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,

x ⊞ y := g⁻¹(g(x) + g(y))

x ⊡ y := 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:

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

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:

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

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

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

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:

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

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

Reciprocal multiplication is defined as:

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

Reciprocal division is defined then as:

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

Reciprocal exponentiation is defined as:

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

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