We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not. My question is why.

As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation where
$H_0(a,b) = b+1$ (successor op)
$H_1(a,b) = a+b$ (addition op)
$H_2(a,b) = ab $ (multiplication op)
$H_3(a,b) = a^b$ (exponentiation op)
$H_4(a,b) = a\uparrow \uparrow b$ (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )

Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$

Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?


  • 3
    $\begingroup$ If you're familiar with the construction $a^b=\exp(b\ln\;a)$, $a\ln\;b\neq b\ln\;a$ most of the time, so... on the other hand, $2^3$ and $3^2$ are so tantalizingly close! $\endgroup$ Apr 28 '11 at 10:13
  • $\begingroup$ On the same page you quote, there are commutative hyper operators as well. $\endgroup$ Apr 28 '11 at 10:14
  • $\begingroup$ @Raskolnikov Yes, but those commutative operators are quite different from the standard arithmetic operators that I am asking about. Their existance does not answer my question (at least not in a way that I understand) $\endgroup$
    – rob levin
    Apr 28 '11 at 10:24
  • $\begingroup$ @J.M. hi yes, indeed this is correct but unfortunately doesn't help my intuition about why the symmetry breaks as the operator order increases $\endgroup$
    – rob levin
    Apr 28 '11 at 10:43
  • 12
    $\begingroup$ Note that H0 isn't commutative: H0(a,b) = b + 1 but H0(b,a) =a + 1. $\endgroup$ Apr 28 '11 at 16:52

14 Answers 14


My personal opinion is that the exponential is not naturally regarded as the next step in the progression from addition to multiplication, so there's no reason to expect it to share properties with the other two.

Notice what happens if you demand that all of your quantities have units. Addition is an operation you do to two quantities with the same units: for example, you can add two distances to get another distance. Multiplication is an operation you do to two quantities with units $a$ and $b$ to get a quantity with units $ab$; for example, you can multiply two distances to get an area.

However, you cannot exponentiate two unital quantities. In computing $a^b$, the quantity $b$ needs to be unitless or else there's no sensible way to assign a value to the result. This tells us that $a$ and $b$ are being treated very differently, so there's no reason to expect that switching them is a physically meaningful operation.

As I've discussed on math.SE before, the exponential has many generalizations in mathematics, and in most of them the base and the exponent are very different objects: in fact, in most of them the base is just $e$.

  • 24
    $\begingroup$ "... is not ... the next step ...", which begs the question, what is? $\endgroup$ Apr 28 '11 at 20:13
  • 4
    $\begingroup$ I don't think addition and multiplication should be regarded as belonging in a progression at all. (Well. There is a theory of objects which act on rings, which I think behave like rings with extra operations, but I'm not familiar with it.) $\endgroup$ Apr 28 '11 at 20:45
  • 6
    $\begingroup$ The difficulty here, though, is that regardless of whether addition, multiplication, and exponentiation can be viewed as a progression in more general settings, we do have a sequence of iteratively-defined operators, which was mentioned in the post, with three of them corresponding to what we call addition, multiplication, and exponentiation, and one may ask why two operators out of that sequence happen to be commutative. $\endgroup$ Nov 17 '12 at 1:54
  • 22
    $\begingroup$ I'm really not so sure I'd agree on the idea that multiplication-as-repeated-addition is just an "accident". This seems to imply that, say, ring theory and "repeated addition" came about entirely separately, and then it was like "HOLY COW! They happen to coincide! Wow, what a nice surprise!". Rather, "repeated addition" came first, ring theory came after, and ring theory was intentionally developed to capture the properties that multiplication has. There's no accident here. $\endgroup$ Nov 17 '12 at 1:57
  • 2
    $\begingroup$ To $($slightly$)$ counter your argument: “nice” properties are usually lost in the process of successive generalizations. For instance, when going from R to C, we lose order; from C to H we lose commutativity, etc. $\endgroup$
    – Lucian
    Nov 26 '15 at 0:14

Here is a thought, which is not a full answer, but too long for a comment.

Addition $a + b$ means something like: Add $1$ to $a$, $b$ times $= 1\cdot b + a$. Commutativity here means that $1\cdot b + a = 1\cdot a + b$. We can see that it's only through the fortunate use of $1$ that this is commutative; $cb + a \not= ca+b$ in the general case. If we defined successorship to be an increment of two, addition would no longer be commutative!

