Relations Their Properties Selected Exercises 10 Which relations

Relations & Their Properties: Selected Exercises

10 Which relations in Exercise 4 are irreflexive? A relation is irreflexive a A (a, a) R. Ex. 4 relations on the set of all people: a) a is taller than b. b) a and b were born on the same day. c) a has the same first name as b. d) a and b have a common grandparent. 2

20 Must an asymmetric relation be antisymmetric? A relation is asymmetric a b ( a. Rb (b, a) R ). 3

20 Must an asymmetric relation be antisymmetric? A relation is asymmetric a b ( a. Rb (b, a) R ). To Prove: ( a b ( a. Rb (b, a) R ) ) ( a b ( (a. Rb b. Ra ) a = b ) ) Proof: 1. Assume R is asymmetric. 2. a b ( ( a, b ) R ( b, a ) R ). 3. a b ( ( a. Rb b. Ra ) a = b ) (step 1. and defn of ) (Implication’s premise is false. ) 4. Therefore, asymmetry implies antisymmetry. 4

20 continued Must an antisymmetric relation be asymmetric? ( a b ( (a. Rb b. Ra ) a = b ) ) a b ( a. Rb (b, a) R ) ? Work on this question in pairs. 5

20 continued Must an antisymmetric relation be asymmetric ? ( a b ( (a. Rb b. Ra ) a = b ) ) a b ( a. Rb (b, a) R ) ? Proof that the implication is false: 1. Let R = { (a, a) }. 2. R is antisymmetric. 3. R is not asymmetric: a. Ra (a, a) R is false. Antisymmetry thus does not imply asymmetry. 6

30 • Let R = { (1, 2), (1, 3), (2, 4), (3, 1) }. • Let S = { (2, 1), (3, 2), (4, 2) }. • What is S R? 1 2 3 4 R S 1 S R 2 3 4 7

40 List the 16 different relations on { 0, 1 }. 8

40 List the 16 different relations on { 0, 1 }. – A relation on {0, 1} is a subset of {0, 1} x {0, 1}. – {0, 1} x {0, 1} = { (0, 0), (0, 1), (1, 0), (1, 1) }. – There are 2|{0, 1} x {0, 1}| = 2|{0, 1}| = 22 x 2 = 24 = 16 such subsets. – They are: 9

50 Let R be a relation on set A. Show: R is antisymmetric R R-1 { ( a, a ) | a A }. To prove: 1. R is antisymmetric R R-1 { ( a, a ) | a A } We prove this by contradiction. 2. R R-1 { ( a, a ) | a A } R is antisymmetric. We prove this by contradiction. 11

50 Prove R is antisymmetric R R-1 { ( a, a ) | a A }. We prove this implication by contradiction: 1. Assume R is antisymmetric: a b ( ( a. Rb b. Ra ) a = b ). 2. Assume it is not the case that R R-1 { ( a, a ) | a A }. 3. a b (a, b) R R-1, where a b. (Step 2) 4. Let (a, b) R R-1, where a b. (Step 3) 5. a. Rb , where a b. (Step 4) 6. a. R-1 b, where a b. (Step 4) 7. b. Ra, where a b. 1) (Step 6 & defn of R 12

50 continued Prove R R-1 { ( a, a ) | a A } R is antisymmetric. 1. Assume R R-1 { ( a, a ) | a A }. 2. Assume R is not antisymmetric: ¬ a b ( ( a. Rb b. Ra ) a = b ) 3. a b ( a. Rb b. Ra a b ) 4. b. R-1 a, where a b. 5. ( b, a ) R R-1 where a b, contradicting step 1. 6. Therefore, R is antisymmetric. (Step 2) (Step 3 & defn. of R-1) (Step 3 & 4) 13