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$$\frac{n}{\infty} + \frac{n}{\infty} +\dots = \frac{\infty}{\infty}$$

You can always break up $\infty/\infty$ into the left hand side, where n is an arbitrary number. However, on the left hand side $\frac{n}{\infty}$ is always equal to $0$. Thus $\frac{\infty}{\infty}$ should always equal $0$.

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    $\begingroup$ so $\lim\limits_{x\to\infty} \frac{10x}{x}=0$? $\endgroup$ – The Count Dec 4 '16 at 19:24
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    $\begingroup$ This is a sincere question, that shows some real thought, so it doesn't deserve the downvotes. $\endgroup$ – littleO Dec 4 '16 at 19:25
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    $\begingroup$ Infinity is not a number. $\endgroup$ – Joffan Dec 4 '16 at 19:25
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    $\begingroup$ I have changed my mind and upvoted. Every time I try to think of a one-sentence refutation of your idea, I end up unsatisfied. $\endgroup$ – The Count Dec 4 '16 at 19:30
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    $\begingroup$ See Extended number line : Arithmetic operations : "The arithmetic operations of $\mathbb {R}$ can be partially extended to $\mathbb {R} \cup \{ \infty \}$ as follows:... " Emphasis added on partially; on the LHS the ratio $\dfrac n {\infty}$ is defined (and thus meaningful) in the extended domain but an infinite sum is not defined in "usual" arithmetic and thus we cannot even try to extend it. $\endgroup$ – Mauro ALLEGRANZA Dec 4 '16 at 20:16

10 Answers 10

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How do you know $0+0+0+0+...... = 0$? If you think about it $0 + 0 + 0 +..... = 0\times\infty = 0\times 1/0 = 1$. (or any other number). We clearly are dealing with a value on which standard arithmetic doesn't apply. (Hence "infinity is not a number".)

So the question becomes what does apply and how do we deal with this? And that is not an easy/simple question. It's not hard... but it's not simple. Bottom line, finite rules of arithmetic do not always apply, and such instincts lead to common traps.

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    $\begingroup$ I liked this answer the best, the 0*1/0 hits the Crux of the matter $\endgroup$ – frogeyedpeas Dec 4 '16 at 19:37
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    $\begingroup$ The problem with this answer is that $0 + 0 + \ldots = 0$ is actually true. $\endgroup$ – Hurkyl Dec 4 '16 at 22:23
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    $\begingroup$ @Hurkyl I don't see the problem. It may be true, but the answer shows that one should be careful when trying to apply familiar arithmetic rules to infinite sums. $\endgroup$ – Servaes Dec 4 '16 at 22:26
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    $\begingroup$ @servaes precisely. It's not enough that 0+0+0+ .... = 0. We must know why. Afterall $0 + 0 + 0+.... = 0 *\infty = 0*1/0 = 1$ so ... one of those basic statements is either wrong or doesn't mean what we think it does. If $1/\infty = 0$ which.... is something people say.... then $1/infty + 1/infty ... = 0+ 0+.....$ which ... is not so clear. $\endgroup$ – fleablood Dec 4 '16 at 22:47
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    $\begingroup$ Morgan Rodgers. Why on earth is that "clear "? It's not clear to me what 0+0+.... means. Or what n/infty means. If infinity can grind something down to nothing why is it clear that it can't build nothing up to something? And it doesn't matter which one is true. What matters is to realize our assumptions and definitions need clarity. Without clarity we don't know 0+0+... is. $\endgroup$ – fleablood Dec 5 '16 at 6:01
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The existing answers are very good; let me give yet another one.

Consider the counter-argument that ${\infty\over\infty}=\infty$: We have $\infty=2^\infty$, so $${\infty\over\infty}={\infty\over 2^\infty}={\infty\over 2}\cdot {1\over 2}\cdot {1\over 2}\cdot {1\over 2}\cdot...$$ But ${\infty\over 2}$ is infinity, and stays infinity no matter how many times we divide it by $2$. So ${\infty\over \infty}=\infty$.

This is exactly the same "shape" as the argument you give, but yields the opposite answer; so something must be wrong with this kind of argument.

The problem lies in the "number" $\infty\over\infty$. It hasn't been precisely defined. Now, some of the time in math it's clear what we mean when we write some complicated expression; however, this isn't one of those times. We need to sit down and precisely define what this thing is.

