# Whats infinity divided by infinity?

This should be a simple question but i just want to make sure. I know from infinity/infinity is undefined.

However if we have 2 equal infinities divided by each other it would be 1?

And if we have an infinity which is twice as big as the other infinity divided by it then it would equal 2?

for example $\frac{1+1+1+\ldots}{2+2+2+\ldots} = \frac12$?

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infinity is not a unit like 1 metre, 1 pound, 1 dollar. –  Rajesh K Singh Aug 11 '12 at 11:58
$\frac{1}{2} = \frac {1+1+1+\dots}{2+2+2+\dots} = \frac {(1+1)+(1+1)+\dots}{2+2+2+\dots} = \frac{2+2+2+\dots}{2+2+2+\dots} = 1$. 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 $$\frac{[(1,1+1,1+1+1,\cdots)]}{[(2,2+2,2+2+2,\cdots)]}=\frac{H}{2H}=\frac{1}{2}‌​.$$ –  sos440 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. –  Ben Crowell 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: 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. –  Ben Crowell Aug 11 '12 at 14:55

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.

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thanks i just wanted to make sure its been a while –  Xitcod13 Aug 11 '12 at 12:20
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). –  Hurkyl Aug 11 '12 at 14:20
Shouldn't $\frac{\infty}{\infty}=1$? or at least $\frac{\infty}{\infty}= \infty$ –  Jesus Christ Aug 29 '12 at 15:52
@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. –  Sharkos May 27 '13 at 7:57

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.

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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

## protected by Chris EagleJul 30 '13 at 23:46

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