# 5.4: The Strong Form of Mathematical Induction

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The strong form of mathematical induction (a.k.a. the principle of complete induction, PCI; also a.k.a. course-of-values induction) is so-called because the hypotheses one uses are stronger. Instead of showing that $$P_k \implies P_{k+1}$$ in the inductive step, we get to assume that all the statements numbered smaller than $$P_{k+1}$$ are true. To make life slightly easier we’ll renumber things a little. The statement that needs to be proved is

$$∀k(P_0 ∧ P_1 ∧ . . . ∧ P_{k−1}) \implies P_k$$.

An outline of a strong inductive proof is:

##### Theorem $$\PageIndex{1}$$

$∀n ∈ \mathbb{N}, P_n$

Proof

(By complete induction)

Basis: (Technically, a PCI proof doesn’t require a basis. We recommend that you show that $$P_0$$ is true anyway.)

Inductive step: (Here we must show that $$∀k, \left( \bigwedge^{k−1}_{i=0} P_i \right) \implies P_k$$ is true.)

Q.E.D.

It’s fairly common that we won’t truly need all of the statements from $$P_0$$ to $$P_{k−1}$$ to be true, but just one of them (and we don’t know a priori which one). The following is a classic result; the proof that all numbers greater than $$1$$ have prime factors.

##### Theorem $$\PageIndex{2}$$

For all natural numbers $$n$$, $$n > 1$$ implies $$n$$ has a prime factor.

Proof

(By strong induction) Consider an arbitrary natural number $$n > 1$$. If $$n$$ is prime then n clearly has a prime factor (itself), so suppose that $$n$$ is not prime. By definition, a composite natural number can be factored, so $$n = a · b$$ for some pair of natural numbers $$a$$ and $$b$$ which are both greater than $$1$$. Since $$a$$ and $$b$$ are factors of n both greater than $$1$$, it follows that $$a < n$$ (it is also true that $$b < n$$ but we don’t need that . . . ). The inductive hypothesis can now be applied to deduce that a has a prime factor $$p$$. Since $$p | a) and \(a | n$$, by transitivity $$p | n$$. Thus $$n$$ has a prime factor.

Q.E.D.

## Exercises:

Give inductive proofs of the following:

##### Exercise $$\PageIndex{1}$$

A “postage stamp problem” is a problem that (typically) asks us to determine what total postage values can be produced using two sorts of stamps. Suppose that you have $$3¢$$ stamps and $$7¢$$ stamps, show (using strong induction) that any postage value $$12¢$$ or higher can be achieved. That is,

$$∀n ∈ \mathbb{N}, n ≥ 12 \implies ∃x, y ∈ \mathbb{N}, n = 3x + 7y$$.

##### Exercise $$\PageIndex{2}$$

Show that any integer postage of $$12¢$$ or more can be made using only $$4¢$$ and $$5¢$$ stamps.

##### Exercise $$\PageIndex{3}$$

The polynomial equation $$x^2 = x + 1$$ has two solutions, $$α = \dfrac{1+\sqrt{5}}{2}$$ and $$β = \dfrac{1− \sqrt{5}}{2}$$. Show that the Fibonacci number $$F_n$$ is less than or equal to $$α_n$$ for all $$n ≥ 0$$.

This page titled 5.4: The Strong Form of Mathematical Induction is shared under a GNU Free Documentation License 1.3 license and was authored, remixed, and/or curated by Joseph Fields.