With multiplication, $a \cdot b $ means add $a$ to itself, $b$ times $= \sum_{i=1}^b a$. It is only through the fact that addition is "hyper-commutative" (i.e. $\sum_{i=1}^b a = \sum_{i=1}^a b$) that multiplication is commutative.

Suppose I define an operation $+_2$ meaning "addition where successorship increases by two" as suggested above. $a+_2 b = 2a + b$. In this case, multiplication loses its commutativity, since $a +_2 a +_2 \dots +_2 a$ $b$ times is not the same as $b +_2 b +_2 \dots$ a times.

So to invert your question: why is it that addition and multiplication are commutative? It seems like it's because of the fortunate choice of successorship.

To stretch it further, we can view multiplication as a weird form of addition. In general, $ab = (a-1)b + b = b +_{a-1} b$. Plugging that in to, say, $b^2$, we would find $b^2=b +_{b-1} b$. So exponentiation is not commutative because its "successor" is not 1. This also explains why exponentiation is commutative for e.g. $2^2$, since $2-1=1$, and so we have a unit successor. It's more complex for $b^b$, but I think the principle holds.

Long story short: it's because "1" is special. Hopefully this is of some interest, even though it's not as "intuitive" as I might like.

EDIT: Perhaps the most interesting point in that long story is: if $H_0\not= +1$ then no operations are commutative.

  • 12
    $\begingroup$ +1 for 'Why aren't + and * non-commutative?" $\endgroup$
    – Mitch
    Apr 28 '11 at 14:54
  • $\begingroup$ Can you define the notion "successor of a (hyper)-operation"? Also, can't we view exponentiation as a weird form of multiplication in exactly the same way? $\endgroup$
    – Stefan
    Apr 28 '11 at 16:17
  • $\begingroup$ @Stefan: my remark was just to say that if $H_0\not=+1$ then no (hyper-)operation is commutative. And indeed, if we shift $+\to+_b$ then this is sort of like moving $H_i\to H_{i+1}$, so it will trickle all the way up the chain. $\endgroup$
    – Xodarap
    Apr 28 '11 at 18:40
  • $\begingroup$ Now I understand what you're saying. Good point! $\endgroup$
    – Stefan
    Apr 28 '11 at 20:43
  • 3
    $\begingroup$ You cannot even write $a+_2a+_2\cdots+_2a$ because $+_2$ isn't associative. Though your statement works both for $((a+_2a)+_2\cdots)+_2a$ and $a+_2(a+_2(\cdots+_2a))$. $\endgroup$
    – firtree
    Dec 16 '15 at 15:10

I think I can dress Qiaochu's answer up in geometric terms. Commutativity of addition and multiplication (in $\bf N$) can be seen as an observation of symmetry of counting blocks that are arranged in certain ways. I recall in elementary school seeing big numbers depicted concretely in our textbooks as piles of blocks.

(Their system involved $a_3$-many $10\times10\times10$ cubes of blocks, $a_2$-many $10\times10$ squares of blocks, $a_1$-many height-$10$ columns of blocks, and $a_0$-many single blocks in order to depict the number with decimal representation $a_3a_2a_1a_0$.)

Let's represent whole numbers by themselves as columns of blocks. Then $a+b$ involves a column with $a$ blocks underneath a column with $b$ blocks, and $b+a$ vice-versa. Both $a+b$ and $b+a$ are represented by the addition operation as a single column. Dissimilarly, we represent $a\times b$ as a rectangle of blocks of dimensions $a$ by $b$.

The symmetry in both of these cases are clear: flip the column $a+b$ upside-down to get $b+a$, and flip the sheet or plane in which the rectangle of $a\times b$ resides over to obtain that of $b\times a$.

Exponentiation, or repeated multiplication, clearly does not manifest the advantage of yielding the same dimension of object after application. That is, $a^b$ should be represented by a $b$-dimensional hypercube of sidelength $a$, whereas $b^a$ a $a$-dimensional hypercube of sidelength $b$. Thus it becomes obvious that no flipping can give us $a^b$ from $b^a$, and the source of symmetry has been taken away.

