If the value of something changes from $\;a\;$ to $\;b\;$, their relative difference can be expressed as a percentage:
$$
\newcommand{\upto}{\mathop\nearrow}
(D0) \;\;\; a \upto b \;=\; (b-a)/a \color{gray}{\times 100\%}
$$
(In this question $\;a\;$, $\;b\;$, and $\;c\;$ are positive (non-zero) real numbers throughout, and $\;a \upto b\;$ is a real number.)
However, percentages are not anti-symmetrical, i.e., the above definition does not satisfy
$$
(P0) \;\;\; a \upto b \;=\; -(b \upto a)
$$
This could be fixed by instead defining
$$
(D1) \;\;\; a \upto b \;=\; (b-a)/(b+a)
$$
(By the way, does this difference measure have a standard name?)
However, neither of the above definitions add up: they don't satisfy the more general property
$$
(P1) \;\;\; a \upto b \;+\; b \upto c \;=\; a \upto c
$$
After some attempts I discovered that
$$
(D2) \;\;\; a \upto b \;=\; \log_q (b/a)
$$
(for an arbitrary base $\;q > 1\;$) does satisfy $(P1)$. And it also satisfies the following other properties of percentage $(D0)$ and of $(D1)$ which seem desirable for any relative difference:
\begin{align}
(P2) \;\; & a \upto a \;=\; 0 \\
(P3) \;\; & \textit{sgn}(a \upto b) \;=\; \textit{sgn}(b - a) \\
(P4) \;\; & (s \times a) \upto (s \times b) \;=\; a \upto b \text{ for all $\;s > 0\;$} \\
(P5) \;\; & a \upto b \to M \text{ for } a \to 0 \text{ for some $\;M\;$ (which may be $\;\infty\;$)} \\
(P6) \;\; & a \upto b \to N \text{ for } b \to \infty \text{ for some $\;N\;$ (which may be $\;\infty\;$)} \\
\end{align}
(I mention $(P2)$ only for completeness; it follows of course from $(P0)$.)
My question: Apart from $(D2)$, what other definitions (if any) satisfy all of the above properties?
Answer
The general (real-valued) solution of axioms P1 and P2 is to take a function $F$ on the set $V$ of values that $a,b,\cdots$ could possibly assume, and define $a \upto b = F(b) - F(a)$.
If $F$ is injective, the converse of P2 is satisfied: $a \upto b = 0$ only for $a=b$.
If $F$ is increasing, P3 is satisfied.
If $F$ is continuous, P5 and P6 are satisfied. (Assume here that the set of values $V$ is an interval, or put the order topology on the set.)
This shows that there is a large family of solutions of the axioms, one for every increasing continuous function on $V$, if you do not assume homogeneity, P4. Homogeneity is a strong requirement that cuts down the space of continuous solutions to the logarithmic ones stated in the question.
Homogeneity of $a \upto b$ is the functional equation $F(sb) - F(sa) = F(b) - F(a)$. Solutions $F$ and $F+c$ are equivalent for constant $c$, and we can assume $F(1)=0$ by adjusting the constant. Taking $a=1$ this is $F(sb)=F(s)+F(b)$ and all continuous solutions (on intervals) are well-known to be multiples of the logarithm.
Assuming that you meant to work with open intervals of values, and $a \upto x$ to be an increasing continuous function of $x$, the not necessarily homogeneous solutions correspond to one parameter groups of homeomorphisms of the interval.
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