# Why is Infinity multiplied by Zero not an easy Zero answer?

I did a bit of math at school and it seems like an easy one - what am I missing?

$$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$

$$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$

(i.e add $0$ to $0$ as many times as you like, result is $0$)

So I thought an infinite number of 0's cannot be anything but 0? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that.

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It is zero in measure theory. – scineram Mar 25 '11 at 10:05
Sum of countable number of 0-s is 0. Multiplication is not easily transformed to addition, unless you know how to add 5 to itself pi times. – Kamil Szot Mar 25 '11 at 12:27
The tag nonstandard analysis needs to be removed, even with minimum amount of knowledge of analysis this is not a question. – Arjang May 27 '13 at 12:02
The software doesn't provide us with a way to link together all the variants of this question that have been posted here over the years, so we could then put this posting into that group of linked-together questions and link to that group of questions whenever it is to be mentioned. – Michael Hardy Jun 13 '14 at 13:06
@MichaelHardy We could always just make a tag specifically for these questions. – Patrick vD Mar 17 at 0:03

The problem is that the laws of addition and multiplication you are using hold for natural numbers, but infinity is not a natural number, so these laws do not apply. If they did, you could use a similar argument that multiplying anything by infinity, no matter how small, gives infinity, thus $\infty \times 0 = \infty$. More sophisticated arguments can also be made, like $\infty \times 0 = \lim_{x \to \infty} (x \times 1/x) = 1$. Clearly all these different values for $\infty \times 0$ mean that $\infty$ cannot be treated like other numbers.

In order to work with infinity, you must first define it. You may think you know what infinity is, but really you don't have a concrete definition. In fact, there are many different definitions of infinity that you could use, each of which result in different behaviors. For example, the real projective line has a concept of infinity such that $1/\infty = 0$, while when talking about infinite sets one uses cardinal numbers (another type of infinity) to represent the sizes of these sets. You must make it clear what infinity you are talking about in order to work with it.

In summary, the expression $\infty \times 0$ using multiplication defined for the natural numbers does not have any meaning, so it cannot be said to be equal to $0$.

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Thanks - that makes sense. Something still feels unintuitive about it though. i.e even if you took a limit approach, surely the limit of the sum of many many 0's is still 0. My next question can be: How many 0's summed together makes a number > 0? :) – Ashley Schroder Mar 25 '11 at 7:39
Any finite number of $0$'s summed together is $0$. Summing infinitely many elements, strictly speaking, does not work. When you use an infinite series, you are secretly actually taking a limit of partial sums. Limits are not simple; calculus is basically the study of limits. When you talk about something like $\lim_{x \to \infty} (x \times 1/x)$, you aren't talking about $\infty \times 0$ per se, but rather what happens to $x \times 1/x$ as $x$ gets arbitrarily large. – Alex Becker Mar 25 '11 at 7:46
@Ashley: If you "sum infinite number of zeros" the answer is "zero". By summing infinite zeros, I assume you mean $\displaystyle \lim_{n \rightarrow \infty} n \times 0$ which is $0$. However, the issue is the interpretation of the indeterminate form $\infty \times 0$. – user17762 Mar 25 '11 at 8:37
@Aschley: Maybe this helps for further clarification: Again you are right that "summing infinite zeros gives zero" if you phrase it like $\sum_{n=1}^\infty 0 = 0$ (since all the partial sums $S_N = \sum_{n=1}^N 0$ are zero and hence, their limit is also zero). – Dirk Mar 25 '11 at 12:56
@Alex An anonymous user tried to get rid of a large chunk of your text (the bit on multiplying infinity with infinity) because they didn't understand it. Dunno if you'd get pinged for this, so thought you might like to know... – user1729 Jun 16 '14 at 9:57

You have to remember that infinity isn't a number. It's more of a concept. When you write

$$n \times 0 = 0 + 0 + 0 +\cdots+ 0 = 0$$

you're doing a finite operation. There's no way to keep adding zero until you reach infinity, because you can't reach infinity. It's this inability to "reach" infinity that makes the operations violate your intuition. Traditional algebra/arithmetic doesn't work on infinity. This is why we use the concept of limits, which is well-defined mathematically and allows us to perform algebra on infinities.

