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

Question

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.

Answer 1

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 $0$.

Answer 2

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.

Answer 3

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.)

Answer 4

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?

Answer 5

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.

Answer 6

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!

Answer 7

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.

Answer 8

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?

Answer 9

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.

Answer 10

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.

Answer 11

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