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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

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

which reduces to

$ 0 \gt b^{-1} \gt a^{-1} $

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  • $\begingroup$ brilliant, thank you. $\endgroup$ Nov 5 '21 at 0:15

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