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karnten07
Feb19-08, 04:14 PM
1. The problem statement, all variables and given/known data
Let F be a field. For any a,b \in F, b\neq0, we write a/b for ab^-1. Prove the following statements for any a, a' \inF and b, b' \in F\{0}:

i.) a/b = a'/b' if and only if ab' =a'b
ii.) a/b +a'/b' = (ab'+a'b)/bb'
iii.) (a/b)(a'/b') = aa'/bb'
iv.) (a/b)/a'/b') = ab'/a'b (if in addition a'\neq0)


2. Relevant equations



3. The attempt at a solution

I'm struggling to understand how i am to prove these statements. What am i to take the dashes to mean, because they are often used to show inverses? So for the first one:

a/b=ab^-1 which = a^-1b = a'/b'?

EnumaElish
Feb19-08, 04:34 PM
My guess is a dash means "alternative value." For example, a = 1, and a' = 2.

karnten07
Feb19-08, 04:35 PM
My guess is a dash means "alternative value." For example, a = 1, and a' = 2.

I also thought that, i will go with that and see what i come up with, thanks

karnten07
Feb19-08, 04:54 PM
My guess is a dash means "alternative value." For example, a = 1, and a' = 2.

Does ab^-1 mean a.b^-1 or (a.b)^-1? I think it might be the former.

If it is, i get:

i) a.b^-1 = a'.b'^-1 when written out fully. So if ab' = a'b, then rearranged gives a= a'b/b' and a' = ab'/b. So inserting them into a.b^-1 = a'.b'^-1 we get:

a'b.b^-1/b' = ab.b^-1/b

and then we get indentity elements leaving a'/b' = a/b

Is this right?