Please help me with this number theory problem that involves

Please help me with this number theory problem that involves equivalence relation proofs. Please be detailed and show all work.

Thank you.

#2. Let T denote the ordered pairs of integers (a, b), with b non-zero. For (a, b) and (c, d) in T, define (a, b) (c, d) to mean adabc. Prove: i. is an equivalence relation. ii. If (a, b) (a\', b) and (c, d) (c\', d\'), then (ad+bc, bod) (a d\'+b\'c\', b\' d\'). iii. If (a, b) (a\', b) and (c, d) (c\', d\'), then (ac, bd) (a c\', b\'d\'.

Solution

(a,b)~(c,d) means ad=bc

1) equivalance relation

reflexive ---- (a,b)~(a,b) since ab=ba ( integers are commutative wrt multiplication )

symetry ----- suppose (a,b) ~(c,d) then ad=bc

since ad=da and bc=cb

(c,d)~ (a,b)  as cb= da

Transitive --------- suppose (a,b) ~(c,d) and (c,d)~(p,q)

then ad=bc ,cq=dp

we will try to show that (a,b) ~(p,q) that is aq=bp

divide ad=bc and dp= cq ,we get a/p=b/q --------.> aq=bp

hence (a,b) ~(p,q) and relation is transitive

~ is equivalance relation.

2) (a,b) ~(a\',b\') and (c,d)~(c\',d\') then ab\'=ba\' and cd\'=dc\'

show (ad+bc,bd)~(a\'d\'+b\'c\',b\'d\') that is   (ad+bc)(b\'d\')=bd((a\'d\'+b\'c\')

take left side of the above equation (ad+bc)(b\'d\')= adb\'d\'+bcb\'d\'= ab\'dd\'+cd\'bb\' = ba\'dd\'+dc\'bb\' = bd(a\'d\'+b\'c\') right side

hence  (ad+bc,bd)~(a\'d\'+b\'c\',b\'d\')

3)(a,b)~(a\',b\') and (c,d)~(c\',d\') then ab\'=ba\' and cd\'=dc\'

show (ac,bd)~(a\'c\',b\'d\') that is acb\'d\'=bda\'c\'

take left side acb\'d\' =ab\'cd\'=ba\'dc\' =bda\'c\' right side hence  (ac,bd)~(a\'c\',b\'d\')