Exam 1 Q2 use "Prove by mathematical induction that". There is no difference in what is required.

What must appear, in the reports' and guides' own words:

On the mark codes. VCAA publishes no glossary for M1, A1, H1 and A1* in the Assessment Guides. The conventional reading, consistent with every usage in the four published guides, is: M = method mark, A = answer/accuracy mark, H = a further intermediate mark within a multi-mark part, and the asterisk in A1* marks a part whose answer was given in the question, which is why every A1* in the corpus is annotated "ans. given", "answer given" or "working required". This is an inference from usage, not a published definition.

The four-mark and three-mark shapes, side by side:

3.2 "Use proof by contradiction to show / prove that …"

[SAMPLE] Exam 1 Q4 and Q5 are the only exemplars; there is no report, so there is no VCAA commentary. What must appear is fixed by the logic and by the model proof VCAA printed in 2024 NHT Exam 2 Section A Q1:

If the statement is an implication ("if P then Q"), the assumption is P AND not-Q — the hypothesis is retained. This is the difference between contradiction and contrapositive and it is the one place candidates reliably err.

3.3 "State the contrapositive of …" / "The contrapositive of this statement is"

Never yet set as a written question; four live multiple-choice instances (§2.11).

In a written form, the answer is one sentence, and the mark would hang on both operations being performed. Given 2024 Exam 2 Section A Q1, where 20% chose the option that performed only one of the two operations, the risk is real.

3.4 "Give a counter-example to show that …"

Never yet set as a written question; one live multiple-choice instance (§2.12). The written form would require:

One counter-example is enough, and more than one wastes time — the mirror image of the over-proving trap in 2008 Exam 1 Q8c (§2.16).

3.5 "Show that …" and "Prove that …" — the general rule

This is the highest-frequency wording in the whole subject and VCAA has restated the rule in its general comments almost every year. The rule, in the reports' own words, across fourteen years:

Three operational rules follow and they are worth more marks per minute than anything else in this document.

1. Never write the target on the first line. [RPT16 E2]: "Some students incorrectly started with the given expression with no explanation of its origin."
2. Work one side to the other, and say which. [RPT07 E2]: "Very few students worked separately on the right hand side to show that it became the left hand side."
3. Substitution is not proof. [RPT19 E2]: "rather than 'verify'". [RPT10 E1]: "just substituting the value and showing that the coefficient of j was zero was not sufficient."

3.6 "Hence …" — the companion instruction

[RPT06 E2]: "The paper also contained six 'hence' type questions. With this type of question, students must use the result obtained in the previous question part to answer the subsequent part of a question. To gain full marks, this instruction must be [followed]." [RPT09 E2]: "It needs to be emphasised that for this type of question students must use a previously established result to answer the question at hand."

The live example: 2009 Exam 1 Q8b (31%), where [RPT09 E1] records "Most students realised that the given result from Question 8a. should be used". And [RPT06 E2] on Q5c (37%): "A large number of students ignored the 'hence' requirement of this question and did not use the answer for cos(π/8)."

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4. The separators

Definition. A separator is a question part where pct ≤ 50 — fewer than half the state earned full marks. Percentages are VCAA's own published figures via [QJSON].

4.1 Separators tagged Discrete mathematics

The area is four years old and [QJSON] holds only 12 questions tagged Discrete mathematics. Five are separators:

The other seven tagged questions are above the line (2013 Exam 1 Q3b 75%, 2024 Exam 1 Q2 65%, 2024 Exam 2 Section A Q1 72%, 2023 Exam 2 Section A Q1 85%, 2025 Exam 2 Section A Q1 93%) or carry no percentage (2024 NHT Exam 2 Section A Q1, 2025 NHT Exam 2 Section A Q1).

4.2 Extended separator list — every proof-flavoured question in the archive, regardless of assigned area

This section deliberately breaks the topic tag. Five instances is not a usable drill set, so the net is widened to every archive question whose task is to establish a stated result — a proof, a vector geometric proof, or a "show that" whose object is a general relation — identified by searching the paper and report text rather than the topic field, and filtered to pct ≤ 50. Questions marked ★ are the five from §4.1.

38 separators. Five carry the Discrete mathematics tag; the other 33 are proof-flavoured questions filed under Space and measurement, Algebra, Calculus or Functions.

