Pancyprian Core Math study materials
Pancyprian Core Math study resources

Pancyprian Core Math: Mathematical Reasoning and Problem Solving

Pancyprian Core Math · Work through your exam, one topic at a time.

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Mathematical Reasoning and Problem Solving

Mathematical Reasoning and Problem Solving is the domain where the Pancyprian Core Mathematics paper tests how you argue rather than what you compute. Items ask you to judge an implication, negate a quantified statement, select a valid proof strategy, or model a written situation - and full credit requires the reasoning to be written out, not merely implied.

Propositional logic

  • A proposition is a statement that is either true or false, never both.
  • p ⇒ q is false only when p is true and q is false; it is automatically true whenever p is false.
  • The contrapositive (not q) ⇒ (not p) is logically equivalent to p ⇒ q; the converse q ⇒ p is not.
  • The negation of "for all x, P(x)" is "there exists x such that not P(x)", and vice versa.
  • "p is necessary for q" means q ⇒ p; "p is sufficient for q" means p ⇒ q.

Methods of proof

  • Direct proof: assume the hypothesis and derive the conclusion step by step.
  • Proof by contradiction: assume the negation of the claim and derive an impossibility.
  • Disproof by counterexample: a single counterexample destroys a universal claim but never establishes one.
  • Induction: establish the base case, assume the statement for n = k, then derive it for n = k + 1 using that assumption.

Sequences and modelling

  • Arithmetic: aₙ = a₁ + (n − 1)d and Sₙ = (n/2)(a₁ + aₙ).
  • Geometric: aₙ = a₁rⁿ⁻¹ and Sₙ = a₁(1 − rⁿ)/(1 − r) for r ≠ 1.
  • In a word problem, define the unknown, translate the constraint into an equation, solve, and check the result against the units and the context.

Worked example

Prove by induction that 1 + 3 + 5 + … + (2n − 1) = n² for every positive integer n. Base case n = 1: the left side is 1 and the right side is 1² = 1, so the claim holds. Assume it holds for n = k, i.e. the sum of the first k odd numbers equals k². For n = k + 1 the sum gains the next odd number, 2k + 1, giving k² + 2k + 1 = (k + 1)². The claim therefore holds for k + 1, and by the principle of induction for all n ≥ 1.

Traps that cost marks

  • Proving the converse rather than the original statement, or assuming the conclusion you are asked to prove.
  • Skipping the base case of an induction, or never using the inductive hypothesis in the step.
  • Offering one confirming example as proof of a general claim.
  • Confusing a necessary condition with a sufficient one.