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Unformatted text preview: Math 55: Discrete Mathematics, Fall 2008 Homework 13 Solutions * 8.5: 16. Reflexive: if ( c,d ) = ( a,b ), the relation becomes ab = ba , which is true. Symmetric: the relation ad = bc is unchanged if we switch ( a,b ) with ( c,d ). Transitive: suppose ( a,b ) R ( c,d ) and ( c,d ) R ( e,f ). Then ad = bc and cf = de . Multiplying the first equation by f and the second by b , we get adf = bcf = bde . Since the letters stand for positive integers, we can cancel d to get af = be , thus ( a,b ) R ( e,f ). 24 (a) Not an equivalence relation, because not symmetric. (b) Equivalence relation. * [5 pts each part] 40 (a) [(1 , 2)] is the set of pairs ( c, 2 c ) for positive integers c . (b) For each positive rational number r , there is an equivalence class consisting of all pairs ( a,b ) of positive integers such that b/a = r . Every class is of this form. (This exercise hints at a deeper idea: the actual definition of the rational number system represents a rational number as an equivalence class of fractions...
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This note was uploaded on 08/16/2010 for the course MATH 55 taught by Professor Strain during the Fall '08 term at University of California, Berkeley.
 Fall '08
 STRAIN
 Math

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