This HL mini-lesson covers Proof: direct proof, disproof by counterexample, proof by contradiction, and proof by mathematical induction — the rigorous heart of the Analysis & Approaches course.
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A proof is a sequence of logically valid steps from accepted facts (axioms and earlier results) to a conclusion — establishing a statement for certain, not just by example.
Notation: ⇒ means "implies"; ⇔ means "if and only if"; ∀ means "for all"; ∃ means "there exists".
Assume the hypothesis and deduce the conclusion algebraically. For example, "the sum of two even numbers is even": 2m + 2n = 2(m + n), which is even. ∎
A universal statement ("for all n …") is disproved by a single failing case.
"n² − n + 41 is prime for all n" fails at n = 41, where n² − n + 41 = 41² = 1681 = 41 × 41, not prime.
Assume the statement is false, then derive a logical contradiction; the assumption must therefore be wrong.
Classic: √2 is irrational — assume √2 = p/q in lowest terms, and show p and q share a factor 2, a contradiction.
Tap a statement, then how you would prove or disprove it.
To prove P(n) for all integers n ≥ 1:
By the domino principle, P(n) then holds for every n ≥ 1.
Prove Σr=1n r = n(n+1)/2.
Base: n = 1 gives 1 = 1(2)/2 = 1. ✓ Step: assume Σ = k(k+1)/2; then adding (k+1) gives k(k+1)/2 + (k+1) = (k+1)(k+2)/2 = P(k+1). ∎
Check at n = 50: 50·51/2 = 1275.
Tap a statement on the left, then its matching partner on the right.
Verifying many cases is not a proof. In induction you must actually use the assumption P(k) to reach P(k+1), and you must state the base case.
Warning: induction proves results for the positive integers, not for all real numbers.
Universal claim with a formula in n → induction. Something is impossible / irrational → contradiction. "For all …" that looks doubtful → hunt for a counterexample. Straightforward algebraic claim → direct proof.
Exam habit: finish an induction with the concluding statement invoking the principle of induction.
Direct: assume the hypothesis, deduce the conclusion
Counterexample: one failing case disproves "for all"
Contradiction: assume false, derive a contradiction (e.g. √2 irrational)
Induction: base case P(1) + step P(k) ⇒ P(k+1)
Rigour: examples are not proof; state the base case
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You've worked through Proof for IB Diploma Mathematics: Analysis & Approaches HL. 🎉
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