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Prove equivalence of different forms for additive inverses of fractions

by RoRi

Prove that if b \neq 0 then

    \[ - \left( \frac{a}{b} \right) = \frac{-a}{b} = \frac{a}{-b}. \]

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

    \begin{align*} - \left(\frac{a}{b} \right) &= -(ab^{-1}) &(\text{Def of } b^{-1}) \\ &= -(b^{-1}a) \\ &= \left( (-b)^{-1} a \right) &(\text{Thm I.12}) \\ &= a(-b^{-1}) &(\text{Thm I.12}) \\ &= \frac{a}{-b} &(\text{Def of } (-b)^{-1}). \end{align*}

Then for the other equality, similarly, we have

    \begin{align*} -\left( \frac{a}{b} \right) &= (-a)b^{-1} \\ &= \frac{-a}{b}. \qquad \blacksquare \end{align*}

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

  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.

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