Infinity is not a number , thus we cannot perform the usual arithmetic operations that we do with real numbers

This is the usual reason given when asked why we can't perform the usual arithmetic operations with infinities. However the terminology used in these types of arguments are very loose and aren't well defined mathematically. For example, there are multiple types of infinities.

I will attempt to formalize this question in the context of Real Analysis below.

Let's assume for the purposes of this question that we are only dealing with the Real Field, $\mathbb{R}$, the set of Real Numbers (Scalars), satisfying the Axioms of Addition and Multiplication.

Correct me if I'm wrong, but infinities of all types are not elements of the Real Field ($\infty \not\in \mathbb{R}$), therefore they do not obey the Axioms of Addition and Multiplication (hence they do not obey the usual 'arithmetic operations', that Real Numbers do).

Some of the results of these operations (not limited to infinities) are not well defined, and leads us to the 7 Indeterminate Forms in Real Analysis:

  1. $\frac{0}{0}$
  2. $\frac{\infty}{\infty}$
  3. $0 \cdot \infty$
  4. $\infty - \infty$
  5. $0^{0}$
  6. $\infty^{0}$
  7. $1^{\infty}$

But why then are the following results true :

$$\text{Given} \ a \in \mathbb{R}$$ $$(+\infty)\cdot( -a) = -\infty $$ $$(-\infty)\cdot( -a) = +\infty$$ $$(+\infty)\cdot( +\infty) = +\infty$$ $$(-\infty)\cdot( +\infty) = -\infty$$ $$(-\infty)^{2k} = +\infty, \ \text{where} \ k\in \mathbb{Z} \geq 1$$

In these cases infinities behave the way we expect elements of the Real Field to, with regards to whether the product of the two multiplicands is either positive or negative. Why is that so?

I have framed this question within the context of Real Analysis, however if there are interesting observations from other areas such as Complex Analysis or Abstract Algebra regarding operations with infinity, I would be highly interested in reading about it.

  • 1
    $\begingroup$ Note that there is a distinction between undefined operations and indeterminate forms. For example, we have the indeterminate form $0^0$ (which just says that $a_n\to 0$ and $b_n\to 0$ does not tell you much about $a_n^{b_n}$) in spite of $0^0=1$. $\endgroup$ – Hagen von Eitzen May 18 '16 at 11:51

Strictly speaking we can define the symbols to work however we want. We choose to define the symbols to work that way because it matches up with the behavior of limits. For example, if $a>0$ and $f(x) \to \infty$ as $x \to c$ then $af(x) \to \infty$ as $x \to c$ while $-af(x) \to -\infty$ as $x \to c$. (Note that $f(x) \to \infty$ and similar expressions are written using the symbol $\infty$ but their definitions can be stated entirely in terms of real numbers.) We generally leave the "indeterminate forms" undefined because there is no way to define them that is consistent with all the ways that they would appear in limits. (One exception: some people define $0^0=1$ and $0 \log 0 = 0$, even though $f(x,y)=x^y$ and $g(x,y)=x \log y$ cannot be continuous at $(0,0)$.)

Also, you should make sure to understand that here we are talking about infinity in the sense of the extended real numbers. There are other notions: infinite cardinal numbers, infinite ordinal numbers, and complex infinity come to mind.


But why then are the following results true : (...)

Logic of these "results" can be explained the following way:

1) In addition to usual operations on $\mathbb{R}$ $+$ and $\cdot$ let's introduce symbols $+_L$ and $\cdot_L$, when $a +_L b = c$, by definition, is a short for statement "$\forall (a_n), (b_n), (a_n \to a, b_n \to b) \implies (\exists \lim(a_n+b_n) = c)$", the same for $\cdot_L$.

2) Note that for $a,b,c \in \mathbb{R}$, $a+b=c \iff a +_L b = c$. Thus, when we work in $\mathbb{R}$ only, we can use simply $+$ instead of $+_L$ without getting ambiguous results.

3) $+_L$ and $\cdot_L$ have meaning for some symbols other than elements of $\mathbb{R}$: for $a \in \mathbb{R}, a > 0$, $-\infty \cdot_L -a = +\infty$ is true. As per 2), we can write simply $(-\infty)\cdot(-a) = +\infty$.


Infinity can be a number, and then we can interpret the usual algebraic operations as applying to them. For example, to illustrate your first principle, for each positive infinite hyperreal $H$ and appreciable number $b<0$ we will have that $H \cdot b\;$ is an infinite negative hyperreal. The so-called "indeterminate forms" are also a lot less mysterious than they used to be: for example, the ratio $\frac{\epsilon}{\delta}$ of two nonzero infinitesimals could be infinitesimal, appreciable, or infinite depending on the specific infinitesimals chosen. This gives a clear meaning to the indeterminate form $\frac{0}{0}$.


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