02/21/2008, 09:34 PM

This isn't really related to zeration per se, but some time ago I searched for an operator ♦ over which addition is distributive (i.e., a+(b♦c) = (a+b)♦(a+c)). (I guess my motivation was that just as exponentiation distributes over multiplication, and multiplication distributes over addition, addition should also distribute over an operation that may be regarded to be, in some sense, "before" addition.)

Eventually, I came upon the definition: . As you can verify, .

This operator is associative and also commutative, and is defined for all positive reals. If you admit complex numbers, it is defined for negative reals as well (although it will be complicated by the question of which branch of logarithm should be chosen). The hypothetical identity element is -∞ (because , which then reduces a♦(-∞) to a; also, ln 0 = -∞ by the dual argument). Interestingly enough, Rubtsov's zeration operation also features an identity at -∞. This property is in fact, a consequence of the distributive law: if addition distributes over any operation #, then the identity e of the # operation must satisfy e+x=e for all x. (The derivation of this is left as an exercise for the reader). This can be seen in exponentiation as well: the multiplicative identity 1 acts as a "zero" for exponentiation: 1^x = 1 for all x.

The only problem is that my operator does not satisfy a♦a♦...♦a (n times) = a+n. The closest one can get is to switch from base e to base 2, in which case this operation satisfies a♦a♦...♦a (2^n times) = a+n.

But anyway, the whole point of this overly-long post is that, in going from addition to zeration, something has to go. Rubtsov's zeration is non-associative and also discontinuous at one point. My operation is both associative and continuous over the positive reals, but then its n'th iteration is not equal to addition by n, which property Rubtsov's zeration does satisfy. So, in deciding which operation comes "before" addition, it's a question of which properties we want to keep, 'cos we can't have them all.

Eventually, I came upon the definition: . As you can verify, .

This operator is associative and also commutative, and is defined for all positive reals. If you admit complex numbers, it is defined for negative reals as well (although it will be complicated by the question of which branch of logarithm should be chosen). The hypothetical identity element is -∞ (because , which then reduces a♦(-∞) to a; also, ln 0 = -∞ by the dual argument). Interestingly enough, Rubtsov's zeration operation also features an identity at -∞. This property is in fact, a consequence of the distributive law: if addition distributes over any operation #, then the identity e of the # operation must satisfy e+x=e for all x. (The derivation of this is left as an exercise for the reader). This can be seen in exponentiation as well: the multiplicative identity 1 acts as a "zero" for exponentiation: 1^x = 1 for all x.

The only problem is that my operator does not satisfy a♦a♦...♦a (n times) = a+n. The closest one can get is to switch from base e to base 2, in which case this operation satisfies a♦a♦...♦a (2^n times) = a+n.

But anyway, the whole point of this overly-long post is that, in going from addition to zeration, something has to go. Rubtsov's zeration is non-associative and also discontinuous at one point. My operation is both associative and continuous over the positive reals, but then its n'th iteration is not equal to addition by n, which property Rubtsov's zeration does satisfy. So, in deciding which operation comes "before" addition, it's a question of which properties we want to keep, 'cos we can't have them all.