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The rule for multiplying rational numbers is this:

$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$

Can the rule be proven or is it meant to be taken as a given?

Edit: Where $b\neq 0$ and $d\neq 0$.

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    @Sara Didier's answer (below) makes it clear that neither b, nor$d$in your original equation can equal 0, since if we suppose b=0, in bx=a, then a=0. But, there does not exist a unique number x such that 0x=0, since 00=0, 01=0, and so on. So, we can infer that$b$cannot equal 0 (and similarly d). anon's comment referenced the original post, which didn't imply this, though I see that you've put it in... so the comments I've made about anon's comment no longer apply.2011-08-29

5 Answers 5

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One knows that $a/b$ may be defined as the number $x$ such that $bx=a$, and that $c/d$ may be defined as the number $y$ such that $dy=c$.

If one wants the multiplication on these objects to be associative and commutative as it is on the integers, one should ask that $ac=(bx)(dy)=(bd)(xy)$ hence that the object $xy$ fits the definition of $(ac)/(bd)$.

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    @Sara, thanks. My pleasure.2011-08-27
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$\rm x\: =\: \dfrac{a}b,\ \ y\: =\: \dfrac{c}d\ \ \Rightarrow\ \ b\ x\: =\: a,\:\ d\ y\: =\: c\ \ \Rightarrow\ \ b\:d\ x\:y\: =\: a\:c\ \ \Rightarrow\ \ x\:y\: =\: \dfrac{a\:c}{b\:d}$

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    What I meant was, given the associativity and commutativity of multiplication in whatever structure (in this case, presumably, the integers) $a$, $b$, $c$, and $d$ live in.2011-08-18
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An approach to make it more clear might be to seperate each rational into a product of its parts ie. $\frac {a}{b}= \frac{a}{1} \cdot \frac{1}{b}$ and $\frac {c}{d}= \frac{c}{1} \cdot \frac{1}{d}$ then use the commutative property to group the "numerator" fractions and the "denominator" fractions seperately: $\frac{a}{b}\cdot \frac{c}{d}=(\frac{a}{1} \cdot \frac{c}{1})\cdot (\frac{1}{b} \cdot \frac{1}{d})=\frac{ac}{1}\cdot\frac{1}{bd}=\frac{ac}{bd}$.

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    @Michael I mean it in the spirit of the comment by anon above, I should have been more clear.2011-08-17
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Because the rationals are a field, it is a given. A field has associative properties defined on it.

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It is more or less taken as a given. It fits the intuition and the construction of the rational numbers (which contains the definition of this multiplication) was generalized to arbitrary commutative rings (+ the choice of a multiplicative subset). This construction is called the localization (see wikipedia).

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    @Mic In the general localization construction for commutative rings, the "pair" representation of fractions needn't be a "given". Rather, more naturally, the pairs $\rm\:(a,b)\:$ representing $\rm\:a/b = a\:b^{-1}\:$ can be *derived as a normal forms* of terms in the natural algebraic presentation in terms of generators and relations. Namely, construct $\rm\:S^{-1}\:R\:$ by adjoining to $\rm\:R\:$ inverses $\rm\:x_s = s^{-1}$ for all $\rm\:s\in S$, i.e. work in $\rm\:R[x_s,x_t,\ldots]/(s\:x_s-1,\:t\:x_t-1,\ldots)\:$ See my [answer here](http://bit.ly/nm9BUV) for more, including references.2011-08-24