In conclusion, commutativity of operations on counting numbers which can be represented via these visual constructions can be seen as a flipping-symmetry which is a property requiring a constant dimension. Thus, dimensional-breaking is the cause of symmetry-breaking.

One may of course fight back with, "well, there is no flipping symmetry, but why isn't there any type of symmetry at all that yields commutativity?" In a sense, "most" binary operations are noncommutative, so believing a "random" binary operation is noncommutative is the natural tentative position to take until evidence to the contrary arises. The evidence to the contrary (flipping symmetry) is strikingly obvious in the cases of addition and multiplication, but not exponentiation.

  • 1
    $\begingroup$ Very nice idea! +1 $\endgroup$
    – Pedro Tamaroff
    May 26 '13 at 23:59

There's a pretty good heuristic that comes from set theory. If you have two sets, $X,Y$, then:

$X\times Y=$ all ordered pairs $(x,y):x\in X, y\in Y$

$Y^X=$ all functions $f:X\rightarrow Y$

Now if $X$ and $Y$ happen to be finite sets, with $|X|=m$ and $|Y|=n$, then we have $|X\times Y|=mn$ and $|Y^X|=n^m$ as usual.

So it should be clear that $|X\times Y|=|Y\times X|$, while $|Y^X|\ne|X^Y|$, for finite sets.

Now these rules also happen to go through for infinite sets as well, which takes more work to show.

Finally, there is a quite natural set isomorphism between $X\times Y$ and $Y\times X$, $ (x,y)\leftrightarrow (y,x)$, but certainly not between $Y^X$ and $X^Y$; i.e. you cannot just switch a domain set with its range and expect to have a set isomorphism on the new set of functions (e.g. let one of the sets have just one member and the other set more than one member).

I realize there are many details to fill in here to make this argument fully rigorous, but I think it shows why there is no reason to think that exponentiation should commute given the set-theoretic analogue of functions between sets.

  • 4
    $\begingroup$ Just to add to this, you can view addition as disjoint union, and there is also a natural bijection between $A \amalg B$ and $B \amalg A$. $\endgroup$
    – aws
    May 6 '13 at 12:00
  • 1
    $\begingroup$ @aws, yes, exactly, thanks for the addition! $\endgroup$
    – user452
    May 6 '13 at 12:02
  • $\begingroup$ @aws what is this symbol? $\amalg$ $\endgroup$
    – MphLee
    May 6 '13 at 14:39
  • 2
    $\begingroup$ @MphLee It is disjoint union. In set theory it is usually implemented as $A \amalg B := \{ ( 0, a ) | a \in A \} \cup \{ ( 1, b ) | b \in B \}$. $\endgroup$
    – aws
    May 6 '13 at 14:52
  • 1
    $\begingroup$ @MphLee It is sometimes written as $+$, but I think $\amalg$ is the more usual notation. $\endgroup$
    – aws
    May 6 '13 at 15:37

When I first read your question, I expected that it must mean that addition would possess some obscure property that multiplication lacks, after all, both the additive structure and multiplicative structure are abelian groups, so you'd expect something like this to just generalize. But after some thinking, I realized that this wasn't the case, and instead that the problem is that we aren't generalizing properly.

For if we define "applying an operator $f$, $n$ times, ie $f^n$" as the recursive procedure $ f^n(x) = \begin{cases} x & \text{if n = 0} \\ f^1(f^{n - 1}(x)) & \text{otherwise} \end{cases} $

Then this definition actually uses addition, so if we'd want to generalize this procedure properly, we'd need to change our definition of "applying an operator $n$ times" as well. And indeed $a^n$ does equal $(a^2)^{n / 2}$, which induces a better generalization of commutativity.

  • $\begingroup$ → Lieven: "Then this definition actually uses addition" where do you see this? I see the function composition and recursion in use in your building of $ f^{n-1} \rightarrow f^n $ but no $ + $. $\endgroup$
    – dan
    Aug 1 '13 at 19:44
  • $\begingroup$ @danielAzuelos: $n = 1 + (n-1)$ is addition $\endgroup$
    – ziggurism
    Oct 30 '17 at 17:00
  • 1
    $\begingroup$ Was the first branch meant to be $\text{if n = 1}$? Otherwise we get $f^1(x) = f^1(x)$. $\endgroup$
    – Jan Tojnar
    Apr 8 '19 at 7:32

Edit. Okay, it turns out this idea has been studied before. See here.

