# 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 $$\frac{a}{m} \times \frac{b}{n}$$.

# Example

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.

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 $$2$$):

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

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.

## Intuition

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

intuition

Parents:

• Mathematics

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