Specialist Mathematics: should I choose it?

What will I learn?

How will I be assessed?

Progression and companion subjects

No complete companion or exclusion review is attached to this profile. Check the official course-entry conditions, language eligibility and your school’s sequence before enrolling. Links between resource hubs show related learning; they do not establish a compulsory prerequisite.

Will I find it hard?

Difficulty depends on your experience, interests, teaching and assessment. These are our reflection prompts, not an official difficulty rating.

• How do you feel about explaining a method and working through unfamiliar multi-step problems?
• Try the sample without looking at a worked answer, then check the steps you need to practise.

Does it scale up or down?

Use applicable SATAC evidence. Vocational certificate credit is not automatically an ATAR input.

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Try a little of the subject

Proof by mathematical induction

1. Why induction proves infinitely many statements Mathematical induction proves a proposition P(n) for every positive integer n by joining two logically different tasks. The initial statement verifies P(1). The inductive step proves the implication P(k) ⇒ P(k+1) for an arbitrary positive integer k. If both are established, P(1) gives P(2), then P(2) gives P(3), and so on. The conclusion is therefore about all positive integers even though the written proof has finitely many lines. Checking P(1), P(2) and P(3) only supplies examples; it never proves the universal claim. A formal proof should name the proposition before using it. For example, for 1+3+⋯+(2n−1)=n², write P(n): 1+3+⋯+(2n−1)=n². The statement starts at n=1, so the initial statement is 1=1². In the inductive step assume P(k): 1+3+⋯+(2k−1)=k². The next sum contains exactly one new term, 2(k+1)−1=2k+1. Hence 1+3+⋯+(2k−1)+(2k+1)=k²+2k+1=(k+1)². This is P(k+1), so the identity holds for every positive integer n. The phrase ‘assume P(k)’ is local to the implication; it does not assume the theorem for every n. The value k must remain arbitrary. Substituting k=4 proves only the transition from the fourth case to the fifth. Likewise, proving P(k+1) without using P(k) may establish a separate algebraic fact, but it does not display the required chain unless that fact independently proves every case. Close the proof by naming both completed parts and the domain: since P(1) is true and P(k) implies P(k+1), P(n) is true for all positive integers n. When reading a proposed proof, test the implication separately from the first case. A flawless transition cannot begin the chain if the initial statement is false, and a true first case cannot carry the chain without a general transition. This separation is the quickest way to

Original practice question: Written task 49: Prove by mathematical induction that 1+3+⋯+(2n−1)=n² for every positive integer n.

This is ATARMAxxing’s original sample learning material. It is not an official assessment task or a prediction of your school’s workload.

Explore the resources

Open the Specialist Mathematics subject hub and free preview (opens in a new tab)

• Mathematics — SACE Stage 1 (opens in a new tab)

Core subject information and planning are free. Full study resources are optional.