7 Mathematical Induction. For the natural numbers N a the following are equivalent:

Transcription

1 7 Mathematical Induction Definition: N a = {n N n a}. For the natural numbers N a the following are equivalent: 1. Ordinary axiom of induction: IfS N a such that (a) a S (b) n N a (n S n +1 S) then S = N a. 2. Strong axiom of induction: IfS N a such that (a) a S (b) n N a (a, a +1,a+2,...,n S n +1 S) then S = N a. 3. (N a, ) is a well ordered set. Note: In most cases we have a =0ora =1. 94

2 Example: Prove n i = n(n +1). 2 This can be solved without induction. Gauss solved it by adding the first n positive integers twice as follows: 1 + n = n n 1=n n 2=n +1. n + 1 = n +1 Thus we get a total of n(n +1). Hence n i = n(n +1). 2 However it can also be done by induction as follows: 95

3 Let S = {n n i = n(n +1) } S. 2. Suppose n S, i.e., Then n+1 i = ( n = = = Hence n +1 S. Thus S = N.2 n i = i)+n +1 n(n +1). 2 n(n +1) + n + 1 by ind. hyp. 2 n(n +1) 2(n +1) (n +1)(n +2). 2 From now on we do these problems a little less formally. 96

4 Example: Prove n i(i!) = (n +1)! 1. Note: 0!=1. Proof: Trueforn =0since0=1 1. Suppose true for n. Then n+1 i(i!) = (n +1)(n +1)!+ n i(i!) = (n +1)(n +1)!+(n +1)! 1 by ind. hyp. = (n +2)(n +1)! 1 = (n +2)! 1. 2 Example: Prove that all integers n 2 can be written as a product of primes. Proof: By Strong Induction. True for n = 2. Suppose true for k, 2 k<n. If n itself is prime, it can be written as a product of 1 prime, namely itself. Otherwise n = a b, where2 a, b < n. By induction, a and b are products of primes. Hence n = a b is a product of primes. 2 97

5 Example: Consider a 2 n by 2 n chessboard. Such a chessboard is called defective if one square is missing. We claim it is possible to tile such a chessboard with L shaped figures. Proof: By induction. True for n =1. Suppose it is true for n = k and consider a 2 k+1 by 2 k+1 defective chessboard. First place an L shaped figure at the center, as shown. If the L shaped figure is placed correctly, there are now four 2 k by 2 k defective chessboards. By the inductive hypothesis, each one can be tiled with L shaped figures. Hence the entire chessboard can be tiled with such figures. 2 98

6 Example: Suppose f : S T,where S = T = n. Then f is 1-1 iff f is onto. We will only prove half of the theorem. Proof: ( ) By induction. If n = 1 then each set has one element and the theorem is true. Suppose the result holds for sets of size n and let f : S T, where S = T = n + 1. We may assume there exists a S, b T such that f(a) =b. Consider f S {a}, i.e., the function f restricted to S {a}. Since f is 1-1, f does not map any point of S {a} to b. Hence f S {a} : S {a} T {b}. Since f is 1-1, clearly f S {a} is 1-1. Now by the inductive hypothesis, f S {a} is onto. Since f(a) =b, f is clearly onto. ( ) Also proved by induction. 2 99

7 Example: Let R AxA. Then for all integers s, t, R s+t = R s R t. Note: We will use a proof that relies on the fact that composition of relations is associative. Proof: Lets be any fixed integer. The proof is by induction on t. The result is true by definition if t = 1. Suppose it holds for t = k, i.e., R s+k = R s R k.then R s+k+1 = R s+k R = (R s R k )R by ind. hyp. = R s (R k R) = R s R k

8 Example: Let (A, ) be a poset. If the length of a longest chain in A is n, thena can be partitioned into n disjoint antichains. Proof: By induction on n. If n = 1, then no two elements of A are related, and hence A itself is an antichain. Suppose the theorem holds for n and that the longest chain in A has length n +1. LetM be the set of all maximal elements of A. Clearly M is an antichain. Now consider (A M), ). The longest chain in A M has length n. By induction, A M can be partitioned into n disjoint antichains. These antichains, together with M, partition A into n + 1 disjoint antichains. 101

9 We now show that if we take strong induction as an axiom, we can prove that N is well ordered, i.e., that every nonempty subset of N has a least element. Theorem 7.1 (N, ) is well-ordered. Proof: Let S N,S φ. We show that S has a least element. Suppose otherwise. We will reach a contradiction. Claim: N S c We prove the claim by induction. First note that 0 S c ;otherwise 0 S and 0 is the least element of S, a contradiction. Suppose 0, 1, 2,...,n 1 S c.thenn S c ;otherwisen S and n is the least element of S. Thus N S c, establishing the claim. However since N S c, S = φ. Thisisacontradiction. Hence we conclude that S has a least element

10 Some Fallacious Proofs Theorem 7.2 n, n = n +1 Proof : Suppose this is true for n = k. Then k +1 = (k)+1 = (k + 1) + 1 by ind. hyp. = k +2 Fallacy: No basis step. Theorem 7.3 In any group of people, if one person has blue eyes, then everyone has blue eyes. Proof : Let S denote the set of people. The proof is by induction on S = n. The theorem holds for n = 1. Suppose the theorem is true for n and let S = {s 1,s 2,...,s n,s n+1 }. We are assuming that there exists s i S such that s i has blue eyes. We must show that all people in S have blue eyes. Case 1: i n. Consider A = {s 1,s 2,...,s n }. By the inductive hypothesis, everyone in A has blue eyes. In particular, s n has blue eyes. Now consider B = {s 2,s 3,...,s n+1 }.Sinces n has blue eyes, the inductive hypothesis says that everyone in B has blue eyes. 103

11 Thus everyone in S has blue eyes. Case 2: i = n +1. Consider A = {s 2,s 3,...,s n+1 }. By the inductive hypothesis, everyone in A has blue eyes. In particular, s 2 has blue eyes. Now consider B = {s 1,s 2,...,s n }.Sinces 2 has blue eyes, we again conclude by the inductive hypothesis that everyone in B has blue eyes. Thus everyone in S has blue eyes. Fallacy: The inductive argument does not show that n =1 n = 2, i.e., it does not show that the result holds for a group of two people. This is because A B = φ when n =1. Theorem 7.4 n 1, ifa and b are positive integers with max {a, b} = n, then a = b. Proof : By induction on n. The result holds for n =1, i.e., if max {a, b} =1,thena = b =1. Suppose it holds for n, i.e., if max {a, b} = n, thena = b. Now suppose max {a, b} = n +1. Case 1: a 1 b 1. Then a b. Hence a =max{a, b} = n +1. Thus a 1=n and max {a 1,b 1} = n. 104

12 By induction, a 1=b 1. Hence a = b. Case 2: b 1 >a 1. Same argument. Fallacy: In the proof we use the inductive hypothesis to conclude max {a 1,b 1} = n a 1=b 1. However we can only use the inductive hypothesis if a 1and b 1 are positive integers. However this need not be the case, e. g., b =1. 105