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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 \(\PageIndex{1}\)

    \[∀n ∈ \mathbb{N}, P_n\]


    (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.)


    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.


    (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.



    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.

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