Quick question, which seems to irritate some people as it apparently strikes them as a nonsensical question: Why does the distributive law only work in one direction (in $(ℝ,+,*)$)? Why does it work this way:

$$ a * (b + c) = (a * b) + (a * c) $$

But not this way:

$$ a + (b * c) = (a + b) * (a + c) $$

What is it about the way addition and multiplication are constructed/composed that inhibits this form of usage? (in ℝ specifically)

  • 8
    $\begingroup$ If there exists an additive identity, then let $b=c=0$ you would get $a=a*a$ which is unhelpful at best :) $\endgroup$
    – user208649
    Jan 22, 2016 at 1:32
  • 7
    $\begingroup$ Given some set of objects with two operations say we arbitrarily chose $+$ to denote one of the operations and $\cdot$ the other. If your second distributive law were true but not the first we'd just switch the two notations so that the regular distribution rule (the first one) applies. There are some (or at least one that I know of) systems of numbers where both equations hold -- e.g. in Boolean algebra. But such a system would require $x\cdot x=x$ for all $x$ (which is easy to confirm in Boolean algebra). $\endgroup$
    – user137731
    Jan 22, 2016 at 1:35
  • 6
    $\begingroup$ An example of two operations distributing over each other is the operations of union and intersection. $\qquad$ $\endgroup$ Jan 22, 2016 at 1:44
  • 4
    $\begingroup$ There are algebraic structures where both equations hold: see Distributive lattice. $\endgroup$
    – lhf
    Jan 22, 2016 at 1:47
  • 2
    $\begingroup$ Every question of the form "Why X?" should be answered by "Why do you expect not-X?". Why should it be the case that the relationship between addition and multiplication is symmetric in this way? Are you saying that just because A bears relationship R to B, that therefore it's reasonable to expect that B should bear relationship R to A? Do you find it surprising that cats eat mice, and yet mice don't eat cats? $\endgroup$
    – Jack M
    Jan 22, 2016 at 14:12

4 Answers 4


Foundations: Asymmetry in definitions

When mathematics starts at the very basic, we practically define multiplication so that it is distributive.

On the natural numbers $1,2,\dots$ you start with:

$$1\cdot n = n\\ (m+1)\cdot n = (m\cdot n) + n$$

This is the recursive way of saying that $m\cdot n$ is the sum of $m$ copies of $n$, but we already see that the second rule is looking like the distributive law.

From this definition of multiplication, can prove both directions of the distributive law more easily than we can prove that multiplication is commutative.

Example: Matrices

In general, we start with a relatively easier notion of "plus" or addition. For example, adding matrices is very simple:

$$\begin{pmatrix}a_1&b_1\\ c_1&d_1\end{pmatrix}+\begin{pmatrix}a_2&b_2\\ c_2&d_2\end{pmatrix}=\begin{pmatrix}a_1+a_2&b_1+b_2\\ c_1+c_2&d_1+d_2\end{pmatrix} $$

Multiplication is a much uglier and less obvious (when you first learn it) thing. Multiplication is not even commutative in matrices.

$$\begin{pmatrix}a_1&b_1\\ c_1&d_1\end{pmatrix}\cdot\begin{pmatrix}a_2&b_2\\ c_2&d_2\end{pmatrix}=\begin{pmatrix}a_1a_2+b_1c_2&a_1b_2+b_1d_2\\ c_1a_2+d_1c_2&c_1b_2+d_1d_2\end{pmatrix} $$

Geometric meaning

So the deep "why" is in the complexity - multiplication is usually defined in terms of addition.

Multiplication of real numbers is an example of this. If real numbers are thought of geometrically as lengths, then to understand addition of positive reals, you can "stay on a line." The only geometric views of multiplication of reals require a step into a higher dimension - you need to talk about areas, or use parallel lines.

The geometry also just makes more sense for the original distributive law - it means that you can connect two rectangles with at edges of the same size to get a larger rectangle, and the new rectangle has area the sum of the two original rectangles.

There is no such geometric meaning for $a+bc$ - it even fails a basic "units" test. One is the measure of area, the other is a measure of length. It is a wonder that we can turn area back into a length at all to add $a$ to $bc$, but the geometry is all wrong to get $a+b^2=(a+b)(a+c)$.

Distributive Lattices: Algebras with symmetry

There isn't a way to "make" two defined operations both distribute, but there are algebras with two operations where both distribute, such distributive lattices and boolean algebras.

You could create two operations on $\mathbb R,$ $+_m$ and $\cdot_m$ which are defined as:

$$a+_m b = \min(a,b)\\ a\cdot_m b = \max(a,b)$$

This is just treating the order on the reals as a distributive lattice.

For lattices in general, we generally use the operator symbols $\cap$ and $\cup$ or $\land$ and $\lor$. These symbols indicate the symmetry. You'd never see a lattice using $+$ and $\cdot$, because they don't fit the "pattern" of addition and multiplication. (In Boolean algebras, you occasionally see $\cap$ written as $\cdot$, but $+$ is something weirder in that case.)