I basically got this idea from user52541's answer. So no claim to originality.

Anyway. Define a sequence of operations $\langle n \rangle : (\mathbb{R}^+)^2 \rightarrow \mathbb{R^+}$ as follows.

  1. For all $x,y \in \mathbb{R}^+$, define $x\langle 0\rangle y = x+y.$
  2. For all $x,y \in \mathbb{R}^+$ and all $n \in \mathbb{N}$, define $x\langle n+1\rangle y=\exp(\log x\langle n\rangle\log y).$

Then $\langle 0 \rangle$ is addition, and $\langle 1 \rangle$ is multiplication. But $\langle 2 \rangle$ is not exponentiation. Furthermore, we can prove that for all $n$ it holds that $\langle n \rangle$ is both commutative and associative.

Remark. To prove commutativity, the functions $\exp$ and $\log$ don't even need to be inverses of one another. They can just be arbitrary functions. To prove associativity, we need a bit more. In particular, we require that $\exp$ and $\log$ are inverses. However, we're still not really using any of their properties, like $\log(xy) = \log x + \log y$ etc.

Question 1. Do we have existence of inverses at each level (we have negatives at level $0$, reciprocals at level $1$, and WHAT, if anything, at level $2$)?

Question 2. Does a form of distributivity hold?

  • $\begingroup$ As can easily be checked, at level 2 the neutral element is $e$, and the inverse element is $\exp(1/\log x)$. Indeed, be $e_n$ the neutral element at level $n$ ($e_0 = 0$, $e_1=1$, etc.), then $e_{n+1}=\exp e_n$. Also, be $i_n(x)$ the inverse of $i$ at level $n$ ($i_0(x)=-1$, $i_1(x)=1/x$, …), then $i_{n+1}(x)=\exp(i_n(\log x))$. Obviously the range on which it is defined shrinks on each step; I'm not sure whether it can always be extended to all of $\mathbb R$ (multiplication can, of course). $\endgroup$
    – celtschk
    Oct 9 '16 at 14:12
  • $\begingroup$ @goblin I really like your idea. I came up with something similar to build a Hypertype Theory. And I am really surprised that the underlying commutative power law is not used more often. Also, I love your profile! Totally inspiring. $\endgroup$
    – ismael
    Jan 6 '19 at 20:48

I know this question is old, but I want to add my answer in case other people like me stumble across this.

First, think about the transition from addition to multiplication. The distributive laws, along with the structure already existing in addition, imply that 0*x = x*0 = 0 always. Thus, we see that 0 is transitioned from being an identity element of addition (x+0 = 0+x = x) to being a zero element or, as I shall call it, a "null" element of multiplication.

Next, we try to do an analogous transition from multiplication to exponentiation. By analogy, the identity of multiplication - 1 - should become a null element of exponentiation, i.e., 1^x = x^1 = 1. However, our analogy also states that exponentiation is repeated multiplication, which implies that 1 is an identity element of exponentiation, since repeating something 1 time should give us that something again; i.e., 1^x = x^1 = x.

Thus, we have "overloaded" 1 by demanding that it be both a null and an identity. This is impossible, so the discoverer of exponentiation (whoever that may have been) came up with a compromise: 1 would be a null on the left (1^x = 1) but an identity on the right (x^1 = x). This implies that, in general, 1^x is different from x^1, and thus exponentiation is not commutative.

  • $\begingroup$ The formal reason for why the "compromise" is what it is can be found in Gottfried Helms' answer to this question. $\endgroup$
    – Matthew W.
    Jun 14 '13 at 16:03

We start with a commutative operation, addition, and iterate it recursively to arrive at the operation, multiplication as repeated addition, and find that it remains commutative. One might assume that a statement such as "iterated commutative operations are again commutative". In which case one would expect exponentiation which is also an iteration of a commutative operation, to itself be commutative. But exponentiation is not commutative, so the expected statement does not hold. Why not?

Well let's look at the proof of the commutativity of multiplication from the axioms of Peano arithmetic.

