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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 )
etc.

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?

Thanks.

  • 3
    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!2011-04-28
  • 0
    On the same page you quote, there are commutative hyper operators as well.2011-04-28
  • 0
    @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)2011-04-28
  • 0
    @J.M. hi yes, indeed this is correct but unfortunately doesn't help my intuition about why the symmetry breaks as the operator order increases2011-04-28
  • 9
    Note that H0 *isn't* commutative: H0(a,b) = b + 1 but H0(b,a) =a + 1.2011-04-28
  • 0
    Good point; I guess H0 is supposed to be a unary operator, but has been set up as a binary operator for consistency with the others. But yes, it is not commutative as set up in the question.2011-04-28
  • 0
    Already $1^2\ne 2^1$, but this could be a "degenerate case". So here is a little problem: Give a combinatorial proof that $2^3\ne 3^2$.2011-04-29

13 Answers 13