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(My question is similar to this one at a high level, but I am looking for something more rigorous.)

I have started into Michael Spivak's "Calculus" textbook. Problem 3 (v) on page 14 asks for a proof that "$\frac{a}{b} \big/\frac{c}{d} = \frac{ad}{bc}$, if b, c, d $\neq 0$".

The only proof that I can come up with assumes that $(c \cdot d^{-1})^{-1}$ is equal to $c^{-1} \cdot d$, which is true, but I can't prove it.

At this point in the chapter, Spivak has listed these nine basic properties of numbers:

  1. $a + (b + c) = (a + b) + c$
  2. $a + 0 = 0 + a = a$
  3. $a + (-a) = (-a) + a = 0$
  4. $a + b = b + a$
  5. $a \cdot (b \cdot c) = (a \cdot b) \cdot c$
  6. $a \cdot 1 = 1 \cdot a = a$; $1 \neq 0$
  7. $a \cdot a^{-1} = a^{-1} \cdot a = 1$, for $a \neq 0$
  8. $a \cdot b = b \cdot a$
  9. $a \cdot (b + c) = a \cdot b + a \cdot c$

This is actually one of the questions with an answer in the back of the book, where Spivak makes the same assumption I do. So here's my question: How do we know that $(c \cdot d^{-1})^{-1}$ is equal to $c^{-1} \cdot d$ given these properties?

Community
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sam.bishop
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  • Whatever $(c \cdot d^{-1})^{-1}$ is, when we multiply it with $c \cdot d^{-1}$, we should get $1$. So if you multiply $c^{-1} \cdot d$ with $c \cdot d^{-1}$ and get $1$... – pjs36 Nov 15 '15 at 04:53
  • Multiply $(c \cdot d^{-1})^{-1}$ by $c^{-1} \cdot d$. If you get $1$, then either one is the inverse of the other. – Steven Alexis Gregory Nov 15 '15 at 05:17
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    @Normal Human: The axiom of choice??? – Asaf Karagila Nov 15 '15 at 05:32
  • @AsafKaragila Autocomplete fail, I was thinking "axioms" (since the question is based on axiomatic treatment of real numbers). There is no "analysis" here, btw... –  Nov 15 '15 at 05:55

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Here is a hint. First, show that the multiplicative inverse in $\mathbb{R}\setminus{\{0\}}$ (are any field in general) is unique.

Then, show that both $(c\cdot d^{-1})^{-1}$ and $c^{-1}\cdot d$ are multiplicative inverses of $d^{-1}\cdot c$.

Aguila
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  • I know what a field is, but I'm afraid that I don't know what "$\mathbb{R}\setminus{{0}}$" means. I should warn you too that I've been out of school for a decade, and I didn't get a math degree. – sam.bishop Nov 15 '15 at 05:10
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    $\mathbb{R}\backslash{0}$ is just a way of saying "the real numbers, not including 0". So he's saying any (nonzero) real number has a unique inverse. – Alex Mathers Nov 15 '15 at 05:16