When we try to do so, we'll find something surprising: it doesn't follow the usual laws of arithmetic! For example, consider the following "proof" that $1=2$: $${1\cdot{\infty\over\infty}}={\infty\over\infty}={2\infty\over \infty}=2\cdot{\infty\over\infty},\mbox{ so }1=2.$$ Since $1\not=2$, at least one of the steps there has to be nonsense. The obvious candidate is the claim that we can cancel $\infty\over\infty$. This suggests that ${\infty\over\infty}=0$; however, then we have $\infty\cdot {1\over\infty}\not=1$, which ruins one of the basic properties of division!

It gets worse: in both arguments, we need to perform infinitely many operations (either infinitely many sums, or infinitely many products), so we need to define how those work. At first glance it seems like limits can help us there, but again, we'll find that there's no good way to define the infinite operations we need which matches with our intuition.

This all points to the following fact:

In order to make sense of arithmetic involving infinity or infinite operations, we need to make some choices in how we precisely define the various operations on infinity; and no matter how we make these choices, some "obvious" properties of numbers will fail to hold.

Picture trying to fit a carpet into a room that's too small. You can maybe get lots of it to lie flat, but somewhere it's going to fold over, or stick up, or crumple.

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    $\begingroup$ On the subject of making sense of infinite operations, you may be interested in various kinds of infinite summation that let us sum divergent series. For example, do you think $1-1+1-1+1-1+...={1\over 2}$? Then Cesaro summation might be right for you! $\endgroup$ – Noah Schweber Dec 4 '16 at 20:07
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    $\begingroup$ @SimpleArt No, that kind of regrouping isn't allowed in Cesaro summation - see the wikipedia page. Basically, any method for summing divergent series is going to have to be "fragile". $\endgroup$ – Noah Schweber Dec 5 '16 at 2:03
  • $\begingroup$ Yes, I know, but many of those viewing this post do not. I was pointing out that perhaps you might want to reconsider leaving that comment there. $\endgroup$ – Simply Beautiful Art Dec 5 '16 at 2:04
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    $\begingroup$ @SimpleArt I disagree. The point of that comment was precisely that there are many ways to sum a non-convergent series. I think it does make sense in this context, because the ambiguity of infinite operations is indeed one of the problems with the OP's argument. I am leaving the comment. $\endgroup$ – Noah Schweber Dec 5 '16 at 2:24
  • $\begingroup$ Fine, as you wish. I just think summing divergent series is completely different from the OP's context, but I am not the judge of that. Cheers anyways for the nice answer. (Also deleted comments that would probably confused others) $\endgroup$ – Simply Beautiful Art Dec 5 '16 at 14:20
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The problem is that before you can even consider a proof of a statement, it has to be clear what that statement means. I have no idea what $\infty/\infty=0$ could possibly mean.

The original answer ended here.

Since not everyone was happy with that answer, let me elaborate. Since $\infty$ is not a number, I just do not know what $\infty/\infty$ should mean. Now you could respond that you are using formulas like $\infty+\infty=\infty$ and have been told that that one is true. Well, it is a useful short-hand, but what people really mean when they write this is the following:

If $(a_n)$ and $(b_n)$ are sequences of real numbers such that $\lim_{n\to\infty} a_n=+\infty$ and $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} a_n+b_n=+\infty$.

Now this is a perfectly fine statement, and it happens to be also true. Another perfectly fine statement (but a wrong one) would be:

If $(a_n)$ and $(b_n)$ are sequences of real numbers such that $\lim_{n\to\infty} a_n=+\infty$ and $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} a_n/b_n=0$.

Now maybe one could understand $\infty/\infty=0$ in this way, but you would have to say so. And now you also see that it would be hard to see how the “proof” in this question relates to this. Maybe one could make the following weaker statement:

If $(b_n)$ be a sequence of real numbers such that $\lim_{n\to\infty} b_n=+\infty$, then $\lim_{n\to\infty} n/b_n=0$.

