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.

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1 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.

2 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.

3 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.

4 The proof that q 0 and r are unique is left as an exercise.

5 Well-Ordering Principle Theorem: The Well-Ordering Principle (WOP) and the Principle of Mathematical Induction (PMI) are equivalent.

6 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

7 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.

8 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

9 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

10 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

11 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)}.

12 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

13 Domain and Range Example: Let X = Y = R, and consider the relation R from X to Y defined by x R y iff x y That is, R = { (x, y) R R x 2 } 16 + y

14 Domain and Range Example: Let X = Y = R, and consider the relation R from X to Y defined by x R y iff x y That is, R = { (x, y) R R x 2 } 16 + y Dom(R) = [ 4, 4] Rng(R) = [ 3, 3] 6

15 Binary Relation on a Set A (binary) relation on a set S is a relation from the set S to S. 7

16 Binary Relation on a Set Examples: Let S = R. The following are examples of binary relations on S. R 1 = {(x, y) x = y}, R 2 = {(x, y) x < y}, R 3 = {(x, y) x y}, R 4 = {(x, y) x > y}, R 5 = {(x, y) x y}, R 6 = {(x, y) x = y or x = y}, R 7 = {(x, y) x = y + 1}, R 8 = {(x, y) x + y 3}. 8

17 Identity Relation The identity relation on S is the relation from S to itself given by I S = {(x, x) x S}. Example: Let S = {a, b, c}. The identity relation on S is the relation I S = {(a, a), (b, b), (c, c)} 9

18 Inverse Relation If R is a relation from A to B, then the inverse of R, denoted R 1, is the relation R 1 = {(y, x) (x, y) R} Example: Consider the relation R = {(2, 4), (2, 6), (2, 8), (2, 10), (3, 6), (3, 9), (5, 10)}. The inverse of R is the relation R 1 = {(4, 2), (6, 2), (8, 2), (10, 2), (6, 3), (9, 3), (10, 5)}. 10

19 Inverse Relation Theorem: If R is a relation from A to B, then (a) Rng(R 1 ) = Dom(R). (b) Dom(R 1 ) = Rng(R). 11

20 Proof:

21 Inverse Relation Example Let R be the relation on R given by x R y iff y = e x. The inverse of R is the relation given by x R 1 y iff x = e y iff y = ln x. 12

22 Inverse Relation Example Let R be the relation on R given by x R y iff y = e x. The inverse of R is the relation given by x R 1 y iff x = e y iff y = ln x. The previous theorem gives Dom(R 1 ) = Rng(R) = (0, ). Rng(R 1 ) = Dom(R) = (, ). 13

23 Composition of Relations Let R be a relation from A to B, and let S be a relation from B to C. The composition of R and S, denoted S R, is a relation from A to C defined by S R = {(a, c) b B ( (a, b) R and (b, c) S )} 14

24 Composition of Relations Example: Consider the sets A = {1, 2, 3, 4, 5} B = {p, q, r, s, t} C = {x, y, z, w} Let R be the relation from A to B given by R = {(1, p), (1, q), (2, q), (3, r), (4, s)}. Let S be the relation from B to C given by S = {(p, x), (q, x), (q, y), (s, z), (t, z)}. 15

25 Composition of Relations Example: Consider the sets A = {1, 2, 3, 4, 5} B = {p, q, r, s, t} C = {x, y, z, w} Let R be the relation from A to B given by R = {(1, p), (1, q), (2, q), (3, r), (4, s)}. Let S be the relation from B to C given by S = {(p, x), (q, x), (q, y), (s, z), (t, z)}. The composition S R is the relation from A to C given by S R = {(1, x), (1, y), (2, x), (2, y), (4, z)}. 16

26 Composition of Relations Example: Consider the relations R = {(x, y) R R y = x + 1} S = {(x, y) R R y = x 2 }. Determine the following: S R = R S = 17

27 Composition of Relations Theorem: Suppose A, B, C, and D are sets. Let R be a relation from A to B, S be a relation from B to C, and T be a relation from C to D. (a) (R 1 ) 1 = R. (b) T (S R) = (T S) R. (c) I B R = R and R I A = R. (d) (S R) 1 = R 1 S 1 18