As per my knowledge and what was taught in school,

$a\times b$ is $a$ times $b$ or $b$ times $a$

Obviously this is commutative as $a$ times $b$ and $b$ times $a$ are same thing. On the other hand there are multiplications like vector multiplication and matrix multiplication that are not commutative.

What does multiplication mean in general, for these? Or should they even be called multiplication?

  • 21
    $\begingroup$ Much of the time it's tradition. But if you have an operation which is associative and distributes over addition (whatever "addition" means), then it's definitely worthy of the name "multiplication". $\endgroup$
    – Arthur
    Nov 21, 2018 at 8:30
  • $\begingroup$ According to inc.com/bill-murphy-jr/…, kids can learn the material from grade 1 better if they start grade when at the age of 8. Maybe by starting so late, they could avoid having damaging past knowledge and be introduced to the concept of a natural number as a finite ordinal and completely avoid teaching them any concept of cardinal numbers until they're ready for it later. That's because it's much harder to learn to add by creating two groups each of which has a certain number of objects then $\endgroup$
    – Timothy
    Nov 21, 2018 at 17:54
  • $\begingroup$ counting the objects in the union than directly from the inductive definition of natural number addition. Also, they should be taught only the bijective unary numeral system in grade 1 and how to add and multiply using it. They should also be guided on how to write a formal proof in a certain weak system of pure number theory and taught how to master that skill. They should later be tested on their ability to write a proof that natural number addition is associative and commutative and that natural number multiplication is also associative and commutative and distributes over addition and $\endgroup$
    – Timothy
    Nov 21, 2018 at 18:00
  • 5
    $\begingroup$ @Timothy: If you're going to write four paragraphs in a sequence of comments you should post an answer instead. $\endgroup$
    – user856
    Nov 22, 2018 at 7:41
  • 1
    $\begingroup$ @Rahul I think I'm fine with doing neither next time I can't think of an answer I'm sure doesn't have wrong information. Maybe comments on the question are only for discussing how the question should be written. Sometimes, other people had a comment to me that was too long so they wrote it as 2 comments. $\endgroup$
    – Timothy
    Nov 22, 2018 at 18:34

7 Answers 7


Terms in mathematics do not necessarily have absolute, universal definitions. The context is very important. It is common for a term to have similar but not identical meanings in multiple contexts but it is also common for the meaning to differ considerably. It can even differ from author to author in the same context. To be sure of the meaning, you need to check the author's definition.

It might be tempting to demand that multiplication be commutative and another term be used when it isn't but that would break some nice patterns such as the real numbers, to the complex numbers, to the quaternions.

In day to day life, multiplication is commutative but only because it deals only with real numbers. As you go deeper into maths, you will need to unlearn this assumption. It is very frequent that something called multiplication is not commutative.

  • 1
    $\begingroup$ (+1) It's a good answer, but my main reason for upvoting are the first two sentences. $\endgroup$ Nov 21, 2018 at 8:42
  • $\begingroup$ @JoséCarlosSantos Thanks. I considered adding some examples e.g. whether a ring has a multiplicative identity and two very different meanings of field. However, I suspected that these might not help. $\endgroup$
    – badjohn
    Nov 21, 2018 at 14:08
  • 4
    $\begingroup$ Matrices are probably a more commonly encountered example of non-commutative multiplication than quaternions. $\endgroup$ Nov 21, 2018 at 19:14
  • $\begingroup$ @HenningMakholm, Yes and it seems that the OP first encountered non-commutative multiplication with matrices. I mentioned quaternions because the progression $\mathbb{R}$ to $\mathbb{C}$ to $\mathbb{H}$ was neater. $\endgroup$
    – badjohn
    Nov 21, 2018 at 20:19
  • $\begingroup$ @Broman The "it" in that quote referred to maths in day to day life. Very few people use complex numbers in day to day life. Various structures in addition to the reals have commutative multiplication but are rarely encountered outside mathematics (or serious science). $\mathbb{Z}_n$ is also commutative but not often used in day to day life. There are 12 and 24 hour clocks but multiplying two times is not common. $\endgroup$
    – badjohn
    Nov 21, 2018 at 20:22

In school you might have also learned that multiplication or addition are defined for "numbers", but vectors, matrices, and functions are not numbers, why should they be allowed to be added or multiplied to begin with?

It turns out that sounds and letters form words which we use in context to convey information, usually in a concise manner.

