Addition: [(a,b)] + [(c,d)] = [(ad+bc,bd)]
Multiplication: [(a,b)] ´ [(c,d)] = [(ac,bd)]

Before we do anything else, we first need to ask whether these definitions make sense. [(a,b)] is an equivalence class and will have many different representations, for example if (a,b) ~ (e,f) and (c,d) ~ (g,h), then [(a,b)] = [(e,f)] and [(c,d)] = [(g,h)]. Then we need to know that [(a,b)] + [(c,d)]  and [(e,f)] + [(g,h)]  give the same answer, since they are the sums of the same equivalence classes. That means we need to know that if (a,b) ~ (e,f) and (c,d) ~ (g,h), then (ad+bc,bd) ~ (eh+fg,fh). To check this we must show (ad+bc)fh = (eh+fg)bd. Now

(ad+bc)fh	= adfh + bcfh		(distributive law)
	= bedh + bcfh		( af = be since (a,b) ~ (e,f) )
	= bedh + bfdg		( ch = dg since (c,d) ~ (g,h) )
	= (eh+fg)bd

and we have shown that addition is "well-defined." You are asked to prove that multiplication is also well-defined on the homework.

We have now defined an algebraic structure with set D^2*/~ and operations + an ´. To simplify the notation, we will call this structure F. Our next task is to show that F is a field. This means checking that all nine field laws are satisfied. Because of limitations of time, space, and my typing ability, I’m only going to go over a sampling of the laws. You can check that the other laws are also satisfied on your own.

Law 1 (associate law of addition): For all x,y,z Î F, (x + y)nbsp;+ z = x + (y +&nsp;z)

Proof: Let x = [(a,b)], y = [(c,d)], and z = [e,f)]. Then

(x + y) + z	= ( [(a,b)] + [(c,d)] ) + [(e,f)]
	= [(ad + bc,bd)] + [(e,f)]
	= [((ad + bc)f + bde,bdf)]
	= [(adf + bcf + bde,bdf)]
	= [((df)a + (cf + de)b,(df)b)]
	= [(a,b)] + [(cf + de),df)]
	= [(a,b)] + ( [(c,d)] + [(e,f)] )
	= x + (y + z)

Law 3 (additive identity): Let 0 Î F be the element 0 = [(0,1)] where 0 is the additive identity of D and 1 is the multiplicative identity of D. It is easy to check that then 0 = [(0,d)] for all non-zero d Î D since (0,1) ~ (0,d). We now check that 0 is an additive identity for F. Let x = [(a,b)] Î F. Then

x + 0	= [(a,b)] + [(0,1)]
	= [(a´ 1+b´ 0,b´ 1)]
	= [(a,b)]
	= x

Law 7 (multiplicative identity): Let 1 Î F be the element 1 = [(1,1)] where 1 is the multiplicative identity for D. It is easy to check that 1 = [(d,d)] for all non-zero d Î D, since (1,1) ~ (d,d). We now check that 1 is a multiplicative identity for F. Let x = [(a,b)] Î F. Then

x ´ 1	= [(a,b)] ´ [(1,1)]
	= [(a´ 1,b´ 1)]
	= [(a,b)]
	= x

Law 8 (multiplicative inverses): Let x = [(a,b)] be a non-zero element of F. Then a ¹  0 and we can define x - ¹ = [(b,a)]. We now check

x ´ x - 1	= [(a,b)] ´ [(b,a)]
	= [(ab,ab)]
	= 1

The proofs of the other laws are all just simple (but tedious) algebraic manipulations. So now we have built a field of fractions from a general domain. The final topic, to be addressed in the next lecture, is to show that the field of fractions includes the original domain. But there is a problem here. It is nonsense to write D Ì F. That would say every element of the set D is an element of the set F. But the elements of F are equivalence classes of pairs of elements of D, so no element of D could possibly be an element of F. We solve this problem by introducing the notion of isomorphic structures. Domains D and E are isomorphic if there is a function f: D ® E such that f is 1-1 (i.e. f(a) ¹ f(b) if a ¹ b), f is onto (i.e. for every e Î E, there is a d Î D with f(d) = e), and for every a,b Î D, f(a+b) = f(a) + f(b) and f(a´b) = f(a)´f(b). We denote isomorphic structures by D @ E. This is the same as the congruence symbol in geometry, and has a very similar meaning. Congruent triangles are different triangles, but have exactly the same geometric structure. In the same way, isomorphic domains may be different sets, but have identical algebraic structure. Since we can’t possibly have D Ì F, we will show D @ E Ì F, where E = {x Î F: x = [(a,1)]}.