GIVEN: To construct the rationals Q from the integers Z, let S={(a,b):a,b in Z a

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] ).

Much Appreciated!!!

Explanation / Answer

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]

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.