One divided by infinity is not zero?

Question

I know that $\frac{1}{\infty}$ is undefined. But my question is - can we say that $\frac{1}{\infty}\neq0$ ?

I've got some idea how to explain that: Let's say we have a random-number generator that generates numbers in interval $(0, \infty)$. What is the probability that it generates 5? $\frac{1}{\infty}$. So it cannot equals zero because if it was zero, it wouldn't generate any number (every number has the same probability - $\frac{1}{\infty}$).

Am I right? Or is that wrong idea?

Thanks

There is something I don't understand. You start with "$1/\infty$ is undefined" and then ask whether something can be said about $1/\infty$ (which is undefined...). How can we say something about a thing that isn't defined? — user37238, Jul 24 '15 at 13:52
"Undefined" isn't another number. It means that the expression doesn't have any value. You can't say the reciprocal of infinity is not equal to zero, because it is undefined, since as pointed out already, infinity is not a number. — Cataline, Jul 24 '15 at 13:53
Just because something can happen doesn't mean it has non-zero probability. (It's also impossible to generate a uniform probability distribution over the reals, by the way.) — fkraiem, Jul 24 '15 at 14:00
Zero probability is not the same as impossible. $1/\infty$ is undefined, so you have to define what you mean, and you seem to be equating it with zero in a sense. That's not a completely wrong interpretation of it as long as you have a deep enough understanding of the underlying structure. — jdods, Jul 24 '15 at 14:02
When we say $\frac{1}{\infty}$ is "undefined", we mean it makes as much sense to say as, for example, $\frac{1}{\mathrm{cow}}$. You shouldn't be performing any operations with undefined objects — Theo Bendit, Jul 24 '15 at 14:03
You'll see the infinity symbol mostly on the right-hand side of the last equation, where the computation stops. It represents an idea, and shouldn't be used in further symbolic manipulation. — Todd Smith, Nov 09 '15 at 00:36

score 23 · Accepted Answer · answered Jul 24 '15 at 14:02

It is worth noting that since $\infty$ is not a number, the expression $\frac{1}{\infty}$ is not meaningful. That is, you cannot evaluate this expression. It is not a number. You can think of it in much the same way as $\frac{1}{0}$ --- an expression without any interpretation.

The only means that we have to talk about expressions involving $\infty$ is through the concept of limits. The shorthand $$ \lim_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0 $$ is sometimes used in a careless way by students, but the middle expression in this string of equations isn't strictly meaningful. It is only a device used to remind you that we are looking at what happens as we divide 1 by increasingly large numbers.

So, I suppose that in some sense, you can say that $\frac{1}{\infty} \neq 0$, as the left-hand side here isn't any number at all.

More info: Dirac function is not really a function, it's functional. But you can treat it as function, especially if you are physicists, especially if you are Dirac. — Ooker, Jul 24 '15 at 16:52

score 6 · Answer 2 · answered Jul 24 '15 at 14:02

6

Your argument makes use of stochastics. If you want to do things the mathematically exact way, you would first need to define a probability measure on $(0,\infty)$. Assuming that you choose a continuous distribution, then yes, the Probability that you hit $5$ is $0$ - but there are infinitely many numbers, so there is no contradiction. (Keep in mind that $0 \cdot \infty$ is not well defined, and unless in the case of $\frac1{\infty}$ there is no heuristical way to define it)

Check out Measure Theory: https://en.wikipedia.org/wiki/Measure_%28mathematics%29

answered Jul 24 '15 at 14:02

supinf

13,433

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uncountable many numbers! :-) – Ant Jul 24 '15 at 14:08
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This is the best answer so far. All the others are caught up in the notational issues of $1/\infty = 0$, but you address the probability aspect of the question, which is (I think) the more significant. – Tim Pederick Jul 24 '15 at 15:12

Ivo Terek · Answer 3 · 2015-07-24T14:00:15.167

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Operations in the real line are maps: \begin{align} +&\colon \Bbb R \times \Bbb R \to \Bbb R \\ -&\colon \Bbb R \times \Bbb R \to \Bbb R \\ \cdot &\colon \Bbb R \times \Bbb R \to \Bbb R \\ \div &\colon \Bbb R \times (\Bbb R \setminus \{0\})\to \Bbb R \end{align}

But $\infty \not\in \Bbb R$, so there is no sense trying to do things like $1/\infty$ in this context. Any pair on which $\infty $ appears is not on the domain of the functions above.

edited Jul 24 '15 at 14:00

answered Jul 24 '15 at 13:54

Ivo Terek

77,665

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There may be a problem with division on $\mathbb{R}\times \mathbb{R}$, no? – user37238 Jul 24 '15 at 13:56
Yes, I was too fast in copy and pasting, fixed – Ivo Terek Jul 24 '15 at 13:57

score 2 · Answer 4 · answered Jul 24 '15 at 15:53

$\infty$ is not a real number (it's not in $\mathbb R$). It's not even an imaginary or complex number. Basically, $\infty$ isn't one of the things people usually call "number," and, as such, $+$, $-$, $\times$, and $\div$ are undefined for it.

