according to Fraleigh,
in the Field of Quotients of An Integral Domain,
( Let D be an integral domain and form the Cartesian product D x D = {(a,b)} | a,b $\in$ D}
and S = {(a,b) | a,b $\in$ D, b is not 0},
let (a,b) , (c,d) are equivalent iff ad= bc
and define F tobe the set of all equivalence classes [(a,b)] for (a,b) $\in$ S.)
[(-a,b)] is an additive inverse for [(a,b)] in F.
But, [(-a,b)] + [(a,b)] = [ (ab-ba, b$^2$ ] = [ (0, b$^2$ ]
how is this equal to [(0.1])?
Also, I'm not sure how to show the distributive laws hold in F.
I tried to show [(a,b)] ( [(c,d)] + [(e,f)] ) = [(ac, bd)] + [(ae, bf)]
but they are clearly different, since
the left side equals to [(adf+bcf+bde,bdf)]
while the right side equls to [(acbf+bdae, bdbdf)]
Give me some help please!
Best Answer
The definition of $(a,b)\sim(c,d)$ you were given here for domains was probably $ad=bc$. When two pairs are similar, the classes are the same: $[(a,b)]=[(c,d)]$.
So, is $(0,s)\sim(0,1)$ for any $s\in D\setminus \{0\}$?
Another good thing to establish is that $(a,b)\sim(at,bt)$ for any $b,t\neq 0$, and any $a$, giving you the equality $[(a,b)]=[(at,bt)]$.
For distributivity, you need to check your work. It looks like there were some weird mistakes. You should have $acf+ade$ in the "numerator" of the left-hand side. You also have one too many $d$'s in the "denominator" of the right-hand side.
After you've fixed the mistakes, the "another good thing" I mentioned should carry you home.