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We have the binary operation addition on numbers. It has an additive identity ( 0 ) and it is commutative.

Multiplication is simply repeated addition. It is a binary operation on numbers.
Its identity ( multiplicative ) is 1 and it is also commutative.

Then the function ^(x,y) = x^y is repeated multiplication, it is again a binary operation but it is not commutative. Why ?
Also, it has a right identity ( 1 ) , but it seems not to have a left identity ( is it true ? ). Why ?

I'm wondering if theres a reason that things change drastically even if we followed the same pattern ( making a new operation out of repeating the last one ).

Thanks in advance.


marked as duplicate by Claude Leibovici, Peter Taylor, Davide Giraudo, kingW3, N. F. Taussig Mar 16 '15 at 11:34

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  • $\begingroup$ If you want a commutative version of exponentiation, look up commutative hyperoperations (really -- they are very interesting). $\endgroup$ – user137731 Mar 16 '15 at 2:57
  • $\begingroup$ It is not a trivial matter to formally prove that addition and multiplication are commutative. You simply cannot cannot prove that exponentiation is commutative. It is sufficient to give a simple counter example: e.g. $0^1\ne 1^0$ from whatever formal definition you settle on. $\endgroup$ – Dan Christensen Mar 16 '15 at 17:16
  • $\begingroup$ As for the formal definition of exponentiation, you have two options (ask your instructor which to use in your course) Option A (with $0^0=1$): 1. $n^0=1$ 2. $n^{m+1}=n^m\times n$ Option B (with $0^0$ undefined): 1. $n^0=1$ for $n\neq 0$ 2. $0^1=0$ 3. $n^{m+1}=n^m\times n$ $\endgroup$ – Dan Christensen Mar 16 '15 at 17:26

How many cases do we have where iterating one commutative associative operation on the integers produces another commutative associative operation? Exactly one, namely the step from addition to multiplication.

One example does not constitute a "pattern".

You can only begin to expect that perhaps a pattern continues if you have seen the pattern repeat between two different cases.


Is $3^5=3*3*3*3*3=5*5*5=5^3$?

Some things aren't deep. They just... are. In this case,$\text{^}(x,y)\neq \text{^}(y,x)$ because of definition. What more is there to say. This isn't something to ponder.

  • $\begingroup$ But why? It is a bit curious, I think. When we apply the repetition procedure for the commutative associative addition, we get a commutative associative multiplication. Why doesn't it stay so when we do the next step? $\endgroup$ – Berci Mar 16 '15 at 2:43
  • $\begingroup$ First slot of $\text{^}(\cdot,\cdot)$ says "what", second says "how many". This is like comparing apples to oranges. $\endgroup$ – Squirtle Mar 16 '15 at 2:46

The most direct answer to your question is that exponentiation simply doesn't have the properties that addition and multiplication do. The "niceties" of commutativity and the existence of identities simply doesn't logically hold any more.

For a heartier answer, let's consider exponentiation in a more general setting. Let G be an number system that is associative, commutative, has a zero, and has (additive) inverses (if you want fancy terminology this makes G an Abelian group). Then we write $ng=g+g+...+g$ for $n\in\mathbb{N}$ and $g\in G$. This looks an awful lot like multiplication, but really, it has been defined in a way more philosophically linked to exponentiation. We have not defined multiplication as a new operation, we have just defined a notational shortcut, using out old friend $\mathbb{N}$, to describe repeating the group operation.

Now, from our current knowledge of what G can be, we know we have a left identity for this repeated addition, and that identity is the natural number 1. But we do not have a right identity, because we multiply naturals on the left and elements of G on the right. For an even more direct example, take E to be the set of even numbers. E fits all our requirements for G, yet when we multiply naturals on the left and even numbers on the right, there is no right identity.

So from our axioms of G we can deduce that $1g=g$ and $n(mg)=(nm)g=(mn)g$. These particular properties show up again when we deal with repeated multiplication. In this case they look like $g^1=g$ and $(g^m)^n=g^{mn}=g^{nm}$. These are some properties that exponentiation and multiplication actually do have in common, they just look a little different now. Multiplication arises naturally out of repeating things (although the pattern breaks down after going past exponentiation). So we shouldn't really be surprised that we loose some nice properties after we go through a couple levels of repeating, we should just be glad that we could extend our nice properties to multiplication to begin with!

Really, the commutativity of multiplication doesn't come from the fact that it is repeated addition. It comes from the nice properties of addition over the integers, and from the fact that we can take the integers to be G itself.

  • $\begingroup$ I see a lot of fancy words, for something not so fancy. Can't we all just agree... there's nothing deep here. This is a verbose discussion for "first slot says 'what', second says 'how many' " $\endgroup$ – Squirtle Mar 16 '15 at 2:54
  • $\begingroup$ Yeah I kinda like that. However, you could say precisely the same thing about multiplication! The first number says "what" gets added, the second number says "how many" times it gets added. Yet this doesn't imply an asymmetry, so the asymmetry of exponentiation does sound curious. $\endgroup$ – Clay Thomas Mar 16 '15 at 2:56

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