5.4: The Strong Form of Mathematical Induction
- Page ID
- 19391
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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:
\[∀n ∈ \mathbb{N}, P_n\]
- 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.
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:
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,
\(∀n ∈ \mathbb{N}, n ≥ 12 \implies ∃x, y ∈ \mathbb{N}, n = 3x + 7y\).
Show that any integer postage of \(12¢\) or more can be made using only \(4¢\) and \(5¢\) stamps.
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 \(n ≥ 0\).