One could then say that the “proof” in your question is to be understood as follows: \begin{multline}\lim_{n\to\infty} \frac{n}{b_n} =\lim_{n\to\infty}\underbrace{\left(\frac1{b_n}+\cdots+\frac1{b_n}\right)}_{\text{$n$ summands}} =\\=\lim_{n\to\infty}\underbrace{\left(\lim_{n\to\infty} \frac1{b_n}+\cdots+\lim_{n\to\infty}\frac1{b_n}\right)}_{\text{$n$ summands}} =\\=\lim_{n\to\infty}\underbrace{(0+\cdots+0)}_{\text{$n$ summands}}=\lim_{n\to\infty}0=0. \end{multline} Now here at least every expression has a defined meaning and we can ask: Where is this wrong? Well, it turns out that the second equality is wrong, and the reason why is again boring: There is just no reason that this equality should hold. It is not enough that it somehow looks nice, we would have to cite some theorem that we have proved earlier to justify this equality, and there is none. (Of course we can pinpoint the wrong equation by plugging in a counterexample. For example for $b_n$=n you can evaluate each expression and get $1=1=0=0=0=0$.)

Anyway, my point is that before we can ask why something is wrong it first has to make minimal sense. And the proof in your question is, as they say, “not even wrong”, because it is unclear what the statement is.

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    $\begingroup$ I find this answer unsatisfying. I get what is being said and I agree with it, but it means exactly what the OP says it means: dividing an arbitrarily large number by itself is 0. Makes no sense, but it is what we are working with as a hypothesis. $\endgroup$ – The Count Dec 4 '16 at 20:05
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    $\begingroup$ I think this answer as "satisfying". You say " exactly what the OP says it means: dividing an arbitrarily large number by itself is 0" but I don't so what "an arbitrarily large number" is supposed to be. Infinity isn't an arbitrarily large number, is it? $\endgroup$ – fleablood Dec 4 '16 at 21:53
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    $\begingroup$ When I write $\infty + \infty = \infty$, I really mean to do arithmetic on extended real numbers. $\endgroup$ – Hurkyl Dec 4 '16 at 23:25
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    $\begingroup$ @Hurkyl, sure, one can also mean that, but then one has to say that one means that, and one has to say what the definitions are, unless they are clear. $\endgroup$ – Carsten S Dec 4 '16 at 23:27
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First of all $\infty$ is not a real number, so it's unclear what you mean by $n/\infty$ in the first place.

But, let's suppose that you extend the real numbers by introducing a new number "$\infty$" which has some expected properties, such as $n/\infty = 0$ for any real number $n$. You would still need to define $\infty/\infty$ before you can refer to it, otherwise we don't even know what "$\infty/\infty$" means. You could define it to be $0$ if you liked, but somebody else could equally well argue that $\infty/\infty$ should be defined to be equal to $10$ (for example) because $$ \lim_{x \to \infty} \frac{10x}{x} = 10. $$ Another person could argue that $\infty/\infty$ should be defined to be equal to $\infty$, because $$ \lim_{x\to\infty} \frac{x^2}{x} = \infty. $$ There is no value that we could assign to $\infty/\infty$ that stands out as being more valid than any other possible value. So, we simply leave $\infty/\infty$ undefined.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Dec 6 '16 at 14:33
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If $\frac\infty\infty=0$, then we should have $\infty\cdot 0 = \infty$.

But OP's main equation $\frac{n}{\infty}+\frac{n}{\infty}+\frac{n}{\infty}+\cdots=\frac\infty\infty$ (which is supposed to equal $0$) suggests that $\infty\cdot 0 =0$.

How can $\infty\cdot 0$ equal both $0$ and $\infty$?

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  • $\begingroup$ Isn't multiplying infinities undefined? I tried to see how I can simplify the infinity/infinity problem by breaking it apart without breaking any rules. $\endgroup$ – Goldname Dec 5 '16 at 3:21
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    $\begingroup$ I don't see why (in this hypothetical situation) we can't look at OP's equation as saying infinity times 0 is $\frac\infty\infty$, since we're adding $0$ (in the form of $\frac{n}{\infty}$) to itself infinitely many times and getting a result of $\frac\infty\infty$. $\endgroup$ – paw88789 Dec 5 '16 at 3:24
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    $\begingroup$ So we no longer require $\frac{a}{b}=c$ to be equivalent to $bc=a$? This principle seems important to me if one wants to have a meaningful conversation about division. $\endgroup$ – paw88789 Dec 5 '16 at 3:36
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    $\begingroup$ @paw88789: This shows that the OP's arithmetic leads to a contradiction, but it doesn't enlighten him at all, because it doesn't explain why that contradiction arises (i.e. the deep reasons that cause it). To me, this post is not an answer. $\endgroup$ – Alex M. Dec 5 '16 at 11:18
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    $\begingroup$ @Goldname, I find funny the way you refer to yourself as "the OP", instead of just "me". $\endgroup$ – Alex M. Dec 5 '16 at 11:19
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When you learned how to extend arithmetic from the natural numbers to the integers to the rational numbers to the real numbers and to the complex numbers, one of the main motivations was to preserve the laws of arithmetic.