What the list says. The median proof-flavoured separator sits around 34%, and the distribution is heavily weighted towards the bottom: nine of the 38 sit at or below 21%. Compare [01SD] §1.5's figure for Functions, relations and graphs — a 44% median full-mark rate across all its written parts. Proof-flavoured questions are harder than the hardest content area, and they are hard for reasons that have nothing to do with the mathematics. Every one of the top ten hardest is a question where the candidate knew the mathematics and failed to write an argument.

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5. What makes a hard one hard

The reports are unusually consistent on this. Six failure modes, each with the evidence.

5.1 The inductive step's logic — differentiating, not multiplying

The single named failure in the archive's hardest induction. [RPT23 E1] on 2023 Exam 1 Q8 (21%):

"Students were then required to differentiate f^{(k)}(x) with respect to x to show that the n = k + 1 case followed. A number of students either did not differentiate the function or differentiated incorrectly. Many students appeared to be thinking of index laws and assumed that f^{(k+1)}(x) was equal to f^{(k)}(x) × f(x)."

The general form of the error: not knowing what operation takes you from case k to case k+1. For a sum it is "add the next term"; for a product it is "multiply by the next factor"; for a derivative it is "differentiate once more"; for a recurrence it is "apply the recurrence". Identify that operation before writing anything, because the entire step is that operation followed by substitution of the assumption.

The 2023 mark distribution proves the point: 39% of the state scored exactly 2 out of 4. They wrote the base case and the assumption, then stalled at the step. The scheme was known; the step was not.

5.2 Stating what is assumed — and not assuming what is to be proved

[RPT25 E1] on 2025 Exam 1 Q7 (29%) names two of its three errors here:

"Misstating the assumption. For example, 'Suppose the proposition is true for n = k. Then …'"

"Assuming equality at the beginning of the inductive step."

The second is the deeper one and it is the commonest way a competent student loses two marks. The wrong shape is:

Required: 1/2 + 1/4 + … + 1/2^{k+1} = 1 − 1/2^{k+1}
LHS = RHS
… manipulate both sides until they match …
∴ true for n = k+1

That argument begins by asserting the thing to be proved. The right shape starts from one side only:

LHS at n = k+1 = (1/2 + 1/4 + … + 1/2^k) + 1/2^{k+1}
              = (1 − 1/2^k) + 1/2^{k+1}     [by the assumption]
              = 1 − 1/2^{k+1}
              = RHS at n = k+1

[GUIDE] makes this explicit for 2025 Exam 1 Q7: the third mark is "A1 – must see use of assumption". A proof that manipulates both sides symmetrically never visibly uses the assumption, and cannot earn that mark even if every line is true.

The same defect appears outside induction. [RPT16 E2] on Q3c (40%): "Some students incorrectly started with the given expression with no explanation of its origin." [RPT14 E1] on Q6a: "many students did not know how a verification or proof should be set out. Some arguments were not convincing, and some eventually showed that a = a or similar."

a = a is the signature of a circular argument. If your last line is a tautology, you have proved nothing.

5.3 Quantifiers

Three distinct quantifier errors are documented or structurally implied.

• The assumption must quantify k existentially, and the conclusion universally. [RPT23 E1]'s model answer reads: "Assume that the statement is true for some n = k …" then "Therefore, by the principle of mathematical induction, the statement is true for all n …" [RPT24 NHT E1] uses the identical pair. The switch from some to all is the content of the induction principle; a proof that never makes it has not invoked the principle.
• The domain determines the base case. n ∈ N starts at 1; n ∈ Z⁺ starts at 1; n ≥ 5 starts at 5. [SAMPLE] Exam 1 Q2 makes this a scored sub-question.
• Negating a quantified implication. [SAMPLE] Exam 2 Section A Q1's distractors B and C are the negation of the statement, offered as fake contrapositives. Negating "For all n, P(n) ⟹ Q(n)" gives "There exists n such that P(n) and not Q(n)" — an existential conjunction, not an implication. In a proof by contradiction of a universally quantified statement, that existential is exactly what you assume.

5.4 Concluding sentences

Every published model answer ends with a sentence, and [GUIDE] attaches a mark to the final line.