In school, for example, most of your mathematical education is about the real numbers, maybe a bit about the complex numbers. Maybe you even learned to derive a function.

But did it ever occur to you that there are functions from $\Bbb R$ to itself such that for every $a<b$, the image of the function on $(a,b)$ is the entire set of real numbers?

If all functions you dealt with in school were differentiable (or at least almost everywhere), why is that even a function? What does it even mean for something to be a function?

Well. These are questions that mathematicians dealt with a long time ago, and decided that we should stick to definitions. So in mathematics we have explicit definitions, and we give them names so we don't have to repeat the definition each time. The phrase "Let $H$ be a Hilbert space over $\Bbb C$" packs in those eight words an immense amount of knowledge, that usually takes a long time to learn, for example.

Sometimes, out of convenience, and out of the sheer love of generalizations, mathematicians take a word which has a "common meaning", and decide that it is good enough to be used in a different context and to mean something else. Germ, stalk, filter, sheaf, quiver, graph, are all words that take a cue from the natural sense of the word, and give it an explicit definition in context.

(And I haven't even talked about things which have little to no relation to their real world meaning, e.g. a mouse in set theory.)

Multiplication is a word we use in context, and the context is usually an associative binary operation on some set. This set can be of functions, matrices, sets, etc. If we require the operation to have a neutral element and admit inverses, we have a group; if we adjoin another operator which is also associative, admits inverses, and commutative and posit some distributivity laws, we get a ring; or a semi-ring; or so on and so forth.

But it is convenient to talk about multiplication, because it's a word, and in most cases the pedagogical development helps us grasp why in generalizations we might want to omit commutativity from this operation.


Terminology for mathematical structures is often built off of, and analogized to, terminology for "normal" mathematical structures such as integers, rational numbers, and real numbers. For vectors over real numbers, we have addition already defined for the coordinates. Applying this operation and adding components termwise results in a meaningful operation, and the natural terminology is to refer to that as simply "addition". Multiplying termwise result in an operation that isn't as meaningful (for one thing, this operation, unlike termwise addition, is dependent on the coordinate system). The cross product, on the other hand, is a meaningful operation, and it interacts with termwise addition in a manner similar to how real multiplication interacts with real addition. For instance, $(a+b)\times c = a \times c + b \times c$ (distributive property).

For matrices, we again have termwise addition being a meaningful operation. Matrices represent linear operators, and the definition of linearity includes many of the properties of multiplication, such as distribution: A(u+v) = A(u) + A(v). Thus, it's natural to treat application of a linear operator as "multiplying" a vector by a matrix, and from there it's natural to define matrix multiplication as composition of the linear operators: (A*B)(v) = (A(B(v)).

  • 2
    $\begingroup$ +1, indeed the common feature of operations called "multiplication" seems to be that they are binary operations that distribute (on both sides) over "addition". $\endgroup$ Nov 21, 2018 at 19:18
  • 1
    $\begingroup$ @HenningMakholm: I don't think that's quite the whole truth. For example, people talk about group "multiplication" even though there's only one operation in that setting. What does seem true is that there are lots of interesting examples of noncommutative operations which distribute over commutative operations and few (or no?) examples of commutative operations which distribute over noncommutative operations, so the tendency is to analogize commutative things with addition and noncommutative things with multiplication. $\endgroup$
    – Micah
    Nov 21, 2018 at 22:08
  • $\begingroup$ @Micah: Hmm, perhaps I'm projecting here, because the word "multiply" about arbirary group operations has never sat well me. (I'm cool with multiplicative notation, of course, and speaking about "product" and "factors" doesn't sound half as jarring as "multiply" either). $\endgroup$ Nov 22, 2018 at 1:21
  • 1
    $\begingroup$ @HenningMakholm: Eh, fair enough. After I wrote my comment it occurred to me that you may be historically correct: if you're Galois and all the groups you like to think about consist of field automorphisms, group multiplication really does distribute over addition... $\endgroup$
    – Micah
    Nov 22, 2018 at 4:32

The mathematical concept below the question is that of operation.

In a very general setting, if X is a set, than an operation on X is just a function

$$X\times X\to X$$

Usually denoted with multiplicative notations like $\cdot$ or $*$. This means that an operation takes two elements of X and give as a result an element of $X$ (exactly as when you take two numbers, say $3$ ant $5$ and the result is $5*3=15$)

Examples of oparations are the usual operations on real numbers: plus, minus, division (defined on non-zero reals), multiplications, exponentiation.

