**9. Final Remarks**

Non-Newtonian calculus, and the non-Diophantine arithmetics behind it, are as simple as the undergraduate arithmetic and calculus we were taught at schools. Their conceptual potential is immense but they remain largely unexplored and unappreciated. Apparently, physicists in general do not feel any need of going beyond standard Diophantine arithmetic operations, in spite of the fact that the two greatest revolutions of the 20th century physics were, in their essence, arithmetic (i.e., relativistic addition of velocities and quantum mechanical addition of probabilities). It is thus intriguing that two of the most controversial issues of modern science—dark energy and Bell's theorem—reveal new aspects when reformulated in generalized arithmetic terms.

One should not be surprised that those who study generalizations of Boltzmann–Gibbs statistics are naturally more inclined to accept non-aprioric rules of physical arithmetic. Anyway, the very concept of non-extensivity, the core of many studies on generalized entropies, is implicitly linked with generalized forms of addition, multiplication, and differentiation [54,59–61].

**Funding:** This research received no external funding.

**Acknowledgments:** I am indebted to Jan Naudts for comments.

**Conflicts of Interest:** The author declares no conflict of interest.

#### **Appendix A. Proof of (39)**

Let us treat this as an exercise in non-Newtonian calculus. Begin with the three diagrams

$$\begin{array}{ccccc} \mathbb{X} & \stackrel{A}{\longrightarrow} & \mathbb{Y} & \mathbb{X} & \stackrel{B}{\longrightarrow} & \mathbb{Y} & \mathbb{X} & \stackrel{A\oplus\chi B}{\longrightarrow} \mathbb{Y} \\ f\mathbf{x} & & & \Big\downarrow f\mathbf{v} & \Big\downarrow & f\mathbf{x} & & \Big\downarrow f\mathbf{v} & \cdot & f\mathbf{x} \\ \mathbb{R} & \stackrel{A}{\longrightarrow} & \mathbb{R} & & \mathbb{R} & \stackrel{B}{\longrightarrow} & \mathbb{R} & \mathbb{R} & \stackrel{A+\emptyset}{\longrightarrow} \mathbb{R} \end{array} \tag{A1}$$

Indeed,

$$\begin{array}{rcl} A \oplus\_{\mathbb{Y}} B(\mathbf{x}) = A(\mathbf{x}) \oplus\_{\mathbb{Y}} B(\mathbf{x}) &=& f\_{\mathbb{Y}}^{-1} \left( f\_{\mathbb{Y}} \circ A(\mathbf{x}) + f\_{\mathbb{Y}} \circ B(\mathbf{x}) \right) = f\_{\mathbb{Y}}^{-1} \left( \vec{A} \circ f\_{\mathbb{X}}(\mathbf{x}) + \vec{B} \circ f\_{\mathbb{X}}(\mathbf{x}) \right) \\ &= f\_{\mathbb{Y}}^{-1} \left( (\vec{A} + \vec{B}) \circ f\_{\mathbb{X}}(\mathbf{x}) \right) \end{array} \tag{A2}$$

By the definition (17) of the limit,

$$\begin{split} \lim\_{\mathbf{x'} \to \mathbf{x}} A(\mathbf{x'}) \oplus\_{\mathbf{Y}} B(\mathbf{x'}) &= \lim\_{\mathbf{x'} \to \mathbf{x}} (A \oplus\_{\mathbf{Y}} B)(\mathbf{x'}) = f\_{\mathbf{Y}}^{-1} \left( \lim\_{r \to f\_{\mathbf{X}}(\mathbf{x})} (A \oplus\_{\mathbf{Y}} B)(r) \right) \\ &= f\_{\mathbf{Y}}^{-1} \left( f\_{\mathbf{Y}} \circ f\_{\mathbf{Y}}^{-1} \left( \lim\_{r \to f\_{\mathbf{X}}(\mathbf{x})} \bar{A}(r) \right) + f\_{\mathbf{Y}} \circ f\_{\mathbf{Y}}^{-1} \left( \lim\_{r \to f\_{\mathbf{X}}(\mathbf{x})} B(r) \right) \right) \\ &= f\_{\mathbf{Y}}^{-1} \left( \lim\_{r \to f\_{\mathbf{X}}(\mathbf{x})} \bar{A}(r) \right) \oplus\_{\mathbf{Y}} f\_{\mathbf{Y}}^{-1} \left( \lim\_{r \to f\_{\mathbf{X}}(\mathbf{x})} \bar{B}(r) \right) \\ &= \left( \lim\_{\mathbf{x'} \to \mathbf{x}} A(\mathbf{x'}) \right) \oplus\_{\mathbf{Y}} \left( \lim\_{\mathbf{x'} \to \mathbf{x}} B(\mathbf{x'}) \right). \end{split} \tag{A3}$$

#### **Appendix B. Proof of Lemma 1**

Ref. (84) may be regarded as a definition of *h*. If *h*(−*x*) = −*h*(*x*) then

$$\begin{array}{rcl} \lg(1-p) + \lg(p) & = & \frac{1}{2} + h\left(1 - p - \frac{1}{2}\right) + \frac{1}{2} + h\left(p - \frac{1}{2}\right) \\ & & \cdot & ./\ 1 & \cdot & ./ & 1 \end{array} \tag{A4}$$

$$l = \left\lfloor 1 + h\left(\frac{1}{2} - p\right) + h\left(p - \frac{1}{2}\right) \right\rfloor \tag{A5}$$

$$=\quad 1 - h\left(p - \frac{1}{2}\right) + h\left(p - \frac{1}{2}\right) = 1\tag{A6}$$

Now let *g*(1 − *p*) + *g*(*p*) = 1. Then

$$\mathbf{1}\_{\!\!\!-1} = \underset{\!\!\!-1}{\operatorname{g}}(\mathbf{1} - \mathbf{p}) + \underset{\!\!\!-1}{\operatorname{g}}(\mathbf{p}) \tag{A7}$$

$$=\frac{1}{2} + h\left(1 - p - \frac{1}{2}\right) + \frac{1}{2} + h\left(p - \frac{1}{2}\right) \tag{A8}$$

$$=\quad 1+h\left(\frac{1}{2}-p\right)+h\left(p-\frac{1}{2}\right).\tag{A9}$$

Denoting *x* = *p* − 1/2 we find *h*(−*x*) = −*h*(*x*).