The situation with the extended real numbers (i.e. the real numbers along with $\pm \infty$) is different — the goal is not to preserve the laws of arithmetic, the goal is to have continuity.

Consequently, arguments that involve naively applying the ordinary laws arithmetic to extended real numbers are quite unreliable.

This is compounded by the fact the argument involves infinite summation. Infinite sums can fail to satisfy many of the laws that finite sums obey, so that's a second source of unreliability.

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    $\begingroup$ +1, but it might be helpful to the OP to explain your second paragraph. $\endgroup$ – Noah Schweber Dec 4 '16 at 22:50
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I think an issue with your reasoning as well is that to "split up infinity" you would have to have an infinite sum of finite terms. You can see with Riemann integration that an infinite sum of 0s is not necessarily 0. Of course none of this is rigorous but I think it should help you intuition a little.

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Infinity is a fine number (and so is its negative counterpart), and is surprisingly well-behaved if you unpack it carefully.

Without going into too much detail about what that means, just note that Cantor showed that there are many types of infinities with different characteristics.

This means that we have to be clear about what sort of infinity we're talking about.

Most of the existing answers implicitly interpret infinity as some number implied to be the result of some unbounded number of arithmetic accumulations, which makes sense given the framing of your question. So what kind of infinity is reached in that fashion? $\omega$!

$\omega$ is a number that, informally, is always bigger than any real number you can produce using finite arithmetic steps. It's a member of the hyperreals, symbolized as $\mathbb{R}^*$. The nice thing about the hyperreals is, anything you can state in first-order logic (i.e., no tricks with arguments on sets) about the reals is still true. This is known as the transfer principle.

In the hyperreals are also infinitesimals $\epsilon$, which are the reciprocals of the infinitudes. They work just as you'd expect: a real number divided by an infinitude is an infinitesimal. So the hyperreal number line looks like this (Wikipedia):

hyperreal number line

With these tools, we note that each of the terms in your series are infinitesimal:

$$\frac{n}{\infty} = \epsilon$$

You have an infinitude of them, so the series can be rewritten as:

$$\epsilon + \epsilon + \dots = \omega\epsilon$$

This is an indeterminate form in the hyperreals (see chapter 1 of Keisler), just as you would expect. The reason for this is because $\frac{n}{\infty}$ is not zero, and in the first place, it wouldn't make sense for a infinitely divisible real number to be zero in that case.

So you can have infinity and infinitesimals as your equation depicted, but you still have to acknowledge indeterminates if you do so.

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    $\begingroup$ $\infty$ is greater than every real number, and consequently it's greater than every hyperreal number too (in particular, $\omega \neq \infty$, no matter which unlimited number you choose for $\omega$). Also, there is a difference between an infinite sum and a (hyper)finite sum of $\omega$ terms. Also, this topic is unrelated to Cantor's work on cardinality and ordinal numbers. $\endgroup$ – Hurkyl Dec 4 '16 at 23:19
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    $\begingroup$ This isn't really accurate; $\omega$ is not an element of the hyperreals. It is an element of the surreals, but that's a very different object. In particular, your answer suggests that there is only one kind of infinity in the hyperreals, and similarly only one kind of infinitesimal. This is false; e.g. if $\epsilon$ is infinitesimal, $\epsilon^2$ is "even more" infinitesimal, and this is crucial to the structure of the hyperreals. Indeed, there is no distinguished infinite/infinitesimal element of the hyperreals. $\endgroup$ – Noah Schweber Dec 4 '16 at 23:23
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    $\begingroup$ The hyperreals are a perfectly valid framework for analysis, but they're more complicated than you're answer suggests; meanwhile, the ordinals (Cantor's transfinites), where $\omega$ lives, are a much worse setting for arithmetic. The surreals are somewhat a mixture of the two, but transfer issues there are much more complicated; they lack the reflexive ease of the hyperreals. $\endgroup$ – Noah Schweber Dec 4 '16 at 23:26
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    $\begingroup$ @bright-star Agh, I didn't know that - that's awful. I'm slightly arguing over notation, just because of how well-established the set-theoretic meaning of "$\omega$" is, and how easily it is to get confused when thinking about ordinals and hyperreals at the same time. I didn't realize wikipedia conflated the two; my apologies. My main point, though, is about the hyperreals being more complicated than is suggested by that picture (again: dang you, wikipedia!). $\endgroup$ – Noah Schweber Dec 4 '16 at 23:29
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    $\begingroup$ @bright-star: Transfer principle: all standard concepts, including the extended real numbers $\pm \infty$ (or whatever hacks one prefers to use them as notation without actually using them as mathematical objects) transfer and apply equally well to the nonstandard model. $\endgroup$ – Hurkyl Dec 4 '16 at 23:31
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You cannot use an "infinite" distributive law with division by the symbol $\infty$. In other words the statement $$\frac{\sum_{n=1}^\infty a_n}{\infty} = \sum_{n=1}^\infty \frac{a_n}{\infty}$$ is not valid if $\sum_n a_n$ is a divergent series. For the left-hand side cannot have any meaning when that sequence diverges, whereas the right-hand side is the zero series (provided that each $a_n$ is an ordinary finite number) which has the sum $0$ for any reasonable summation definition.

