Q: Where do the weird rules for rational numbers come from? (Dealing with fractions)

The original question was:  Why is it when we multiply fractions we multiply the numerators across  and the denominators across?  Whereas when we divide we don’t do the same?  Who came up with these rules and why do they work the way they do?


Physicist: For all of these rules, start with addition and then extend and extend and extend, while doing the least damage possible.  At first blush I thought this would be a short and straightforward question, but it really isn’t.  There are wrong ways to construct the rules of arithmetic, but if you get technical there’s not necessarily a right way.  That said, as far as basic arithmetic goes (no highfalutin calculus, or set theory, or anything) there’s really just one “right” way.

Rational numbers (“fractions”) have been in use for thousands of years and they, and their rules, have been independently invented at least dozens of times.  More recently (19th century) a bunch of really A-type mathematicians got together to “put mathematics on a more rigorous footing“.  Those mathematicians (there were many) are responsible for unpleasant statements like “there are exactly as many prime numbers as rational numbers“, and are a big reason behind why modern mathematicians always look just a little pained when they speak.  It was these paragons of compulsivity that established how the rules are stated, and gave everything definite names like “associative”, “commutative”, etc.

Constructing the rational numbers and all of their behavior starts with an innocent statement; that “\frac{1}{A} is the unique number such that A\left(\frac{1}{A}\right) = \left(\frac{1}{A}\right)A = 1“.  This one definition is the headwaters for all of the properties of rational numbers that follow.  If ever you’re totally stuck, and don’t know how to handle fractions, keep this single and tremendously important definition in mind.  Everything else follows (but maybe not immediately).  Just a quick warning for those of you expecting any of the rest of this post be about history or something interesting; everything that follows is a dry-as-bones, utterly literal, derivation of the arithmetic rules for fractions and rational numbers from the ground up.  Definitely boring, but worth seeing once.

Starting with integers, along with regular addition, subtraction, and multiplication (basic arithmetic tools), and the definition of “\frac{1}{A}“, you’ll find that all of the weird properties of rational numbers come tumbling out.

Often \frac{1}{A} will be multiplied by another integer.  To save room we write: B\left(\frac{1}{A}\right) = \frac{B}{A}.  This is strictly a convention; a standard defined notation that’s agreed upon.

For example: \frac{3}{4} = 3\left(\frac{1}{4}\right).

For example: \frac{x+2}{6} = (x+2)\frac{1}{6}.

Already we can say things like: \left(\frac{3}{4}\right)4 = 3\left(\frac{1}{4}\right)4 = 3\cdot 1 = 3.  We can even do a little better and say B\left(\frac{C}{A}\right) = BC\left(\frac{1}{A}\right) = \frac{BC}{A}.

For example: 7\cdot\frac{3}{4} = 7\cdot3\cdot\frac{1}{4} = 21\cdot\frac{1}{4} = \frac{21}{4}.

How do fractions multiply?  It would be nice to know: \left(\frac{1}{A}\right)\left(\frac{1}{B}\right) = ?.  But check this out: \left(\frac{1}{A}\right)\left(\frac{1}{B}\right)AB = \left(\frac{1}{A}\right)A\left(\frac{1}{B}\right)B = 1\cdot 1 = 1.  This means that \left(\frac{1}{A}\right)\left(\frac{1}{B}\right) does the exact same thing that \frac{1}{AB} does, and (since this is all there is in the definition) in fact is the same.  So, \frac{1}{A}\cdot\frac{1}{B} = \frac{1}{AB}.

For example: \frac{1}{5}\cdot\frac{1}{3}\cdot15 = \frac{1}{5}\cdot\frac{1}{3}\cdot3\cdot5 = \left(\frac{1}{5}\cdot5\right)\left(\frac{1}{3}\cdot3\right) = 1.  But keep in mind that “\frac{1}{15} is the unique number such that 15\left(\frac{1}{15}\right) = \left(\frac{1}{15}\right)15 = 1“, and since \frac{1}{5}\cdot\frac{1}{3} has been shown to have the same property we know that \frac{1}{5}\cdot\frac{1}{3} = \frac{1}{15}.  This may seem anal-retentive and unnecessary, but that’s how math is done.

