What is it? How do you define multiplication?

All of us had to memorize the multiplication table in elementary school, but how did they come up with it if so many people claim that it is wrong to think of it as repeated addition?

Voodoo magic?

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    $\begingroup$ Who claims that it's wrong to think of it as repeated addition? $\endgroup$ – joriki Sep 14 '11 at 13:30
  • $\begingroup$ "There are differences amongst educationalists as to which number should normally be considered as the number of copies and whether multiplication should even be introduced as repeated addition." - Wikipedia $\endgroup$ – Curiosity Sep 14 '11 at 13:31
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    $\begingroup$ Whether it's wrong to think of it as something or whether that's the best way to introduce it to children are two quite separate questions. As to which number should be considered the number of copies, that doesn't make any sense since multiplication is commutative. $\endgroup$ – joriki Sep 14 '11 at 13:32
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    $\begingroup$ What I'm trying to ask is... If 3 times 0.3 is the same as $0.3 + 0.3 + 0.3$, what is 3 0.3 times? How to think about it? Especially when we have 0.3 * 0.2, This is basically a fraction multiplied with a fraction... But how to interpret the operation of multiplication here? As scaling? Between 0 and 1 it could be thought as division... Am I making any sense? I am just quite confused. $\endgroup$ – Curiosity Sep 14 '11 at 13:35
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    $\begingroup$ Keith Devlin has written a few columns at the MAA site in recent years explaining why he thinks it's a bad idea to introduce multiplication to children as repeated addition. See maa.org/devlin/devlin_01_11.html and the other essays cited there. $\endgroup$ – Gerry Myerson Sep 14 '11 at 13:40

The first thing you should consider is that there are, in some sense, different "hierarchies" of numbers. At each stage, we enlarge the class of numbers, and try to do so in a way that leaves everything we could do before still the same, but that now we can do more.

A common way of doing this is to start with the natural numbers (sometimes called "counting numbers" or positive integers). We start with $1$, $2$, $3,\ldots$.

Here, we do define multiplication as repeated addition. For example, one way to define multiplication is to assume we know how to add, and then define it by saying: $$\begin{align*} n\times 1 &= n\\ n\times (k+1) &= (n\times k) + n % need to edit at least 6 characters \end{align*}$$ Using mathematical induction, we can show this defines multiplication for all positive integers, and that it has the usual properties we know (commutative, so that $n\times k = k\times n$ for all positive integer $n$ and $k$, distributes over the sum, is associative, etc).

Then we have two choices for "expanding our universe of numbers": we can now define negative integers, by considering things that would help us solve all equations of the form $a+x=b$ with $a$ and $b$ positive integers; or we can introduce positive rationals (fractions) by considering all things that would help us solve all equations of the form $ax = b$. Let's do the latter, since that came first historically.

So, we had the positive integers, and we knew how to add and multiply them. Now we are going to have more numbers: now, for every pair of positive integers $a$ and $b$, we will have a number "$\frac{a}{b}$", which is a number that satisfies the property that $$b\times\left(\frac{a}{b}\right) = a.$$ We also say that $\frac{a}{b}$ is "the same fraction" as $\frac{c}{d}$ if and only if $ad=bc$ (here we are comparing products of positive integers, so that's fine).

We also notice that our old positive integers can also be considered fractions: the positive integer $a$ is a solution to $1x = a$, so $a$ corresponds to the fraction $\frac{a}{1}$.

Now, how do we add two of these numbers? Since $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sx=r$, then $\frac{a}{b}+\frac{r}{s}$ represents the solution to something; to what? A bit of algebra will tell you that it is the solution to precisely $(bs)x = (as+br)$. So we define $$\frac{a}{b}+\frac{r}{s} = \frac{as+br}{bs}.$$ There's a bit of work that needs to be done to ensure that if you write the fractions differently, the answer comes out the same (if $\frac{c}{d}=\frac{a}{b}$, and if $\frac{t}{u}=\frac{r}{s}$, does $\frac{cu+td}{du} = \frac{as+br}{bs}$? Yes). And we also notice that if we add positive integers as if they were fractions, we get the same answer we did before: $$\frac{a}{1} + \frac{c}{1} = \frac{a1+c1}{1} = \frac{a+c}{1}.$$ That's good; it means we are enlarging our universe, not changing it.