However, there are ways to get around this. Define the set $\overline{\mathbb R}$ to be $\mathbb R\cup\{-\infty,\infty\}$, where $\infty$ and $-\infty$ are just meaningless symbols. Then define what $+$, $-$, $\times$, and $\div$ mean in this set. Really, you could define them to be almost anything, as long as you don't end up with contradictions, but a reasonable definition is outlined here. In the system constructed there, $\frac1\infty=0$. However, it seems kind of artificial, since we've basically just added a symbol, called it "infinity," and made it work like we hope an actual infinity would work.

Another thing is called "potential infinity." Consider the following equation: $$\lim_{x\to\infty}\frac1x=0$$ This is read as "the limit as $x$ goes to infinity of $\frac1x$ equals $0$." What this means is, as $x$ gets larger and larger, $\frac1x$ gets closer and closer to $0$. That's why this is called a "potential" infinity — it asks what happens as $x$ gets closer to infinity, without really getting there. This "limiting" operation happens entirely within $\mathbb R$, the set of real numbers. The above equation is as close to your question as you can get with limits.

Then there are things called "cardinalities," which are too complicated for me to get into. (The basic idea is that they measure sizes of sets, sets which might be infinite.) Division is not defined for cardinalities.

By the way: A set $S\subseteq\mathbb R$ is called a "neighborhood" of a point $x$ if there is an interval $(a,b)$ such that $x\in(a,b)\subseteq S$. If we want to extend the notion of "neighborhood" to $\overline{\mathbb R}$, we could define the neighborhoods of $\infty$ to be sets such that there is an $a$ such that $(a,\infty]\subseteq S$, and we could define the neighborhoods of $-\infty$ to be sets such that there is a $b$ such that $[-\infty,b)\subseteq S$. — Akiva Weinberger, Jul 24 '15 at 16:03
(The notation $[\infty,a)$ doesn't make sense in $\mathbb R$ but it does it $\overline{\mathbb R}$.) — Akiva Weinberger, Jul 24 '15 at 16:03

score 0 · Answer 5 · answered Jul 24 '15 at 13:57

In some contexts it makes sense to say $1/0=\infty$ and $1/\infty=0$. If you're working in a context in which there is just one $\infty$ that can be approached in either the positive or the negative direction, then it makes sense. This makes sense for the values of (but not the arguments to) the tangent and secant functions. For rational functions it makes sense for both arguments and values.

score 0 · Answer 6 · answered Jul 24 '15 at 15:31

First: Can $1/\infty = 0$, or $1/\infty\ne0$? Neither is true, according to ordinary algebraic conventions, but there are two ways in which the former can be correct.

Notational abuse. We can't use $\infty$ like a number in our expressions, because it isn't a number in any of the usual senses. But let's be honest, everyone knows what you mean when you say $1/\infty$; you mean $\lim_{n\to\infty}\frac{1}{n}$, which does equal zero.
Choose-your-own-conventions. I find that one of the cool things about maths is that you can come up with your own system of definitions, axioms, and notations, and as long as it's consistent, it's maths. (And if people take notice and find it useful, then it's good maths. That's why we have things like imaginary numbers: "You can't take the square root of a negative number." "Okay, but what if we could?")

I'm not aware of any commonly-accepted algebra in which infinity is an ordinary quantity and can be operated upon (as opposed to a geometry in which this is the case: see projective geometry). But that doesn't mean that one couldn't exist... and if it did, I'm pretty sure that $1/\infty$ would be zero rather than non-zero.

Second: If random variable $X$ has infinite range, how can $P(X=x)$ be non-zero for any $x$? This is actually a very important question and does not at all depend on the matter of $1/\infty$.

What it boils down to is this: yes, $P(X=5)=0$. However, the probability that $X$ is in any range of values (e.g. $P(4.99\lt x\lt5.01)$) is non-zero, and you can make those bounds arbitrarily close to 5.

Wikipedia's probability density function article has a pretty good explanation. To quote:

Suppose a species of bacteria typically lives 4 to 6 hours. What is the probability that a bacterium lives exactly 5 hours? The answer is actually 0%. Lots of bacteria live for approximately 5 hours, but there is negligible chance that any given bacterium dies at exactly 5.0000000000... hours.

jdods · Answer 7 · 2015-07-24T14:17:56.737

If we are to understand "$=$" as a binary relation of identity and "$\neq$ as a binary relation of non-identity, then in most contexts, formally, $1/\infty$ is not identical to zero. Therefore it would make logical sense to write $1/\infty \neq 0$ as long as we "define what we mean by $1/\infty$". This is more in a strictly logical sense than in a mathematical sense.

In order to interpret $1/\infty \neq 0$ mathematically, you have to carefully describe the underlying structure of the statement.

If we are to only consider "$=$" and "$\neq$" for numbers or algebraic expressions involving numbers, then $1/\infty \neq 0$ doesn't make sense because the left side isn't a number. However, if you define $1/\infty$ appropriately (e.g. as a limiting procedure), then the statement makes sense and can be evaluated as false.

One divided by infinity is not zero?

7 Answers7