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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 $d\neq 0$.

Sara
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    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. – anon Aug 17 '11 at 21:53
  • Besides commutation and association of multiplication in the rationals, you need that b^(-1)d^(-1)=(bd)^(-1). – Doug Spoonwood Aug 18 '11 at 04:01
  • @Doug, If the composition law is associative, b^(-1)d^(-1)=(db)^(-1). If it is commutative, db=bd. If it is both, b^(-1)d^(-1)=(bd)^(-1). So it seems nothing besides commutativity and associativity is needed. – Did Aug 19 '11 at 08:24
  • @Dider I don't see how association by itself implies b^(-1)d^(-1)=(db)^(-1). Do you have a proof of that? If we have association, the existence of inverses, and the existence of an identity, then b^(-1)d^(-1)=(db)^(-1). But, I simply don't see how association by itself implies that b^(-1)d^(-1)=(db)^(-1), since the notion of an inverse does not get defined by that of association (and no "^" does not associate if you thought it did). – Doug Spoonwood Aug 19 '11 at 21:10
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    @Doug, the very use of the notation $b^{-1}$ presupposes the existence of the identity and of an inverse for $b$. – Gerry Myerson Aug 20 '11 at 12:18
  • @Gerry No, it doesn't. The notation b^(-1) may just come as a way of writing 1/b. Now, 0^(-1) comes as a way of writing 1/0. But, I haven't presupposed the existence of an inverse for 0, since I have NOT made any claim that says or implies that 0(1/0)=1. Again, anon said "Once you know that multiplication is commutative and associative (and therefore work with multiplicative inverses too) these steps are justified." Well, for all* rational numbers commutation and association holds. But, those steps he gave don't come as justified, since 1/b doesn't have meaning for all b. – Doug Spoonwood Aug 20 '11 at 13:35
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    @Doug, I think your quarrel is with OP. You should tell Sara that what she calls a rule isn't actually a rule, since neither side of her equation has a meaning for all $b$ and $d$. – Gerry Myerson Aug 21 '11 at 00:10
  • @Gerry I've tried to edit the question. I don't think I've quarreled with Sara, but rather explained what she said. The post-as-it-stands-now isn't a rule for multiplying all rational numbers a, b, c, d, since say b=0 and d=0, which do qualify as rational numbers. We simply don't know that (a/b)(c/d)=(ac)/(bd), even though multiplication is defined for all rational numbers. Again, my "quarrel" lies with the claim that "if multiplication is associative and commutative, then anon's steps hold." Multiplication associates and commutes on the set of all rational numbers, but his steps don't. – Doug Spoonwood Aug 21 '11 at 00:21
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    @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. – Gerry Myerson Aug 21 '11 at 01:28
  • @Gerry If the restrictions exist, then MORE THAN commutation and association imply anon's steps, since in such a case we have inverses and the identity. So, I can consistently object to his statement "Once you know that multiplication is commutative and associative (and therefore work with multiplicative inverses too) these steps are justified." Look, write out the proof for b^(-1)d^(-1)=(bd)^(-1) with a proof analysis citing every step along the way. You'll have the definition of an inverse and the identity there. So, multiplication and commutation together do not suffice. – Doug Spoonwood Aug 21 '11 at 02:23
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    @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. – anon Aug 21 '11 at 04:34
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    @Doug, you're not listening. I give up. – Gerry Myerson Aug 21 '11 at 08:12
  • @anon Your last statement basically came as my point. – Doug Spoonwood Aug 21 '11 at 14:18
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    Welcome back, Sara. Some answers have been posted since you went away - do you have anything to say about any of them? Clarifications to ask for? Maybe you like one answer enough to accept it? – Gerry Myerson Aug 27 '11 at 06:49
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    Why is Didier's answer not attracting any comments as anon's? It appears to me that the answers are the same as both rely on multiplication being associative and commutative. – Sara Aug 27 '11 at 08:19
  • @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. – Doug Spoonwood Aug 29 '11 at 20:16

5 Answers5

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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 $(ac)/(bd)$.

Did
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$$\rm x\: =\: \dfrac{a}b,\ \ y\: =\: \dfrac{c}d\ \ \Rightarrow\ \ b\ x\: =\: a,\:\ d\ y\: =\: c\ \ \Rightarrow\ \ b\:d\ x\:y\: =\: a\:c\ \ \Rightarrow\ \ x\:y\: =\: \dfrac{a\:c}{b\:d}$$

Bill Dubuque
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  • Taking as given the associativity and commutativity of multiplication. – Gerry Myerson Aug 18 '11 at 01:39
  • @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. – Bill Dubuque Aug 18 '11 at 01:57
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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. – Gerry Myerson Aug 18 '11 at 02:03
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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: $\frac{a}{b}\cdot \frac{c}{d}=(\frac{a}{1} \cdot \frac{c}{1})\cdot (\frac{1}{b} \cdot \frac{1}{d})=\frac{ac}{1}\cdot\frac{1}{bd}=\frac{ac}{bd}$.

  • But how do you know that $\frac ab = \frac a1\cdot\frac1b$ before you've proved the thing that the question is about? – Michael Hardy Aug 17 '11 at 22:12
  • @Michael I mean it in the spirit of the comment by anon above, I should have been more clear. –  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.

MathApprentice
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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).

Louis La Brocante
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    Maybe in some abstract-algebra contexts it is "taken as given", but the question was about rational numbers, not about fields of fractions. – Michael Hardy Aug 17 '11 at 22:11
  • @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. – Bill Dubuque Aug 24 '11 at 15:27