1. Strong Induction: Selected Exercises Goal Explain & illustrate proof construction of a variety of theorems using strong induction

2. Copyright © Peter Cappello2 Strong Induction Domain of discussion is the positive integers, Z +. Strong Induction: If 1.p( 1 ) 2.  k ( [ p( 1 )  p( 2 )  …  p( k ) ]  p( k + 1 ) ) then  n p( n ).

3. Copyright © Peter Cappello3 Example: Fundamental Theorem of Arithmetic Let the universe of discourse be N. Let P( n ) denote “n is a product of primes.” Prove that  n  2 P( n ). Basis P( 2 ): Since 2 is prime, 2 is a product of primes. Assume P( 2 ),..., P( n ). (Strong induction hypothesis)

4. Copyright © Peter Cappello4 Show P( n + 1 ): –Case n + 1 is a prime: It is a product of 1 prime: itself. –Case n + 1 is not a prime: 1. n + 1 = ab, such that 1 < a  n, 1 < b  n 2.P( a ): a = p 1 p 2... p k, where the p i s are primes. (S.I.H.) 3.P( b ): b = q 1 q 2... q l, where the q i s are primes. (S.I.H.) 4. n + 1 = ( p 1 p 2... p k )( q 1 q 2... q l ).

5. Copyright © Peter Cappello5 Exercise 10 –A chocolate bar consists of n squares arranged in a rectangle. –The bar can be broken into smaller rectangles of squares with a vertical or horizontal break. –How many breaks suffice to break the bar into n squares? Consider an example bar consisting of 4 x 3 squares.

6. Copyright © Peter Cappello6 Exercise 10 continued Use strong induction to prove that n - 1 breaks suffice. Basis n = 1: 1 – 1 = 0 breaks suffice. Assume for bars of up to n squares that n – 1 breaks suffice. Show for bars of n + 1 squares that n breaks suffice. 1.Break the bar horizontally, if there is > 1 row, or vertically, if there is only 1 row of squares.

7. Copyright © Peter Cappello7 Exercise 10 continued 2. Let piece 1 have s 1 squares & piece 2 have s 2 squares. 3. s 1 + s 2 = n + 1. 4. s 1 - 1 breaks suffice to break piece 1 into squares. (SIH) 5. s 2 - 1 breaks suffice to break piece 2 into squares. (SIH) 6. 1 + ( s 1 – 1 ) + ( s 2 – 1 ) = n breaks suffice.

8. Copyright © Peter Cappello8 Exercise 30 Find the flaw with the following “proof” that a n = 1,  n ≥ 0, when a  0 is real. Proof: Basis n = 0: a 0 = 1. Inductive step: Assume that a j = 1, 0  j  k. Show a k+1 = 1: a k+1 = ( a k. a k ) / a k-1 = 1. 1 / 1 = 1.

9. Copyright © Peter Cappello9 Exercise 30 continued The proof fails for n = 1. Why?

10. Copyright © Peter Cappello 201110 Exercise 18 Show: If a convex polygon P with consecutive vertices v 1, v 2, …, v n is triangulated into n – 2 triangles (Δ), the triangles can be numbered 1, 2, …, n – 2 so that v i is a vertex of Δ i, for i = 1, 2, …, n – 2. 1 2 34 5 6

11. Copyright © Peter Cappello 201111 Exercise 18 Show: If a convex polygon P with consecutive vertices v 1, v 2, …, v n is triangulated into n – 2 triangles (Δ), the triangles can be numbered 1, 2, …, n – 2 so that v i is a vertex of Δ i, for i = 1, 2, …, n – 2. 1 2 34 5 6 1 2 3 4

12. Exercise What is the P( n ) that is claimed? For what n is it claimed to be true? What is a recursive formulation of P( n )? Copyright © Peter Cappello12

