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Unformatted text preview: Solutions to Homework 7 Problem #1 (3 points) Reflexive: ( a 1 , a 2 ) R ( a 1 , a 2 ) since E 1 and E 2 are themselves reflexive and therefore a 1 E 1 a 1 and a 2 E 2 a 2 . Symmetric: ( a 1 , a 2 ) R ( b 1 , b 2 ) ⇔ a 1 E 1 b 1 and a 2 E 2 b 2 ⇔ b 1 E 1 a 1 and b 2 E 2 a 2 (since E 1 and E 2 are themselves symmetric) ⇔ ( b 1 , b 2 ) R ( a 1 , a 2 ) Transitive: Assume that ( a 1 , a 2 ) R ( b 1 , b 2 and ( a 1 , a 2 ) R ( c 1 , c 2 ). This is true if and only if: a 1 E 1 b 1 and a 2 E 2 b 2 and a 1 E 1 c 1 and a 2 E 2 c 2 ⇔ b 1 E 1 c 1 and b 2 E 2 c 2 (by the transitivity of E 1 and E 2 ) ⇔ ( b 1 , b 2 ) R ( c 1 , c 2 ) Therefore, R is an equivalence relation. Problem #2 (3 points) This is not a partial ordering since it is not antisymmetric. Consider ( a, b ) = (1 , 4) and ( c, d ) = (2 , 3). Then, ( a, b ) R ( c, d ) since 1 ≤ 2. But also, ( c, d ) R ( a, b ) since 3 ≤ 4. But, ( a, b ) 6 = ( c, d )....
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This note was uploaded on 01/31/2011 for the course MATH 300 taught by Professor Ctw during the Spring '08 term at Rutgers.
 Spring '08
 ctw
 Math

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