# Why is $\infty \cdot 0$ not clearly equal to $0$?

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.

• It is zero in measure theory. Commented Mar 25, 2011 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. Commented Mar 25, 2011 at 12:27
• It is true that a sum of $0$s is $0$. I don't know why that means somebody is missing something. $\qquad$ Commented May 27, 2016 at 18:53
• @MichaelHardy: I thought at first that indeterminate-forms would cover it, but that seems not quite good enough. We need a way to say something like applying intuition about finite expressions to infinite expressions leads to difficulties; think what would be the formal consequences of your suggestion. A finite sum of $0$s is $0$; an infinite sum needs a definition that is shewn to be workable; who says it means somebody is missing something? Commented Jul 11, 2018 at 21:19
• Zero added to itself infinity times is the same as 0 . Infinity which is the same as 0 . (0/0) or 0 . (1/0) or 0 . (2/0)… which is the same as infinity or an undefined number or indeed any number Commented Jun 2 at 8:26

The problem is that the laws of addition and multiplication you are using hold for real 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 $$0$$.

• 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. Commented Mar 25, 2011 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
Commented Mar 25, 2011 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
Commented Mar 25, 2011 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... Commented Jun 16, 2014 at 9:57
• @Jiminion: So what was Ramanujan's answer? Commented Oct 26, 2016 at 17:44

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.

• There are infinite sums that converge, so your answer is wrong. Commented May 27, 2016 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. Commented May 28, 2016 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. Commented May 29, 2016 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. Commented May 30, 2016 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.)

• 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
Commented Sep 1, 2011 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 $-\infty$. Commented Jun 8, 2013 at 5:43
• So you're saying 0 is not 0? How about "You can't limit infinity."? Commented May 5, 2014 at 14:34
• I'm not agree. You write lim(f(x)×g(x)) and then say it's the same as lim(n×0). It's totally not the same. And you also should prove that lim(f(x)×g(x)) = lim(f(x))×lim(g(x)). However if we have f(x)×g(x)=1 and f(x)×g(x)=2 you will have different results (1 and 2 in these cases). But sure n×0=0. Commented Oct 20, 2016 at 16:22

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?

• 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. Commented May 27, 2016 at 15:44
• @AndreasK. they confuse $0$ with $\lim_{x \to 0} (x)$ in all their "proofs". $0 \times \infty$ actually is $0 \times \lim_{x \to \infty} (x)$ which is clearly $0$. Commented Oct 28, 2020 at 19:45
• The issue is that $0\times\lim_{x\rightarrow\infty}(x)$ is not clearly $0$. The manipulation $a\times\lim\limits_{x\rightarrow\infty}f(x)=\lim\limits_{x\rightarrow\infty}(a\times f(x))$ is only sensible assuming both limits exist. Commented Dec 31, 2022 at 20:59

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.

The symbol $\infty$ was originally introduced by Wallis in the 17th century. He used it to denote a specific infinite number, and went on to consider partitions of intervals into $\infty$ parts of width $\frac{1}{\infty}$ 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.

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?

• 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. Commented May 27, 2013 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. Commented May 27, 2013 at 12:41
• [...] because multiplying anything by the constant zero sequence is the constant zero sequence. Without showboating with non-standard analysis. Commented May 27, 2013 at 12:45
• Yes. It is obvious that the OP which asked the possibly most naive question in mathematics about infinity meant it in the context of non-standard analysis. That makes a lot of sense. Commented May 27, 2013 at 13:34
• Don't pretend to know what I believe in, and what are my beliefs regarding one thing or another. It's not only a huge mistake on your side, and an obvious lie to yourself, it's also extremely insulting to me. Commented May 27, 2013 at 13:44

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!

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.

The problem with most of the undefined numbers is that they can't be given a unique value. There are infinitely many numbers satisfying those properties. Start with $\frac{0}{0}$. Let's say it equals $x$, which means $0\cdot x=0$. So, for each value of $x$, that equation is satisfied, so we can't get a unique $x$. Another one is $\frac{\infty}{\infty}$ which is equal to $\frac{\frac{1}{\infty}}{\frac{1}{\infty}}$=$\frac{0}{0}$, so for similar reasons, this can't be uniquely defined either.

Now, another one is $1^{\infty}$. Intuitively , you might say that multiplying 1 by itself an infinite number of times would still produce 1. But, let's say $1^{\infty}=x$, so $1=x^{\frac{1}{\infty}}$ or $1=x^0$. Since every $x$ satisfies this too, so we can't get a unique $x$ in this case either.

Another one is $0\cdot \infty$. Let's say $0\cdot \infty=x$ or $\frac{x}{\infty}=0$ or $x\cdot 0=0$ because $\frac{1}{\infty}\rightarrow 0$. So, we can't get a unique $x$ in this case either.

In fact, $$\infty\cdot 0=\infty\cdot \frac{1}{\infty} (\text{because } \frac{1}{\infty}\rightarrow 0) =\frac{\infty}{\infty} =\frac{0}{0}=\text{undefined}$$ (because all of these can't be given a unique value). So, it seems that all of these indeterminate forms are basically the same thing. But I'm not sure if $1^{\infty}$ can also be proved to be equal to $\frac{0}{0}$ or not.

• This is not really an answer. Just a list of indeterminate forms. Not to mention that $\infty$ is not a number, and $0$ is in fact a well-defined number. Commented Mar 6, 2017 at 4:32
• You should never say something is "equal to 'undefined' "; rather you should say that it is undefined. $\qquad$ Commented Oct 2, 2017 at 20:11

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.

• The approach I like for infinite sums is to not regard them as having a special value, but instead treat map infinite sums of reals to reals, by recognizing certain forms of infinite sums, like Σ2ⁿ (n∈ℤ), as equivalent to zero and deriving the rest algebraically from that. For example, a sum like 1+2+4+8... would be zero minus (1/2+1/4+1/8+1/16...), i.e. -1, but math on such sums would work as it should. Adding 1+2+4+8... to itself would yield 2+4+8+16... which is zero minus (1+1/2+1/4+1/8...), i.e. -2. If one takes this approach, multiplying any infinity of this form by zero would yield 0. Commented Jun 21, 2018 at 22:06

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

• 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 $0$. Commented May 27, 2016 at 15:55
• @Andyk : You see to make assumptions that are too strong about the meaning of the question. It said "someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy". In other words, there is a lot of vagueness in the question, but there is some evidence that something other than what you're talking about may have been intended. Commented Oct 2, 2017 at 20:12