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
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
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
This is not exercise 10 :'(
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Being a little pedantic, you should first prove that $-(b^{-1}) = (-b)^{-1}$ in order to justify the transition from $-(b^{-1}a)$ to $((-b)^{-1}a)$.
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.
This is not exercise 9 !!!
Ha. Good point! I did Exercise #10 twice it seems. Fixed now.
this is not excercise 10 :(