How about products? If $\frac{a}{b}$ represents the solution to $bx=a$, and $\frac{r}{s}$ represents the solution to $sy=r$, their product will be the solution to $(bs)z = ar$. So we define $$\frac{a}{b}\times\frac{r}{s} = \frac{ar}{bs}.$$ And then we notice that it extends the definition of multiplication for integers, since $\frac{a}{1}\times\frac{b}{1} = \frac{a\times b}{1}$. And we check to see that multiplication and addition still have the properties we want (commutativity, associativity, etc).

(There are other ways to figure out what multiplication of fractions "should be", on the basis of what we want it to do. For example, we want multiplication to extend multiplication of integers, so $\frac{a}{1}\times\frac{b}{1}$ should be $\frac{ab}{1}$; and we want it to distribute over the sum, so we want $$\frac{a}{1} = \frac{a}{1}\times \frac{1}{1} = \frac{a}{1}\times\left(\underbrace{\frac{1}{b}+\frac{1}{b}+\cdots+\frac{1}{b}}_{b\text{ summands}}\right) = \underbrace{\left(\frac{a}{1}\times\frac{1}{b}\right) + \cdots + \left(\frac{a}{1}\times\frac{1}{b}\right)}_{b\text{ summands}}.$$ So $\frac{a}{1}\times \frac{1}{b}$ should be a fraction which, when added to itself $b$ times, equals $a$; that is, a solution to $bx=a$; that is, $\frac{a}{b}$. And so on).

Then we move on from the positive rationals (fractions) to the positive reals. This is more complicated, as it involves "filling in gaps" between rationals. It is very technical. But what it turns out is that for every real number you can find a sequence of rationals $q_1,q_2,q_3,\ldots$ that get progressively closer to each other and to $r$ (we say the sequence "converges to $r$"); it won't hurt too much if you think of the $q_i$ as being progressive decimal approximations to $r$ (they don't have to be, and ahead of time you don't have any notion of decimal approximation, but you can think of it that way for our purpose). So then the way we define multiplication of real numbers $r$ and $s$ is to find a sequence of rationals $q_1,q_2,q_3,\ldots$ giving the approximation to $r$, and one $p_1,p_2,p_3,\ldots$ giving the approximation to $s$, and we define $r\times s$ to be whatever it is that the sequence $$p_1\times q_1,\ p_2\times q_2,\ p_3\times q_3,\ \ldots$$ approximates. This ensures that if you take rational numbers and multiply them as if they were reals you get the same thing as if you multiply them as rationals, and likewise for integers.

So multiplication of positive reals is really a series of approximations made up by multiplying rationals; and multiplication of rationals is really a way to codify solutions to certain equations with integers; and it is only multiplication of (positive) integers that really corresponds to "repeated addition".

Finally, once you have the positive reals, you can introduce the negative real numbers. We consider solutions to equations of the form $a+x=b$ with $a$ and $b$ positive real numbers. Some of these already have solutions, some don't. This gives us "zero" and "negative reals". We then extend the definition of multiplication to "zero" and to "the negative reals" in a way that makes sense relative to this definition. Turns out we need to make $0\times r = 0$ for all $r$, and have to respect the "rules of signs" to make sure everything still works. So we define it that way to make sure everything works and what we had before still works exactly the same.

