# Speculative- Associativity and Information Loss

In Axler's Linear Algebra Done Right, the following problem (1.B.6) was posed:

Let $$\infty$$ and $$-\infty$$ denote two distinct objects, neither of which is in $$\mathbb{R}$$. Define an addition and scalar multiplication on $$\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$$ as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for $$t \in \mathbb{R}$$ define \begin{align*} & t \infty = \begin{cases} -\infty & \text{ if t < 0} \\ 0 & \text{ if t = 0} \\ \infty & \text{ if t > 0} \end{cases} \; \; \; \; \; t(-\infty) = \begin{cases} \infty & \text{ if t < 0} \\ 0 & \text{ if t = 0} \\ -\infty & \text{ if t > 0} \end{cases} \\ & t + \infty = \infty + t = \infty, \; \; \; \; \; t + (-\infty) = (-\infty) + t = -\infty, \\ & \infty + \infty = \infty, \; \; \; \; (-\infty) + (-\infty) = -\infty, \; \; \; \; \infty + (-\infty) = 0. \end{align*} Is $$\mathbb{R} \cup \{\infty\} \cup \{-\infty\}$$ a vector space over $$\mathbb{R}$$? Explain.

Now, it came out that this is not a vector space because it fails on the associative and distributive properties. And, looking into why it is that associativity fails, I wondered whether associativity is a property that stipulates the retention of previous operations in some basic sense. Here is what I mean.

Take $$x$$, $$y$$, and $$z$$ to be elements of this set where $$x$$ $$\in$$ $$\mathbb{R}$$, $$y = ∞$$ and $$z = −∞$$. The one grouping leaves $$x$$ as the result of the addition, and the other leaves $$0$$ as the result. The discrepancy, from what I can see, is due to the fact that addition with either $$∞$$ or $$−∞$$ (on one hand) and a real number (on the other) completely "loses the information" that a real number was involved in the operation. Perhaps this type of information loss is allowed only for the additive identity?

So, I understand that one way to frame associativity is that grouping should not matter, but in what ways is it also a stipulation that no information loss must occur? Are those the same? What is wrong with looking at it this way?

We know that $$ab=a$$, implies $$b=e_{G}$$ in group $$G$$ (Cancellation law). Now as you mentioned $$\infty +t = \infty$$, which implies that $$t=0$$ for any $$t\in \mathbb{R}$$. Impossible! You’re absolutely right we lose some information in addition if we talk a a little bit friendly.