These two operators are very different from the usual $+,\cdot$. There is no identity for either operation (although you could add $+\infty$ and $-\infty$ for identities,) and you have $a\cdot_m a = a+_ma=a$.

Deep dive into Algebra: Rings of endomorphisms

This will really only make sense if you have taken some algebra.

Given any Abelian group $(A,+)$, the set $R=\mathrm{End}(A)$ of homomorphisms $A\to A$ forms a ring with identity by defining $(f+g)(a)=f(a)+g(a)$ and $(f\cdot g)(a)==(f\circ g)(a)=f(g(a))$.

It turns out, every ring $R$ is basically a subset of such an endomorphism ring. The distributive law is what makes that possible:

$$R\hookrightarrow End(R,+)$$

That is the most trivial "representation" of $R$, but, for example, in the case of matrices, there are simpler abelian groups to use.

So, let's say we only have the notion of addition and an order ($<$) on the real numbers.

It turns out, for each $r\in\mathbb R$ there is a unique $f_r\in\mathrm{End}(\mathbb R)$ which has $f_r(1)=r$ and is "nice" with respect to the notion of between-ness - if $c$ is between $a$ and $b$ (inclusive) then $f_r(c)$ is between $f_r(a),f_r(b)$ (inclusive.) (This "between-ness condition can be seen as requiring $f_r$ to be continuous.)

Thus there is a map:

$$\mathbb R\hookrightarrow \mathrm{End}(\mathbb R,+)\\r\mapsto f_r$$

Now we can define multiplication in $\mathbb R$ as $r\cdot s=f_r(s)$. It takes some effort to prove that this works, but the more remarkable thing is not that multiplication is distributive - that is built in by the nature of how we defined endomorphisms. The remarkable thing is that multiplication is associative and commutative.

[It's easier to define multiplication on the integers and rationals, because you don't require the order part - if $A=(\mathbb Q,+)$ or $(\mathbb Z,+)$ then there is exactly one $f_p\in\mathrm{End}(A)$ which satisfies $f_p(1)=p$.]

This notion of endomorphisms is actually pretty useful. For example, a vector space can be defined as an abelian group and a field $k$ with a ring homomorphism: $$k\hookrightarrow \mathrm{End}(A)$$

And the same for $R$-modules, although in that case the map $R\to \mathrm{End}(A)$ is not required to be an inclusion.

Units of measure

One last thing to note is to wonder what happens when units are in use.

You can only add two numbers if they are in the same "units." You don't add $1$ inch to $2$ meters. You have to convert units in that case. You certainly can't add $1$ meter per second $2$ meters.

However, you can multiply any two numbers with any units, you just get a different unit. for example: $1\text{ m/s}\times 3\text{ s}=3m$.

So, in the equation $a(b+c)=ab+ac$, you get the same units and are "allowed" to do this operation if $b$ and $c$ have the same units.

But if you had $a+bc=(a+b)(a+c)$, then the right side would require $a,b,c$ and $bc$ to all have the same units. That essentially means that they can't have units, unless you can find a unit $u$ such that $u^2$ is the same unit as $u$.

  • $\begingroup$ Thanks, that's a lot of input which is what I came here for. I did though wonder more about the exact workings of + and * in the body of real numbers. Definitely gonna check out boolean algebra and latices, they were pointed out multiple times now. :-) $\endgroup$
    – John Smith
    Jan 22, 2016 at 2:03

My way of interpreting the distributive law in an image:

My way of interpreting the distributive law in an image

The reason why people get annoyed with stuff like this is because they see the operations $+$ and $\times$ as symbols, not actual processes.

Standard index laws when using $\times$ allows us to relate multiplication with dimensions.

Standard addition laws regarding 'like terms' means we can't just simplify two entities from these 'separate' dimensions.

From image: "The output of the * operator can be visualised by adding dimensions (as per the index laws, x^2 * x^3 = x^5" "The output of the + operator can be visualised as lumping two of the same dimension objects together." "If you imagine x=a=b=c and apply the distributive law, you'll end up with non-similar terms. i.e. you can't group x^2 and x terms"

  • $\begingroup$ You shouldn't post answers (or questions for that matter) as images. Among other reasons, this makes it very hard to search for them and (almost) impossible to read them for visually impaired people. Images are perfectly fine as a support for an answer, though. Could you please crop the two images and type out the rest of the answer? $\endgroup$
    – A.P.
    Jan 22, 2016 at 14:48
  • 1
    $\begingroup$ Sorry, will do. I just joined SE when I posted my answer $\endgroup$ Jan 24, 2016 at 23:54

Update: The previous version of this answer was somewhat misleading, suggesting that there might be non-trivial rings where the sum distributes over the product.

