DEFINITIONS: 1. FACTORIAL: n! = n (n 1) (n 2) ! = 1. n! = n (n 1)! = n (n 1) (n 2)! (n + 1)! = (n + 1) n! = (n + 1) n (n 1)!

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1 MATH 2000 NOTES ON INDUCTION DEFINITIONS: 1. FACTORIAL: n! = n (n 1) (n 2) ! = 1 n! = n (n 1)! = n (n 1) (n 2)! (n + 1)! = (n + 1) n! = (n + 1) n (n 1)! 2. SUMMATION NOTATION: f(i) = f(1) + f(2) + f(3) f(n 1) + f(n) f(i) = f(1) and 2 f(i) = f(2) and i=2 f(i) = f(n) i=n c b 1 f(i) = f(i) + i=a i=a c f(i) i=b n+1 f(i) = f(i) + n+1 i=n+1 f(i) Examples: (3i + 1) = = 4 2 (3i + 1) = ( ) + ( ) = = 11 (3i + 1) = ( ) + ( ) + ( ) (3n + 1) = (3n + 1) n+1 (3i + 1) = ( ) + ( ) + ( ) (3n + 1) + (3 (n + 1) + 1) = (3n + 1) + (3n + 4) 1

2 3. WELL ORDERED : A set S of real numbers is well ordered if every non-empty subset of S has a minimum element. Examples: S = { 6, 0, 15, 102} S is well ordered. S = (0, 1) is not well ordered since it has no minimum element. S = [0, 1] is not well ordered since (0, 1) [0, 1] Q, Z and R are not well ordered. 4. Well ordering axiom: The set N is well ordered. 5. The Principle of Mathematical Induction, abbreviated to PMI. THE INDUCTION PRINCIPLE (PMI): For each n N, let P (n) be a statement. If a) P (1) is true and b) k N, P (k) P (k + 1) is true, then n N, P (n) is true. Condition a), that P (1) is true, is called the basis step or base case of the induction. Condition b) is called the inductive step. Some analogies to describe induction: - dominoes: setting up a line of dominoes so that each one will trigger the next is the inductive step; then tipping over the first domino is the base step: together, the two steps result in all the dominoes being knocked over. - climbing an infinite ladder: you need to be able to step onto the bottom rung (the base step) and then be able to climb from any one rung to the next (the inductive step). Two main uses of induction: - to make inductive definitions - to prove statements about N. 2

3 INDUCTIVE DEFINITIONS: We can use the induction property to define a function on the set N of all natural numbers. Example: The factorial function can be defined inductively by giving a base case and an inductive step: a) 1! = 1, b) n! = n (n 1)!. Example: The odd natural numbers can be inductively defined by: a) 1 is odd; b) for all n, if n is odd then n + 2 is odd. Example: The sum of the numbers from 1 to n can be defined inductively by: a) i = 1 b) n 1 i = n + i Note: induction is closely related to the idea of RECURSION used in computer science. The main difference is the point of view: in induction we start with a base case (usually n = 1) and work our way inductively up the chain of natural numbers; in a recursive calculation of something like 20!, we recursively calculate 19! then 18! then... until we reach a stopping case of 1!. STRONG INDUCTION: There is a variation of the basic principle called the Principle of Strong Induction. In this version we use not just the claim for n, but the claim for all numbers from 1 to n, to prove the claim for n + 1. Example: The fibonacci numbers 1, 2, 3, 5, 8, 13, 21,... can be defined by the rule that each number is the sum of the two previous ones. Inductively: f(n + 1) = f(n) + f(n 1). 3

4 PROOFS BY INDUCTION: Standard method to prove a statement about all natural numbers: - show that P (1) is true - usually very simple to do! - show that k N; P (k) P k + 1) is true - this is a for-all-if-then proof! - conclude that P (n) is true n N. We will look at proofs by induction of 3 basic kinds: summation formulas; divisibility statements; order relationships. EXAMPLE: Prove that n N, (3n 2) = n(3n 1)/2. OR (3i 2) = n(3n 1)/2. PROOF BY INDUCTION: a) Base case: Check that P (1) is true. For n = 1, (3i 2) = = 1 and n(3n 1)/2 = 1(3(1) 1)/2 = 1(2)/2 = 1. So P (1) is true. b) Inductive Step: Show that for any k N, P (k) P (k + 1) is true. ASSUME: that P (k) is true, i.e. that (3k 2) = k(3k 1)/2. GOAL: Show that P (k + 1) is true, i.e. that (3k 2) + [3(k + 1) 2] = (k + 1)[3(k + 1) 1]/ (3k 2) + [3(k + 1) 2] = k(3k 1)/2 + (3k + 1), by the induction hypothesis, = k(3k 1)/2 + 2(3k + 1)/2, common denominator, = (3k 2 k + 6k + 2)/2 = (3k 2 + 5k + 2)/2 = (k + 1)(3k + 2)/2 = (k + 1)[3(k + 1) 1]/2. Therefore by induction, (3n 2) = n(3n 1)/2, n N. 4

