You need to check that P satisfies the conditions in the definition of an

equivalence relation:

1. Reflexivity: aPa

2. Symmetry: if aPb then bPa

3. Transitivity: if aPb and bPc then aPc.

a. Let a=(x,y), y!=0, then xy = yx so aPa, and condition 1 is satisfied.

b. Let a=(x,y) b=(u,v), y, v !=0, and aPb. Therefore xv=yu so yu = xv,

and as ordinary multiplication is comutative uy=vx, and so bPa and

condition 2 is satisfied.

c. Let a=(x,y), b=(u,v), c=(z,w), y, v, w !=0, and aPb and bPc.

Then we have xv = yu and uw = vz. and as y!=0 we can write

u = xv/y,

so:

(xv/y)w = vz,

hence:

xw = yz

which is aPc, so condition 3 is satisfied and hence P is an equivalence

relation on R X (R - {0} ).

For the second part observe that if a=(x,y) is in R X (R - {0} ), then

aPb, where b=(x/y,1). So every element of the set is in an equivalence

class with an element of the form (r,1). Now suppose there are two distinct

elements of this form in the same equivalence class say (r1,1) and (r2,1),

with r1 != r2, then:

r1.1 = 1.r2

so we are forced to conclude that r1=r2 which is a contradiction.

hence we have proven that every element in the set is eqi=uivalent to a

unique element of the form (r,1) which may be taken to be representative of

the equivalence class.

RonL