• [RPT23 E1]: "Therefore, by the principle of mathematical induction, the statement is true for all n …"
• [RPT24 NHT E1]: "Therefore, by the principle of mathematical induction, the proposition is true for all n …"
• [RPT25 NHT E1]: "Therefore, by mathematical induction, …"
• [GUIDE] 2025 E1 Q7: final mark is A1* and the guide's printed solution includes "By the principle of mathematical induction, P(n) is true for n ∈ N."

Outside induction the same applies. [RPT22 E1] on 2022 Exam 1 Q6bii (47%): "Some algebraic errors were made and incorrect conclusions drawn." A dot product of zero with no sentence saying what it means is an incomplete answer to "determine whether OP is perpendicular to QP".

The rule: every proof ends with a sentence in English that names what has been established. It costs one line and it is worth a mark.

5.5 Verification mistaken for proof

The most frequently repeated complaint in the entire twenty-year archive.

• [RPT06 E2] on Q2c (16%): "Few managed to show ∠ABC and ∠ADC were supplementary. Most resorted instead to finding numerical approximations of the two angles in an attempt to show they were supplementary."
• [RPT06 E2] on Q2d (20%): "only a minority attempted to use a double angle formula to show the given result. A large number again resorted to finding numerical approximations to the two angles to try to prove ∠APC = 2∠ADC."
• [RPT14 E1] on Q8b (40%): "Several students simply substituted a few values in for θ and then asserted that the result was [established]."
• [RPT19 E2] on Q3bii: "students are expected to explicitly show that the given information leads to the required conclusion rather than 'verify' that the given values of z are solutions of the equation."
• [RPT22 E2] on Q2ai (34%): "A number of students apparently used a CAS to solve the given equation and then substituted their answers, again using CAS to verify the given result."
• [RPT10 E1] on Q8a: "many students were unable to show that the given value was the second occurrence; just substituting the value and showing that the coefficient of j was zero was not sufficient."
• [RPT23 E2] on Q2fii (7%): "most students did not 'show that' the required result arose through a series of logical steps."

Checking a claim at some values is evidence. Producing the claim by valid steps is proof. The one place where checking values is the method is [SAMPLE] Exam 1 Q2a — "Show that n₀ = 5" — because that sub-question asks for a threshold, which is a finite search.

A related and equally penalised move is the unconvincing derivation that nonetheless lands on the given answer. [RPT11 E1]: "As often happens in a 'show that' question, a number of students were unable to do appropriate working but still managed to 'show' the result." [RPT12 E1] on Q9c (35%): "some students were unable to do any convincing algebra, yet still managed to obtain dx/dy". Assessors read for the derivation, not the destination.

5.6 Necessary versus sufficient — proving more or less than is required

Two failure directions, both expensive.

Proving too little is the necessary-condition error: f''(a) = 0 does not establish a point of inflection (2006 Exam 2 Section B Q4cii, 8%; 2025 Exam 2 Section A Q2, 48%).

Proving too much wastes time that a 40-mark/60-minute paper does not have. [RPT08 E1] on Q8c (36%, one mark): "Many wasted time showing that the opposite pairs of sides had equal length or were parallel. Some students tried to show that all four angles were right angles … but then wasted time showing that adjacent sides were unequal in length, not realising that a square is a type of rectangle."

Assuming a convenient special case is the third variant and the most insidious, because the algebra looks fine. [RPT10 E2] on Q1b (44%): "Others attempted to show that opposite sides were equal using simplifying assumptions such as |a| = |b| = 1/2, believing that a and b were orthogonal unit vectors." And on Q1dii (49%): "it was common to see |b| = |a| = 1 used for their scalar product to give zero."

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6. A worked method sheet

Six skeletons. Each is reproducible under time pressure and loses no mark to presentation. Write them in this order, on separate lines, every time.

6.1 Induction — the universal skeleton

Let P(n) be the proposition that  ⟨statement⟩,  for all n ∈ ⟨domain⟩.

BASE CASE.  n = ⟨first value in the domain⟩.
   LHS = ⟨evaluate⟩
   RHS = ⟨evaluate⟩
   LHS = RHS, so P(⟨first value⟩) is true.

ASSUMPTION.  Assume P(k) is true for some k ∈ ⟨domain⟩ (with k ≥ ⟨threshold⟩ if any).
   That is,  ⟨the statement with n replaced by k⟩.