The law $\frac{a+b}{d}=\frac{a}{d}+\frac{b}{d}$ is true for real or complex numbers $a$, $b$ and $d$ provided $d\ne 0$, but it does not quite generalize to the form you use.

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Euler once heavily looked at the ratio you mentioned (albeit archaically in today's terms) although as it's reciprocal (0/0). What he found is that it can only be referred to as an "indeterminate form". What this means in a literal and logical sense is that if you declare $\frac 00 = a$ to be true, then you find that $\frac 00$ equals every number through simple algebraic manipulation. The end result is that 0/0 equals every number and every number equals every other number. This is obviously false, so you reach a contradiction.

Therefore, what stops you is that giving it any single value is an issue. The "trick" is to give it a different value depending on the context. This is the concept of limits and various ideas of the value a function approaches rather than the value the function has.

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  • $\begingroup$ From my experience most of those tricks I know of is adding/subtracting infinity, or doing multiplication/division with two infinites or 0, i.e. infinity/infinity = infinity(1/infinity). Infinity was divided by infinity when factoring it out. Thus, I avoided those but I still end up with this incorrect proof. $\endgroup$ – Goldname Dec 5 '16 at 3:25
  • $\begingroup$ @Goldname have you learned the concept of limits?What you have is a set of two limits.One is a limit of the sum as it increases without bound.One is the limit as some divisor increases without bound.The rate at which they increase determines the ratio of one infinity to the other(yes,there are infinities of varying degrees!).So, in reality yes your function is equal to the left hand side.However,your ability to separate it in that manner depends upon the particular function.In essence, you have selected a particular form of infinity or series.If you haven't taken Calculus google the "limit". $\endgroup$ – The Great Duck Dec 5 '16 at 4:35
  • $\begingroup$ @Goldname unfortunately, it's a paradox and infinity is a number which does not truly exist. Saying infinity equals "n+n+n+n+n+n..." directly implies you have infinity of the linearly grown sense. What if I propose that infinity is a^n as n increases forever for some integer a? One grows faster. Divide the two infinities and you get an ideal ratio of their heights based upon their rate as it continues infinitely. In fact, I would guess it gives 0 because the exponential pulls it downward. $\endgroup$ – The Great Duck Dec 5 '16 at 4:39
  • $\begingroup$ I am currently evaluating the limit of the divisor first, so whether or not it is linearly increasing should result in an infinite sum of 0 $\endgroup$ – Goldname Dec 5 '16 at 4:40
  • $\begingroup$ @Goldname you don't evaluate one limit first... It's one single limit of the entire rational function f(x)/g(x). If you wish to discuss the function that is a limit over a limit, then if either limit diverges (goes to infinity) it diverges... period. The point of the form infinity/infinity is that it is not the true value of the limit. It's the value you get if you directly plug in the "value" of infinity. Algebraic manipulation of f and g reveal the true value of the limit. So once again, it all depends on f and g. $\endgroup$ – The Great Duck Dec 5 '16 at 4:43

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