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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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    $\begingroup$ Well, $$\frac{a}{b}\cdot\frac{c}{d}=a\cdot b^{-1}\cdot c\cdot d^{-1}=(a\cdot c)\cdot(b\cdot d)^{-1}=\frac{a\cdot c}{b\cdot d}.$$ Once you know that multiplication is commutative and associative (and therefore work with multiplicative inverses too) these steps are justified. $\endgroup$ – anon Aug 17 '11 at 21:53
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    $\begingroup$ @Doug, the very use of the notation $b^{-1}$ presupposes the existence of the identity and of an inverse for $b$. $\endgroup$ – Gerry Myerson Aug 20 '11 at 12:18
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    $\begingroup$ @Doug, Sara says $(a/b)(c/d)=ac/bd$, you say we don't know that $(a/b)(c/d)=ac/bd$, so of course you are disagreeing with Sara. anon has (silently, implicitly, perhaps unconsciously) added the hypothesis that $b$ and $d$ are not zero (which is the only way to make sense of Sara's question) and has taken it from there. If you have no quarrel with Sara then you, too, are implicitly accepting the restrictions on the numbers, and you can't object to what anon did. $\endgroup$ – Gerry Myerson Aug 21 '11 at 1:28
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    $\begingroup$ @Doug: It's unclear to me what you're talking about in a number of your statements. My steps, though I didn't show them all for sake of brevity, do make valid use of commutativity and associativity. And yes, I'm taking the existence of inverses and the identity for granted - you are correct in pointing out that in an arbitrary algebraic structure these are also necessary for the argument to work. $\endgroup$ – anon Aug 21 '11 at 4:34
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    $\begingroup$ @Doug, you're not listening. I give up. $\endgroup$ – Gerry Myerson Aug 21 '11 at 8:12
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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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    $\begingroup$ I like this answer because the author explains the logic behind each step. $\endgroup$ – Sara Aug 27 '11 at 8:20
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    $\begingroup$ @Sara, thanks. My pleasure. $\endgroup$ – Did Aug 27 '11 at 14:00
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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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  • $\begingroup$ Taking as given the associativity and commutativity of multiplication. $\endgroup$ – Gerry Myerson Aug 18 '11 at 1:39
  • $\begingroup$ @Gerry It's not "given" but, rather, trivially deduced, since in the above algebraic approach the fraction field is a constructed as a quotient ring of an associative and commutative ring, hence trivially inherits those properties. See my answer here. $\endgroup$ – Bill Dubuque Aug 18 '11 at 1:57
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson Aug 18 '11 at 2:03
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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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  • $\begingroup$ But how do you know that $\frac ab = \frac a1\cdot\frac1b$ before you've proved the thing that the question is about? $\endgroup$ – Michael Hardy Aug 17 '11 at 22:12
  • $\begingroup$ @Michael I mean it in the spirit of the comment by anon above, I should have been more clear. $\endgroup$ – user12998 Aug 17 '11 at 22:38
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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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    $\begingroup$ Maybe in some abstract-algebra contexts it is "taken as given", but the question was about rational numbers, not about fields of fractions. $\endgroup$ – Michael Hardy Aug 17 '11 at 22:11
  • $\begingroup$ @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 for more, including references. $\endgroup$ – Bill Dubuque Aug 24 '11 at 15:27

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