strong mathematical induction - h-scpeople.hsc.edu/faculty-staff/robbk/math262/lectures/spring...

Strong Mathematical InductionLecture 23Section 5.4

Robb T. Koether

Hampden-Sydney College

Mon, Feb 24, 2014

Robb T. Koether (Hampden-Sydney College) Strong Mathematical Induction Mon, Feb 24, 2014 1 / 34

1 The Principle

2 The Method

3 ExamplesPrime FactorizationThe Checkerboard PuzzleTreesBinary Strings

4 Assignment

Outline

1 The Principle

2 The Method

4 Assignment

The Principle of Strong Mathematical Induction

Let P(n) be a predicate defined for integers n. Let a be an integer.If it is true that

P(a), P(a + 1), . . . , P(b) are true, andFor all integers k ≥ b, if P(a), P(a + 1), . . . , P(k) are true, thenP(k + 1) is true,

then it follows that P(n) is true for all n ≥ a.

Outline

1 The Principle

2 The Method

4 Assignment

The Method

The basic step.Show that P(a), P(a + 1), . . . , P(b) are true.

The inductive step.Show that for all integers k ≥ b, if P(a), P(a + 1), . . . , P(k) aretrue, then P(k + 1) is true,

The conclusion.Conclude that P(n) is true for all n ≥ a.

The Method

The power of the method is that we are allowed to assume thatthe statement is true for all integers less than or equal to k , notjust k itself.Depending on the nature of the statement, this can be atremendous advantage.

Outline

1 The Principle

2 The Method

4 Assignment

Outline

1 The Principle

2 The Method

4 Assignment

Prime Factorization

TheoremEvery integer n ≥ 2 can be factored into a product of primes.

Proof.The basis step.

Let n = 2.2 is prime, so the statement is true.

Prime Factorization

Proof.The inductive step.

Suppose that the statement is true for all n ≤ k for some integerk ≥ 2.That is, suppose that every integer n from 2 through k factors into aproduct of primes, for some integer k ≥ 2.Consider the integer k + 1.Either it factors or it does not factor.If it does not factor, then it is prime and we are done.

Prime Factorization

Proof.So suppose that it does factor, say n = rs for some integers r and swith 2 ≤ r < k + 1 and 2 ≤ s < k + 1.Then, by the induction hypothesis, r and s factor into products ofprimes.Therefore, k + 1 factors into a product of primes.

Therefore, all integers n ≥ 2 factor into a product of primes.

Outline

1 The Principle

2 The Method

4 Assignment

The Generalized Checkboard Puzzle

TheoremGiven a checkboard with an even number of squares, if we remove anytwo squares of opposite color, the remaining squares can be coveredwith 1× 2 and 2× 1 tiles.

A Lemma

LemmaFor any integers m and n, if the diagonally opposite corners of anm× n checkerboard are of opposite colors, then either m is even and nis odd or m is odd and n is even.

Proof.If m and n are both odd, then each row and column will end withthe same color that it started with.If m and n are both even, then each row and column will end withthe opposite color that it started with.Either way, diagonally opposite corners would be the same color.Therefore, if diagonally opposite corners are of opposite colors,then one of m and n must be even and the other one odd.

Inductive Proof of the Theorem

Proof.Let m = 2 and n = 1 and remove two squares of opposite from a 2× 1checkerboard.

Proof.There are no squares left, so they can be covered (vacuously) by 2× 1and 1× 2 tiles.

Proof.Let m and n be two integers, at least one of which is even, andsuppose that the theorem is true for all smaller checkerboards with aneven number of squares

Proof.Remove any two squares of opposite colors.

Proof.Consider the “bounding rectangle” of the removed squares.

Proof.It either equals the original checkerboard or it is smaller.

Proof.Case 1: Suppose it is smaller. Then, by the induction hypothesis, it canbe tiled.

Proof.We must show that the remainder of the checkerboard can also betiled.

Proof.We may assume that the original checkerboard has an even number ofcolumns. (Why?)

Proof.The smaller checkerboard must have one even dimension and one odddimension.

