# 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 $\frac{1+1+1+\ldots}{2+2+2+\ldots} = 2$?

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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 $\frac{\infty}{\infty}= \infty$ – Gustavo Bandeira Aug 29 '12 at 15:52