REQUIRED.  Show P(k+1) is true, i.e. that ⟨the statement with n replaced by k+1⟩.

STEP.  ⟨LHS at n = k+1⟩
       = ⟨apply the defining operation once: add the next term / multiply by the
          next factor / differentiate once more / apply the recurrence⟩
       = ⟨substitute the assumption — write "by the assumption" here⟩
       = ⟨algebra⟩
       = ⟨RHS at n = k+1⟩
   So P(k+1) is true.

CONCLUSION.  P(1) is true and P(k) ⟹ P(k+1); therefore, by the principle of
   mathematical induction, P(n) is true for all n ∈ ⟨domain⟩.

Three non-negotiables. (i) LHS and RHS of the base case are evaluated separately and then compared. (ii) The step starts from one side only. (iii) The words "by the assumption" appear somewhere — [GUIDE] pays a mark for visible use of the assumption.

Divisibility variant. Replace the assumption line with "Assume ⟨expression in k⟩ = d·m for some m ∈ Z", and end the step with "= d(⟨integer expression⟩), which is divisible by d since ⟨expression⟩ ∈ Z."

Inequality variant. Carry the threshold into the assumption (for some k ≥ 5), and chain inequalities in one direction only, justifying each link.

6.2 Proof by contradiction

Suppose, for the sake of contradiction, that ⟨negation of the conclusion⟩
   ⟨— and, if the statement is an implication, retain the hypothesis: "and ⟨P⟩"⟩.

Then  ⟨valid deduction⟩
      ⟨valid deduction⟩
      ⟨… down to something impossible⟩

But ⟨the impossible thing⟩ is false, because ⟨reason⟩.
This is a contradiction.
Therefore ⟨the original statement⟩.

Negation table to memorise.

6.3 Proof by contrapositive

The contrapositive of "if P then Q" is "if not Q then not P".
These are logically equivalent, so it suffices to prove the contrapositive.

Assume ⟨not Q⟩.
Then ⟨valid deductions⟩
Therefore ⟨not P⟩.
Hence, by contraposition, if P then Q.

To state a contrapositive (the multiple-choice form): swap, then negate both halves, and keep any leading quantifier. Do both operations. Twenty per cent of the 2024 cohort did one.

6.4 Disproof by counter-example

The statement is false.
Counter-example: let ⟨specific object⟩.
Then ⟨hypothesis⟩ holds, because ⟨check⟩.
But ⟨conclusion⟩ fails, because ⟨check⟩.
Therefore the statement is false.

One example. Check both halves. Stop.

6.5 Direct proof with quantifiers

Let n be an arbitrary ⟨odd integer / element of the domain⟩.
Then n = ⟨2k + 1⟩ for some k ∈ Z.
Substituting:  ⟨expression⟩ = ⟨expand⟩ = ⟨factor out the structure you need⟩
             = 2(⟨integer expression⟩),  which is even since ⟨expression⟩ ∈ Z.
Since n was arbitrary, this holds for all ⟨odd integers⟩.

[GUIDE] for 2024 Exam 1 Q2 pays three marks for exactly three moves: parametrise (M1), substitute (M1), conclude (A1*).

6.6 Vector proof of a geometric result

Set up.  Let ⟨position vectors⟩ = a, b, … as given.
         ⟨Write every vector you will need using the triangle rule, e.g. AC = OC − OA.⟩

Claim.   ⟨What is to be shown, restated.⟩

Work.    ⟨Compute symbolically — never substitute numbers or assume |a| = 1.⟩
         ⟨Use whatever relation the stem supplies: |a| = |b| for a rhombus,
          the circle equation for a semicircle, a midpoint definition, etc.⟩

Conclude. Since ⟨dot product = 0 / the two vectors are equal / … ⟩,
          ⟨the named geometric property⟩ holds.

The minimal sufficient conditions, memorised:

6.7 "Show that" — the general skeleton

⟨Start from the given information or the left-hand side. Write it down.⟩
= ⟨step, with the rule named if it is a formula-sheet identity⟩
= ⟨step⟩
= ⟨step⟩
= ⟨the target, exactly as printed in the question⟩

as required.

Never write the target first. Never manipulate both sides towards each other. Never substitute numbers to check. If the question says "hence", the previous part's result must physically appear in your working.

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7. Quick reference — the whole area on one page