1. Relations & Their Properties: Selected Exercises

2. 2 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.

3. 3 20 Must an asymmetric relation be antisymmetric? A relation is asymmetric   a  b ( aRb  (b, a)  R ).

4. 4 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. and defn of  ) 3.  a  b ( ( aRb  bRa )  a = b ) (Implication’s premise is false.) 4.Therefore, asymmetry implies antisymmetry.

5. 5 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.

6. 6 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.

7. 7 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 RS 1 2 3 4 S  R

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

9. 9 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}||{0,1}| = 2 2x2 = 2 4 = 16 such subsets. –They are:

10. 10 1.  2.{ (0,0) } 3.{ (0,1) } 4.{ (1,0) } 5.{ (1,1) } 6.{ (0,0), (0,1) } 7.{ (0,0), (1,0) } 8.{ (0,0), (1,1) } 9.{ (0,1), (1,0) } 10.{ (0,1), (1,1) } 11.{ (1,0), (1,1) } 12.{ (0,0), (0,1), (1,0) } 13.{ (0,0), (0,1), (1,1) } 14.{ (0,0), (1,1), (1,0) } 15.{ (1,1), (0,1), (1,0) } 16.{ (0,0), (0,1), (1,0), (1,1) }

11. 11 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.

12. 12 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 ( ( aRb  bRa )  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.aRb, where a  b. (Step 4) 6.aR -1 b, where a  b. (Step 4) 7.bRa, where a  b. (Step 6 & defn of R -1 ) 8.R is not antisymmetric, contradicting step 1. (Steps 5 & 7) 9.Thus, R is antisymmetric  R  R -1  { ( a, a ) | a  A }.

13. 13 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 ( ( aRb  bRa )  a = b ) 3.  a  b ( aRb  bRa  a  b ) (Step 2) 4.bR -1 a, where a  b. (Step 3 & defn. of R -1 ) 5.( b, a )  R  R -1 where a  b, contradicting step 1. (Step 3 & 4) 6.Therefore, R is antisymmetric.