Recursion Lecture 9 Justin Stevens Justin Stevens Recursion (Lecture 9) 1 / 26 Outline 1 Strong Induction Induction Strong Induction 2 Recursion 3 Parting Shots Justin Stevens Recursion (Lecture 9) 2 / 26 Base Case: Show P(1). Inductive Step: Show P(k) implies P(k + 1) for any integer k ≥ 1. The assumption we make is known as the induction hypothesis. Induction Theorem. Suppose we have a statement P that we wish to show is true for all positive integers at least 1. We can prove this in two steps: Justin Stevens Recursion (Lecture 9) 3 / 26 Induction Theorem. Suppose we have a statement P that we wish to show is true for all positive integers at least 1. We can prove this in two steps: Base Case: Show P(1). Inductive Step: Show P(k) implies P(k + 1) for any integer k ≥ 1. The assumption we make is known as the induction hypothesis. Justin Stevens Recursion (Lecture 9) 3 / 26 h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 ! k2 = (k + 1)2 + k + 1 4 (k + 1)(k + 2)2 = . 2 This is the sum of cubes formula for n = k + 1, hence the identity holds. Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Adding the next cube to both sides of our assumption gives: Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Justin Stevens Recursion (Lecture 9) 4 / 26 h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 ! k2 = (k + 1)2 + k + 1 4 (k + 1)(k + 2)2 = . 2 This is the sum of cubes formula for n = k + 1, hence the identity holds. Adding the next cube to both sides of our assumption gives: Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Justin Stevens Recursion (Lecture 9) 4 / 26 ! k2 = (k + 1)2 + k + 1 4 (k + 1)(k + 2)2 = . 2 This is the sum of cubes formula for n = k + 1, hence the identity holds. Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Adding the next cube to both sides of our assumption gives: h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 Justin Stevens Recursion (Lecture 9) 4 / 26 (k + 1)(k + 2)2 = . 2 This is the sum of cubes formula for n = k + 1, hence the identity holds. Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Adding the next cube to both sides of our assumption gives: h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 ! k2 = (k + 1)2 + k + 1 4 Justin Stevens Recursion (Lecture 9) 4 / 26 This is the sum of cubes formula for n = k + 1, hence the identity holds. Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Adding the next cube to both sides of our assumption gives: h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 ! k2 = (k + 1)2 + k + 1 4 (k + 1)(k + 2)2 = . 2 Justin Stevens Recursion (Lecture 9) 4 / 26 Sum of Cubes Identity 2 3 3 3 3 h n(n+1) i Example. Prove the identity 1 + 2 + 3 + ··· + n = 2 . Proof. When n = 1, 1 = 12. We assume the formula is true for n = k: k(k + 1)2 13 + 23 + 33 + ··· + k3 = . (Hypothesis) 2 Adding the next cube to both sides of our assumption gives: h i k(k + 1)2 13 + 23 + 33 + ··· + k3 + (k + 1)3 = + (k + 1)3 2 ! k2 = (k + 1)2 + k + 1 4 (k + 1)(k + 2)2 = . 2 This is the sum of cubes formula for n = k + 1, hence the identity holds. Justin Stevens Recursion (Lecture 9) 4 / 26 Proof Without Words of Sum of Cubes Figure 1: Alan L. Fry, Mathematical Association of America (1985) Justin Stevens Recursion (Lecture 9) 5 / 26 Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5). k+1 Then Ni is a k + 1-digit number divisible by 5 all of whose digits are odd. n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Justin Stevens Recursion (Lecture 9) 6 / 26 k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5). k+1 Then Ni is a k + 1-digit number divisible by 5 all of whose digits are odd. n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. Justin Stevens Recursion (Lecture 9) 6 / 26 k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5). k+1 Then Ni is a k + 1-digit number divisible by 5 all of whose digits are odd. n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: Justin Stevens Recursion (Lecture 9) 6 / 26 From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5). k+1 Then Ni is a k + 1-digit number divisible by 5 all of whose digits are odd. n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . Justin Stevens Recursion (Lecture 9) 6 / 26 k+1 Then Ni is a k + 1-digit number divisible by 5 all of whose digits are odd. n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5). Justin Stevens Recursion (Lecture 9) 6 / 26 n Digit Number Divisible by 5n Example. (USAMO) Prove that for every positive integer n, there exists an n-digit number divisible by 5n all of whose digits are odd. Solution. We begin by showing the first 4 base cases: 5, 75, 375, 9375. k k Assume that the number N = a1a2a3 ··· ak is divisible by 5 , so N = 5 A. Consider the k + 1-digit number for i ∈ {1, 3, 5, 7, 9}: k k k k Ni = ia1a2a3 ··· ak = i · 10 + 5 A = 5 i · 2 + A . From the reduced residue system, there exists i with i · 2k + A ≡ 0 (mod 5).