Now, if you want to use operations in mathematics you usualy requires properties that are useful in calculations. Here the most common properties that an operation can have.

$1)$ Associativity. That means that $(a*b)*c=a*(b*c)$ and allow you to omit parenthesis and write $a*b*c$. The usual summation and multiplications on real numbers are associative. Subtraction, division, and exponentiation are not. For example: $$(5-2)-2=3-2=1\neq 5=5-(2-2)$$ $$ (9/3)/3=3/3=1\neq 9=9/1=9/(3/3)$$ $$ 2^{(3^2)}=2^9=512\neq 64= 8^2=(2^3)^2$$ A famous examples of non-associative multiplication often used in mathematics is that of Cayley Octonions. A useful example of associative operations is the composition of functions. Let $A$ be any set and $X$ be the set of all functions from $A$ to $A$, i.e. $X=\{f:A\to A\}$. The composition of two function $f,g$ is the function $f*g$ defined by $f*g(a)=f(g(a))$. Clearly $f*(g*h)(a)=f(g(h(a)))=(f*g)*h(a)$.

$2)$ Commutativity. That means that $a*b=b*a$. Usual sum and multiplications are commutative. Matrix product is not commutative, the composition of functions is not commutative in general (a rotation composed with a translation is not the same as a translation first, and then the rotation), vector product is not commutative. Non commutativity of composition of functions is on the basis of Heisemberg uncertainty principle . The Hamilton Quaternions are a useful structure used in math. They have a multiplication which is associative but non commutative.

$3)$ Existence of Neutral element. This means that there is an element e in $X$ so that $x*e=e*x=x$ for any $x$ of $X$. For summation the netural element is $0$, for multiplication is $1$. If you consider $X$ as the set of even integers numbers, than the usual multiplication is well-defined on $X$, but the neutral element does not exists in $X$ (it would be $1$, which is not in $X$).

$4)$ Existence of Inverse. In case there is neutral element $e\in X$, this means that for any $x\in X$ there exists $y$ so that $xy=yx=e$. Usually $y$ is denoted by $x^{-1}$. The inverse for usual sum is $-x$, the inverse for usual multiplication is $1/x$ (which exists only for non-zero elements). In the realm of matrices, there are many matrices that have no inverse, for instane the matrix $\begin{pmatrix} 1 & 1\\1&1\end{pmatrix}$.

Such properties are important because they make an operation user-friendly. For example: is it true that if $a,b\neq 0$ then $a*b\neq0$? This seems kind of obviuos, but it depends on the properties of the operation. For example $\begin{pmatrix} 1 & 0\\3&0\end{pmatrix}\begin{pmatrix} 0 & 0\\1&2\end{pmatrix}=\begin{pmatrix} 0 & 0\\0&0\end{pmatrix}$ but both $\begin{pmatrix} 1 & 0\\3&0\end{pmatrix}$ and $\begin{pmatrix} 0 & 0\\1&2\end{pmatrix}$ are different from zero.

Concluding, I would say that when a mathematician hear the word multiplication, immediately think to an associative operation, usually (but not always) with neutral element, sometimes commutative.

  • 1
    $\begingroup$ Another example of non-associative multiplication is IEEE floating point numbers (where intermediate results can underflow to zero or not, depending on the order of evaluation. small * small * big can be zero if the first two terms underflow, or small if the second two terms multiply to 1.) $\endgroup$ Nov 21, 2018 at 21:12
  • 1
    $\begingroup$ Not to mention the cross product in $\mathbb R^3$. $\endgroup$ Nov 21, 2018 at 21:41
  • $\begingroup$ @MartinBonner: An interesting example, in that it is a non-associative approximation to an associative operation. I see in Wikipedia, incidentally, that it may be subnormal rather than zero, though I do not see exactly when. $\endgroup$
    – PJTraill
    Nov 21, 2018 at 21:46

a∗b is a times b or b times a

That is a theorem not a definition, thank you very much. Of course the easiest proof is geometric: the former is an a x b rectangle, and you rotate it and you get a b x a rectangle. To prove it from axioms is a bit trickier because you have to answer the question if multiplication is not commutative as an axiom then what do we prove it from? The answer is some combination of mathematical induction, which is the idea that if something works for all numbers "and so on" then it works for all numbers, but it only applies in the integers because other number systems don't really have an "and so on."

