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Unformatted text preview: EECS 203: DISCRETE MATHEMATICS Homework 9 Solutions 1. (9 points) If R V V is a binary relation over V then R- 1 (the inverse) is defined as: ( x,y ) R if and only if ( y,x ) R- 1 . Prove or disprove the following: (a) If R is transitive & reflexive then R R- 1 is an equivalence relation. (b) If R is a partial order then R- 1 is a partial order. (c) ( R R- 1 ) * is an equivalence relation. Solution : (a) R is reflexive, so is R- 1 by its definition. Since R is transitive x,y,z (( x,y ) R ( y,z ) R ) = ( x,z ) R (( y,x ) R- 1 ( z,y ) R- 1 ) = ( z,x ) R- 1 hence R- 1 is also transitive. Since both R and R- 1 are transitive and reflexive, so is R R- 1 . Consider ( x,y ) R R- 1 = ( x,y ) R ( x,y ) R- 1 = ( x,y ) R ( y,x ) R (by the definition of R- 1 ) = ( y,x ) R- 1 ( y,x ) R (by the definition of R and R- 1 ) = ( y,x ) R R- 1 which proves that R R- 1 is also symmetric, hence an equivalence relation. (b) As shown above, if R is reflexive and transitive then so is R- 1 . Also, since R is anti symmetric x,y ( x,y ) R ( y,x ) R = ( x = y ) ( y,x ) R- 1 ( x,y ) R- 1 = ( x = y ) which proves that R- 1 is also anti symmetric and hence a partial order. (c) No, if an element v is not related to any others then there is no path from v to v of any length. Consider the following example R= 1 1 1 Here R R- 1 = R, and (R R- 1 ) * = 1 1 1 1 which is not reflexive....
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- Fall '07