Now we’ve got the tools to define multiplication in general: \frac{A}{B}\cdot\frac{C}{D} = A\frac{1}{B}C\frac{1}{D} = AC\left(\frac{1}{B}\frac{1}{D}\right) = AC\frac{1}{BD} = \frac{AC}{BD}.  In other words “fractions multiply across”.

For example: \frac{3}{7}\cdot\frac{4}{5} = \frac{3\cdot4}{7\cdot5} = \frac{12}{35}.  Deriving the rules is often complex, but using them isn’t.

Now, again using nothing new, \frac{AC}{BC} = \frac{A}{B}\cdot\frac{C}{C} = \frac{A}{B}\cdot 1 = \frac{A}{B}.  This means two things: fractions can be reduced, and multiplying the top and the bottom by the same thing does nothing.

For example: \frac{9}{6} = \frac{3\cdot3}{2\cdot3} = \frac{3}{2}\cdot\frac{3}{3} = \frac{3}{2}.

It would be great if the addition and subtraction of fractions follow all of the same rules that hold for integers.  As a mathematician, the way you make this happen is to declare that it’s true, and then check for inconsistencies.  The rule most important for defining addition and subtraction of fractions is the “distributive law”, which says that “A(B+C) = AB+AC” (this doesn’t lead to any new inconsistencies).

First, \frac{A}{B} + \frac{C}{B} = \frac{1}{B}A + \frac{1}{B}C = \frac{1}{B}(A +C) = \frac{A+C}{B}.  So now, if the fractions have the same denominator, then they can be added together.

For example: \frac{3}{7} + \frac{5}{7} = \frac{3+5}{7} = \frac{8}{7}

But what about fractions with different denominators?  The trick is to not have them: \frac{A}{B} + \frac{C}{D} = \frac{AD}{BD} + \frac{BC}{BD} = AD\frac{1}{BD} + BC\frac{1}{BD} = (AD + BC)\frac{1}{BD} = \frac{AD+BC}{BD}

For example: \frac{2}{3}+\frac{5}{7}=\frac{2}{3}\cdot\frac{7}{7}+\frac{3}{3}\cdot\frac{5}{7}=\frac{2\cdot7}{3\cdot7}+\frac{3\cdot5}{3\cdot7}=\frac{14}{21}+\frac{15}{21}=\frac{14+15}{21}=\frac{29}{21}

Huzzah!  We can multiply, add, and reduce fractions with ease and impunity!

But what about dividing fractions?  Well, here we have to tread lightly and describe exactly what is meant by “dividing by a fraction”.  So, in a perfectly reasonable extension of the original definition, “\frac{1}{\left(\frac{A}{B}\right)} is the unique number such that \frac{A}{B}\left(\frac{1}{\left(\frac{A}{B}\right)}\right) = \left(\frac{1}{\left(\frac{A}{B}\right)}\right)\frac{A}{B} = 1“.

But check this out: \frac{B}{A}\cdot\frac{A}{B} = \frac{B\cdot A}{A\cdot B} = 1.  Since \frac{B}{A} does exactly what \frac{1}{\left(\frac{A}{B}\right)} is defined to do, we can say that \frac{1}{\left(\frac{A}{B}\right)} = \frac{B}{A}.

Another way to see this is to say \frac{1}{\left(\frac{A}{B}\right)} = \frac{1}{\left(\frac{A}{B}\right)}\cdot\frac{\left(\frac{B}{A}\right)}{\left(\frac{B}{A}\right)} = \frac{1\cdot\frac{B}{A}}{\frac{A}{B}\cdot\frac{B}{A}} = \frac{\frac{B}{A}}{\frac{AB}{BA}} = \frac{B}{A}\cdot\frac{1}{\frac{1}{1}} = \frac{B}{A}\cdot\frac{1}{1}= \frac{B}{A}

For example: \frac{\left(\frac{3}{4}\right)}{\left(\frac{7}{2}\right)} = \frac{3}{4}\cdot\frac{1}{\left(\frac{7}{2}\right)} = \frac{3}{4}\cdot\frac{2}{7} = \frac{3\cdot2}{4\cdot7} = \frac{6}{28}.