Suppose that $(A,+,*)$ is a ring where $+$ distributes over $*$, i.e. where $$ a + b*c = (a+b)*(a+c) \quad \text{for every } a,b,c \in A. \tag{1} \label{eq:dist} $$ By definition of ring, there must be an element $0 \in A$ that acts as the identity for $+$, therefore $$ a = a + 0 = (a+0)*(a+0) = a*a \quad \text{for every } a \in A $$ and a ring where every element is idempotent is called a Boolean ring. On the other hand, since $A$ is a ring we know that $0*a = 0$ for every $a \in A$, so $\eqref{eq:dist}$ implies that we must also have $$ a = a + 0*a = (a + 0)*(a+a) = a*a + a*a = a+a $$ and summing to both sides the additive inverse of $a$ — which must exist for $A$ to be a ring — gives $$ 0 = a. $$ Therefore the only ring where $\eqref{eq:dist}$ can hold is the trivial ring, $A = \{0\}$.

In case this wasn't clear, $(\Bbb{R},+,*)$ is a ring, and clearly $\Bbb{R} \neq \{0\}$.

On the other hand, rings are not the only kind of algebraic structure comprised of a set with two operations defined on it, and for each of those we can define a notion of distributivity.

For example, it can be proved that in a lattice $(L,\vee,\wedge)$ if one operation distributes over the other, then the converse must also hold, too. In other words, we have that $$ a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c) \quad \text{for every } a,b,c \in L $$ if and only if $$ a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c) \quad \text{for every } a,b,c \in L $$ and in this case the lattice is simply said to be distributive. In his answer Thomas Andrews gave an example of distributive lattice structure on $\Bbb{R}$.

  • $\begingroup$ I see. But is there somesort of term for this condition? I know that there are all kind of names for operator groups (like all the morphisms). Can there maybe a statement be made about when operators can be double distributive and when they can not? $\endgroup$
    – John Smith
    Jan 22, 2016 at 2:09
  • 1
    $\begingroup$ @JohnSmith I'm sorry, it was late for me when I wrote my answer, which wasn't complete. I updated my answer to show that for rings (as in the case you had in mind) double distributivity isn't a useful condition. On the other hand, double distributivity is possible for other structures. $\endgroup$
    – A.P.
    Jan 22, 2016 at 14:42

Not very "exact", but you could see it this way:

$$a * (b + c) = \underbrace{(b + c) + \dots + (b + c)}_{a \text{ times}} = \underbrace{b + \dots + b}_{a \text{ times}} + \underbrace{c + \dots + c}_{a \text{ times}} = (a * b) + (a * c)$$

But this is not going to work:$$a + (b * c) = a + \underbrace{c \dots + c}_{b \text{ times}} \neq\underbrace{ (a + c) + \dots + (a + c)}_{(a + b) \text{ times}} = (a + b) * (a + c)$$

  • 1
    $\begingroup$ Thanks, I thought about this too. But this would require multiplication to be a "shortcut" operator for the addition operator, wouldn't it? The way I understood it is that multiplication can be simulated by repeated addition, but isn't actually repeated addition. (i.e. exists seperately) $\endgroup$
    – John Smith
    Jan 22, 2016 at 1:47
  • 1
    $\begingroup$ Well. in $\mathbb{N}$ multiplication is exactly repeated addition. And also things like $2.5 * 2$ can be seen as $2 + 2 + 1/2$. This intuition stops obv at negative or real numbers. $\endgroup$
    – adjan
    Jan 22, 2016 at 1:52
  • 1
    $\begingroup$ I'm not sure at what level you are asking this question, but at some point we need to derive the laws of arithmetic (which includes the distributive law) - the 'standard' approach taught in first year undergrad course would be via the Peano Axioms which gives us a basis for the laws of arithmetic on the natural numbers. One of these laws would be the distributive law of * over +, there is no law of + over * as it cannot be derived from the Peano Axioms, or simply shown by a counter example. Yes, for a basically simple question, its not an easy answer. $\endgroup$
    – user247608
    Jan 22, 2016 at 1:52
  • $\begingroup$ I mean, it would make it easier if multiplication was exactly just repeated addition. I find that multiplication is much more complex though, as there are multiple ways to recreate it (repeated addition being one of them). (Definitely gonna check out the derivation of arithmetic laws though) $\endgroup$
    – John Smith
    Jan 22, 2016 at 1:58
  • 1
    $\begingroup$ Whichever way you create multiplication, its got to obey the distributive law. I guess there's a bunch of equivalent theories to generate the laws of arithmetic as we know them. If you take a look at the Peano approach, it would be adequate to understand the theory just for the naturals so you -can see why distributivity is baked-in for * over + but not for + over *. You could take a look at Boolean algebra and see why from the definition of AND and OR that AND is distributive over OR and vice versa - all a matter of the definition, I guess that also goes for the algebra of sets too. $\endgroup$
    – user247608
    Jan 22, 2016 at 2:18

You must log in to answer this question.

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