# Does this algebra have a name?

I thought of viewing $$\bar{\mathbb{R}}$$ as a relative subalgebra of a total algebra $$\mathbb{R}^\ast := \bar{\mathbb{R}} \cup \{\ast\}$$, wherein an output of an operation, if not already defined in $$\bar{\mathbb{R}}$$, is set equal to $$\ast$$. In a sense, $$\ast$$ acts as the definition for those extended real sums or products that were previously undefined, while also absorbing everything it touches algebraically. In this construction $$\ast$$ is not equal to any extended real number.

I'm not claiming this is an original idea. I think this is a typical approach of viewing a partial algebra as a relative subalgebra of a total algebra. Nonetheless, I liked it. Before I get into why, my question is, does the total algebra I'm calling $$\mathbb{R}^\ast$$ have a name?

Definitions... Addition in $$\bar{\mathbb{R}}$$ is defined for all extended real pairs except $$(\pm\infty,\mp\infty)$$. By keeping these definitions, and further defining $$\pm\infty + \mp\infty = \ast$$ and also $$a+b = \ast$$ whenever $$\ast \in \{a,b\}$$, we obtain the definition of addition on $$\mathbb{R}^\ast$$. Similarly, multiplication in $$\bar{\mathbb{R}}$$ is defined for all extended real pairs (I let $$\pm\infty\cdot 0 = 0$$), so by further defining $$ab = \ast$$ whenever $$\ast \in \{a,b\}$$ we get the definition of multiplication in $$\mathbb{R}^\ast$$. To extend negation we just let $$- \ast = \ast$$.

I even went so far as to give $$\ast$$ the name "undefined", so I could use the phrases "$$a$$ is defined" and "$$a$$ is undefined" to respectively mean that $$a \neq \ast$$ and $$a = \ast$$. For instance, to say the sum $$a+b$$ is undefined is to say $$a+b = \ast$$.

It turns out that addition and multiplication in $$\mathbb{R}^\ast$$ are associative and commutative, $$0$$ and $$1$$ act as additive and multiplicative identities, respectively, and negation is an involution. Moreover, $$\mathbb{R}^\ast$$ is real-cancellative in the sense that $$x +(-x) = 0$$ only holds for real $$x$$.

Distributivity is a little more delicate. We'll say that a sum $$b+c$$ is heterogeneous in case $$(b,c) \;\; \in \;\; [-\infty,0)\times(0,+\infty] \;\;\; \cup \;\;\;(0,+\infty] \times [-\infty,0).$$ So a defined heterogeneous sum is one that is heterogeneous and equal to an extended real number, whereas an undefined heterogeneous sum is one that is heterogeneous and equal to $$\ast$$. For any $$a,b,c \in \mathbb{R}^\ast$$, $$a(b+c) = ab + ac$$ so long as $$a \neq \pm\infty$$ when $$b+c$$ is a defined heterogeneous sum, and $$a \neq 0$$ when $$b+c$$ is an undefined heterogeneous sum.

Why I like it... I know there is nothing special about $$\mathbb{R}^\ast$$. I'm just using new symbols to say old things. I guess this is the way math has always gone. But it does make it easier for me. I'll only ever care about algebraic expressions whose terms are in $$\bar{\mathbb{R}}$$, but it's nice to know I don't have to worry about checking for infinities in said expressions - if they don't evaluate to an extended real number, they'll at least evaluate to something ($$\ast$$).

There is similar ease in considering algebraic expressions involving extended real functions on some domain $$X$$. For extended real functions $$e,f,g,h$$, it may be that $$d = f(g+h) + e$$ is not extended-real for some $$x$$, but that is no problem; $$d(x) = \ast$$ for such $$x$$, and $$d$$ at least exists in some setting. You can also use indicator-like functions $$\ast_E(x) = \begin{cases}\ast, & x \in E, \\ 0, & x \notin E \end{cases}$$ to adjust expressions, and even call these functions "adjustors". For instance, for any functions $$f,g,h : X \to \mathbb{R}^\ast$$, $$h(f+g) + \ast_A = hf+hg + \ast_B$$ and $$f + (-f) = \ast_C$$ where $$A = \{x \in X: \,h(x) = \pm\infty \;\text{and} \; f(x) + g(x) \; \text{defined heterogenous sum}\}$$ $$B = \{x \in X: \,h(x) =0 \;\text{and} \; f(x) + g(x) \; \text{undefined heterogenous sum}\}$$ $$C = \{x \in X \;: \; f(x) \; \text{not real}\}.$$ Thus you could say things like $$\mathbb{R}^\ast$$ function spaces are adjusted-cancellative and adjusted-distributive.

But I ask again, does $$\mathbb{R}^\ast$$ have a name? Visually, I picture $$\mathbb{R}^\ast$$ a coproduct of topological spaces: a closed line segment with $$\ast$$ hovering off to the side like a satellite. Is there anything like this construction for the complex numbers? Whereby the complex numbers are visualized as an open circle, and to close that circle a pair of antipodal "infinities" are defined as the endpoints of each diameter through the circle, so that each such infinity is distinct; and then hovering off to the side of the closed circle is the single point $$\ast$$. You could call this space $$\mathbb{C}^\ast$$ if you're comfortable doing so.

• Is not it trivial? What is $0\cdot*$ and $*^0$? Commented Sep 9, 2023 at 2:02
• I think one person called your $*$ "nullity", though their presentation wasn't well-received. Commented Sep 9, 2023 at 2:04
• @MarkS. in a wheel $0/0=\bot$ is a defined value, this is standard terminology. en.wikipedia.org/wiki/Wheel_theory Commented Sep 9, 2023 at 2:06
• To be clear, the purpose of extending $\bar{\mathbb{R}}$ to $\bar{\mathbb{R}} \cup \{\ast\}$ is in no way to "make sense" of $\infty - \infty$. I don't need it to make sense. The only reason for the extension is to rid myself of the frustration of checking cases to see when expressions involving extended real numbers do or don't evaluate to extended real numbers. I don't want to be afraid of writing something like $(a+b)c -d +ef$ just because I don't know beforehand that it is defined in $\bar{\mathbb{R}}$. I'd like to write it down first, manipulate it a bit, then see if it was defined.
– joeb
Commented Sep 9, 2023 at 2:22
• $0 \cdot \ast = \ast$. That may not be nice, but I wanted to make sure $\ast$ absorbs everything it touches.
– joeb
Commented Sep 9, 2023 at 2:25

In that algebraic system, $$0/0=\bot$$ is called "nullity". It looks like all your rules are satisfied.
Particularly, $$\infty-\infty=\bot$$, $$0\bot=\bot$$.