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Mathematics · Ch 2 — Principle of Mathematical Induction

The Principle of Mathematical Induction

2.3

The Principle of Mathematical Induction

Statement of the Principle

Let P(n)P(n) be a mathematical statement involving a natural number nn. The Principle of Mathematical Induction says P(n)P(n) is true for every natural number nn provided both of the following hold:

  1. Base case (basis step): P(1)P(1) is true.
  2. Inductive step: for every positive integer kk, if P(k)P(k) is true then P(k+1)P(k+1) is also true, i.e. P(k)⇒P(k+1)P(k) \Rightarrow P(k+1).

When both conditions are verified, we conclude P(n)P(n) is true for all n∈Nn \in \mathbb{N}.

  • The base case is a single, concrete check — confirm the statement for the starting value of nn (usually n=1n=1, though a statement that only holds from a later point, say n≥4n \ge 4, instead has its base case at P(4)P(4)).
  • The inductive step is conditional: it does not assert that P(k)P(k) is actually true. It only proves the implication "if P(k)P(k) holds, then P(k+1)P(k+1) holds." The assumption "P(k)P(k) is true," made purely to carry out this argument, is called the inductive hypothesis.

Together, the base case supplies the first domino, and the inductive step supplies the falling-tile rule that carries truth indefinitely forward — exactly the mechanism motivated in Section 4.2.


Worked Illustration: Sum of the First nn Odd Numbers

Observe the pattern:

1=12,1+3=22,1+3+5=32,1+3+5+7=42,…1 = 1^2,\quad 1+3 = 2^2,\quad 1+3+5 = 3^2,\quad 1+3+5+7 = 4^2, \ldots

This suggests the general claim

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

Base case. P(1)P(1) says 1=121 = 1^2, which is true.

Inductive step. Assume the inductive hypothesis P(k)P(k) holds, i.e.

1+3+5+⋯+(2k−1)=k2....(1)1 + 3 + 5 + \cdots + (2k-1) = k^2. \qquad \text{...(1)}

We must show P(k+1)P(k+1) follows. Add the next odd number, 2(k+1)−1=2k+12(k+1)-1 = 2k+1, to both sides of (1): …