13. Copyright © Peter Cappello 201113 Exercice18 Proof Basis n = 3: True, as shown. Assume proposition for polygons of k vertices, where 3  k  n - 1. Show proposition for polygons with n vertices. 1.For n > 3, every triangulation of P includes a diagonal from v n or v n-1. (If not, P is not triangulated.) 2.Case: A diagonal connects v n to v k : 1 23 1 1 2 34 5 6

14. Copyright © Peter Cappello 201114 18 Proof continued Subdivide P into 2 smaller polygons, P 1 & P 2, defined by this diagonal: P 1 has vertices v 1, v 2, …, v k, v n. Renumber v n of P 1 as v k+1 ; (k – 1 triangles when triangulated) 1 2 3 4

15. Copyright © Peter Cappello 201115 18 Proof continued Subdivide P into 2 smaller polygons, P 1 & P 2, defined by this diagonal: P 1 has vertices v 1, v 2, …, v k, v n. Renumber v n of P 1 as v k+1 ; (k – 1 triangles when triangulated) Triangulate P 1 with triangles properly numbered 1, …, k – 1. (SIH) 1 2 3 1 2 4

16. Copyright © Peter Cappello 201116 18 Proof continued P 2 has vertices v k, v k+1, …, v n. P 2 has n – k + 1 vertices; (n – k – 1 triangles) For each v i of P 2, renumber vertices v’ i = v i – k +1. 4= 6 - 2 1= 3 – 22= 4 - 2 3= 5 - 2

17. Copyright © Peter Cappello 201117 18 Proof continued P 2 has vertices v k, v k+1, …, v n. For each v i of P 2, renumber vertices v’ i = v i – k +1. P 2 has n – k + 1 vertices; (n – k – 1 triangles) Triangulate it with triangles properly numbered 1, …, n – k – 1. (SIH) 1 2 4 12 3

18. Copyright © Peter Cappello 201118 18 Proof continued P 2 has vertices v k, v k+1, …, v n. For each v i of P 2, renumber vertices v’ i = v i – k +1. P 2 has n – k + 1 vertices; (n – k – 1 triangles) Triangulate it with triangles properly numbered 1, …, n – k – 1. (SIH) Add k – 1 to each triangle number: k, k + 1, …, n – 2. 3 4 4 12 3

19. Copyright © Peter Cappello 201119 18 Proof continued P 2 has vertices v k, v k+1, …, v n. For each v i of P 2, renumber vertices v’ i = v i – k +1. P 2 has n – k + 1 vertices; (n – k – 1 triangles) Triangulate it with triangles properly numbered 1, …, n – k – 1. (SIH) Add k – 1 to each triangle number: k, k + 1, …, n – 2. Original vertex v i now participates in Δ i, i = k, …, n – 2. 1 2 34 5 6 1 2 3 4

20. Copyright © Peter Cappello20 18 Proof continued 3. Case: A diagonal connects v n-1 to v k : This case is handled similarly. Do this as an exercise. This constructive proof is essentially a recursive algorithm for computing these triangles & their numbers.

21. Copyright © Peter Cappello21 End

22. Copyright © Peter Cappello 201122 40 Well-ordering principle: Every nonempty set of nonnegative integers has a least element. Use the well-ordering principle to show that [x, y  R  x < y]   r  Q, x < r < y.

23. 23 40 Proof 1.y – x > 0. 2.1/(y – x) > 0. 3.Let a  Z, a > 1/(y – x). 4.y – x > 1/a. 5.Choose the least positive j such that └ x ┘ + j/a > x. (WOP) 6. Let r = └ x ┘ + j/a. 7. r  Q. ( └ x ┘, j, a  Z) 8. x < r. (defn of r) 9. └ x ┘ + (j – 1)/a < x. (choice of j) 10. r – 1/a < x. (defn of r.) 11. r – (y – x) < x. (step 10 & 4) 12. r < y.

24. Copyright © Peter Cappello24 Characters    ≥ ≡ ~ ┌ ┐ └ ┘        ≈      Ω Θ     Σ        