5 EXAMPLE: Prove n N, 4 5 n 1. PROOF BY INDUCTION: a) Base case: Check that P (1) is true. For n = 1, 5 n 1 = 4, which is divisible by 4. So P (1) is true. b) Inductive Step: Show that for any k N, P (k) P (k + 1) is true. ASSUME: that P (k) is true, i.e. that 4 5 k 1 OR 5 k 1 = 4m, m Z GOAL: Show that P (k + 1)is true, i.e. that 4 5 k k+1 1 = 5(5 k ) 1 = (4 + 1)(5 k ) 1 = 4(5 k ) + 5 k 1 = 4(5 k ) + (5 k 1) = 4(5 k ) + 4m by induction hypothesis = 4(5 k + m) where 5 k + m Z 4 5 k+1 1 Therefore by induction, 4 5 n 1, n N. 5

6 EXAMPLE: Prove that n N, (3i 1) = n(3n + 1)/2. PROOF BY INDUCTION: a) Base case: Check that P (1) is true. For n = 1, So P (1) is true. (3i 1) = 2 and n(3n+1)/2 = (1 4)/2 = 2. b) Inductive Step: Show that for any k N, P (k) P (k + 1) is true. ASSUME: that P (k) is true, i.e. that k (3i 1) = k(3k + 1)/2. k+1 GOAL: Show that P (k + 1)is true, i.e. that (3i 1) = (k + 1)(3(k + 1) + 1)/2. k+1 (3i 1) = k (3i 1) + k+1 i=k+1 (3i 1) = k(3k + 1)/2 + (3(k + 1) 1), by the induction hypothesis, = k(3k + 1)/2 + (3k + 2), = k(3k + 1)/2 + 2(3k + 2)/2, common denominator, = (3k 2 + k + 6k + 4)/2 = (3k 2 + 7k + 4)/2 = (k + 1)(3k + 4)/2 = (k + 1)(3(k + 1) + 1)/2. Therefore by induction, (3i 1) = n(3n + 1)/2, n N. 6

7 EXAMPLE: Prove that n N, 3 5 n 2 n. PROOF BY INDUCTION: a) Base case: Check that P (1) is true. For n = 1, 5 n 2 n = 5 2 = 3, which is divisible by 3. So P (1) is true. b) Inductive Step: Show that for any k N, P (k) P (k + 1) is true. ASSUME: that P (k) is true, i.e. that 3 5 k 2 k OR 5 k 2 k = 3m, m Z GOAL: Show that P (k + 1) is true, i.e. that 3 5 k+1 2 k+1. 5 k+1 2 k+1 = 5(5 k ) 2(2 k ) = (3 + 2)(5 k ) 2(2 k ) = 3(5 k ) + 2(5 k ) 2(2 k ) = 3(5 k ) + 2(5 k 2 k ) = 3(5 k ) + 2(3m) by induction hypothesis = 3(5 k + 2m) where 5 k + 2m Z 3 5 k+1 2 k+1. Therefore by induction, 3 5 n 2 n, n N. 7

8 EXAMPLE: Prove that n 2, n N, n > n 2 + n. PROOF BY EXTENDED INDUCTION: a) Base case: NOTE THE BASE CASE HERE IS n = 2 Check that P (2) is true. For n = 2, n = 9 and n 2 + n = 6, and 9 > 6. So P (2) is true. b) Inductive Step: Show that for any k N, k 2, P (k) P (k + 1) is true. ASSUME: that P (k) is true, i.e. that k > k 2 + k. GOAL: Show that P (k + 1) is true, i.e. that (k + 1) > (k + 1) 2 + (k + 1). (k + 1) = k 3 + 3k 2 + 3k , = (k 3 + 1) + (3k 2 + 3k + 1), regrouping > (k 2 + k) + (3k 2 + 3k + 1), by induction hypothesis, = 4k 2 + 3k + (k + 1), regrouping > 4k 2 + 3k + 2 since k 2, k + 1 > 2, > k 2 + 3k + 2, since 4k 2 > k 2, = (k + 1)(k + 2), = (k + 1)((k + 1) + 1), = (k + 1) 2 + (k + 1). So (k + 1) > (k + 1) 2 + (k + 1). Therefore by induction, n > n 2 + n, n N, n 2. 8

9 EXAMPLE: Prove that n 2, n N, PROOF BY EXTENDED INDUCTION: n A i = n A i. a) Base case: NOTE THE BASE CASE HERE IS n = 2 Check that P (2) is true. For n = 2, So P (2) is true. 2 A i = A 1 A 2 = A 1 A 2, by DeMorgan s 2 = A i b) Inductive Step: Show that for any k N, k 2, P (k) P (k + 1) is true. k ASSUME: that P (k) is true, i.e. that A i = k A i. GOAL: Show that P (k + 1) is true, i.e. that k+1 A i = k+1 A i. k+1 A i = = = = k A i A k+1 k A i A k+1, by DeMorgan s k A i A k+1, by induction hypothesis, k+1 A i. n Therefore by induction, A i = n A i, n N, n 2. 9

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