Complete Induction and the Well- Ordering Principle Complete Induction as a Rule of Inference In mathematical proofs, complete induction (PCI) is a rule of inference of the form P (a) P (a + 1) P (b) k b ([P (a) P (a + 1) P (k)] P (k + 1)) n a (P (n)) where a and b are positive integers with a b, and P is any propositional function with domain n a.
Complete Induction To prove that a propositional function P (n) is true for all positive integers n a, we complete two steps: 1. (Base Case) Verify the truth of P (a) P (a + 1) P (b) 2. (Inductive Step) Prove the conditional statement [P (a) P (a + 1) P (k)] P (k + 1) is true for all positive integers k b. To complete the inductive step, we assume P (m) is true for all m = a, a+1, a+2,..., k, then show P (k + 1) must also be true for all k b.
Complete Induction Special Case: a = b = 1 To prove that a propositional function P (n) is true for all positive integers n, we complete two steps: 1. (Base Case) Verify that P (1) is true. 2. (Inductive Step) Prove the conditional statement [P (1) P (2) P (3) P (k)] P (k + 1) is true for all positive integers k.
Complete Induction Special Case: a = b = 1 Example: Prove that every positive integer n has a binary expansion of the form n = a 0 + a 1 2 + a 2 2 2 + + a j 2 j, where j is a nonnegative integer, a j = 1, and a i {0, 1} for all i.
Complete Induction Proof: Let P (n) be the proposition: n = a 0 + a 1 2 + a 2 2 2 + + a j 2 j, where j is a nonnegative integer, a j = 1, and a i {0, 1} for all i. We want to show P (n) is true for all n 1. Base Step: P (1) is true, since 1 = a 0 2 0 where j = 0 and a 0 = 1. Inductive Step: Assume P (m) is true for all positive integers m k. Then, consider the integer k + 1. We have two cases:
Case 1: (k+1 is even) If k + 1 is even, then k + 1 = 2m where m k. Therefore, P (m) is true and we have k + 1 = 2m = 2(a 0 + a 1 2 + a 2 2 2 + + a j 2 j ) = 0 + a 0 2 + a 1 2 2 + a 2 2 3 + + a j 2 j+1 = â 0 + â 1 2 + â 2 2 2 + â 3 2 3 + + â j+1 2 j+1 where â j+1 = a j = 1, â 0 = 0, and â i = a i 1 {0, 1} for all 1 i j. Therefore, P (k + 1) is true in this case. Case 2: (k+1 is odd) If k + 1 is odd, then k = 2m where m < k.
Therefore, P (m) is true and we have k + 1 = 1 + 2m = 1 + 2(a 0 + a 1 2 + a 2 2 2 + + a j 2 j ) = 1 + a 0 2 + a 1 2 2 + a 2 2 3 + + a j 2 j+1 = â 0 + â 1 2 + â 2 2 2 + â 3 2 3 + + â j+1 2 j+1 where â j+1 = a j = 1, â 0 = 1, and â i = a i 1 {0, 1} for all 1 i j. Therefore, P (k + 1) is true in either case. This completes the inductive step. Therefore, it follows by complete induction that P (n) is true for all n 1.
Complete Induction Example: Prove that every positive integer n 2 can be expressed as a product of prime numbers.
Complete Induction Proof: Let P (n) be the proposition: n = p 1 p 2 p j, where j 1 and p i is prime for all j. We want to show P (n) is true for all n 2. We will use complete induction corresponding to the case a = b = 2. Base Step: P (2) is true, since is a prime number. 2 = p 1 Inductive Step: Assume P (m) is true for all integers m = a, a + 1,..., k. That is, assume all integers m = 2, 3,..., k have a prime factorization.
Then, for k b = 2, we have two cases. Case 1: (k+1 is prime) If k +1 = p 1 is prime, then P (k +1) is true. Case 2: (k+1 is composite) If k + 1 is composite, then there exist integers c and d with 1 < c d < k + 1, such that k + 1 = c d Then, 2 c k and 2 d k, and it follows by the induction hypothesis that c = p 1 p 2 p j d = q 1 q 2 q l where p i and q i are prime for all i. Thus, k + 1 = (p 1 p 2 p j ) (q 1 q 2 q l )
and we conclude that P (k + 1) is true. This completes the inductive step, and it follows by complete induction that P (n) is true for all n 2.