Denote the successor of $n$ as $s(n).$ Multiplication is defined recursively via $s(m)\cdot n = m\cdot n + n$ with a base case of $0\cdot n=0.$

We wish to show that $m\cdot n = n\cdot m$ for all $m,n\in\mathbb{N}.$ We proceed by induction on $m$ nested inside an induction on $n$. Let $S$ be the set of naturals that commute with everything in $\mathbb{N}$. Clearly $0\in S$. Suppose $n\in S$. Let $T$ be the set of all naturals that commute with $S(n)$. Clearly $0\in T$. Suppose $m\in T.$ Then we have

$$ s(m)\cdot s(n) = m\cdot s(n) + s(n)\qquad\text{(def of mult)}\\ = s(n)\cdot m + s(n)\qquad\text{(ind hyp 2)}\\ = (n\cdot m + m) + s(n)\qquad\text{(def of mult)}\\ = n\cdot m + (m + s(n))\qquad\text{(add assoc)}\\ = n\cdot m + s(m + n)\qquad\text{(def of add)}\\ = n\cdot m + s(n + m)\qquad\text{(add comm)}\\ = n\cdot m + (n + s(m))\qquad\text{(def of add)}\\ = (n\cdot m + n) + s(m)\qquad\text{(add assoc)}\\ = (m\cdot n + n) + s(m)\qquad\text{(ind hyp 1)}\\ = s(m)\cdot n + s(m)\qquad\text{(def of mult)}\\ = n\cdot s(m) + s(m)\qquad\text{(ind hyp 1)}\\ = s(n)\cdot s(m)\qquad\text{(def of mult)}. $$

So $s(m)\in T$ and therefore $T=\mathbb{N}.$ Therefore $s(n)\in S$, and so $S=\mathbb{N}$.

Now let's repeat the proof, line for line, but changing mutatis mutandi the addition to multiplication and the multiplication to exponentiation, to see where it fails.

Define exponentiation recursively as $m^{s(n)} = m^n\cdot m,$ with base case $m^0 = 1.$

$$ s(n)^{s(m)} = s(n)^m \cdot s(n)\\ = m^{s(n)} \cdot s(n)\\ = (m^n \cdot m) \cdot s(n)\\ = m^n \cdot (m \cdot s(n))\\ = m^n \cdot \color{red}{s(m \cdot n)}\\ = m^n \cdot s(n \cdot m)\\ = m^n \cdot \color{red}{(n \cdot s(m))}\\ = (m^n \cdot n) \cdot s(m)\\ = (n^m \cdot n) \cdot s(m)\\ =n^{s(m)} \cdot s(m)\\ = s(m)^n \cdot s(m)\\ = s(m)^{s(n)}. $$

The two steps in red are the invalid steps (besides also the base case fails), which assume that the successor of a product is a product with the successor of one of the factors, which fails: $\color{red}{s(m\cdot n) \neq m\cdot s(n)}$. So this is the reason that the formal proof of commutativity of multiplication fails to carry over to a proof of commutativity of exponentiation. This is the difference between addition and multiplication: the successor operation commutes with addition but not multiplication. $$(m+1)+n=(m+n)+1,$$ but $$(m+1)\cdot n\neq (m\cdot n)+1$$.

Multiplication distributes over addition and the successor operation, rather than commuting with it.

To say it another way, the question shouldn't be "why isn't exponentiation commutative?". The better question is "why is multiplication commutative?" Addition is commutative because successor is associative, and multiplication is commutative because, recursively viewed, it's just addition, and addition commutes with successor. For no other operations, certainly no higher iterates, should we expect this to be true.

Note that this is essentially the same answer as Lieven gave: if we use a multiplicative recursion instead of additive, then parallels between iterated of addition and multiplication do exist. But I wanted to see an answer that pinpoints the precise step where the proof breaks down, hence my submission.


This might be slightly circular since the definition of exponentiation and logarithm comes after addition and multiplication, but consider: If we perform a certain unary operator on the two arguments, perform addition on them, and then take the inverse, we get $e^{\log a + \log b} = ab$. If we do the same with multiplication, $e^{\log a \log b} = a^{\log b} = b^{\log a}$, which is not exponentiation.


A question "why..." in number theory is difficult to answer if it is not just an algebraic but a qualitative one, but let's give a try to improve at least intuition.

For me it is more intuitive to see the (hyper-)operators as (possibly repetatively/iteratively applied) operation at a startvalue, say $x_0$ , having a "base"-parameter $b$ and an iteration-counter $h$.

