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# Prove equivalence of different forms for additive inverses of fractions

Prove that if then Proof. We can use the first exercise of this section (Section I.3.3, Exercise #1) and the previous exercise (Section I.3.3, Exercise #8) to compute Then for the other equality, similarly, we have ### 7 comments

1. José Vanderson says:

The answer to exercise 10:

(ab^-1) – (cd^-1) = 1.[(ab^-1) – (cd^-1)] , (Substitute 1 by (bd/bd))
= (bd/bd).[(ab^-1) – (cd^-1)]
=[(adbb^-1) – (cbdd^-1)] / bd
= (ad-cb)/bd

2. Jon says:

This is not exercise 10 :'(

3. Daniel Nagase says:

[latextpage]
Being a little pedantic, you should first prove that $-(b^{-1}) = (-b)^{-1}$ in order to justify the transition from $-(b^{-1}a)$ to $((-b)^{-1}a)$.

• Jon says:

wasn’t that proved already in exercise 8? I mean, we can go ahead and generalize saying that if (ab)^{-1} = a^{-1}b^{-1} then (a)^{-1} = a^{-1} , right? Correct me if I’m mistaken please.

4. Pablo says:

This is not exercise 9 !!!

• RoRi says:

Ha. Good point! I did Exercise #10 twice it seems. Fixed now.

• this is not excercise 10 :(