Multiplication of rational numbers (Math 0)

We’ve seen how to add and sub­tract pairs of ra­tio­nal num­bers. But the nat­u­ral num­bers have an­other op­er­a­tion on them: mul­ti­pli­ca­tion.

Re­mem­ber, a given ra­tio­nal num­ber rep­re­sents what we get when we cut an ap­ple into pieces all of the same size, then take some num­ber notePos­si­bly more than we ac­tu­ally made, and pos­si­bly nega­tive! of the lit­tle pieces. The product of \(\frac{a}{m}\) and \(\frac{b}{n}\) noteRe­call that \(\frac{a}{m}\) is “$a$ copies of the lit­tle-piece we get when we cut an ap­ple into \(m\) equal pieces. is what we call “$\frac{a}{m}$ mul­ti­plied by \(\frac{b}{n}\)”, and it an­swers the ques­tion “What hap­pens if we do the pro­ce­dure that would make \(\frac{b}{n}\), but in­stead of start­ing by cut­ting one ap­ple into \(n\) pieces, we started by cut­ting \(\frac{a}{m}\) ap­ples into \(n\) pieces?”.

We write the product of \(\frac{a}{m}\) and \(\frac{b}{n}\) as \(\frac{a}{m} \times \frac{b}{n}\).


It’s hope­fully easy to see that \(1 \times \frac{b}{n} = \frac{b}{n}\). In­deed, the defi­ni­tion is “what do we get if we would make \(\frac{b}{n}\), but in­stead of start­ing by cut­ting one ap­ple, we started by cut­ting \(1\) ap­ple?”; but that’s just the same! It’s like say­ing “What if, in­stead of putting bread around my sand­wich filling, I tried putting bread?”—I haven’t ac­tu­ally changed any­thing, and I’ll still get the same old sand­wich no­teor \(\frac{b}{n}\) out at the end.

How about \(2 \times \frac{3}{5}\)? (Strictly speak­ing, I should prob­a­bly be writ­ing \(\frac{2}{1}\) in­stead of \(2\), but this way saves a bit of writ­ing.\(\frac{2}{1}\) means “two copies of the thing I get when I cut an ap­ple into one piece”; but an ap­ple cut into one piece is just that ap­ple, so \(\frac{2}{1}\) just means two ap­ples.) Well, that says “in­stead of cut­ting one ap­ple, we cut two ap­ples” into \(\frac{3}{5}\)-sized pieces.

From now on, my pic­tures of ap­ples will get even worse: rather than be­ing cir­cles, they’ll now be squares. It just makes the di­a­grams eas­ier to un­der­stand.

Two times three-fifths

In the pic­ture, we have two ap­ples (squares) which I’ve drawn next to each other, sep­a­rated by a dashed line. Then I’ve taken \(\frac{3}{5}\) of the whole shape (shaded in red): that is, to the group of two ap­ples I have done the pro­ce­dure that would cre­ate \(\frac{3}{5}\) if it were done to one ap­ple alone.

No­tice, though, that this di­vides neatly into \(\frac{3}{5}\) of the left-hand ap­ple, and \(\frac{3}{5}\) of the right-hand ap­ple. So the red-shaded area comes to \(\frac{3}{5} + \frac{3}{5}\), which you already know how to calcu­late: it is \(\frac{6}{5}\).

Gen­eral in­te­ger times fraction

Can you work out, from the case of \(2 \times \frac{3}{5}\) above, what \(m \times \frac{a}{n}\) is, where \(m\) is an in­te­ger?

It is \(\frac{a \times m}{n}\).

In­deed, the pro­ce­dure to get \(\frac{a}{n}\) is: we split \(1\) into \(n\) equal pieces, and then take \(a\) of them. So the pro­ce­dure to get \(m \times \frac{a}{n}\) is: we split \(m\) into \(n\) equal pieces, and then take \(a\) of them.

But each of the pieces we’ve just made by split­ting \(m\)—that is, those de­mar­cated by the longer solid lines in the \(2 \times \frac{3}{5}\) di­a­gram above—can be viewed as be­ing \(m\) copies of what we get by split­ting \(1\). (In the di­a­gram above, we have \(2\) copies of that which we get by split­ting \(1\): namely the two copies in­di­cated by the dashed line.) So we can view the sec­ond pro­ce­dure as: we split \(1\) into \(n\) equal pieces <div><div>noteIn the di­a­gram above, there are \(5\) such equal pieces, and right now we’re look­ing only at one square, not at both squares joined to­gether.%, and then take \(a\) of them noteIn the di­a­gram above, \(a\) is \(3\): this has given us the red shaded bit of one of the squares., and then do this \(m\) times. noteIn the di­a­gram above, \(m\) is \(2\): we’re fi­nally look­ing at the two squares joined to­gether into a rec­t­an­gle.

This pro­duces \(a \times m\) pieces, each of size \(\frac{1}{n}\), and hence the ra­tio­nal num­ber \(\frac{a \times m}{n}\). %

You should check that you get the right an­swer for a differ­ent ex­am­ple: \(-5 \times \frac{2}{3}\).

This is “do the pro­ce­dure that makes \(\frac{2}{3}\), but in­stead of start­ing with \(1\), start with \(-5\)”.

So we take five anti-ap­ples, and di­vide them into thirds (ob­tain­ing \(15\) anti-chunks of size \(\frac{1}{3}\) each, grouped as five groups of three); and then we take two chunks out of each group of three, ob­tain­ing \(10\) anti-chunks of ap­ple in to­tal.

