What is infinity divided by infinity?

Question

This should be a simple question but I just want to make sure.

I know $\infty/\infty$ is undefined. However, if we have 2 equal infinities divided by each other, would it be 1?

And if we have an infinity divided by another half-as-big infinity, would we get 2? For example $\frac{1+1+1+\ldots}{2+2+2+\ldots}=\frac12$?

$\frac{1}{2} = \frac {1+1+1+\dots}{2+2+2+\dots} = \frac {(1+1)+(1+1)+\dots}{2+2+2+\dots} = \frac{2+2+2+\dots}{2+2+2+\dots} = 1$. Hence such division is undefined. Any value could be assigned to it. — Karolis Juodelė, Aug 11 '12 at 12:02
In the realm of hyperreal numbers, we can speak of *infinitely large numbers*. Since the field of hyperreal numbers is really a field, even for an infinitely large number $H$, $2H$ makes sense, yielding $2H/H = 2$. In your example, let us say $H$ is the hyperreal number corresponding to the sequence $$(1,1+1,1+1+1,\cdots)$$ via ultrapower construction of the hyperreal field. Then the sequence $$(2,2+2,2+2+2,\cdots)$$ corresponds to $2H$, yielding $$\frac{[(1,1+1,1+1+1,\cdots)]}{[(2,2+2,2+2+2,\cdots)]}=\frac{H}{2H}=\frac{1}{2}.$$ — Sangchul Lee, Aug 11 '12 at 12:30
@sos440: In NSA, infinite numbers don't have specifiable sizes, and you can't uniquely identify a sum like $1+1+1+\ldots$ with a specific hyperreal. Hyperreals can be defined as equivalence classes of sequences under an ultrafilter. Since ultrafilters can't be explicitly constructed, you can't, in general, take infinite sums $\sum a_i$ and $\sum b_i$ and say whether they refer to the same hyperreal. More correct if you used Conway's surreal numbers. In the surreals, it would be natural to associate $1+1+\ldots$ with $\omega$, although there is still an ambiguity as pointed out by Karolis. — , Aug 11 '12 at 14:50
Two points that I think a freshman calc student needs to absorb: (1) Things we would write as $\infty/\infty$ are called indeterminate forms, and calculus offers specific techniques for studying them. (2) Is infinity is a number? See this question: http://math.stackexchange.com/questions/36289/is-infinity-a-number There are many different number systems. Some of them have infinite quantities and some don't. In many of them, there are different sizes of infinity, in which case $\infty$ isn't just one number, it's many, so $\infty/\infty$ depends on the sizes of the $\infty$s. — , Aug 11 '12 at 14:55
Related, I think: __[Double Think about Numerosity](https://math.stackexchange.com/questions/778265/double-think-about-numerosity)__ — Han de Bruijn, Jun 09 '21 at 13:56

score 24 · Accepted Answer · answered Aug 11 '12 at 12:06

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Essentially, you gave the answer yourself: "infinity over infinity" is not defined just because it should be the result of limiting processes of different nature. I.e., since such a definition would be given for the sake of completeness and coherence with the fact "the limiting ratio is the ratio of the limits", your

$$ \frac{1 + 1 + \cdots}{2 + 2 + \cdots} = \lim_{n \to \infty} \frac{n}{2n} = \frac{1}{2} $$

and, say (this is my choice)

$$ \frac{1 + 1 + 1 + \cdots}{1 + 2 + 3 + \cdots} = \lim_{n \to \infty} \frac{n}{n(n+1)/2} = 0 $$

would have to be equal (as they commonly define $\infty/\infty$), which does not happen.

answered Aug 11 '12 at 12:06

Filip Chindea

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thanks i just wanted to make sure its been a while – Xitcod13 Aug 11 '12 at 12:20
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These equations, of course, assumes that you *actually mean* a limiting process of that sort, and that the number of terms on the top and the bottom accrue at the same rate (more or less). – Aug 11 '12 at 14:20
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Shouldn't $\frac{\infty}{\infty}=1$? or at least $\frac{\infty}{\infty}= \infty$ – Red Banana Aug 29 '12 at 15:52
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@Gustavo: The whole point is to answer that. The answer is no. Your idea is motivated entirely by guesswork and some vague similarity to $n/n$ for normal $n$. However in standard analysis there is no such thing as $\infty$ in the same field as $1,6,85.45,\cdots$. You can't use the normal rules. Nonetheless, compare this to $0/0$ to get some sense of what's going on. This isn't defined either, and sense made only by letting the fraction gradually approach this. – not all wrong May 27 '13 at 07:57
@BillyRubina No, because $\frac{∞}{∞} = x$ reduces to $∞ = ∞x$, which is true for all positive values of $x$. – Sapphire_Brick May 30 '21 at 05:37

score 18 · Answer 2 · answered Aug 22 '15 at 15:50

I will quote the following from Prime obsession by John Derbyshire, to answer your question.

Nonmathematical people sometimes ask me, “You know math, huh? Tell me something I’ve always wondered, What is infinity divided by infinity?” I can only reply, “The words you just uttered do not make sense. That was not a mathematical sentence. You spoke of ‘infinity’ as if it were a number. It’s not. You may as well ask, ‘What is truth divided by beauty?’ I have no clue. I only know how to divide numbers. ‘Infinity,’ ‘truth,’ ‘beauty’—those are not numbers.”

Not all that convincing, since there are many systems including infinite numbers, so perhaps the answer should be “it depends which infinity you take”. Other answers elaborate this. — PJTraill, Jun 04 '20 at 09:57

Mikhail Katz · Answer 3 · 2017-09-26T12:16:04.540

To elaborate a bit on the comment by sos440, there are at least two approaches to the issue of infinity/infinity in calculus:

(1) $\frac \infty\infty$ as an indeterminate form. In this approach, one is interested in the asymptotic behavior of the ratio of two expressions, which are both "increasing without bound" as their common parameter "tends" to its limiting values;

(2) in an enriched number system containing both infinite numbers and infinitesimals, such as the hyperreals, one can avoid discussing things like indeterminate forms and tending, and treat the question purely algebraically: for example, if $H$ and $K$ are both infinite numbers, then the ratio $\frac H K$ can be infinitesimal, infinite, or finite appreciable, depending on the relative size of $H$ and $K$.

One advantage of approach (2) is that it allows one to discuss indeterminate forms in concrete fashion and distinguish several cases depending on the nature of numerator and denominator: infinitesimal, infinite, or appreciable finite, before discussing the technical notion of limit which tends to be confusing to beginners.

Note 1 (in response to user Xitcod13): Here an infinitesimal number, in a number system $E$ extending $\mathbb{R}$, is a number smaller than every positive real $r\in\mathbb{R}$. An appreciable number is a number bigger in absolute value than some positive real. A number is finite if it is smaller in absolute value than some positive real.

thanks great addition to the answer. I think you should elaborate when infinitesimal , and appreciable finite means. It might be clear from context to some but not to others. — Xitcod13, May 27 '13 at 16:57

What is infinity divided by infinity?

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