Well-Ordering Principle Axiom: Every nonempty subset of Z + has a least element. That is, if S Z + and S, then S has a smallest element.
Well-Ordering Principle Example: Use well-ordering property to prove the division algorithm: If a is an integer and d is a positive integer, then there exist (unique) integers q and r with 0 r < d such that a = dq + r.
Well-Ordering Principle Proof: Consider the set S = {n Z + n = a dq +1 where q Z} This set is nonempty because dq can be made as large as desired (by taking q < 0 with q sufficiently large). By the wellordering principle, S has a least element n 0 = (a dq 0 ) + 1 where q 0 Z. We claim n 0 d. If not, then n 1 = (a d(q 0 + 1)) + 1 = n 0 d > 0, which implies n 1 S and n 1 < n 0. This contradicts the assumption that n 0 is the least element of S. Therefore, 1 n 0 d. This proves a = dq 0 + r where r = n 0 1 is an integer such that 0 r < d.
The proof that q 0 and r are unique is left as an exercise.
Well-Ordering Principle Theorem: The Well-Ordering Principle (WOP) and the Principle of Mathematical Induction (PMI) are equivalent.
WOP PMI Proof: Assume WOP holds. We want to prove PMI holds. Let S Z + such that (i) 1 S, (ii) k S k + 1 S for all k Z +. We want to prove S = Z +. Assume for the sake of contradiction that S Z +. Then, T = Z + S is a non-empty subset of Z +. Therefore, by the well-ordering principle, T has a least element m. That is, there exists m T such that m n for all n T. Since 1 S, we know m 2. Therefore m 1 is a positive integer. Clearly m 1 / T (otherwise m 1 would be the least element
of T ). This means m 1 S, and by assumption (ii) (m 1) S (m 1) + 1 S. Therefore, m S which contradicts the fact that m T. We conclude that S = Z +. Therefore PMI holds.
Relations Binary Relation Let A and B be sets. A (binary) relation from A to B is a subset of A B. Notation Let R A B be a relation from A to B. If (a, b) R, we write a R b. 1
Binary Relation Example: Consider the sets A = {2, 3, 5, 7} and B = {4, 6, 8, 9, 10}. Let R be the relation from A to B defined by a R b iff b is divisible by a. Then R consists of the ordered pairs {(2, 4), (2, 6), (2, 8), (2, 10), (3, 6), (3, 9), (5, 10)}. 2
Domain and Range The domain of a relation R from A to B is the set Dom(R) = {x A y (y B and x R y)}. The range of the relation R from A to B is the set Rng(R) = {y B x (x A and x R y)}. 3
Domain and Range Example: Consider the sets A = {2, 3, 5, 7} and B = {4, 6, 8, 9, 10}. Let R be the relation from A to B defined by a R b iff b is divisible by a. Then R consists of the ordered pairs {(2, 4), (2, 6), (2, 8), (2, 10), (3, 6), (3, 9), (5, 10)}.
Domain and Range Example: Consider the sets A = {2, 3, 5, 7} and B = {4, 6, 8, 9, 10}. Let R be the relation from A to B defined by a R b iff b is divisible by a. Then R consists of the ordered pairs {(2, 4), (2, 6), (2, 8), (2, 10), (3, 6), (3, 9), (5, 10)}. Dom(R) = {2, 3, 5} Rng(R) = {4, 6, 8, 9, 10} 4
Domain and Range Example: Let X = Y = R, and consider the relation R from X to Y defined by x R y iff x 2 16 + y2 9 1. That is, R = { (x, y) R R x 2 } 16 + y2 9 1. 5
Domain and Range Example: Let X = Y = R, and consider the relation R from X to Y defined by x R y iff x 2 16 + y2 9 1. That is, R = { (x, y) R R x 2 } 16 + y2 9 1. Dom(R) = [ 4, 4] Rng(R) = [ 3, 3] 6