So \(-5 \times \frac{2}{3} = \frac{-10}{3}\), in ac­cor­dance with the rule of \(n \times \frac{a}{n} = \frac{a \times m}{n}\). <div><div>

Gen­eral frac­tion times fraction

ex­am­ples and pic­tures in­stant rule

Order doesn’t matter

No­tice that while it was fairly ob­vi­ous that or­der doesn’t mat­ter dur­ing ad­di­tion (that is, \(\frac{a}{m} + \frac{b}{n} = \frac{b}{n} + \frac{a}{m}\)), be­cause it’s sim­ply “putting two things next to each other and count­ing up what you’ve got”, it’s not all that ob­vi­ous that the product of two frac­tions should be in­de­pen­dent of the or­der we mul­ti­plied in. How­ever, you should check, from the gen­eral ex­pres­sion above, that it ac­tu­ally is in­de­pen­dent of the or­der.

Why is this? Why should it be that “do the pro­ce­dure that made \(\frac{b}{n}\), but start­ing from \(\frac{a}{m}\) in­stead of \(1\)” and “do the pro­ce­dure that made \(\frac{a}{m}\), but start­ing from \(\frac{b}{n}\) in­stead of \(1\)” give the same an­swer?

Well, re­mem­ber the di­a­gram we had for \(2 \times \frac{3}{5}\) (re­mem­ber­ing that that is “do the pro­ce­dure that would make \(\frac{3}{5}\), but in­stead of do­ing it to \(1\), we do it to \(2\)):

Two times three-fifths

What would we get if we ro­tated this di­a­gram by a quar­ter-turn?

Two times three-fifths, rotated

But wait! The shaded bit is just what we get when we do the pro­ce­dure that makes \(2\) (namely “put two copies of the shape next to each other”), but in­stead of do­ing it on the sin­gle (up­per-most) square, we do it to the ver­sion of the num­ber \(\frac{3}{5}\) that is rep­re­sented by the shaded bit of the up­per-most square! And that is ex­actly what we would do to get \(\frac{3}{5} \times 2\).

In gen­eral, \(\frac{a}{m} \times \frac{b}{n}\) is the same as \(\frac{b}{n} \times \frac{a}{m}\), be­cause the two just “come from the same di­a­gram, ro­tated by a quar­ter-turn”. They are mea­sur­ing the same amount of stuff, be­cause the amount of stuff in a di­a­gram doesn’t change sim­ply be­cause we ro­tated it.

Another example

We’ll do \(\frac{-5}{7} \times \frac{2}{3}\).

this example

Med­i­ta­tion: why the no­ta­tion makes sense

At this point, a di­gres­sion is in or­der. We have already seen the no­ta­tion \(\frac{a}{n}\) for “take an ap­ple; di­vide it into \(n\) pieces, each \(\frac{1}{n}\)-sized; and then take \(a\) of the chunks”. In the lan­guage of mul­ti­pli­ca­tion that we’ve now seen, that is “do what we would do to make \(a\), but do it start­ing from a \(\frac{1}{n}\)-chunk in­stead of \(1\)”. That is, \(\frac{a}{n}\) is just \(\frac{1}{n} \times a\).

And we can do that in a differ­ent way: we can take \(a\) ap­ples, di­vide each into \(n\) chunks, and then just draw one of the chunks from each ap­ple. In the lan­guage of mul­ti­pli­ca­tion, that is just “do what we would do to make a \(\frac{1}{n}\)-chunk, but do it to \(a\) in­stead of \(1\)”. That is, \(\frac{a}{n} = a \times \frac{1}{n}\).

Re­call­ing that \(a\) is just \(\frac{a}{1}\), our no­ta­tion \(\frac{a}{n}\) is sim­ply the same as \(\frac{a}{1} \times \frac{1}{n}\), as an in­stance of the “in­stant rule” \(\frac{a}{1} \times \frac{1}{n} = \frac{a \times 1}{1 \times n} = \frac{a}{n}\).

In­verses: putting things in reverse

Re­mem­ber that we had “anti-ap­ples” as a way of mak­ing noth­ing ($0$) by adding to some quan­tity of ap­ples. In a similar vein, we can “in­vert” mul­ti­pli­ca­tion.

When­ever \(a\) is not \(0\), we can find a ra­tio­nal num­ber \(\frac{c}{d}\) such that \(\frac{a}{b} \times \frac{c}{d} = 1\). (No­tice that we’ve got \(1\) as our “base point” now, rather than the \(0\) that ad­di­tion had.)

In­deed, us­ing the in­stant rule, we see that \(\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}\), so to make \(1\) we want \(a \times c\) to be the same as \(b \times d\).

But we can do that: if we let \(c = b\) and \(d = a\), we get the right thing, namely \(\frac{a \times b}{b \times a} = \frac{a \times b}{a \times b} = \frac{1}{1} = 1\).

So \(\frac{b}{a}\) works as an in­verse to \(\frac{a}{b}\). And this is why we needed \(a\) not to be \(0\): be­cause \(\frac{b}{a}\) isn’t ac­tu­ally a ra­tio­nal num­ber un­less \(a\) is nonzero.


We’ve seen how this defi­ni­tion fol­lows from the in­stant rule. Where does it ac­tu­ally come from, though?



  • Mathematics

    Math­e­mat­ics is the study of num­bers and other ideal ob­jects that can be de­scribed by ax­ioms.