# Prove that from $0 > a > b$ follows $0 > b^{-1}>a^{-1}$ [duplicate]

As the title already suggests, I'm trying to prove the following statement:

0 > a > b follows 0 > $$b^{-1}$$ > $$a^{-1}$$

My approach seems to lead me nowhere, as I've ended up (sort of?) disproving it:

0 > a | a*1/a = 1
$$0 * a^{-1} * a > a$$ | /a
$$0 *a^{-1} > a/a$$
$$\rightarrow 0 > 1$$ Nonsense

The problem requires you to only use the ordered field axioms

• $a$ is negative, every time you multiply both sides by it, the inequality is supposed to flip. Nov 4 '21 at 23:13

Since $$0 \gt a \gt b$$ both $$a$$ and $$b$$ are negative, so their product is positive.
Divide the inequality by $$ab$$, the inequality sign remains the same. So
$$\dfrac{0}{ab} \gt \dfrac{a}{ab} \gt \dfrac{b}{ab}$$
$$0 \gt b^{-1} \gt a^{-1}$$