Question

^{2. Recall that the set Q of rational numbers consists of equivalence classes of elements of Z × Z\{0} under the equivalence relation R defined by: (a, b)R(c, d) ⇐⇒ ad = bc. We write [a, b] for the equivalence class of the element (a, b). Using this setup, do the following problems: 2A. Show that the following definition of multiplication of elements of Q makes sense (i.e. is “well-defined”): [a, b] · [r, s] = [ar, bs]. (Recall this means that we must check that the definition gives the same answer no matter which representative of the equivalence class we use to compute the product.) [This is the same as problem 19 of section 4.2.]}

Answer 1

The multiplication on is defined by, [a,b].[r,s][ar,bs] for [a,b],[r,s] are equivalence classes on Q.

Now to check the well-definedness of the multiplication, we are to check if [a,b]=[c,d].....(1) and [r,s]=[p,q].....(2) then [a,b].[r,s]=[c,d].[p,q] that is [ar,bs]=[cp,dq] for [a,b], [c,d], [p,q] and [r,s] belonging to Z xZ\{0}.

Now, [a,b]={(x,y): (a,b)R(x,y) holds}, for (a,b),(x,y) belongs to Z x Z\{0}.

Since [a,b]=[c,d] ( from (1)) then (a,b)R(x,y)....(3) holds and (c,d)R(x,y)....(4) holds.

Since R is an equivalence relation then it is symmetric then from (4) we've (x,y)R(c,d) holds....(5).

Again since R is equivalence relation then it is transitive also. Then from (3) and (5) we've (a,b)R(c,d) holds so ad=bc...(6).

Again from (2) we'll have (r,s)R(p,q) holds so rq=ps.....(7).

Now multiplying equation (6) and (7) we get,

ar dq=bs cp

Or, (ar, bs)R(cp, dq) holds

This gives [ar, bs]=[cp, dq].

Since [a,b], [c,d], [p,q] and [r,s] are arbitrary equivalence classes on Q.

So the multiplication of equvalance classes on Q is well defined.