Relations Sections 8.1 & 8.5 Based on Rosen and slides by K. - PDF document

Relations Sections 8.1 & 8.5 Based on Rosen and slides by K. Busch 1 Relations and Their Properties A binary relation R from set to set A B is a subset of Cartesian product A  B   Example: UW students UW courses A B R  {( , ) | is enrolled in } a b a b  B  Example: { 0 , 1 , 2 } { b , } A a R  {( 0 , ), ( 0 , ), ( 1 , ), ( 2 , )} a b a b 2 1

A  A relation on set is a subset of A A Example:  A relation on set : { 1 , 2 , 3 , 4 } A  {( 1 , 1 ), ( 1 , 2 ), ( 2 , 1 ), ( 2 , 2 ), ( 3 , 4 ), ( 4 , 1 ), ( 4 , 4 )} R 3 More Examples Relations over integers:   {( , ) | } R a b a b     {( , ) | or } R a b a b a b    {( , ) | (mod )} for positive integer 1 R a b a b m m    {( , ) | 1 } R a b b a (Actually a function) 2

Functions as Relations Relation over integers Z    {( , ) | 1 } R a b b a    Function from Z to Z ( ) 1 f a b a  : f Z Z Function from A to B assigns exactly one element from B to each input from A i.e., a functions is a restricted type of relation where every a in A is in exactly one ordered pair (a,b). Reflexive relation on set : R A    , ( , ) a A a a R  Example: { 1 , 2 , 3 , 4 } A  {( 1 , 1 ), ( 1 , 2 ), ( 2 , 1 ), ( 2 , 2 ), ( 3 , 4 ), ( 3 , 3 ), ( 4 , 3 ), ( 4 , 4 )} R 6 3

Symmetric relation : R    ( , ) ( , ) a b R b a R  Example: { 1 , 2 , 3 , 4 } A  {( 1 , 1 ), ( 1 , 2 ), ( 2 , 1 ), ( 2 , 2 ), ( 3 , 4 ), ( 4 , 3 ), ( 4 , 4 )} R 7 Antisymmetric relation : R      ( , ) ( , ) a b R b a R a b  Example: { 1 , 2 , 3 , 4 } A  {( 1 , 1 ), ( 1 , 2 ), ( 2 , 2 ), ( 3 , 4 ), ( 4 , 4 )} R 8 4

Transitive relation : R      ( , ) ( , ) ( , ) a b R b c R a c R  Example: { 1 , 2 , 3 , 4 } A  {( 1 , 1 ), ( 1 , 2 ), ( 2 , 3 ), ( 3 , 4 )( 1 , 3 ), ( 1 , 4 ), ( 2 , 4 )} R 9 Combining Relations 1  {( 1 , 1 ), ( 2 , 2 ), ( 3 , 3 )} R 2  {( 1 , 1 ), ( 1 , 2 ), ( 1 , 3 ), ( 1 , 4 )} R   {( 1 , 1 ), ( 1 , 2 ), ( 1 , 3 ), ( 1 , 4 ), ( 2 , 2 ), ( 3 , 3 )} R R 1 2   {( 1 , 1 )} R R 1 2   {( 2 , 2 ), ( 3 , 3 )) R R 1 2 10 5

Composite relation: S  R        ( , ) : ( , ) ( , ) a b S R x a x R x b S      Note:  ( , ) ( , ) ( , ) a b R b c S a c S R Example:  {( 1 , 1 ), ( 1 , 4 ), ( 2 , 3 ), ( 3 , 1 ), ( 3 , 4 )} R  {( 1 , 0 ), ( 2 , 0 ), ( 3 , 1 ), ( 3 , 2 ), ( 4 , 1 )} S  S  {( 1 , 0 ), ( 1 , 1 ), ( 2 , 1 ), ( 2 , 2 ), ( 3 , 0 ), ( 3 , 1 )} R 11 Power of relation: n R  1 R   1 n n  R R R R  Example: {( 1 , 1 ), ( 2 , 1 ), ( 3 , 2 ), ( 4 , 3 )} R   2  {( 1 , 1 ), ( 2 , 1 ), ( 3 , 1 )( 4 , 2 )} R R R   3 2  {( 1 , 1 ), ( 2 , 1 ), ( 3 , 1 )( 4 , 1 )} R R R   4 3 3  R R R R 12 6

Theorem: A relation is transitive R R n  if and only if R for all   1 , 2 , 3 , n R  Proof: 2 1. If part: R 2. Only if part: use induction 13 R  1. If part: We will show that if 2 R then is transitive R R  Assumption: 2 R  Definition of power: 2  R R R  ( , ) a c R Definition of composition:       ( , ) ( , ) ( , ) a b R b c R a c R R Therefore, is transitive R 14 7

2. Only if part: We will show that if is transitive R R n   then for all 1 R n Proof by induction on n  Inductive basis: 1 n   1 It trivially holds R R R 15 Inductive hypothesis: R k  Assume that R  k  for all 1 n 16 8

 1  Inductive step: We will prove R n R Take arbitrary   n 1 ( , ) a b R We will show  ( , ) a b R 17   n 1 ( , ) a b R definition of power n   ( , ) a b R R definition of composition     n : ( , ) ( , ) x a x R x b R R n  inductive hypothesis R     : ( , ) ( , ) x a x R x b R is transitive R  ( , ) a b R End of Proof 18 9