Relations & Their Properties: Selected Exercises

Copyright © Peter Cappello2 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)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.

Copyright © Peter Cappello3 Exercise 20 Must an asymmetric relation be antisymmetric? A relation is asymmetric   a  b ( aRb  (b, a)  R ).

Copyright © Peter Cappello4 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: 1.Assume R is asymmetric. 2.  a  b ( ( a, b )  R  ( b, a )  R ). (step 1. & defn of  ) 3.  a  b ( ( aRb  bRa )  a = b ) (implication premise is false.) 4.Therefore, asymmetry implies antisymmetry.

Copyright © Peter Cappello5 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 Cappello6 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: 1.Let R = { (a, a) }. 2.R is antisymmetric. 3.R is not asymmetric: aRa  (a, a)  R is false. Antisymmetry thus does not imply asymmetry.

Copyright © Peter Cappello7 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? RS S  R

Copyright © Peter Cappello8 Exercise 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.

Copyright © Peter Cappello9 Exercise 50 Prove R is antisymmetric  R  R -1  { ( a, a ) | a  A }. 1.Proceeding by contradiction, we assume that: 1.R is antisymmetric:  a  b ( ( aRb  bRa )  a = b ). 2.It is not the case that R  R -1  { ( a, a ) | a  A }. 2.  a  b (a, b)  R  R -1, where a  b. (Step 1.2) 3.Let (a, b)  R  R -1, where a  b. (Step 2) 4.aRb, where a  b. (Step 3) 5.aR -1 b, where a  b. (Step 3) 6.bRa, where a  b. (Step 5 & defn of R -1 ) 7.R is not antisymmetric, contradicting step 1. (Steps 4 & 6) 8.Thus, R is antisymmetric  R  R -1  { ( a, a ) | a  A }.

Copyright © Peter Cappello10 Exercise 50 continued Prove R  R -1  { ( a, a ) | a  A }  R is antisymmetric. 1. Proceeding by contradiction, we assume that: 1.R  R -1  { ( a, a ) | a  A }. 2.R is not antisymmetric: ¬  a  b ( ( aRb  bRa )  a = b ) 2. Assume  a  b ( aRb  bRa  a  b ) (Step 1.2) 3.bR -1 a, where a  b. (Step 2s & defn. of R -1 ) 4.( b, a )  R  R -1 where a  b, contradicting step 1. (Step 2 & 3) 5.Therefore, R is antisymmetric.