# Reciprocal arithmetic

This revised post follows up on reciprocal (or harmonic or parallel) addition mentioned in a previous post here.

Reciprocal arithmetic is based on an automorphism that interchanges the zero with the infinite and the greater-than-one with the less-than-one: 0 ↔ ∞ and x ↔ 1/x. So zero becomes the new inaccessible number and infinity becomes the new additive unit. That is,

xy := g−1(g(x) + g(y))

xy := g−1(g(x) ∙ g(y))

etc., where g(x) = 1/x with x ≠ 0.

The identity element is ∞ if it is included, so that x ⊡ ∞ = x and x ⊡ ∞ = ∞. The additive inverse of x is −x, and the multiplicative inverse is x−1.

Reciprocal arithmetic is isomorphic to the ordinary numerator operations exchanged with their denominator counterparts. It is like counting down from a maximum, in which an increment of one reduces the amount slightly.

Reciprocal addition (also known as parallel addition) is defined as a power operation:

$O_{-1}(a_{1},a_{2},...,a_{n})\equiv&space;\left&space;(&space;\sum_{i=1}^{n}a_{i}^{-1}&space;\right&space;)^{-1}&space;,\,\,&space;a_{i}&space;\neq&space;0$

with the understanding that (1/0) → ∞ and (1/∞) → 0. An intermediate value may be zero but not a final value.

Simple reciprocal addition is thus defined as:

$x&space;\boxplus&space;y&space;=&space;\left&space;(&space;\frac{1}{x}&space;+&space;\frac{1}{y}&space;\right&space;)^{-1}&space;\equiv&space;\frac{xy}{x+y};&space;\,\,&space;x,&space;y&space;\neq&space;0$

The reciprocal additive unit is infinity instead of zero: x ⊞ ∞ = ∞. Reciprocal increment is:

$x&space;\boxplus&space;y&space;=&space;\left&space;(&space;\frac{1}{x}&space;+&space;1&space;\right&space;)^{-1}&space;\equiv&space;\frac{x}{x+1};&space;\,\,&space;x&space;\neq&space;0$

Reciprocal addition is commutative, associative, and distributes with multiplication. See Kent E. Erickson, “A New Operation for Analyzing Series-Paralled Networks,” IRE Trans. on Circuit Theory, March 1959, pp.124-126, See also here and here.

Reciprocal subtraction is defined as:

$x&space;\boxminus&space;y&space;=&space;x&space;\boxplus&space;(-y)&space;=&space;\left&space;(&space;\frac{1}{x}&space;-&space;\frac{1}{y}&space;\right&space;)^{-1}&space;\equiv&space;\frac{xy}{y-x};&space;\;&space;x,y\neq&space;0$

$x\boxminus&space;y&space;>0&space;\Leftrightarrow&space;y>x;\;&space;x\boxminus&space;y&space;<0&space;\Leftrightarrow&space;y

Reciprocal multiplication is defined as:

$x&space;\boxtimes&space;y&space;=&space;\left&space;(&space;\frac{1}{x}&space;\times&space;\frac{1}{y}&space;\right&space;)^{-1}&space;\equiv&space;xy;&space;\,\,&space;x,&space;y&space;\neq&space;0$

Reciprocal division is defined then as:

$x&space;\:&space;\Box&space;\:&space;y&space;=&space;\left&space;(&space;\frac{1}{x}&space;\div&space;\frac{1}{y}&space;\right&space;)^{-1}&space;\equiv&space;\frac{x}{y};&space;\,\,&space;x,&space;y&space;\neq&space;0$

Reciprocal exponentiation is defined as:

$x&space;\curlywedge&space;y&space;=&space;\left(\frac{1}{x}&space;\right)^{(-(\frac{1}{y})^{-1})}&space;\equiv&space;\left(&space;\frac{1}{x}&space;\right)&space;^{-y};&space;\,\,&space;x,&space;y&space;\neq&space;0$