HW9-Solutions

# HW9-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 9...

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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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HW9-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 9...

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