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Deveno
Deveno is offline
#4
Feb27-12, 12:43 AM
Sci Advisor
P: 906

Apostol's Field Axioms


i'm not sure which "axiom" you're talking about, is it:

1/(xy) = (1/x)(1/y)?

one doesn't need "uniqueness" of an inverse to show that, just "existence":

(xy)[(1/x)(1/y)] = x(y[(1/x)(1/y)] (associativity)

= x(y[(1/y)(1/x)] (commutativity)

= x([y(1/y)](1/x)] (associativity)

= x((1)(1/x)) (using y(1/y) = 1, since 1/y is "a" inverse for y)

= x(1/x) (multiplicative identity property of 1)

= 1 (since 1/x is "a" inverse for x).

thus 1/(xy) is "one" inverse for xy (it might be, that there are others, but at the very least we know for sure that (xy)(1/(xy)) = 1 if x(1/x) = 1, and y(1/y) = 1).

what is going on is this:

the actual process of computing a/b + c/d goes like this:

1. a/b + c/d = (a/b)(1) + (c/d)(1) =
2. (a/b)(d/d) + (c/d)(b/b) =
3. (ad/(bd)) + (bc/(bd)) =
4. (ad + bc)/(bd)

1-->2 is obvious
2-->3 uses commutativity and associativity of multiplication

(which is why writing 1/x = x-1 makes it clearer)

3-->4 uses the distributive law

for the above to work, we just need the existence of a-1 and b-1, we don't need uniqueness (the possibility of division).

of course, uniqueness is easy to prove given existence:

suppose ab = ba = 1, and ca = ac = 1:

then c = c(1) = c(ab) = (ca)b = (1)b = b.