### **Appendix C. Proof of Lemma 2**

*g*(*p*1) + ··· + *g*(*pn*) = 1 must hold for any choice of probabilities. Setting *p*<sup>1</sup> = *p*, *p*<sup>2</sup> = 1 − *p*, we find

$$\mathbf{g}(p) + \mathbf{g}(1-p) + (n-2)\mathbf{g}(0) = 1,\tag{A10}$$

If *g*(0) = 0 then, by Lemma 1, *g*(*p*) = 1/2 + *h*(*p* − 1/2), with antisymmetric *h*. Returning to arbitrary *pk*, we get

$$11 = \frac{n}{2} + \sum\_{k=1}^{n-1} h\left(p\_k - \frac{1}{2}\right) + h\left(1 - \sum\_{k=1}^{n-1} p\_k - \frac{1}{2}\right). \tag{A11}$$

By antisymmetry of *h*,

$$1 - \frac{n}{2} - \sum\_{k=2}^{n-1} h\left(p\_k - \frac{1}{2}\right) = h\left(p\_1 - \frac{1}{2}\right) - h\left(p\_1 - \frac{1}{2} + \sum\_{k=2}^{n-1} p\_k\right),\tag{A12}$$

which implies that the right-hand side of (A12) is independent of *p*<sup>1</sup> for any choice of *p*2, ... , *pn*−1. In other words, the difference *h*(*x*) − *h*(*x* + *p*) is independent of *x* for any 0 ≤ *p* ≤ 1/2 − *x*, so *h*(*x*) = *ax*. *g*(0) = 0 implies *h*(1/2) = 1/2, *a* = 1, and *g*(*p*) = *p* for any *p*.

Now let *g*(0) > 0. Normalization

$$\lg(1) + (n-1)\lg(0) = 1\tag{A13}$$

combined with (A10), imply

$$
\lg(p) + \lg(1-p) = \lg(0) + \lg(1) > 0. \tag{A14}
$$

Accordingly, *G*(*p*) = *g*(*p*)/(*g*(0) + *g*(1)) satisfies *G*(*p*) + *G*(1 − *p*) = 1, so that

$$G(p) = \frac{1}{2} + H\left(p - \frac{1}{2}\right),\tag{A15}$$

where *H*(−*x*) = −*H*(*x*). Returning to

$$\mathcal{g}(p) = \left(\mathcal{g}(0) + \mathcal{g}(1)\right) \left[\frac{1}{2} + H\left(p - \frac{1}{2}\right)\right],\tag{A16}$$

we find

$$\frac{1}{g(0) + g(1)} \quad = \frac{n}{2} + \sum\_{k=1}^{n-1} H\left(p\_k - \frac{1}{2}\right) + H\left(1 - \sum\_{k=1}^{n-1} p\_k - \frac{1}{2}\right). \tag{A17}$$

and *H*(*x*) = *ax* by the same argument as before. Now,

$$\mathbf{g}(p) = \left(\mathbf{g}(0) + \mathbf{g}(1)\right) \frac{1 - a + 2ap}{2} \tag{A18}$$

Summing over all the probabilities,

$$1 = \sum\_{k=1}^{n} \lg(p\_k) = \left(\lg(0) + \lg(1)\right) \frac{n - an + 2a}{2},\tag{A19}$$

we get

$$\lg(p) \quad = \quad \frac{1 - a + 2ap}{n + (2 - n)a} \, \text{(A20)}$$

$$\lg(0) = \begin{array}{c c} 1-a \\ \hline n + (2-n)a' \end{array} \tag{A21}$$

$$\mathcal{g}(1) \;=\; \frac{1+a}{n+(2-n)a}.\tag{A22}$$

For *a* = 1 we reconstruct the case *g*(0) = 0, *g*(*p*) = *p*. *g*(0) > 0 and *g*(1) ≥ 0 imply either

$$1 - a > 0, \quad 1 + a \ge 0, \quad n + (2 - n)a > 0,\tag{A23}$$

or

$$1 - a < 0, \quad 1 + a \le 0, \quad n + (2 - n)a < 0,\tag{A24}$$

but (A24) is inconsistent. The first two inequalities of (A23) imply −1 ≤ *a* < 1, but then *n* + (2 − *n*)*a* > 0 is fulfilled automatically for *n* ≥ 3. Non-negativity of *g*(*p*) requires 0 ≤ 1 − *a* + 2*ap* for all 0 ≤ *p* ≤ 1. For positive *a* the affine function *p* → 1 − *a* + 2*ap* is minimal at *p* = 0, implying 0 < *a* ≤ 1. For negative *a* the map *p* → 1 − *a* + 2*ap* is minimal at *p* = 1, so −1 ≤ *a* < 0. Finally, −1 ≤ *a* ≤ 1 covers all the cases. The case *a* = 0 implies *g*(*pk*) = 1/*n*, which is possible, but uninteresting for non-Newtonian applications as such a *g* is not one-to-one.
