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Mathematics · Ch 4 — Combinatorics and Mathematical Induction

Mathematical induction

4.6

Mathematical induction

Consider the sum of the first nn positive odd numbers: 1=1, 1+3=4, 1+3+5=9, 1+3+5+7=16,…1=1,\ 1+3=4,\ 1+3+5=9,\ 1+3+5+7=16,\ldots — the right-hand sides are exactly the perfect squares 1,4,9,16,…1,4,9,16,\ldots, suggesting the conjecture

1+3+5+⋯+(2n−1)=n2.1+3+5+\cdots+(2n-1) = n^2.

A pattern noticed on a handful of cases is not yet a proof for every nn — that is exactly what the Principle of Mathematical Induction supplies. It applies to statements P(n)P(n) phrased in terms of a positive integer nn, and mirrors climbing an infinite staircase: if you can (a) climb the first step, and (b) climb from any given step to the next, then you can climb every step. Formally, to prove P(n)P(n) true for all nn:

  • Step 1 (initial/base step). Verify P(1)P(1) is true.
  • Step 2 (inductive step). Assume P(k)P(k) is true for some positive integer kk (the inductive hypothesis), and show that this forces P(k+1)P(k+1) to be true as well.
  • Step 3 (conclusion). If Steps 1 and 2 both hold, then P(n)P(n) is true for every positive integer nn.
Note

The inductive step never independently re-proves P(k+1)P(k+1) — it must genuinely use the assumption P(k)P(k) (usually by isolating the sum/expression up to the kkth term inside the (k+1)(k+1)th expression and substituting the assumed formula for it) to reach P(k+1)P(k+1). Skipping that link breaks the proof. …