Proof.Case 1-A: Suppose that it has an even number of rows and an oddnumber of columns.

OddEve

Proof.Then we can tile the checkerboard as shown.

OddEve

Proof.Case 1-B: Suppose that it has an odd number of rows and an evennumber of columns.

EvenOdd

Proof.Case 2: Suppose the “smaller” checkerboard equals the originalcheckerboard. Then the two removed squares must be in diagonallyopposite corners.

Proof.There is a gap in the previous proof. Where is it?

Outline

1 The Principle

2 The Method

4 Assignment

Definition (Tree)A graph is a tree with at least one vertex if it is connected and containsno cycles.

A Tree Not a Tree Not a Tree

TheoremIf a connected graph with n ≥ 1 vertices is a tree, then it has exactlyn − 1 edges.

Proof.When n = 1, there is only one vertex.If there were an edge, then it would connect that vertex to itself,creating a cycle.Therefore, there are 0 edges and the statement is true whenn = 1.

Proof.Suppose that the statement is true for all n ≤ k for some k ≥ 1.

Let G be a graph with k + 1 vertices

Select any vertex V in the graph

Let m be the number of edges incident to V

m must be at least 1. Why?

Remove V and the incident edges

There are m separate component graphs Gi . Why?

Each component Gi is a tree. Why?

9 vertices8 edges

7 vertices6 edges

8 vertices7 edges

5 vertices4 edges

Each component Gi has ki vertices and ki − 1 edges

Proof.The total number of edges in the components is

(k1 − 1) + · · ·+ (km − 1) = (k1 + · · · km)−m= k −m.

Now add back in the 1 vertex and m edges that we removed, andwe have k + 1 vertices and k edges.Therefore, the statement is true when n = k + 1.Therefore, the statement is true for all n ≥ 1.

It is possible to give induction proof based on standard induction.The basic case is the same as before.

Proof.Suppose that the statement is true when n = k for some integerk ≥ 1.That is, suppose that any tree with k vertices has exactly k − 1edges.Let G be a tree with k + 1 edges.G must have a vertex of index 1.That is, there must be a vertex that is incident to only 1 edge.Remove that vertex and the incident edge, creating the graph G′.

Proof.The graph G′ is a tree (why?) and it has k vertices.So G′ has exactly k − 1 edges.Thus, G has exactly k edges.So the statement is true when n = k + 1.Therefore, it is true for all n ≥ 1.

Outline

1 The Principle

2 The Method

4 Assignment

Binary Strings

TheoremLet S be the set of all binary strings with an equal number of 0’s and1’s. Then every string x ∈ S is of the form

x = 0s1, where s ∈ S,x = 1s0, where s ∈ S, orx = st, where s, t ∈ S.

Binary Strings

Proof.Let x ∈ S have length n.Clearly, if x begins with 0 and ends with 1, or begins with 1 andends with 0, then it must be in the form 0s1 or 1s0 for some s ∈ S.So, suppose that x begins and ends with 0 or begins and endswith 1.Without loss of generality, assume that x begins and ends with 0.

Binary Strings

Proof.Let x = x1x2 . . . xn−1xn, where x1 = 0 and xn = 0.Let ki be the number of 0’s minus the number of 1’s in the first idigits.Clearly, k0 = 0, k1 = 1, kn−1 = −1, kn = 0.For all i , the change from ki to ki+1 is either +1 or −1, dependingon whether xi+1 = 0 or xi+1 = 1.Therefore, for some j , with 2 ≤ i ≤ n − 2, we must have kj = 0.Then let s = x1 . . . xj and t = xj+1 . . . xn.Then s, t ∈ S and x = st .

Outline

1 The Principle

2 The Method

4 Assignment

Collected

CollectedSec. 4.8: 16.Sec. 5.1: 15, 44.Sec. 5.2: 14, 26.Sec. 5.3: 10, 18.

Assignment

AssignmentRead Section 5.4, pages 268 - 276.Exercises 1, 6, 7, 8, 10, 11, 12, 17, page 277.

strong mathematical induction - h-scpeople.hsc.edu/faculty-staff/robbk/math262/lectures/spring...

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