[strbean]

I think this is the novel bit:

> This includes constants such as e, pi, and i; arithmetic operations including addition, subtraction, multiplication, division, and exponentiation as well as the usual transcendental and algebraic functions.

[SideQuark]

And those come from the infinite series needed to compute exp and ln. They’re just as much work either way. The exp and ln way are vastly costlier for every op, including simply adding 1 and 2.

[krick]

It's not about being costly or not, this is completely irrelevant to the point being made. eml is just some abstract function, that maps ℝ² to ℝ. Same as every other mathematical function it is only really defined by the infinite set of correspondences from one value to some other value. It is NOT exp(x) - ln(y), same as exp is not a series (as you wrongfully stated in another comment). exp can be expressed (and/or defined) as a series to a mathematician familiar with a notion of series, and eml can be expressed as exp(y) - ln(y) to a mathematician familiar with exp and ln. They can also be expressed/defined multiple other ways.

I am not claiming this is better than 1/(x-y) in any way (I have no idea, maybe it isn't if you look closely enough), but you are simply arguing against the wrong thing. Author didn't claim eml to be computationally efficient (it even feels weird to say that, since computational efficiency is not a trait of a mathematical function, but of a computer architecture implementing some program) or anything else, only that (eml, 1) are enough to produce every number and function that (admittedly, somewhat vaguely defined) a scientific calculator can produce.

However, I want to point out that it's weird 1/(x-y) didn't appear on that graph in Figure 1, since if it's as powerful as eml, it should have all the same connections as eml, and it's a pity Odrzywołek's paper misses it.

[SideQuark]

His paper misses infinitely many such functions.

[krick]

Oh, glad you are still here. Because I kept wondering about 1/(x-y), and came to the conclusion it actually cannot do nearly as much as eml. So maybe you could confirm if I understood your assumptions correctly and help me to sort it out overall.

In your original post you were kinda hand-wavy about what we have except for x # y := 1/(x-y), but your examples make it clear you also assume 0 exists. Then it's pretty obvious how to get: identity function, reciprocity, negation, substraction & addition. But I effectively couldn't get anywhere past that. In fact, I got myself convinced that it's provably impossible to define (e.g.) multiplication as a tree of compositions of # and 0.

So here's my interpretation of what you actually meant:

1. I suppose, you assumed we already have ℕ and can sample anything from it. Meaning, you don't need to define 5, you just assume it's there. Well, to level things out, (#, 0, 1) are enough to recover ℤ, so I assume you assumed at least these three. Is that right?

2. Then, I suppose you assumed that since we have addition, multiplication simply follows from here. I mean at this point we clearly have f(x) = 3x, or 4x, or 5x, … so you decided that the multiplication is solved. Because I couldn't find how to express f(x, y) = x⋅y, and as far, as I can tell, it's actually impossible. If I'm wrong, please show me x⋅y defined as a sequence of compositions of (#, 0, 1).

3. This way (assuming №2) we get (ℚ, +, -, ⋅, /). Then, I suppose, you assume we can just define exp(x) as its Taylor series, so we also have all roots, trig functions, etc., and then we obviously have all numbers from ℝ, that are values of such functions acting on ℚ. Exactly as we do in any calculus / real analysis book, with limits and all that jazz.

If that's what you actually meant, I'm afraid you completely missed the point, and 1/(x-y) in fact isn't nearly as good as eml for the purposes of Odrzywołek's paper. Now, I didn't actually verify his results, so I just take them for granted (they are totally believable though, since it's easy to se how), but what he claims is that we can use eml essentially as a gate, like Sheffer stroke in logic, and express "everything else" just as a sequence of such gates and constant 1 (and "everything else" here is what I listed in №3). No words, limits, sets and other familiar mathematical formalism, just one operation and one constant, and "induction" is only used to get all of ℕ, everything else is defined as a finite tree of (eml, 1).

[SideQuark]

All of these are standard fare in abstract algebra classes, and I didn’t care to write it all out. Once you have the “inverse” operations - and reciprocal, the entire structure follows, for a large set of objects, whether N or Q or R or C or finite fields or division rings, and a host of other structures. So I only wrote - and 1/x

Then, subtraction is (x#y)#0 = x-y. Reciprocal is x#0 = 1/x. Addition follows from x+y=x-((x-x)-y). This used the additive identity 0.

Multiplication follows from

x^2= x-1/(1/x + 1/(1-x)), so we can square things. Then -2xy = (x-y)^2 -x^2 - y^2 is constructible. Then we can divide by -2 via x/-2 = 1/((0-1/x)-1/x), and there’s multiplication. In terms of #, this expression only needed the constant 1, which is the multiplicative identity.

Now mult and reciprocal give x * 1/y = x/y, division.

Any nontrivial ring needs additive and multiplicative identities 1!=0, which are the only constants needed above. If you assume this is Q or R or C, it may be possible to derive one from the other, not sure. But if you’re in these fields, you know 0 and 1 exist.

Then any element of Q is a finite set of ops. R can be constructed in whatever way you want: Dedekind cuts, Cauchy sequences, whatever usual constructions. Or assume R exists, and compute in it via the f(x,y).

This also works over finite fields (eml does not), division rings, even infinite fields of positive characteristic, function fields (think elements are ratio of polynomials), basically any algebraic object with the 4 ops.

[applicative]

You are taking `exp` and `ln` to require infinite series, because you aren't taking his operator as primitive.