Then we can also reflect the left- or right-associativity in the completed table:

$\displaystyle \qquad \begin{eqnarray} G_{1,L}(x_0,b,h) &=& ((((x_0) \underset{\text{h times}}{ \underbrace{+ b) + b) \ldots + b)} } &\quad= \quad& G_{1,R}(x_0,b,h) &=& \underset{\text{h times}}{ \underbrace{ (b + \ldots (b + (b +} } (x_0)))) \\ G_{2,L}(x_0,b,h) &=& ((((x_0) \underset{\text{h times}}{ \underbrace{\cdot b) \cdot b) \ldots \cdot b)} } &\quad =^{1)} \quad& G_{2,R}(x_0,b,h) &=& \underset{\text{h times}}{ \underbrace{ (b \cdot \ldots (b \cdot (b \cdot} } (x_0)))) \\ G_{3,L}(x_0,b,h) &=&(((( x_0) \underset{\text{h times}}{ \underbrace{ \uparrow b) \uparrow b ) \ldots \uparrow b)} } &\quad \quad & G_{3,R}(x_0,b,h) &=& \underset{\text{h times}}{ \underbrace{ (b \uparrow \ldots (b \uparrow (b \uparrow} } (x_0 )))) \end{eqnarray}$
$\qquad \qquad ^{1)}$ only over certain (for instance:scalar) fields

Here we have symmetry in the left- or right-associativity $G_{1,L} = G_{1,R} (=G_1)$ and over a scalar field $G_{2,L} = G_{2,R} (=G_2) $

In $G_1$ we have even the interchangeability of the iterator and the base $G_1(x_0,b,h) = G_1(x_0,h,b)$

We can now look at elements with special values in $x_0$ and $b$ to find that the operation is "neutral" ($b=0$ for the addition and $b=1$ for the multiplication and the "power" ($G_{3,L}$)) and becomes "absorbing" for some elements ($b=0$,$x_0=0$ for multiplication and $x_0=1$ for "power" ($G_{3,L}$) and $b=1$ for "exponentiation" ($G_{3,R}$)) and also at the absorbing effects of $\infty$ and of fixpoints in $G_{3,R}$ etc. ...

If I look at all this and even a bit further then the miracle of the "broken symmetry" after the two operations addition and multiplication becomes less mysterious and only one small between other changes of properties (although it is surely remarkable...) .

An additional remark: although I expand here on a hierarchical scheme based on the iteration of operations, I find it worth to remember what Quiaochu Yuan says, that the different operations should be seen in some other view focused at some qualitative(?)/topological(?) properties, for instance the exponential map in the complex plane.


I totally agree that this "breaking of symmetry" happens and that it is because we do not understand addition and multiplication completely. There are many types of rings, also e.g. power-associative rings, etc. but that's not the point. What exactly makes it possible to so easily interchange the "operation" with the "quantity" under the operator for addition?

For your suggestion:

I'd like to see an answer that either: defines a "new" exponentiation (by generalizing differently) which is both associative and commutative

I thought for ordinal numbers exponentiation is associative, it isn't. But there might be mathematical constructs, Jacobsthal's exponentiation(?), for which it is the case.


Because nobody had found any real-world application for a commutative exponentiation operator, until someone looked hard enough at the log-normal probability distribution, which can be found in countless natural and artificial processes, in physics, life sciences, social sciences, and economy. The Missing Operation article gives a simple introduction to this topic.


Maybe this is what you are looking for :


  • $\begingroup$ You might want to lookup tetration if you are not familiar with it. $\endgroup$
    – mick
    May 11 '13 at 21:34

For any two positive real numbers, their product to the base 2 can be described as the exponential to the base 2 of the sum of their logs to the base 2. We could define another function $f$ for all positive integers $x$ and $y$ by $f(x, y) = 2^{2^{\log_2(\log_2x) + \log_2(\log_2y)}}$. This operation is commutative and distributes over multiplication. However, it is a different function than exponentation. The exponential to the base 2 of a product is not the product of the exponentials to the base 2. The exponential to the base 2 of a sum is the product of the exponentials to the base 2. On the other hand, multiplication distributes over addition. This shows that exponentation is not the double exponentation to the base 2 of the sum of the double logs to the base 2. Thus, exponentation doesn't necessarily have to be commutative.


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