The HSC marker's pen has three jobs: tick the opening line (you named the method), tick each logical jump (your algebra justified the next claim), and tick the closing line (you actually proved what was asked). This final lesson distils every technique from Module 11 into one workflow, shows you how the marking criteria are written, and rehearses the three most common pitfalls that cost top-band students whole proofs in the exam.
Start with the warm-up and record your first reasoning.
List, from memory, the five proof techniques you have studied in Module 11 and the one situation each is best suited to. No looking back, this is a snapshot of what you've actually internalised.
Orient yourself to the lesson goals, language, and core approach.
Every HSC proof rewards two habits: name the method and structure in the opening line, then justify every algebraic jump in words. Marks are awarded for visible logical structure, not just for arriving at the right algebra.
The name-and-justify standard: (1) the opening line declares the method ("We prove by induction…", "Suppose, for a contradiction…"), (2) each line of algebra carries a brief reason ("since $k$ is even", "by the inductive hypothesis"), (3) the closing line states what has been proved.
Open · Justify · Close, three places where marks are won or lost.
Work through the concept and complete its check.
The NESA marker uses a published rubric. For a 3-mark induction question the typical allocation is:
For a 2- or 3-mark contradiction/contrapositive question, marks are typically allocated as:
The hook revisited: A student with correct base case and correct inductive algebra but no closing line typically scores 2 out of 3 they have lost the conclusion mark. The fix is one sentence: "Therefore, by the principle of mathematical induction, $P(n)$ holds for all integers $n \geq 1$."
Induction mark split: 1 base, 1 step, 1 conclusion · "Therefore, by the principle of mathematical induction…", memorise this exact phrase · Each algebraic line should have a brief reason in words · Missing the closing line is the single biggest source of dropped marks on induction questions
Pause, copy the 1-1-1 mark split, the exact closing-line phrasing, and the rule that each algebraic step needs a brief written reason into your book.
Quick check: Which line is required to earn the conclusion mark on a 3-mark induction question?
Build the next proof move and complete its check.
We just saw that HSC induction marks split 1-1-1 across base, step and conclusion, and that the closing line "by the principle of mathematical induction…" is a mandatory mark-gate. That raises a question: how do you decide which proof method to reach for first in an exam? This card answers it → a trigger-phrase table mapping prompt language to technique, with contradiction as fallback not default.
The full Module 11 toolkit, mapped to its trigger phrase in the prompt:
Trigger phrases map to techniques, memorise the table on this card · Default to the most direct method that the prompt allows · Contradiction is a fallback, not a default · The opening line is half the structure mark, write it deliberately
Pause, copy the trigger-phrase table (prompt language → technique), the rule that direct proof is the default, and that contradiction is a fallback, into your book.
Did you get this? True or false: a counterexample is sufficient to disprove a claim of the form "for all positive integers $n$, $P(n)$ holds".
Follow the worked examples, then complete the check.
Worked examples · 3 in a row, reveal as you go
Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} r(r+1) = \dfrac{n(n+1)(n+2)}{3}$ for all positive integers $n$. Lay out the three mark-bearing components clearly.
Prove that $\log_2 3$ is irrational.
Determine whether the following statement is true or false: "For all positive integers $n$, the expression $n^2 - n + 11$ is prime." Justify your answer fully.
Fill the gap: A counterexample to the universal claim "for all positive integers $n$, $n^2 + n + 17$ is prime" is $n =$ , since $n^2 + n + 17 = 256 + 16 + 17 = 289 = 17^2$ is not prime.
Correct the common traps and test your understanding.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: in a proof by mathematical induction, the inductive step is allowed to begin by writing the target form of $P(k+1)$ as its first line.
Complete the practice activities and revisit your opening thinking.
Activities · practice with the ideas
Write a complete induction proof, base, step, conclusion, for $\displaystyle\sum_{r=1}^{n} r^2 = \dfrac{n(n+1)(2n+1)}{6}$. Mark the three mark-bearing lines in the margin.
A claim states: "for all integers $n \geq 1$, $n^2 - 79n + 1601$ is prime." Find a counterexample and present it in the format expected by the HSC marker.
For each prompt, state which method is best and write the opening line only:
(a) "Prove that the sum of two odd integers is even."
(b) "Prove that there are no positive integer solutions to $x^2 - y^2 = 1$ with $y \geq 1$."
(c) "Prove that $\displaystyle\sum_{r=1}^{n} (2r-1) = n^2$ for all positive integers $n$."
(d) "Disprove: for all positive integers $n$, $2^n - 1$ is prime."
Critique the following student response: "Prove $\sqrt{5}$ is irrational. Suppose $\sqrt{5} = a/b$. Then $a^2 = 5b^2$. So $a$ is a multiple of $5$. Therefore $\sqrt{5}$ is irrational." Identify all missing components and rewrite a complete proof.
Across Module 11 you have met: direct proof, contrapositive, contradiction, induction, counterexample. Write a one-paragraph reflection (3–5 sentences) on which method you find hardest to choose correctly, and one rule of thumb that helps.