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    $\begingroup$ @Curiosity: The properties of addition and multiplication for positive integers usually need to be proven by induction in one way or another, if you want to really do it from scratch. Above, you require generalized commutativity and generalized associativity to justify the step from $\sum(a+b)$ to $(\sum a)+(\sum b)$, and you would need to prove several things beyond the definition of multiplication I give to establish the first step. Assuming you had all those results in hand, then that would work, but one has to be careful not have a circular argument. $\endgroup$ – Arturo Magidin Sep 15 '11 at 13:30
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    $\begingroup$ @Doug: The difference is that the number of times addition is done implicitly in multiplication (according to the interpretation on the table here) is indefinite, whereas an addition table shows the results of adding exactly two numbers together. The fact that addition between two numbers is done in each step of the process - and that addition tables exhaust the minimal examples of addition - are not inconsistent with the fact that they are used in a broader scheme. If you put code inside a FOR loop while programming it doesn't mean you altered the original snippet of code. $\endgroup$ – anon Sep 15 '11 at 19:45
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    $\begingroup$ @Doug: Since he explicitly distinguishes between the original axiomatic framework (where you can't define recursion, or use the essence of "FOR loops") and the outside meta-schema (where we actually understand repeated iteration), it's clear he's pointing out that an indefinite number of operations makes a function ill-defined in the former while not the latter. There's obviously an issue of whether distinguishing hierarchies and teaching rigid formalism is healthy / useful pedagogically at an elementary level, but saying repeated operations can't ultimately be understood is just absurd. $\endgroup$ – anon Sep 15 '11 at 20:30
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    $\begingroup$ I doubt anyone could have given a better answer to this question. Bravo! $\endgroup$ – Mathemagician1234 Sep 15 '11 at 23:23
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    $\begingroup$ @Doug: Indeed, human beings can't mentally visualize more than say 6 or 7 copies of an object simultaneously, but that doesn't mean, for example, the very concept of whole number addition isn't understood, and understood for arbitrarily large numbers, so your point is patently disingenuous. Moreover, the idea that repeated operations cannot be understood is utterly absurd, because we would then not have product ($\prod$) and summation ($\sum$) notation and it would throw out a large plurality of all of mathematics. You wouldn't have a computer if not for well-defined repeated operations. $\endgroup$ – anon Sep 15 '11 at 23:26

At Willie Wong's suggestion, I post my comment as an answer.

Keith Devlin has written a few columns at the MAA site in recent years explaining why he thinks it's a bad idea to introduce multiplication to children as repeated addition. See this and the other essays cited there.

  • $\begingroup$ Oh, man. I totally disagree with this but I'll see what he has to say. $\endgroup$ – anon Sep 14 '11 at 23:02
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    $\begingroup$ The link in the answer is now broken, but here's a currently working one: What Exactly is Multiplication? $\endgroup$ – Rahul Sep 13 '14 at 5:39

Multiplication of positive integers is defined as repeated addition. However, this doesn't help us define multiplication by a negative number, or a rational, or anything else -- you can't add three to itself minus four times, or half of a time. This means we have to start from whole-number multiplication and then "fill in the gaps" to define multiplication for larger classes of numbers.

  • $\begingroup$ How did we fill the gaps? That's my question, exactly. It's quite confusing for me. $\endgroup$ – Curiosity Sep 14 '11 at 13:36
  • $\begingroup$ @Curiosity: Something like this: If $n \cdot x$ is $x + \dots + x$ ($n$ times) for $n$ a positive integer, then one declares $(-n) \cdot x$ to be $-(n \cdot x)$, and $(n/m)\cdot x$ to be $(n \cdot x)/m$. And so on... $\endgroup$ – Hans Lundmark Sep 14 '11 at 14:12
  • $\begingroup$ No, it's not, even if you exclude 0 from the natural numbers. Consider (2*1). There doesn't exist the sum of just one number 2. Maybe you think you can say "well, that's just the sum of 1 and 1". But, see, multiplication commutes, so the other way should also work if "multiplication is repeated addition". Also, if "multiplication is repeated addition", what does (1*1)? I simply have no idea how to perform an addition with just one element. (0*0), (0*1), and (0*x) seem even worse. $\endgroup$ – Doug Spoonwood Sep 14 '11 at 19:52
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    $\begingroup$ @Doug why do you claim that 'there doesn't exist the sum of just one number 2'? I think when you tell people that a.b is calculated (for naturals) by 'taking a copies of b and adding them all together', they'll know exactly what to do when a is 1, and I don't think the notion is nearly so confusing (even for children) as you're making it out to be. $\endgroup$ – Steven Stadnicki Sep 15 '11 at 0:49
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    $\begingroup$ @Doug: You take people too literally. Obviously what user3296 wants is to be able to speak of a number of instances without an assumption about singular (1), plural (1+), or even a presence at all (0) (in other words, "a number" - which is the full concept), but the English language doesn't have a graceful way to do this, so people generally have become accustomed to understanding phrases such as "$n$ apples" to allow singular or empty cases depending on context, seeing as how often that is the intention behind the utterance, despite the literal grammar indicating otherwise. $\endgroup$ – anon Sep 15 '11 at 20:44

I gave a long response to this question as part of an answer to another question. The basic point I want to get across is that repeated addition does not generalize in the way that multiplication generalizes; for example you'd be hard-pressed to think of multiplication of complex numbers or matrices as repeated addition. That is because what they really are are compositions of functions.

