Relations & Their Properties: Selected Exercises

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 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.

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

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

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.

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.

R S S R 1 1 2 2 3 3 4 4 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?

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

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}| • = 22x2 • = 24 • = 16 such subsets. • They are:

 • { (0,0) } • { (0,1) } • { (1,0) } • { (1,1) } • { (0,0), (0,1) } • { (0,0), (1,0) } • { (0,0), (1,1) } • { (0,1), (1,0) } • { (0,1), (1,1) } • { (1,0), (1,1) } • { (0,0), (0,1), (1,0) } • { (0,0), (0,1), (1,1) } • { (0,0), (1,1), (1,0) } • { (1,1), (0,1), (1,0) } • { (0,0), (0,1), (1,0), (1,1) }

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. We prove this by contradiction.

50 • Prove R is antisymmetric R  R-1 { ( a, a ) | a  A }. • We prove this implication by contradiction: • Assume R is antisymmetric: a b ( ( aRb  bRa )  a = b ). • Assume it is not the case thatR  R-1 { ( a, a ) | a  A }. • a  b (a, b)  R  R-1, where a  b. (Step 2) • Let (a, b)  R  R-1, where a  b.(Step 3) • aRb , where a  b.(Step 4) • aR-1b, where a  b. (Step 4) • bRa, where a  b. (Step 6 & defn of R-1) • R is not antisymmetric, contradicting step 1. (Steps 5 & 7) • Thus, R is antisymmetric R  R-1 { ( a, a ) | a  A }.

50 continued Prove R  R-1 { ( a, a ) | a  A }R is antisymmetric. • Assume R  R-1 { ( a, a ) | a  A }. • Assume R is not antisymmetric: ¬a b ( ( aRb  bRa )  a = b ) • a  b ( aRb  bRa  a  b ) (Step 2) • bR-1a, where a  b. (Step 3 & defn. of R-1) • ( b, a )  R  R-1 where a  b, contradicting step 1. (Step 3 & 4) • Therefore, R is antisymmetric.

Relations & Their Properties: Selected Exercises