Odd one out: Three of these are HSC-required components of a complete induction proof. Which one is NOT?
Earlier you considered the student who wrote a correct base case and correct inductive step but no closing line on a 3-mark induction question.
The expected mark is 2/3, base case mark and inductive-step mark, but the conclusion mark is forfeited. NESA's published rubrics treat the closing line "Therefore, by the principle of mathematical induction…" as a substantive marker, not decoration. Memorising and routinely writing this exact phrase is one of the highest-leverage moves in HSC Ext 2: it converts near-perfect proofs into perfect proofs at zero algebraic cost.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. The statement "for every positive integer $n$, $n^2 + 3n + 1$ is prime" is false. Provide a counterexample with full working. (2 marks)
Q2. Prove by mathematical induction that $\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)} = \dfrac{n}{n+1}$ for all positive integers $n$. Lay out base, step, and conclusion as separate components. (3 marks)
Q3. Prove by contradiction that there are no positive integers $a, b, c$ with $a^2 + b^2 = 3c^2$. (Hint: consider remainders modulo $3$.) (3 marks)
Activity answers:
1. Base $n=1$: LHS $=1$, RHS $=\frac{1\cdot 2 \cdot 3}{6}=1$ ✓. Step: assume $\sum_1^k r^2 = \frac{k(k+1)(2k+1)}{6}$. Then $\sum_1^{k+1} r^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{(k+1)[k(2k+1)+6(k+1)]}{6} = \frac{(k+1)(2k^2+7k+6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$ ✓. Conclusion: Therefore, by the principle of mathematical induction, the identity holds for all $n \geq 1$. $\blacksquare$
2. Try $n=80$: $80^2 - 79 \cdot 80 + 1601 = 6400 - 6320 + 1601 = 1681 = 41^2$. Not prime. Counterexample: $n=80$ shows the statement is false. $\blacksquare$
3. (a) Direct: "Let $m = 2j+1$ and $n = 2k+1$." (b) Contradiction: "Suppose positive integers $x, y$ with $y \geq 1$ satisfy $x^2 - y^2 = 1$." (Then $(x-y)(x+y)=1$ forces both factors to be $\pm 1$, contradicting $x, y \geq 1$.) (c) Induction: "Base case $n=1$: LHS $=1$, RHS $=1$ ✓." (d) Counterexample: "$n=4$: $2^4-1 = 15 = 3 \cdot 5$, not prime."
4. Missing: (i) "in lowest terms" / $\gcd(a,b)=1$, (ii) the contradiction is not derived, must show $b$ is also a multiple of $5$, (iii) explicit conclusion. Rewrite: Suppose $\sqrt 5 = a/b$ with $\gcd(a,b)=1$. Then $5b^2 = a^2$, so $5 \mid a^2$ and hence $5 \mid a$. Write $a = 5m$, then $5b^2 = 25m^2$, so $b^2 = 5m^2$ and $5 \mid b$. But $5 \mid a$ and $5 \mid b$ contradicts $\gcd(a,b)=1$. Therefore $\sqrt 5$ is irrational. $\blacksquare$
5. Open-ended reflection, typical answer: "Hardest choice is contradiction vs contrapositive. Rule of thumb: if the body of the proof never uses the assumed $P$, it's really a contrapositive; if it uses both $P$ and $\neg Q$ to chase $\bot$, it's a contradiction."
Q1 (2 marks): $n = 13$: $n^2 + 3n + 1 = 169 + 39 + 1 = 209 = 11 \cdot 19$. Not prime [1]. Therefore the statement is false [1]. $\blacksquare$
Q2 (3 marks): Base $n=1$: LHS $= \frac{1}{1 \cdot 2} = \frac{1}{2}$, RHS $= \frac{1}{2}$ ✓ [1]. Step: assume $\sum_1^k \frac{1}{r(r+1)} = \frac{k}{k+1}$. Then $\sum_1^{k+1} \frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$ ✓ [1]. Conclusion: Therefore, by the principle of mathematical induction, the identity holds for all positive integers $n$ [1]. $\blacksquare$
Q3 (3 marks): Suppose, for a contradiction, positive integers $a, b, c$ satisfy $a^2+b^2 = 3c^2$. Choose the solution with smallest $c$. Squares modulo $3$ are $0$ or $1$, so $a^2 + b^2 \equiv 0 \pmod{3}$ forces $a \equiv b \equiv 0 \pmod 3$ [1]. Write $a = 3a'$, $b = 3b'$: then $9(a'^2 + b'^2) = 3c^2$, so $c^2 = 3(a'^2 + b'^2)$, hence $3 \mid c$. Write $c = 3c'$: then $a'^2 + b'^2 = 3c'^2$ with $c' < c$ [1], contradicting the minimality of $c$. Therefore no such positive integers exist [1]. $\blacksquare$
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review. This is the final lesson of Module 11.
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