In particular, multiplication of real numbers is (in my opinion) better thought of as compositions of scalings of the real line. This generalizes immediately to multiplication of complex numbers, which are compositions of scalings-and-rotations of the plane.

(Another point I want to get across is that it is completely unnecessary, even when dealing with natural numbers, to define multiplication as repeated addition.)

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    $\begingroup$ On the other hand, repeated addition does generalize to the hierarchy multiplication -> exponentiation -> tetration -> .... $\endgroup$ – Charles Sep 15 '11 at 20:23
  • $\begingroup$ @Charles: as I explain in detail in the other question, I disagree that this is a good way to think about either multiplication or exponentiation. $\endgroup$ – Qiaochu Yuan Sep 15 '11 at 21:32
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    $\begingroup$ There aren't really many other options on tetration, though, which was my point. $\endgroup$ – Charles Sep 16 '11 at 3:43

I like to look at the original geometrical sources: A non-negative number is the length of a line segment, the sum of two numbers is the combined lengths of the segments, and the product is the area of the rectangle with the segments as the sides. I know this has its problems (it's surprisingly hard to get the integers from the reals), but I like to look at things from different points of view.

  • $\begingroup$ When you consider such a perspective, you can always console yourself that the ancient Greeks considered numbers from that point-of-view. Euclid's number theory existed in that sort of context, including his proofs. $\endgroup$ – Doug Spoonwood Sep 15 '11 at 23:16
  • $\begingroup$ Additionally, the Archimedean property stated as definition 4 of Euclid's book V is originally due not to Archimedes but to Eudoxus. see en.wikipedia.org/wiki/Eudoxus_of_Cnidus $\endgroup$ – CopyPasteIt Oct 22 '18 at 17:27

To understand multiplication, discover it yourself (see 2nd section below)!

Starting point: The natural numbers $\mathbb N$ with only one binary operation called addition.

Before long, you'll find it both useful and convenient to define multiplication. For example, you store many cans of Campbell's soup in your ultra-organized pantry.

enter image description here

You see that two cans are in the front and you have three rows. Without the use of a calculator, you 'know' that you have six cans. You like this so much that you set up a notation for what it means to multiply two natural numbers $a$ and $b$:

${\displaystyle a\times b=\underbrace {b+\cdots +b} _\text{a-times}}$

You call the 'thing being multiplied', $b$, the multiplicand, and the other 'thing', $a$, the multiplier. You are not bothered at all if either $a$ or $b$ is equal to $0$ - you see the result as a big fat zero (no soup).

You like to think about things in an abstract way, and you discover something that 'knocks your socks off' - multiplication distributes over addition:

$\tag 1 a \times (b + c) = (a \times b) + (a \times c)$

You now think that perhaps multiplication can be viewed as a binary operation in its own right. You then ask if the distributivity law 'gives you back' multiplication as a uniquely defined arithmetic operation. You decide to add the 'law', $\, 1 \times 1 = 1$, and testing some more,

$\quad 2 \times 2 = 2 \times (1 + 1) = $
$\qquad 2 \times 1 + 2 \times 1 = $
$\qquad (1 + 1) \times 1 + (1 + 1) \times 1 =$
$\qquad 1 \times 1 + 1 \times 1 + 1 \times 1 + 1\times 1 = $
$\qquad 1 + 1 + 1 + 1 = 4$

This is a great start! You realize that for your 'laws of arithmetic' it will be both convenient and useful to include multiplication, an operation that is 'tied to the hip' with addition via (1).

The link

What Exactly is Multiplication? \ Keith Devlin

that Gerry Myerson provides contains the quote

multiplication is complex and multi-faceted

and the author suggests that to understand multiplication, you need to form a personal 'mental amalgam' with a heuristic, like 'multiplication is scaling'.


Devlin is profane - do not trust him!

Multiplication is repeated addition ($N = n·u= u+u+u+...$, unit is preserved) , but there is another thing called product.($u×u = u^2$, unit is changed).

Euclid had problems with this issue too, but clearly he was aware of it. Euclid speaks of product $A\times B$ as plane number, and of product $ A \times B \times C$ as solid number.

René Descartes resolved the issue completely.

These terms is often(almost always) misused, because $|A × B| = |A| · |B|$.


protected by Qiaochu Yuan Sep 25 '11 at 19:12

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