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There are infinite sums that converge, so your answer is wrong. – Andyk May 27 at 15:29
@Andyk I fail to see how the existence of convergent series makes my conceptual answer wrong. As I said, you can't add an infinite number of terms together using traditional algebra/arithmetic. You can use limits to compute whether the sequence of partial sums of a series converges to a value, and we define the value of the series to be that value, if it exists. It isn't, however, truly a sum of an infinite number of terms in the traditional algebraic/arithmetic sense, as evidenced by the existence of the Riemann series theorem. – thrnio May 28 at 2:25
Zeno's paradox. The motion is possible because we can cover (add up) infinite number of intervals (each half of the previous) in finite time. So you can reach infinity and add an infinite number of terms. – Andyk May 29 at 19:52
@Andyk Zeno's paradox defends my point. In Zeno's paradox, the argument is that there is always further to go, so at no point will said person reach their destination, or even move at all, but we know that it is possible to move, so by reductio ad absurdum, one of our assumptions must be wrong. Since Zeno's paradox is based on our intuitions about the interaction between addition and infinity, our intuition must be wrong, which was exactly what I stated. – thrnio May 30 at 2:14

As several others have pointed out, $\infty$ is not a number. So you need to deal it with a bit of care.

To clarify your doubt, the way you have written is to look at $n \times 0$ and then let $n \rightarrow \infty$. So it is true that $$\displaystyle \lim_{n \rightarrow \infty}\left( n \times 0 \right)= 0$$

However, when people write $\infty \times 0$ usually it is a shorthand to denote the indeterminate form when some quantity tends to infinity and some other quantity tends to zero in a limiting sense i.e. expressions of the form $$\lim_{x \rightarrow 0} \left( f(x) \times g(x) \right)$$ where $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$.

(Note that $\infty$ is not a number in the conventional sense. It is just a shorthand to denote that something grows unbounded i.e. given any number your function can take a value larger than that number.)

For instance, let $f(x) = \frac{1}{x}$ as $g(x) = x$, then $f(x) \times g(x) = 1$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0} \left( f(x) \times g(x)\right) = \lim_{x \rightarrow 0} 1 = 1$$ However, $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$ and hence in this case, the indeterminate form evaluates to $1$.

The case which resembles what you have written down is when $f(x) = \frac{1}{x}$ and $g(x) = 0$. This again is an indeterminate form since $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0} g(x) = 0$. However in this case, $f(x) \times g(x) = 0$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0} \left( f(x) \times g(x) \right) = 0$$

Yet another example is to look at $f(x) = \frac{1}{x}$ and $g(x) = \sqrt{x}$. This again is an indeterminate form since $\displaystyle \lim_{x \rightarrow 0} f(x) = \infty$ and $\displaystyle \lim_{x \rightarrow 0^+} g(x) = 0$. Note that $f(x) \times g(x) = \frac{1}{\sqrt{x}}$, $\forall x \neq 0$ and hence $$\displaystyle \lim_{x \rightarrow 0^+} (f(x) \times g(x)) = \displaystyle \lim_{x \rightarrow 0^+} \frac{1}{\sqrt{x}} = \infty$$

Hence, you cannot associate a unique value to $\infty \times 0$. It depends on the problem at hand. You can read more about indeterminate form here. (As always with wikipedia, read it just to get a general overall idea.)

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To reply to user 3302, Your wrong. lim[as x->0]f(x)=1/x does not exist, if you approach it from the right you get +infinity and if you approach it from the left you get -infinity, therefore your answer does not exist. – user15414 Sep 1 '11 at 20:13
That depends on what infinity is... if it is the extra point in the one-point compactification of the reals or the complex plane, then there is just $\infty$, no $-\infty$. – dfeuer Jun 8 '13 at 5:43
So you're saying 0 is not 0? How about "You can't limit infinity."? – Cees Timmerman May 5 '14 at 14:34

As pointed in the other answers, the issue is the interpretation of what do you mean by $\infty \cdot 0$. Strictly speaking, $0+0+\cdots+0$ when the number of terms goes to infinity IS 0 (this is just the sum of a series).

But look at this example

\begin{eqnarray} \begin{split} 1&=&1 \\ \frac{1}{2}+\frac{1}{2}&=&1 \\ \frac{1}{3}+\frac{1}{3}+\frac{1}{3} &={}&1 \\ \vdots \\ \frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n} &={}&1 \\ \end{split} \end{eqnarray}

By repeating the process, at every step I get more numbers, each of them closer to zero...This is also what we can understand by $\infty \cdot 0$, but this is $1$, isn't it?

