# From ratios to quotients

Ratios and proportions are symmetric.

A:B ≡ B:A and A:B :: C:D iff C:D :: A:B.

But when ratios are converted to quotients or fractions, they are no longer symmetric. There must be a convention as to which is the denominator and which is the numerator.

In an ordinary fraction or quotient or rate the numerator is above the line (the vinculum) or left of the slash, and the denominator is below the line (the vinculum) or right of the slash. Additionally, in an ordinary quotient or rate the denominator is one.

However, there is also what may be called a reciprocal fraction or reciprocal quotient or reciprocal rate that is the converse: the numerator is below the line (the vinculum) or right of the slash, and the denominator is above the line (the vinculum) or left of the slash. Thus there are two numbers (or a single number with an implied number one) but their meaning may change.

For example, uniform motion is defined as

S1 : S2 :: T1 : T2.

But in reality either distance is measured over a given time or time is measured over a given distance.

Thus we may write

ΔS / ΔT = V, average speed.

But we may equally well write

ΔT / ΔS = W, average pace.

In general, if we look at motion functionally, we may write

ΔS / ΔT = S(ΔT).

And we may equally well write

ΔT / ΔS = T(ΔS).

These rates use ordinary algebra if their denominators are equal:

(ΔS1 / ΔT) + (ΔS2 / ΔT) = (ΔS1 + ΔS2) / ΔT

and

(ΔT1 / ΔS) + (ΔT2 / ΔS) = (ΔT1 + ΔT2) / ΔS

Their difference quotients also use ordinary algebra:

$\frac{\Delta&space;S(T)}{\Delta&space;T}&space;=&space;\frac{S(T+\Delta&space;T)-S(T)}{\Delta&space;T}$

and

$\frac{\Delta&space;T(S)}{\Delta&space;S}=&space;\frac{T(S+\Delta&space;S)-T(S)}{\Delta&space;S}$

For the derivatives take the limits to get

$V(T)=&space;S'(T)=\lim_{\Delta&space;T\rightarrow&space;0}&space;\frac{\Delta&space;S}{\Delta&space;T}=&space;\lim_{\Delta&space;T\rightarrow&space;0}&space;\frac{S(T+\Delta&space;T)-S(T)}{\Delta&space;T}$

and

$W(S)=&space;T'(S)=\lim_{\Delta&space;S\rightarrow&space;0}&space;\frac{\Delta&space;T}{\Delta&space;S}=&space;\lim_{\Delta&space;S\rightarrow&space;0}&space;\frac{T(S+\Delta&space;S)-T(S)}{\Delta&space;S}$

Addition of these derivatives is straightforward:

V1(T) + V2(T) = V3(T)

and

W1(S) + W2(S) = W3(S)

But there are two other options. We can have a speed with a given distance:

ΔS / ΔTn = 1 / (ΔTn / ΔS) = 1/W = V~, average inverse pace,

or a pace with a given time:

ΔT / ΔSn = 1 / (ΔSn / ΔT) = 1/V = W~, average inverse pace.

These rates use reciprocal arithmetic if their numerators are equal:

(ΔS / ΔT1) ⊞ (ΔS / ΔT2) = ((ΔT1 / ΔS) + (ΔT2 / ΔS))−1 = ((ΔT1 + ΔT2) / ΔS)−1 = ΔS / (ΔT1 + ΔT2)

and

(ΔT / ΔC ⊞ (ΔT / ΔS2) = ((ΔS1 / ΔT) + (ΔS2 / ΔT))−1 = ((ΔS1 + ΔS2) / ΔT)−1 = ΔT / (ΔS1 + ΔS2)

Their difference quotients use reciprocal arithmetic:

$\left&space;(\frac{\Delta&space;T(S)}{\Delta&space;S}&space;\right&space;)^{-1}=\frac{\Delta&space;S}{\Delta&space;T(S)}=&space;\frac{\Delta&space;S}{T(S+\Delta&space;S)-T(S)}$

and

$\left&space;(\frac{\Delta&space;S(T)}{\Delta&space;T}&space;\right&space;)^{-1}=\frac{\Delta&space;T}{\Delta&space;S(T)}=&space;\frac{\Delta&space;T}{S(T+\Delta&space;T)-S(T)}$

These lead to the reciprocal derivatives

$\tilde{V}(S)&space;=&space;\grave{}\,&space;T(S)&space;=&space;\left&space;(\lim_{\Delta&space;S\rightarrow&space;0}\frac{\Delta&space;T(S)}{\Delta&space;S}&space;\right&space;)^{-1}&space;=\left&space;(\lim_{\Delta&space;S\rightarrow&space;0}&space;\frac{T(S+\Delta&space;S)-T(S)}{\Delta&space;S}&space;\right&space;)^{-1}$

and

$\tilde{W}(T)&space;=&space;\grave{}\,&space;S(T)&space;=&space;\left&space;(\lim_{\Delta&space;T\rightarrow&space;0}\frac{\Delta&space;S(T)}{\Delta&space;T}&space;\right&space;)^{-1}&space;=\left&space;(\lim_{\Delta&space;T\rightarrow&space;0}&space;\frac{S(T+\Delta&space;T)-S(T)}{\Delta&space;T}&space;\right&space;)^{-1}$

$\tilde{V_{1}}(S)&space;\boxplus&space;\,\tilde{V_{2}}(S)=&space;(1/T_{1}(S)&space;+&space;1/T_{2}(S))^{-1}=(T_{1}(S)+T_{2}(S))^{-1}&space;=\tilde{V_{3}}(S)$
$\tilde{W_{1}}(T)&space;\boxplus&space;\,\tilde{W_{2}}(T)=&space;(1/S_{1}(T)&space;+&space;1/S_{2}(T))^{-1}=(S_{1}(T)+S_{2}(T))^{-1}&space;=\tilde{W_{3}}(T)$