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I'm trying to prove the following:

$${ad\over bc} = {\frac ab \over \frac cd} $$

First, $${\frac ab} = ab^{-1} $$ and $${\frac cd} = cd^{-1} $$ So the compound fraction above equals $$ (ab^{-1})({cd^{-1}})^{-1} $$

Here's where I get stuck. How do I handle the second term? I know $$(ab)^{-1} = (a^{-1})(b^{-1})$$ but what how is a number such as $$(b^{-1})^{-1}$$ simplified? I know it equals b, but how do I reach that conclusion from defining division as the multiplication of reciprocals?

Nate
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    $\dfrac cd = cd^{-1}$, $$;\dfrac{1}{\frac cd} = (cd^{-1})^{-1}$$ – amWhy Dec 31 '14 at 18:22
  • $(b^{-1})^{-1}=b^{-1\times -1}=b$ – Vikram Dec 31 '14 at 18:31
  • Nice fundamental stuff. Also see http://math.stackexchange.com/questions/71157/proof-of-dividing-fractional-expressions/71186#71186 (dealing with the same expression as this, but you took it a lot farther so you deserve a very different answer). – David K Dec 31 '14 at 20:25

2 Answers2

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It sounds like you're doing this informally; it's not unreasonable at all to take $(x^{-1})^{-1} = x$ as axiomatic.

Here's a common little trick with inverses to show they are unique: if we posit that $xy = 1$ and $yz = 1$, then we can calculate:

$$ x = x1 = xyz = 1z = z $$

A more convenient fact that can be proven from this is

If $ax = 1$ has a solution for $x$, then that solution is unique

so to prove that $c = a^{-1}$, all you need to do is to show that $ac = 1$.

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Here is a useful fact: for any number $x$ (a member of a field) and any integers $m$ and $n$ (which may each be positive or negative),

$$ (x^m)^n = x^{mn}.$$

From this you can conclude

$$ (b^{-1})^{-1} = b^{(-1)(-1)} = b^1 = b. $$

Of course then you have to have first established $(x^m)^n = x^{mn},$ including the cases where one or both of $m$ and/or $n$ is negative, so that you can make use of that identity. But it is such a useful identity that it is well worth proving.

David K
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