I am currently teaching a GED math class. While learning about the order of operations, the students asked why does a number next to a parentheses mean multiplication?

I understand the rule that two terms next to each mean multiply those two terms together, regardless as to what those terms look like. But did this notation that we now all understand come from shorthand? Or was it easier to read? Or is there some other reason entirely?

Any insight that you can provide will be appreciated. I would get very angry with teachers in my life who explained things by saying, because that is the way it is. If this is the case, then I at least want to be able to tell them that honestly.

  • $\begingroup$ Omitting the multiplication sign or omitting it and using parenthesis instead are abbreviations of the expression with the multiplication sign. For instance, $a\times b$ maybe be abbreviated by $ab$ or $(a)(b)$ or $a(b)$... All these notations (except for $a\times b$) mean, by definition, $a\times b$. $\endgroup$
    – Git Gud
    Jun 1, 2014 at 18:43
  • 4
    $\begingroup$ It's just shorthand. To be very precise, we have the following operations (amongst others): $+$, $-$, $\cdot$, $\div$. When we first learn how to multiply, we use $\times$ but once you start using variables, this notation is really awful. Handwritten, $6\times x$ looks like $6xx$ which is wholly confusing. For this reason, mathematicians will likely prefer to use $6 \cdot x$ over $6\times x$. However it gets to be pretty tedious to put $\cdot$ everywhere so we omit it. $\endgroup$ Jun 1, 2014 at 18:45
  • 2
    $\begingroup$ @CameronWilliams I disagree that it's "just" shorthand. If it were just shorthand, the standard order of operations would unambiguously still apply, but there is notable precedent for $1/6x$ being $\frac{1}{6\cdot x}$ rather than $\frac{1}{6}x$. It is, in a way, shorthand, but the way multiplication by juxtaposition fits outside the "standard" PEMDAS/GEMS semantics means it's perhaps a little different in common use. $\endgroup$
    – Linear
    Jun 2, 2014 at 3:31
  • 5
    $\begingroup$ @Jsor for me "1/6x" is ambiouous and if I were to guess, I would say "(1/6) times x". I think it depends on where you have been educated. I have completed my education in Finland. $\endgroup$ Jun 2, 2014 at 10:28
  • 1
    $\begingroup$ I'd expect a well-founded answer to refer to the hoistory of mathematical notation (including the origins of $+$, $\times$, $\cdot$ and juxtaposition) $\endgroup$ Jun 2, 2014 at 10:35

5 Answers 5


I would argue a case for how we describe multiples using natural language. When I say, "I have three boxes", I don't need any other words between 'three' and 'boxes'. Similarly, "I have three $x$", in shorthand, becomes simply $3x$.

Contrast this with addition: "I have an apple and two oranges". The 'and' separates objects that are together but are not necessarily of the same form. "I have an $x$ and two $y$s" becomes $x+2y$.

  • 2
    $\begingroup$ Another thing to point out is that we do a lot of multiplication in algebra, so removing the $\times$ sign is short-hand that makes things simpler in the end. @Theophile, excellent answer! +1 $\endgroup$ Jun 1, 2014 at 18:55
  • 8
    $\begingroup$ "Four score and seven years ago" = $4 \times 20 + 7\ \textrm{years ago}.$ $\endgroup$
    – Emily
    Jun 2, 2014 at 16:19
  • $\begingroup$ @Arkamis I think that comment is the best; most compelling answer here. $\endgroup$
    – OJFord
    Jun 3, 2014 at 1:10
  • $\begingroup$ At least in Austria we use a simple dot (⋅) for multiplication (not ×), so that sign (among the others like +, −, etc) is the easiest to miss. I always thought that is the reason. $\endgroup$ Jun 3, 2014 at 9:11

Theophile's answer is greatly intuitive, but what's really behind all this?

I believe in the end it comes down to the precedence of the operations. Say you have expression

$$2 \cdot x \cdot y + 3 \cdot x$$

The meaning of the expression actually is:

enter image description here

It can be clearly seen that the $\cdot$ operation is the "tightest" bond between those operands - so if you want to cluster the operands, i.e. if you want to save writing by omiting some of the operators, you must clearly start at the very bottom level of the syntactic tree. So you write

$$2 x y + 3 x$$

Just imagine what would happen if you try to omit the $+$ operator instead:

$$2 \cdot x \cdot y 3 \cdot x$$

That would mess up the syntax completely.

