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I understand that the addition operator, defined as $F(x,y)=x+y$, and the multiplication operator, defined as $F(x,y)=xy$, are both commutative and associative (meaning, of course, that $F(y,x)=F(x,y)$ and that $F(F(x,y),z)=F(x,F(y,z))=F(F(x,z),y)$). My question is: Why are exponentiation, tetration, pentation, etc. not generally commutative or associative? I understand that exponentiation just works that way, but my question is, is there an intuitive explanation and its mathematical proof?

Cheers, and thanks in advance.

I hope I've explained everything like I intended to. If I messed something up, feel free to edit or redirect me to a different post.

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  • $\begingroup$ I'm not downvoting because I think you are asking honestly, but all the same I have to say that I think this is a non-question. Why should the "based solely on the operator a 'dimension' below" be of any relevance? If you consider other properties than commutativity and associativity, it doesn't even work for multiplication as derived from addition. $\endgroup$ – David Jan 24 '17 at 23:46
  • $\begingroup$ Ah, okay. I suppose so. $\endgroup$ – Drew Christensen Jan 24 '17 at 23:48
  • $\begingroup$ How could it be $2^3 = 2*2*2 = (2*2) + (2*2) = 2 + 2 + 2+ 2 = 1+1+1+1+1+1+1+1+1=8$ while $3^2 = 3*3 = 3+3+3 = 1+1+1+1+1+1+1+1+1=9$. An $(a^n)^m = a^{n*m}$ where as $a^{n^m} = a^{n*n*n*n....}$. These simply are not conceptually the same things at all. $\endgroup$ – fleablood Jan 25 '17 at 0:04
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    $\begingroup$ I guess intuitively I think of this as "order matters". Putting on eating food for breakfast order doesn't matter as adding a piece of food in your mouth doesn't affect whether you can put another in. Putting you socks and shoes on order does matter because once you put something on it affects whether you can put something else on. addition doesn't matter as that's just counting and grouping. multiplication is adding and putting in bags. Once you put in bags you can't undo but as you are going no further it's okay. Exponetion you are grouping the bags and you can't go back. $\endgroup$ – fleablood Jan 25 '17 at 0:13
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To give some insight:

In the Integers we might naïvely define: $x\cdot 1$ as $\underbrace{1+\cdots+1}_{x\text{ times}}$.

Then multiplication would then be : $x\times y = \underbrace{x+\cdots+x}_{y\text{ times}} \\= (1+\cdots+1)+\cdots+(1+\cdots+1) \\= \underbrace{y+\cdots+y}_{x\text{ times}} \\ = y\times x$

Exponentiation however is $x^y = \underbrace{x\times\cdots\times x}_{y\text{ times}} \\ = (1+\cdots+1)\times\cdots\times(1+\cdots+1) \\ \underbrace{\neq\qquad}_{\text{generally}} y^x$

eg: $2^3 = (1+1)(1+1)(1+1) = (1+3+3+1) = 3\times 3-1 = 3^2-1$

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Of course, in the rationals, reals, et cetera, we have to extend the definition of the operators beyond this naïve approach (hence your work in operations).   But if the property does not hold for operations over the integers, then it shall not hold over the extentions.   ( $~8.0\neq 9.0~$ )

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    $\begingroup$ The most trivial of cases: consider $1$. $1^x\ne x^1$, and $1\uparrow\uparrow x\ne x\uparrow\uparrow1$. Same argument for higher order operations. $\endgroup$ – Simply Beautiful Art Jan 25 '17 at 0:12

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