So once you talk about multiplication outside of the integers you lose all these good assurances that multiplication is commutative. In my opinion, that's really it. Multiplication makes sense in a lot of places, and many of those places don't have strong reasons that it would be commutative.


It ought to be mentioned that multiplication in general is an action of a collection of transformations on a collection of objects, and multiplication of elements of a group is in fact a special case. This might have even been inherent in the original notion of multiplication; consider:

$3$ times as many chickens is not the same as "chicken times as many $3$s". (The latter makes no sense.)

Symbolically, if we let "$C$" stand for one chicken, and denote "$k$ times of $X$" by "$k·X$", then we have

$3·(k·C) = (3k)·C$ but not $3·(k·C) = (k·C)·3$ (since "$(k·C)·3$" does not make sense).

What is happening here? Even in many natural languages, numbers have been used in non-commutative syntax (in English, "three chickens" is correct while "chickens three" is wrong), because the multiplier and the multiplied are of different types (one is a countable noun phrase while the other is like a determiner).

In this special case of natural numbers, they actually act as a semi-ring on countable nouns:

$0·x = \text{nothing}$.

$1·x = x$.

$(km)·x = k·(m·x)$ for any natural numbers $k,m$ and countable noun $x$.

$(k+m)·x = (k·x)\text{ and }(m·x)$ for any natural numbers $k,m$ and countable noun $x$.

$k·(x\text{ and }y) = (k·x)\text{ and }(k·y)$ for any natural numbers $k,m$ and countable noun $x$.

(English requires some details like pluralization and articles, but mathematically the multiplication is as stated.)

Observe that multiplication between natural numbers is commutative, but multiplication of natural numbers to countable nouns is not. So for example we have:

$3·(k·C) = (3k)·C = (k3)·C = k·(3·C)$.

And in mathematics you see the same kind of scalar multiplication in vector spaces, where again the term "scalar" (related to "scale") is not accidental. Again, multiplication of scalings to vectors is not commutative, but multiplication between scalings (which is simply composition) is commutative.

More generally, even the transformations that act on the collection of objects may not have commutative composition. For example, operators on a collection $S$, namely functions from $S$ to $S$, act on $S$ via function application:

$(f∘g)(x) = f(g(x))$ for any $f,g : S \to S$ and $x \in S$.

And certainly composition on functions from $S$ to $S$ is not commutative. But maybe the "multiplication" name still stuck on as it did for numbers and scalings. That is why we have square matrix multiplication defined precisely so that it is equivalent to their composition when acting on the vectors.

  • $\begingroup$ Can we say multiplication is precisely an endomorphism on a monoid? For example, $(\mathbb{N}, \times) \cong (\mathrm{End} (\mathbb{N}, +), \circ)$. $\endgroup$
    – user76284
    Jul 23, 2020 at 22:36
  • $\begingroup$ Though perhaps we might want to exclude the example involving $\max$ and $+$... $\endgroup$
    – user76284
    Jul 24, 2020 at 1:48
  • $\begingroup$ @user76284: Hmm, I prefer not to try to express things in terms of higher-order objects unless they seem to be intrinsic. In the case of numbers and reals, they seem to be to be really just a matter of ring action on countable nouns and mass nouns respectively. $\endgroup$
    – user21820
    Jul 24, 2020 at 5:43

Commutativity is not a fundamental part of multiplication. It is just a consequence of how it works in certain situations.

Here is an analogy. When you multiply two positive integers, a and b, you can define multiplication as repetitive additions. If consider a*b, that means that we will do a repetitive addition with a, and we will repeat b times. This is how multiplication is taught in the early years.

Now have a look what happens when we use this on negative integers. Using the above definition, 4 * (-3) would mean that we would repeatedly add the number 4. Nothing strange so far. But we will repeat this -3 times. Hold on, how can we do something a negative amount of times? Why should we call this multiplication when it is not a repetitive addition?

My point here is that when we transitioned from just positive integers to include the negative integers, we lost something that we thought was fundamental. But it was never fundamental.

Let's have a look at matrix multiplication. Multiplication of real numbers could be seen as a special case of matrix multiplication. This special case is where the size of the matrices is 1x1. If you have the 1x1 matrices [a] and [b] and perform matrix multiplication between them, you would get the matrix [ab].


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .