This posting has been Edited. The edited material shall be noted.


[The projectively extended real line](https://en.wikipedia.org/wiki/Projectively_extended_real_line) $\hat {\mathbb R}= \mathbb R \cup \{\infty\}$ is one system which allows division by zero! Yet it has many [undefined](https://en.wikipedia.org/wiki/Projectively_extended_real_line#Arithmetic_operations_that_are_left_undefined) arithmetic operations.


If we weaken the requirement of the operators being binary *functions* to being ternary ***relations***, and so can relate many values to the same arguments; then those undefined operations can be settled in the sense of getting some kind of closure under them.


I'll adopt the notation "$a \* b \to c$" to mean: the operator $\*$ is a ternary relation that sends the ordered pair $(a,b)$ to $c$; for convience, it can be read as: $c$ is a result of $a \* b$.


Now we may coin a closure notion over $\hat {\mathbb R}$ under an operator $\*$ as:


$ \hat {\mathbb R} \text { closed under } \* \iff \forall x,y \in \hat{\mathbb R} \exists z \in \hat {\mathbb R}: x \* y \to z$


I'll revert to the functional expression (i.e. $x \* y = z$) only when $z$ is unique per $(x,y)$


The following is a possible extension:


$\infty + \infty = \infty \\ \infty - \infty \to r \\ \infty \times 0 \to r \\ 0 \times \infty \to r \\ \infty / \infty \to r \\ 0 /0 \to r$


Where $r$ is any element of $\hat{\mathbb R}$.



> 
> [EDIT]: the above was the original suggestion, which was motivated by defining subtraction and division as the reciprocal relation of addition and multiplication respectively, i.e.:
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> 
> $a - b \to c \iff c + b \to a \\ a / b \to c \iff c \times b \to a$
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> 
> But, ($\infty + \infty \to \infty$) doesn't conform to that! We have $r - \infty = \infty$ so this results in $\infty + \infty 
> \to r$, so this to be corrected to ($\infty + \infty \to r$); for all $r \in \hat{\mathbb R}$.
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> 
> Additional motivation is to have multiplication by zero be equivalent with self subtraction.
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> 
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Accordingly $\hat{\mathbb R}$ would be closed under operators "$+,-,\times, /$" in the sense defined here.


Perhaps a similar move can be done for the Riemanian sphere $\mathbb C^\*$, to gain closure under more operators.



> 
> Is this system consistent relative to $\sf ZFC$? More specifically, what are the particulars of what could be viewed to constitute somehwat natural interpretation of this system in $\sf ZFC$?
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> 
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Is this extension of the projectively extended real line, consistent?
This is an account on the particulars of an interpretation of the original system [before the edit] presented in this question in set theory.


First we define an extended kind of rationals to suit adding a rational that is higher than all other rationals, the latter would correspond to $\infty$. The set of all those rationals shall be denoted by $\mathbb Q^\*$


$\text {Define}: r \in \mathbb Q^\* \iff r \subseteq \mathbb R \times \mathbb R \land [\operatorname {image}(r) = \mathbb R \lor \operatorname {preimage}(r) = \mathbb R] \land \\\exists a,b \in \mathbb Z : r= \{\langle a \times x, b \times x \rangle \mid x \in \mathbb R \}$


In the above definition $r$ is meant to represent the rational number $a/b$ for integers $a,b$.


A strict smaller than relation $<$ can be defined on $\mathbb Q^\*$ as:


$r < s \iff \\ r \neq s \land \exists x \exists a \, \exists b \,( a < b \land \langle a, x \rangle \in r \land \langle b, x \rangle \in s)$


As usual elements of $\mathbb Q^\*$ strictly below $0/1$ are negatively signed, those above except $\infty$ are positively signed, while the rest (i.e.; $0, \infty$) are unsigned, this can also be captured in terms of sets as the signed rationals being those which have both their images and preimages being $\mathbb R$.


So we define: $\infty = \{ \langle x,0 \rangle \mid x \in \mathbb R \}$, i.e. the reciprocal of $0/1 = \{\langle 0,x\rangle \mid x \in \mathbb R \}$


**Addition** of any two extended rationals $r,s$ is given by:


$r + s =\{\langle a+b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\}$


This gives: $r + \infty = \infty + r= \infty$ for all $r \in \mathbb Q$


**Subtraction over** $\mathbb Q^\*$:


$r - s \to q \iff \\ q+s = r \lor q= \{\langle a-b,c \rangle \mid \langle a,c \rangle \in r\land \langle b,c \rangle \in s\} $


This gives: $r - \infty = \infty-r = \infty$, for all $r \neq \infty$


and $ \infty - \infty \to r $, forall $r \in \mathbb Q^\*$.


**Multiplication over** $\mathbb Q^\*$:


$r \neq 0, s \neq 0 \\ r \times s = \{\langle k \times x,a \rangle , \langle t \times h, m^2 \rangle \mid (\langle t,m \rangle \in r [s] \land \langle h,m \rangle \in s [r]) \land (\langle k,1 \rangle \in r [s] \land \langle x,a \rangle \in s [r]) \} $


Where: $(z \in r[s] \land u \in s[r])$, is short for: $(z \in r \land u \in s) \lor (z \in s \land u \in r)$


if $r =0 \lor s=0 \implies r \times s \to (r-r)+(s-s) $


So, we get: $r \times \infty = \infty \times r= \infty$, for all $r \neq 0$


and, $0 \times \infty \to r; \infty \times 0 \to r $, forall $r \in \mathbb Q^\*$


Now, that we defined addition and multiplication of the extended rationals, we can define extended reals as Dedekind cuts over $\mathbb Q^\*$


A Dedekind cut shall be defined here as a binary partition on $\mathbb Q^\*$, with one block being an initial segment (i.e.; closed under $<$) of $\mathbb Q^\*$ that is open upwardly. So, for example $\{\mathbb Q^\* \setminus \{\infty\}, \{\infty\}\}$ is a Dedekind cut, and it is taken to represent the real number that corresponds to the extended rational number $\infty$, and so it'll be denoted by "$[\infty]$". The elements of a Dedekind cut are to be termed *initial*, *terminal* abbreviated as *init,term*, the former is the one closed dowardly, the latter is the one closed upwardly.


We can define a total order $<$ on Dedekind cuts themselves, this is given by:


$K < L \iff \operatorname {init}(K) \subsetneq \operatorname {init}(L)$


That said, then clearly $[\infty]$ is strictly greater than all other cuts. Formally this is: $$ r \neq [\infty] \implies r < [\infty]$$, for every extended real $r$.


The set of extended reals to be designated by $\hat {\mathbb R}$


As a terminology if $S$ and $C$ are nonempty subsets of $\mathbb Q^\*$, then:


$S \ \* \ C = \{a \ \* \ b \mid a \in S \land b \in C \} $


Where "$\*$" is some arithmetic operator.


Define: $X=\{S,-\} \iff X=\{S, \mathbb Q^\* \setminus S \}$


**Addition of extended reals:**


$K+L = \{ \operatorname {init}(K) + \operatorname {init}(L), - \} $


This yields: $r + [\infty] = [\infty]$, for all $r \in \hat{\mathbb R}$


**Subtraction of extended reals:**


$ K-L \to X \iff \\ X + L=K \lor \\K < [\infty] \land L=[\infty] \land X=[\infty]$


Another definition is:
$K-L \to X \iff \\ X + L=K \lor \\ X = \{\operatorname {init} (K) - \{x \in \operatorname {term}(L) \mid L \neq [\infty] \to x \neq \infty \}, - \}$


This gives: $ r - [\infty]= [\infty]-r = [\infty] $, for all $r \neq [\infty]$,


and: $[\infty] -[\infty] \to r $, for all $ r \in \hat {\mathbb R}$.


Define: $\operatorname {Comp} (S) = S \cup \{\operatorname {Inf}(S)\} $


$\operatorname {Comp}$ is read as the *completion* set of.


**Multiplication of extended reals:**


$K > 0 \land L > 0: \\K \times L= \{\operatorname {Comp} (\operatorname {term}(K) \times \operatorname {term}(L)) , -\}$


$ K > 0 \land L < 0: K \times L = 0 - (K \times (0 - L))$


$ K < 0 \land L < 0: K \times L = (0-K) \times (0-L)$


$ K = 0 \lor L = 0:\\ K \times L \to X \iff (K-K) + (L-L) \to X$


Accordingly: $r \times [\infty] = [\infty]$, if $r \neq 0$;


and: $0 \times [\infty] \to r, [\infty] \times 0 \to r $; for all $r \in \hat {\mathbb R}$


**Division of extended reals:**


$ K/L \to X \iff X \times L \to K $


Yielding: $[\infty]/r = [\infty], r/[\infty]=0$, for all $r \neq [\infty] $;


and: $r/ 0 = [\infty], [\infty]/[\infty] \to r, 0/0 \to r $, for all $r \in \hat{\mathbb R}$.


This would establish the interpretation of the projectively extended real line in $\sf ZFC$, and also provides an answer to the undetermined expressions mentioned in the Wikipedia page, along the lines mentioned in this question. So, it proves the consistency of this system.