Just a quick note on behalf of whoever grades your tests or homework: Please reduce fractions, \frac{6}{28} = \frac{2\cdot3}{2\cdot14} = \frac{2}{2}\cdot\frac{3}{14} = \frac{3}{14}.

For example: \frac{x+7}{\frac{1}{2}} = (x+7)\frac{1}{\frac{1}{2}} = (x+7)\cdot2 = 2x+14

What about subtraction and negative numbers? In general, in every case, without exception of any kind, whenever you see “-A” you can exchange it with “(-1)A” and exchange “B-A” with “B+(-1)A”.  After that, treat “(-1)” just like any other number or variable*.

So (still using no new rules!), we can say \frac{A}{B} - \frac{C}{D} = \frac{A}{B} + (-1)\frac{C}{D} = \frac{A}{B} + \frac{(-1)C}{D} = \frac{AD}{BD} + \frac{(-1)BC}{BD} = \frac{AD+(-1)BC}{BD} = \frac{AD-BC}{BD}

For example: -\frac{3}{7} + \frac{9}{2} = \frac{-3}{7} + \frac{9}{2} = \frac{-3\cdot2}{7\cdot2} + \frac{9\cdot7}{2\cdot7} = \frac{-3\cdot2+9\cdot7}{2\cdot7} = \frac{-6+63}{14} = \frac{-57}{14} = -\frac{57}{14}

For those of you who’ve read this far, you can show using the same definitions and tricks above that:

\frac{1}{(-A)} = -\frac{1}{A}.

Using the fact that A^n=\overbrace{A\cdot A\cdots A}^{n\textrm{ times}}, you can get \left(\frac{A}{B}\right)^n = \frac{A^n}{B^n}.

Similarly, you can show that \left(\frac{A}{B}\right)^{-n} = \left(\frac{B}{A}\right)^n.

There are big issues with \frac{1}{0}, because it’s defined as the number that, when multiplied by 0, gives 1.  But of course that number doesn’t exist*.  In practice, if you ever see a “1/0”, stop mathing.  And every time you divide by something that could be zero, make a note on the side of the paper.  The short answer to almost every question about “1/0” is “doesn’t”.

For example: If x+1 = x-1, then \frac{x+1}{x-1} = 1 and by the way only when x\ne1.

It’s also worth pointing out that when things are added in the denominator, there’s not much that can be done with it.  So, if you’ve got something like \frac{3}{x+2}, then you’re stuck.  That’s as simplified as it can get.  The one and only thing you can say about \frac{3}{x+2} is that \frac{3}{x+2}\cdot(x+2) = 3 (so long as x\ne-2).

Also, for those of you wondering (this is a bit technical), the rational numbers also inherit their positions in the number line very naturally.  Using only the fact that “if A<B and C>0, then AC<BC” and the fact that we know how to order integers, you can figure out which rational number is bigger than which other rational numbers.  Since \frac{A}{B}<\frac{C}{D}\Leftrightarrow AD<BC, if you can figure out which of AD and BC is bigger (they’re both integers), then you can figure out which of \frac{A}{B} and \frac{C}{D} is bigger.

For example:

\begin{array}{ll}\frac{7}{4}\,[?]\,\frac{5}{3}\\\Rightarrow 3\cdot4\cdot\frac{7}{4}\,[?]\,\frac{5}{3}\cdot3\cdot4\\\Rightarrow 3\cdot7\,[?]\,5\cdot4\\\Rightarrow 21\,[?]\,20\\\Rightarrow 21>20\\\Rightarrow\frac{7}{4}>\frac{5}{3}\end{array}

Even better, we can describe exactly where the rational numbers are!

For example: You can show that \frac{1}{2} is just as far from 1 as it is from 0.  1-\frac{1}{2} = \frac{2}{2}-\frac{1}{2} = \frac{2}{2}+\frac{-1}{2} = \frac{2-1}{2} = \frac{1}{2} and \frac{1}{2}-0 = \frac{1}{2}.

Just to get ahead of the most obvious follow up questions:  There’s a lot of weird emphasis on how exactly mathematical notation is used in text.  The first key to dealing with complicated text-based notation is: don’t.  Writing equations using the symbols found only on a keyboard is something our unfortunate and sadly limited ancestors had to consider.  If you’re reading this now, then you’re in a bigger and better-notated world.

In general, you can always write A/B = A(1/B), and A/B/C = A(1/B)(1/C), and so on.  A little fancier: A/B/C = A\cdot\frac{1}{B}\cdot\frac{1}{C} = \frac{A}{BC}.  If for some horrifying, bizarre reason you find yourself looking at a string of numbers or variables being multiplied together, with no parentheses in sight, just replace “A/B” with “A\frac{1}{B}” and go*.

For example: 2/3\cdot7A/4/2/D = 2\cdot\frac{1}{3}\cdot7\cdot A\cdot\frac{1}{4}\cdot\frac{1}{2}\cdot\frac{1}{D} = \frac{2\cdot7\cdot A}{3\cdot4\cdot2\cdot D} = \frac{7A}{12D}.

That (way too long of a post) all said, if you’re reading this because you’re presently panic-studying the night before a test, and actually needed a short answer, memorize this:

“A/B is A times 1/B.  1/B times B is one.  There’s nothing else to say about 1/B.”

and go to sleep sooner rather than later.


*If your math background is extensive enough to know some exceptions, then… be cool, you know what I mean.

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7 Responses to Q: Where do the weird rules for rational numbers come from? (Dealing with fractions)

  1. [email protected] Hyde says:

    “Physicist”

    Looking at your very last asteric *, it appears to me that your mention of any ‘exceptions’ is worthy of a post in and of it’s self! Will any be forthcoming?!

    From ‘a philosophical perspective’ if the way that fractions are added subtracted multiplied and divided do indeed reflect Reality, as in the dividing of candy bars among kids…! (Who seem [little monsters!] to be more than ‘excellent’ in looking out for ‘mathematical chicanery’ in such matters!) I wonder how can there be any such ‘Exceptions'(?), if so, would the universe not cease to exist at that very moment(?) or would MM’s then melt in any given kids hands?!

    Or maybe would that indicate that there may be some problems with the axiom(s) or definitions underlying operations with fractions…? Or ‘some such’?

    Thanks.

  2. Joe says:

    What are the rules for partial fraction decomposition?

  3. Diego says:

    Well, In the first part it ‘s supposed the abelian (Commutative property) property of the “·” to the new set. We can expect that there will be, but in the finish we must see that every property we have enfose really works.

  4. Eric says:

    @Hyde:
    Whether the last asterisk is worth its own post will depend on what The Physicist meant. In the upper levels of mathematics, people get lazy and stop inventing new symbols for new operations. Thus, there is a lot of ambiguity. For example, I’ve seen both – (the first asterisk) and / (the last asterisk) used as set difference operators. I’ve also seen / used for integer division. Finally, I’ve seen R/I used as notation for quotient ring formed from a ring R and an ideal I.

    @The Physicist
    Wonderful explanation. When I was asked this question a few years ago, I said that when we did things this way, things matched the reality we wanted to model. Your response is much more elegant and satisfying.

  5. Luke says:

    “Why is it when we multiply fractions we multiply the numerators across and the denominators across? Whereas when we divide we don’t do the same?”

    Actually, one can do the same when dividing, but just divide across:

    A/B “divided by” C/D, done by way of multiplying inverses, equals AD/BC. When you divide across, you get A/C “divided by” B/D, and when you multiply these inverses you get AD/CB which is the same as AD/BC.

  6. The Physicist The Physicist says:

    @Eric
    That’s exactly the sort of thing I was thinking of (and didn’t want to mention)!

  7. JosephHyde says:

    I guess that I don’t have to worry about any melted chocolate or crying youngsters…

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