PS: the syntax tree was generated using http://mshang.ca/syntree/

  • 2
    $\begingroup$ I don't think this is "clearly" at all. After a few seconds of feeling into it, your example of 2*x*y3*x looks pretty natural to me. It's just the degree of habit that differs. Knowing a bunch of programming languages, markups and math notations (e.g. hungarian), it is clearly obvious that it is not clearly at all. By the way: en.wikipedia.org/wiki/Weasel_word . Look for "clearly". $\endgroup$
    – phresnel
    Jun 2, 2014 at 13:22
  • $\begingroup$ It can be clearly, and I mean clearly, seen that the $*$ operation is the closest (i.e. tightest) path between adjacent nodes of the syntax tree, compared to the $+$ operation. So if you want to cluster some nodes, you have to start from the bottom levels of the tree and the $*$ is the first operation you will come accross. $\endgroup$
    – Tomas
    Jun 2, 2014 at 14:15
  • $\begingroup$ I meant your second "clearly": if you want to save writing by omiting some of the operators, you must clearly start at the very bottom level of the syntactic tree. This can equally valid be transformed to if you want to save writing by omitting some of the operators, you must clearly start at the very top level of the syntactic tree, leading to 2*x*y3*x, which is very, very clear to me after thinking myself a bit into it. IMHO, yours is not a "natural" reasoning; it is rather based on familiarity with the mode. $\endgroup$
    – phresnel
    Jun 2, 2014 at 14:20
  • 1
    $\begingroup$ The "clearly" word is used quite commonly in mathematics. I don't think it is just a matter of habit. "y3" looks like a cluster because there is nothing in between, while we see from the diagram that it doesn't form a cluster. This seems like a reasonable reason to prefer omitting the multiplication symbol. $\endgroup$
    – gmatht
    Jun 2, 2014 at 14:31
  • 1
    $\begingroup$ Dosen't carry over for tighter-binding operations - most notably exponentiation written as ‘^’. $\endgroup$
    – mudri
    Jun 2, 2014 at 14:40

Another thing to consider and not mentioned is that the multiplication symbol $\times$ also happens to look a lot like the variable $x$, especially when writing them out by hand. It is one reason $\cdot$ is sometimes used for multiplication as well.

  • $\begingroup$ And what's with 't' vs. '+'? I know many whose 't' looks a lot like a '+', i.e. without the "hook". $\endgroup$
    – phresnel
    Jun 2, 2014 at 13:25

While I happen to think this notation is bad1, there's a rather compelling way to think about it: a numeral can be interpreted as a linear operator. This is natural as a "base case" of the way we write linear mappings $\mathbb{R}^n \to \mathbb{R}^m$ as $m\times n$ matrices.

1What I really dislike about juxtaposition multiplication is how it clashes with general (i.e. non-linear) function application. It's very natural to shorten $\sin(x)$ to $\sin x$ – after all, the parens don't really group anything! Yet people will get confused if you write $f\: x$ to notate $f(x)$ for some general function $f$, and possibly mistake it for something like $f(x)\cdot x$; especially in physics, where it's commonplace to omit the function argument and assume you'll put on the "usual" symbol.


Teophile's answer is great.

I would like to add that multiplication is the "natural" mathematical operation, not addition, as some people might think. A group has multiplication, but no addition. Addition is only introduced in rings, once we already have a multiplication.

Also, function concatenation is a multiplication, and mathematicians like to write this without an operator (look e.g. into the Lambda calculus), unless we talk about function application. But even then, mathematicians do not need an additional symbol - note how there is no extra symbol on the right hand side.

$$ (fg)(x) = f(g(x)) $$

(you might more be used to write $(fg)(x)$ as $f\circ g(x)$ ($\circ$ here being function concatenation))

  • 3
    $\begingroup$ There are additive groups, e.g.: $\langle \mathbb Z, + \rangle$ $\endgroup$
    – amWhy
    Jun 2, 2014 at 16:35
  • $\begingroup$ The operation in the group is a multiplication. In a ring, you have a multiplicative and additive subgroup - you use the operation to distinguish. As an operation, the + in your group is a multiplication. It's an addition in the ring. Taking this even further: the 0-element of the ring is the 1-element of your additive group. $\endgroup$
    – Zane
    Jun 2, 2014 at 16:39
  • 2
    $\begingroup$ @Zane In a group you have an operation, that happens to be denoted by $\cdot$ and called "product", but can be an addition (integers with the product of addition) or something else (rotations of a square). $\endgroup$
    – Davidmh
    Jun 2, 2014 at 16:51
  • $\begingroup$ And the dot is not written between adjacent elements, which is what the OP asked about. Call this dot whatever you want, but this is the natural operation I'm talking about. Mathematicians used to called that a multiplication, or, yes, as you say, a product (multiplying two values yields a product). And you're right, this multiplication can be an addition, a rotation, whatever. One is the formal, group-theoretic name, the other is the intuitive interpretation of the operation. Note e.g. an addition can be a subtraction. $\endgroup$
    – Zane
    Jun 2, 2014 at 17:01
  • $\begingroup$ @Rahul: No. The operation of a group is typically called a multiplication (or a product). In the ring, you have two subgroups, and both operations are multiplications in their groups. To distinguish in the ring, they are called differently, and we use different operators. None of them "stops being a multiplication". The addition is a multiplication, or better said "a multiplicative operation". $\endgroup$
    – Zane
    Jun 3, 2014 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.