Is $1/\infty=0{}$?

Question

If $1\div0=\infty$, then $1\div(1\div0)=0$. Does this mean $1\div\infty=0$? Or is it always close to zero?

Please let me know what is correct: either it's zero or always close to zero.

Answer 1

This will depend on what mathematical model you use to model $\infty$. Different models may give different answers depending on what, precisely, "$\infty$" means.

The model you are probably thinking of, however, is most likely one of the following two, which are very similar:

1. the extended real number line, $\bar{\mathbb{R}}$
2. the projective real number line $\hat{\mathbb{R}}$

Both are based on taking the usual real number line and adjoining either two or one, respectively, additional objects, which you could call as numbers, but perhaps better as ideal points: in the first case, $\infty$ (or $+\infty$) and $-\infty$, in the second, only $\infty$. These are explicit objects just like the numbers that we are inserting into the set, by the way, not just abuse of the symbol off of a limit. Then it is made as a definition that $\frac{1}{\infty} = 0$ in both systems (same for $-\infty$). For the first system, the following are defined:

$$a \infty = \infty, a \ne 0$$ $$\frac{a}{\pm \infty} = 0, a \in \mathbb{R}$$ $$a \pm \infty = \pm \infty, a \in \mathbb{R}$$ $$\infty + \infty = \infty$$ $$(-\infty) + (-\infty) = -\infty$$

But no other operations involving $\infty$ are defined, such as $\frac{\infty}{\infty}$ (note the "$a \in \mathbb{R}$" restriction), $\frac{\infty}{0}$, etc. nor is $\frac{a}{0}$. In the second system (projective, where $\infty = -\infty$) $\frac{a}{0} = \infty$ as well (for $a \in \mathbb{R}$) except when $a = 0$, where it is undefined. But also there, $\infty + \infty$ is undefined as well, in particular, the last two rules above do not hold.

Why are these definitions used? The intuition they attempt to capture is that "plugging $\pm \infty$ should give you the limit at infinity" of the operation - that is, in this system we inject a real point "at infinity" into the number system and then ask to "extend by continuity" there as though we were "filling in a removable singularity". Where this fails to happen, the operation involving infinite points is undefined. If $a$ is a fixed real number, then $f(x) = a + x$ approaches $\infty$ as $x \rightarrow \infty$, so we define $a + \infty = \infty$ in the extended real number line. But $\infty - \infty$ is not defined, since the limit $\lim_{(a, b) \rightarrow (\infty, \infty)} a - b$ does not exist: it depends on the precise path by which $a$ and $b$ go toward $\infty$. This is also why "$\infty - \infty$" when not considered as extended real numbers but rather as standing for a form of a limit (so "$\infty$" is now just "a symbol off a limit"), is called as indeterminate form.

Answer 2

The answer depends on how you interpret the expression $\frac{1}{\infty}$.

(1) If you interpret it as the limit of $\frac{1}{n}$ as $n$ tends to infinity then the answer is $0$: $\lim_{n\to\infty}\frac{1}{n}=0$.

(2) On the other hand, if you interpret $\infty$ as a specific infinite number $H$ in a modern theory of infinitesimals such as Robinson's framework then the answer is that the ratio $\frac{1}{H}$ is "close to zero", namely an infinitesimal.

In fact answers (1) and (2) are really the same answer because one can view the limit of a function $f(n)$ as the standard part of $\frac{1}{H}$: $\lim_{n\to\infty}f(n)=\text{st}(f(H))$, and the standard part of an infinitesimal is zero.