GIVEN: To construct the rationals Q from the integers Z, let S={(a,b):a,b in Z a
ID: 2944636 • Letter: G
Question
GIVEN: To construct the rationals Q from the integers Z, let S={(a,b):a,b in Z and b not zero}. The equivalence relation "--" on S is defined by (a,b) -- (c,d) iff ad=bc. Thus, we define the set Q of rationals to be the set of equivalence classes corresponding to "--". The equivalence class determined by the ordered pair (a,b) we denote by [a/b]. Then [a/b] is what we usually think of as the fraction a/b. For a, b, c, d in Z with b and d not zero,we define addition and multiplication in Q by: [a/b]+[c/d] = [(ad+bc)/bd]; [a/b].[c/d]=[(ac)/(bd)].SHOW: Verify that "--" is an equivalence relation on S. Then show that addition and multiplication are well defined (That is, suppose [a/b]=[p/q] and [c/d]=[r/s] and show that [(ad+bc)/bd]=[(ps+qr)/qs] and [ac/bd]=[pr/qs] ).
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Explanation / Answer
GIVEN: To construct the rationals Q from the integers Z, let S={(a,b):a,b in Z and b not zero}. The equivalence relation "--" on S is defined by (a,b) -- (c,d) iff ad=bc. Thus, we define the set Q of rationals to be the set of equivalence classes corresponding to "--". The equivalence class determined by the ordered pair (a,b) we denote by [a/b]. Then [a/b] is what we usually think of as the fraction a/b. For a, b, c, d in Z with b and d not zero,we define addition and multiplication in Q by: [a/b]+[c/d] = [(ad+bc)/bd]; [a/b].[c/d]=[(ac)/(bd)]. SHOW: Verify that "--" is an equivalence relation on S. Then show that addition and multiplication are well defined (That is, suppose [a/b]=[p/q] and [c/d]=[r/s] and show that [(ad+bc)/bd]=[(ps+qr)/qs] and [ac/bd]=[pr/qs] ). WE HAVE THE GIVEN RELATION AS [A,B] IS RELATED TO [C,D] ...IFF......AD=BC. LET US TEST THE 3 CRITERIA FOR EQUIVALNCE 1.REFLEXIVE [A,B] SHALL BE RELATED TO [A,B] IT IS TRUE SINCE WE HAVE C=A...D=B..AND AD=AB BC=BA SINCE MULTIPLICATION OF REAL NUMBERS IS COMMUTATIVE WE HAVE AB=BA AND HENCE AD=BC....OK 2.SYMMETRIC ... IF [A,B] IS RELATED TO [C,D]...THEN [C,D] SHALL BE RELATED TO [A,B]. IT IS OK SINCE WE HAVE AD=BC FROM THE GIVEN RELATION WHICH BY COMMUTATIVE PROPERTY MEANS DA=CB...WHICH MEANS [C,D] IS RELATED TO [A,B]. 3.TRANSITIVE [A,B] IS RELATED TO [C,D] AND [C,D] IS RELATED TO [E,F] TO CHECK IF [A,B] IS RELATED TO [E,F]....THAT IS TO CHECJK IF AF=BE.......................1 WE HAVE ... [A,B] IS RELATED TO [C,D]...SO AD=CB....................2 AND [C,D] IS RELATED TO [E,F]...SO CF=DE..............................3 2*3 GIVES ADCF=CBDE..........................4 SINCE D IS NOT ZERO DIVIDING 4 WITH D AFC=BEC ...................................5 IF C IS ZERO THEN OBVIOUSLY A AND E ARE ZEROS IN WHICH CASE AF=0=BE....OK IF C IS NOT ZERO THEN WE CAN DIVIDE 5 WITH C TO GET AF=BE.....OK... SO THE RELATION IS AN EQUVALENCE RELATION ------------ suppose [a/b]=[p/q] and [c/d]=[r/s] and show that [(ad+bc)/bd]=[(ps+qr)/qs] and [ac/bd]=[pr/qs] ). A/B=P/Q = K SAY A=BK AND P=QK...............................1 C/D=R/S = K' SAY C=DK' AND R=SK'...................................3 [AD+BC]/BD=[BKD+BDK']/BD = K+K'....................4 [PS+RQ]/QS=[QKS+SK'Q]/QS=K+K'.......................5 SO [(ad+bc)/bd]=[(ps+qr)/qs]...............PROVED.... SIMILARLY [ac/bd]= BKDK'/BD=KK'.................................7 [pr/qs] )=QKSK'/QS=KK'...............................8 SO [ac/bd]=[pr/qs]
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