## Types of non-associative arithmetics

on März 21, 2011 in General

When considering the free (additive) groupoid/magma that is generated by 1, we can define a multiplication inductively by:
1 * x = x
(a+b) * x = a*x + b*x
which gives us the plain non-associative arithmetic sometimes denoted by N.

Now we can impose equations on the groupoid. For example the free associative groupoid is the natural numbers (without 0) and the above multiplication is also definable there (and it is the multiplication on the natural numbers).

For any finite set of equations the above multiplication is still well defined, because of the right-distributivity of the multiplication.

This is satisfied by a number of well-known identities in quasigroup theory:

### Commutative Law x+y=y+x

The free monogenetic commutative groupoid can be embedded into the free monogenetic commutative loop.

### Associative Law (x+y)+z=x+(y+z)

The free monogenetic associative groupoid is the natural numbers, it can be embedded into the free associative loop which is the integer numbers.

### Left-Commutative or Left-Permutable Law x+(y+z)=y+(x+z)

The elements of the free left-commutative groupoid P can be represented as recursive multisets (two multisets [a1,...,am] and [b1,...,bn] are equal if m=n and (b1,…,bn) is a permutation of (a1,…,am))

[] is element of P and if a1,…,an are in P then also [a1,...,an] is element of P.
The operations can then be defined as
1=[]
a+[b1,...,bn] = [a,b1,...,bn]

We can not add a right neutral element 0 as this would imply commutativity
x+y=x+(y+0)=y+(x+0)=y+x

The free left-commutative groupoid can be embedded into the free left-commutative quasigroup with left-neutral element 0:

We extend the notation into
{±a1,…,±an}k
and define the operations as:

{}0 = 0 = 1-1
{}1 = [] = 1
{b1,…,bn}1=[b1,...,bn]

±a + {b1,…,bn}k = [±a,{b1,...,bn}(k-1)]
{a1,…,am}p – {b1,…,bn}q = {{a1,…,am}(p-1),-{b1,…,bn}(q-1)}

We gather elements in zero-braces {a1,…,am}0,{b1,…,bn}0 = {a1,…,am,b1,…,bn}0.
in each {a1,…,an}k we cancel two elements with opposite sign.
We gather nested braces by
{{b1,…,bn}k1}k2={b1,…,bn}(k1+k2)

Indeed 0 is not right neutral, for example 1+0=[1,{}(-1)].

An realization of P with function is given by starting with the identity function id(x)=x and then for two functions f, g also adding the function f^g.
Left inverse is the operation f^{1/g} and right inverse the operation ln(f)/ln(g).

[f1,...,fn] = x^(f1(x)…fn(x))
{f1,…,fn} = [f1,...,fn]-1 = log_x(x^{f1…fn})=f1…fn
0 = 1
1+0 = x^(x*log_x(1)) = 1 = 0
{1,{}(-1)} = {{}(-1)}={}(-1)
-1 + 1 = [-1] = x^(1/x)
0+

### (Bi)Symmetric or Medial or Entropic Law (x+y)+(z+w)=(x+z)+(y+w)

The medial law makes the multiplication commutative (!):
(a+b)*(x+y) = a*(x+y) + b*(x+y)
=(a*x + a*y) + (b*x+b*y) by induction & r-distributivity
=(a*x + b*x) + (a*y+b*y) medial law
=(x*a+x*b) + (y*a+y*b) by induction assumption
=x*(a+b) + y*(a+b) by induction and r-distributivity
= (x+y)*(a+b) by r-distributivity

The free monogenetic medial groupoid is neither left nor right cancellative, and hence can not be embedded into a quasigroup.  p. 175

### Commutative Medial Law

The free monogenetic commutative medial quasigroup can be represented by the polynomials (with allowed negative exponents) over the integers.

Addition is given by where x is the polynomial variable.
Accordingly Multiplication is the default multiplication of the polynomials.

Verify distributivity: ### idempotency x+x=x

This just yields the trivial groupoid {1}

### Stein-II-Identiy x+(y+x)=(y+x)+y

 J. Ježek, T. Kepka, Equational theories of medial groupoids, Algebra Universalis 17 (1983), 174-190.