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You never get to the infinity by repeating this process. Limit means that you approach the infinity but never actually get to it because it's not a number and cannot be reached. The expression $\infty \cdot 0$ means strictly $\infty\cdot 0=0+0+\cdots+0=0$ per se. I don't understand why the mathematical community has a difficulty with this. – Andyk May 27 at 15:44

This is just a general comment regarding questions like these:

When encountering definitions and results you think you could disagree with, try to think through the definitions. In this case, how would you define multiplication by $\infty$? More fundamentally, how would you define $\infty$? One approach would be to formally define $\infty$ as a symbol such that $a \infty = \infty$ for all numbers $a$, but then we would have to exclude $0$, and set $0 \infty = 0$. This causes problems however (if we want the distributive law to hold): $0=(a-a)\infty =\infty-\infty$. And how would you define this?

Shortly, trying to do arithmetic with this new symbol causes a lot of problems. Whenever you are dealing with entities which are "unusual" you must define what they are and how they interact with other mathematical objects.

In the same way, we could ask how to define infinitesimals, numbers that are smaller than any other number. Say, a number $\epsilon$ such that $|\epsilon| < a$ for all real non-zero numbers $a$. Such a number does not exist in the real number system, but it is possible to define such a system. (Google non-standard analysis) Now the task is to define how to do arithmetic with infinetesimals... (but this is really off-topic)

To summarize this somewhat unorganized answer: try to think through the consequences of defining $0 \infty = 0$. (that is, try to reduce ad absurdum). Think through what definitions you are using and try to find examples.

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It may also have to do with conflicting definitions: generally speaking, for some number n, n * 0 = 0, and n * infinity = infinity. So what is infinity * 0? It's undefined.

Also, as Alex pointed out, infinity is not a natural number so the same rules don't apply. And he's right, there are different definitions for infinity, so don't think of it as a number with an actual value. For instance, the set of all real numbers and the set of all rational numbers are both infinite, but the set of all real numbers is a 'bigger' infinity. Thus, infinity does not have a concrete value, and it doesn't make sense to treat it as any other natural number.

But it's true that instinctively you would expect the answer '0' so I think it's helpful to look back at axioms and identities for these kinds of problems, and go from there. Unfortunately math is not always intuitive!

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Suppose that in $36n$ independent trials, the probability of success on each trial is $1/(10n)$. What is the probability that the number of successes is $5$?

\begin{align} 36n & \to \infty \\[10pt] \frac{1}{10n} & \to 0 \\[10pt] 36n\cdot\frac{1}{10n} & = 3.6 = \text{expeccted number of successes} \\[10pt] \Pr(\text{number of successes}=5) & \to \frac{e^{-3.6} 3.6^5}{5!} \end{align}

That "$0\cdot\infty$" is an "indeterminate form" means precisely that if you multiply something that approaches $0$ by something that approaches $\infty$, then the product might approach $0$ or $\infty$ or some number between those extremes, depending on what the two factors being multiplied are.

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The symbol $\infty$ was originally introduced by Wallis in the 17th century. He used it to denote a specific number, and went on to consider partitions of intervals into $\infty$ parts in applications such as calculation of areas of plane figures. Such giants as Leibniz, Euler, and Cauchy exploited infinite quantities to obtain results in analysis. More details can be found for instance in the recent article here: http://dx.doi.org/10.1007/s10699-012-9285-8

In an enriched number system containing such infinite numbers, it is correct that an infinite number times $0$ is indeed "an easy Zero answer". One could take such a number system to be the hyperreals, but there are many such number systems. So long as the number system is a field, anything multiplied by $0$ will necessarily give $0$, even if the "anything" is an infinite number.

The customary interpretation of the expression "infinity times zero" is in terms of the so-called "indeterminate forms" (see also Whats infinity divided by infinity?), and then the answer is far from easy and in fact is not zero in general.

However, when interpreted literally, the OP's hunch that $\infty \times 0=0$ is "easy" can be fully justified as above.

For a related question on "infinity times infinitesimal", see infinity times infinitesimal - what happens?

