Why multiplication is commutative, but exponentiation is not?

A definition of exponentiation as an operation made from multiplication seem to be similar to a definition of multiplication made from addition.

I.e., 2*5 is 2 + 2 + 2 .. 2, for the total of five numbers two. Similarly, 2^5 is 2 * 2 * 2 .. 2, for the total of five numbers two.

Yet multiplication is commutative, but similarly defined exponentiation is not. Why?

Re: Why multiplication is commutative, but exponentiation is not?

Well exponentiation is commutative and associative in the sense that $\displaystyle (x^k)^l = x^{(kl)} = (x^l)^k$. But if your asking why $\displaystyle x^y \neq y^x$ (provided that $\displaystyle x \neq y$) it is because you cannot consider the exponent as a number the way you consider the base a number. The exponent is more of a descriptor.

Re: Why multiplication is commutative, but exponentiation is not?

Can you elaborate on why I can't consider the exponent the number?

My concern is what is the difference between making multiplication out of addition and making the exponentiation out of multiplications. Yes, I'm asking why $\displaystyle x^y \neq y^x$ .

Re: Why multiplication is commutative, but exponentiation is not?

One little thing I want to clarify first... when I say exponents are associative and commutative I only mean when they are on the same "level" so $\displaystyle a^{b^{c}} \neq a^{bc}$.

On to your question... Let us assume we are working in the real numbers (hopefully this set suffices your needs). The real numbers are a field, which by definition means that multiplication and addition are commutative. Exponentiation came along later and is a definition of convenience more than anything else. There really is no deep answer to your question, it's just the way exponentiation was defined does not allow for commutativity among the different "levels".

Re: Why multiplication is commutative, but exponentiation is not?

one reason is that 1 and 0 are "badly behaved" with regard to exponentiation.

for example, 1^n = 1, but n^1 = n, and 0^n = 0 (let's just agree for the moment n is not 0), while n^0 = 1.

since for natural numbers, n is defined as +1 done to 0 n times, if 0 and 1 do not commute via exponents, it's unreasonable to expect anything else will, either.

and of course, 0^1 = 0, while 1^0 = 1.

but i am afraid you're drawing the wrong conclusion from this. in actual fact, it's the commutativity of multiplication that is the real mystery. for example, with matrices,

A+B = B+A, but AB usually doesn't equal BA.

Re: Why multiplication is commutative, but exponentiation is not?

Thank you guys.

Let me rephrase a bit. Suppose one has an operation - call it A - and numbers.

Then he creates operation AA in the following manner - defining "x AA y" as "x A x A x .. x" so that the total number of x-s is y.

The question is - can we find out if AA is commutative? Turns out if A is addition, then the answer is yes, and if A is multiplication, the answer is no.

Now what information we need to predict the commutativeness of AA, made out of A? What is different between addition and multiplication, which leads to different results?

Re: Why multiplication is commutative, but exponentiation is not?

Looks like a problem is either somehow incorrectly stated or less well known than I thought :( .

Is there a branch dealing with properties of operations such as commutativity? For example, Cayley-Dickson construction allows creation of algebras with various properties. Are there ideas why particular properties exist or not in each case?