Relations & Their Properties: Selected Exercises

Copyright © Peter Cappello Exercise 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 is taller than b. a and b were born on the same day. a has the same first name as b. a and b have a common grandparent. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 Must an asymmetric relation be antisymmetric? A relation is asymmetric  a b ( aRb  (b, a)  R ). Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 Must an asymmetric relation be antisymmetric? A relation is asymmetric  a b ( aRb  (b, a)  R ). To Prove: (a  b ( aRb  (b, a)  R ) )  (a  b ( (aRb  bRa )  a = b ) ) Proof: Assume R is asymmetric. a  b ( ( a, b )  R  ( b, a )  R ). (step 1. & defn of  ) a  b ( ( aRb  bRa )  a = b ) (implication premise is false.) Therefore, asymmetry implies antisymmetry. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 continued Must an antisymmetric relation be asymmetric? (a b ( ( aRb  bRa )  a = b ) )  a  b ( aRb  ( b, a )  R )? Work on this question in pairs. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 20 continued Must an antisymmetric relation be asymmetric ? (a b ( (aRb  bRa )  a = b ) )  a  b ( aRb  (b, a)  R ) ? Proof that the implication is false: Let R = { (a, a) }. R is antisymmetric. R is not asymmetric: aRa  (a, a)  R is false. Antisymmetry thus does not imply asymmetry. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 30 Let R = { (1, 2), (1, 3), (2, 3), (2, 4), (3, 1) }. Let S = { (2, 1), (3, 1), (3, 2), (4, 2) }. What is S  R? 1 2 3 4 R S S  R Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 50 Let R be a relation on set A. Show: R is antisymmetric  R  R-1  { ( a, a ) | a  A }. To prove: R is antisymmetric  R  R-1  { ( a, a ) | a  A } We prove this by contradiction. R  R-1  { ( a, a ) | a  A }  R is antisymmetric. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 50 Prove R is antisymmetric  R  R-1  { ( a, a ) | a  A }. Proceeding by contradiction, we assume that: R is antisymmetric: a b ( ( aRb  bRa )  a = b ). It is not the case that R  R-1  { ( a, a ) | a  A }. a b (a, b)  R  R-1, where a  b. (Step 1.2) Let (a, b)  R  R-1, where a  b. (Step 2) aRb , where a  b. (Step 3) aR-1b, where a  b. (Step 3) bRa, where a  b. (Step 5 & defn of R-1) R is not antisymmetric, contradicting step 1. (Steps 4 & 6) Thus, R is antisymmetric  R  R-1  { ( a, a ) | a  A }. Copyright © Peter Cappello

Copyright © Peter Cappello Exercise 50 continued Prove R  R-1  { ( a, a ) | a  A }  R is antisymmetric. Proceeding by contradiction, we assume that: R  R-1  { ( a, a ) | a  A }. R is not antisymmetric: ¬a b ( ( aRb  bRa )  a = b ) Assume a b ( aRb  bRa  a  b ) (Step 1.2) bR-1a, where a  b. (Step 2s & defn. of R-1) ( b, a )  R  R-1 where a  b, contradicting step 1. (Step 2 & 3) Therefore, R  R-1  { ( a, a ) | a  A }  R is antisymmetric. Copyright © Peter Cappello