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This answer completely misses the point of the question. In particular, if you consider "infinite numbers", then you should consider multiplying them by infinitesimals, not just by zero. And the result could be transfinite, infinitesimal, standard real, bounded real, or zero if you really did multiply something by zero. – Asaf Karagila May 27 '13 at 12:01
Asaf: the OP wanted to know specifically about the product of infinity by zero. If he wanted infinitesimals he could have asked about them, just as he asked about infinity. I was addressing his question specifically, rather than your interpretation thereof. – Mikhail Katz May 27 '13 at 12:41
No, you're forcing the interpretation through non-standard analysis. Obviously the reason that $\infty\cdot0$ is an indeterminate form as a limit comes from two reasons: (1) $\infty$ is not a real number, therefore the context means that we multiply two sequences; (2) the multiplication of a divergent sequence and a sequence converging to zero can be divergent, zero, or convergent to a finite number. If we talk about sequences, then the appropriate thing is to talk about infinitesimals, not just about transfinite reals. If you want to talk about proper zero, then $0\cdot\infty=0$ always. – Asaf Karagila May 27 '13 at 12:45
[...] because multiplying anything by the constant zero sequence is the constant zero sequence. Without showboating with non-standard analysis. – Asaf Karagila May 27 '13 at 12:45
This is not particularly highlighting non-standard analysis but any number system containing infinite numbers. You are forcing an "indeterminate form" interpretation on the OP's question. My point was precisely that one needn't force that interpretation on his question. – Mikhail Katz May 27 '13 at 12:52

The problem is that there is no way to extend the usual operations of addition and multiplication on $\mathbb R$ so that $\mathbb R \cup \{\infty,-\infty\}$ forms a field. The technical difficulty is actually with adding infinities, not subtracting them. What is $\infty + \infty$? Well, it can't be $\infty$, because then we'd subtract $\infty$ from both sides and get $\infty = 0$. We can't have $\infty + \infty = -\infty$, because then we'd get $3\infty = 0$, so $\infty = 0$. And we can't have $\infty+\infty = r$ for $r$ real, because then we'd have $\infty = r/2$. So there's no way in general to make arithmetic with just a positive and a negative infinity make sense. So if for convenience you want to usea limited sort of arithmetic with infinities, you have to lay out the rules you've chosen to use ahead of time—there's no real standard. However, as some other answers have suggested, it's possible to regain sense by adding lots of infinities, like $2\infty$, $\frac 2 3 \infty$, etc., and the "infinitesimals" to match them, but that goes beyond my own knowledge.

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$$\underbrace{\frac 1 n + \cdots + \frac 1 n}_\text{5n terms} = 5$$ As $n$ approaches $\infty$, each of these terms approaches $0$, but their sum remains equal to $5$, so its limit is $5$.

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This doesn't answer the question. Each term approaches $0$ but never really becomes zero. So this sum correctly equals to $5$. But this is not the same as $0+0+\cdots + 0$ which equals to $0$. – Andyk May 27 at 15:55

There are several good answers here, any of which will shed light on your question. However, there is one other issue at hand that is not being addressed: $0$.

Zero is $\infty$'s opposite and also isn't a number (well, it isn't just a number/integer, although you can usually treat it like one without drastic repercussions - provided you follow the "special rules" governing operations with zero).

You can set up the problem as you attempted:

$$0 \times \infty = \underbrace{0 + 0 + 0 + \cdots.}_{\infty \text{ many times}}$$

But you must realize that zero is actually an infinitesimal, an ever-shrinking quantity.

So what this problem boils down to is a race. Will the terms shrink faster than you can add them together? It depends on how fast $\infty$ is growing and how fast $0$ is shrinking$^*$. Thus

$$0 \times \infty = \lim_{x \to \infty}\frac{1}{x} \times \lim_{x \to \infty}x= \lim_{x \to \infty}\frac{x}{x} = \lim_{x \to \infty}1 = 1$$

AND

$$0 \times \infty = \lim_{x \to \infty}\frac{2}{x} \times \lim_{x \to \infty}x= \lim_{x \to \infty}\frac{2x}{x} = \lim_{x \to \infty}2 = 2$$

because zero shrinks at different rates ($\infty$ grows at the same rate in each expression).

$^*$disregard the "time" components of my statements. $\infty$ is simultaneously every large quantity and $0$ is simultaneously every small quantity. The time component just makes thinking about this stuff easier!

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I've downvoted your answer because you say "Zero is ∞'s opposite and also isn't a number". 0 is a well known intezer. On the contrary $\infty$ is not a number so " 0×∞=0+0+0+⋯ ∞ many times" doesn't make sense. – user103816 Mar 7 '15 at 6:17
You know, you're right that I should not have said that zero isn't an number; it's a very special number! It is the only element of the set of integers that is also in the set of infinitesimals (and also in the set of imaginary numbers, but that's beside the point). In this case, we're more interested in its properties as related to it being an infinitesimal than it being an integer. – Nick G. May 28 '15 at 18:09
How come $0$ is in the set of infinitesimals? – sequence May 11 at 0:23

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