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Can $1 \div 0$ equal $\infty$?

My evidence is:

$1 \div 0.1 = 0.1$

$1 \div 10^{-2} = 100$

$1 \div 10^{-3} = 1000$

$1 \div 10^{-4} = 10^4$

$1 \div 10^{10^{3}} = 10^{1000}$

$1 \div 0 = \infty$

Can $1 \div 0$ equal $\infty$, or does it have to be intermediate?

Lucas Urban
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  • The title does not coincide with the body. And no, division by zero is not allowed, and this has good reasons. – Peter May 20 '21 at 16:28
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    Your "evidence" is saying that $\lim_{n\rightarrow0} \frac{1}{n}=\infty$ for $n>0$, which is fine (even if you've typed it incorrectly). This is not the same as $1/0=\infty$ though. For example, if we instead take those $n<0$ we get $\lim_{n\rightarrow0} \frac{1}{n}=-\infty$. (These are called one-sided limits.) – user1729 May 20 '21 at 16:33
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    Or $\lim_{x \downarrow 0} 1/x = \infty$, which is also fine. – Magdiragdag May 20 '21 at 16:34
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    @Peter Whoopse! Fixed. – user1729 May 20 '21 at 16:37
  • Additionally to the problem that $\infty$ is not a number : As pointed out, the limit does not even exist (even if we allow $\infty$ as a limit which is usually done) since we have a sign-problem. – Peter May 20 '21 at 16:41
  • What your example shows is that if $x$ is a positive number that is getting closer and closer to $0$, then $1/x$ becomes larger and larger. This is not the same as saying that if $x=0$, then $1/x=\infty$. – Joe Lamond May 20 '21 at 16:59
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    Does this answer your question? Division by $0$ – Hans Lundmark May 21 '21 at 05:41

2 Answers2

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Suppose that $\frac{1}{0} = \infty$. Then $0 \cdot \infty = 1$. Also $$ 2 = 1 + 1 = 0 \cdot \infty + 0 \cdot \infty = (0 + 0) \cdot \infty = 0 \cdot \infty = 1 $$ That seems bad. Also $$ 1 = 2-1 = 1 - 1 = 0 $$ That seems worse. Trying to assign a value to $\frac{1}{0}$ will not be consistent with the other axioms of arithmetic.

Matthew Leingang
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It's important to be a little fussy with details, but here's a capsule account.

There are at least two useful ways to "add infinity" to the real number system:

  1. We may add "signed infinities" that respect the ordering, in that $-\infty < a < +\infty$ for all real $a$.
  2. We may add a single "point at infinity" $\infty$, which amounts to saying "$-\infty = +\infty$". This is a useful way to represent the slope of a vertical line, for example, since "vertically up" and "vertically down" are geometrically identical.

If we want to assign "an infinite value" to an expression $a/0$ for some real number $a$, we pick a continuous function $f$ so that $f(0) = a$ and a non-vanishing function $g$ such that $\lim(g, 0) = 0$, and determine (with suitable definitions, see below) whether or not $\lim(f/g, 0)$ is independent of $f$ and $g$.

If the answer is yes, we write "$a/0 = \ell$" as a shorthand. If instead $\lim(f/g, 0)$ depends on the choices of $f$ and $g$, the expression $a/0$ is indeterminate. (Incidentally, choosing the point of evaluation to be $0$ is not significant; we just need to pick some real number, and $0$ is our favorite.)

There are important caveats, as noted by others here:

  • As a colleague puts it to his calculus students, to say $\lim(f, 0) = \infty$ (signed or not) is a Very Special Way of saying the limit of $f$ at $0$ does not exist.
  • To reiterate, $\ell$ is not a real number. If it were, we could for example multiply $a/0 = \ell$ by $0$ obtaining $a = 0$, and then "deduce" falsehoods such as $0 = 1$.

Now, to your question:

If we add signed infinities, "$1/0 = +\infty$" is not correct: Think of $g(x) = x$, and $f(x)/g(x) = 1/x$ has no limit at $0$ (though the one-sided limits at $0$ do "exist": $+\infty$ from the right—as your numerical evidence suggests, and $-\infty$ from the left).

If we add an unsigned infinity, then $a/0 = \infty$ for all non-zero $a$. To assert the same thing without referring to non-real quantities, if $\lim(f, 0) = a \neq 0$ and if $\lim(g, 0) = 0$, then $|f/g|$ grows without bound near $0$. Precisely, "For every real number $M$, there exists a positive real number $\delta$ such that if $0 < |x| < \delta$, then $|f(x)/g(x)| > M$."

Andrew D. Hwang
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