I was surprised to see that 1/0 is undefined. One answer mentions that $1/0$ can be +$\infty$ or -$\infty$ depending on whether $0$ is approached by the left or the right:
But why does this make a difference? Aren't both numbers equal by all measurable accounts?

Also, if zero is neither positive nor negative, why $1/0$ does not equal "unsigned infinity", which would be infinity in its own dimension (like imaginary numbers), and hence, $0$ in the real dimension?

Also, if we keep dividing 1 in parts of zero size, won't we have +$\infty$ number of parts?

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    $\begingroup$ Aren't $+\infty$ and $-\infty$ as far away as possible by all "measurable" accounts? $\endgroup$ – Alex Becker Apr 13 '12 at 2:33
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    $\begingroup$ The answers there do not say that 1/0 = anything, so for you to say that it "can be $\pm \infty$" is not a fair representation of those answers. $\endgroup$ – The Chaz 2.0 Apr 13 '12 at 2:34
  • $\begingroup$ @TheChaz What I understood is that if 1 is divided by something that approaches zero from the left or right, we get negative or positive infinity. I'm asking why it makes a difference. $\endgroup$ – Camilo Martin Apr 13 '12 at 2:37
  • $\begingroup$ @AlexBecker Good point, but it doesn't clear up my doubt. $\endgroup$ – Camilo Martin Apr 13 '12 at 2:39
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    $\begingroup$ @CamiloMartin you can define $\frac{1}{0} = \infty$ if you are willing to require that $\infty = -\infty$ and this space has nice topological properties. It is called the one point compactification of $\mathbb{R}$. If you do this in the complex plane, you get the Riemann sphere, which is a remarkably beautiful space. The basic issue with these definitions is that you can no longer perform arithmetic with infinity which we sometimes want to do. Things still don't work nicely with limits, but then they don't work nicely now so that's not much of an argument. $\endgroup$ – Chris Janjigian Apr 13 '12 at 2:48

Changed to an answer and expanded at asker's request.

The problems with extending the definition of division to make $\frac{1}{0}$ meaningful appear when you treat it like any other real number. The real numbers have certain operations defined on them, notably addition, subtraction, multiplication and (except when the denominator is $0$) division. These satisfy certain properties, such as $b\cdot \frac{a}{b}=a$ and $0\cdot a=0$ for any real numbers $a,b$ such that the expressions are defined. If we want these to still hold after defining $\frac{1}{0}$, we would have to let $1=0\cdot \frac 1 0 = 0$! This is absurd, so whatever we build by defining $\frac{1}{0}$ is very different from the real numbers, and lacks at least one of its fundamental properties.

  • $\begingroup$ Chosen for the concise explanation. $\endgroup$ – Camilo Martin Apr 13 '12 at 3:56
  • $\begingroup$ @Alex Becker You forgot that for $\displaystyle b \cdot \frac{a}{b} = a$, $b$ must $\neq 0$. And if $\displaystyle \frac{x}{0} = 0$ then $\displaystyle \frac{0}{x} = \frac{1}{0} = 0$ which make sense. $\endgroup$ – user93957 Sep 30 '13 at 18:14

Another problem here lies in that there doesn't exist a single object like "unsigned infinity". When we write (2+3)=5, there exists a single, unique number 5 which (2+3) equals. But, if we let (1/0) equal unsigned infinity, we've let it equal both positive infinity and negative infinity which are not the same infinity. Multiple infinities don't just happen because we have both positive and negative numbers. If "0" indicates an infinitesimal number, which may or may not equal the real number 0, then (1/0) and (15/0) will equal different hyperreal infinite numbers.

  • $\begingroup$ I would like to know why someone downvoted you. +1 until a justification for a downvote is given. (Also: can one mantain that $1/0$ and $15/0$ are different infinities? How?) $\endgroup$ – Camilo Martin Apr 13 '12 at 14:34
  • $\begingroup$ @CamiloMartin A lot of what I've written elsewhere has evoked negative responses from others for various reasons. The downvoter might have just downvoted, because they saw I wrote the answer... though perhaps not, I don't know. Since (1/0) and (15/0) are both undefined, given that "0" means the number 0, one can legitimately maintain they equal different infinities. If "0" means an infinitesimal number, which may or may not equal the real number "0", then (1/0) and (15/0) will equal different infinite numbers in a hyperreal number system. $\endgroup$ – Doug Spoonwood Apr 13 '12 at 20:23

I think it is mainly due to the sign (positive or negative) of $1/x$ depending on whether $x$ is positive or negative.

There may be more "deeper" answers than the above, but for most practical purposes that is the reason.

  • $\begingroup$ But if 0 does not have a sign, why the result does not go into some imaginary plane and results in 0 in the real plane, or what about the number of times I could divide anything positive by zero being positive infinity? $\endgroup$ – Camilo Martin Apr 13 '12 at 2:41
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    $\begingroup$ @CamiloMartin No reference is made to the sign of $0$, but rather to the sign of $x$ when we examine the expression $1/x$ for small $x$. $\endgroup$ – Alex Becker Apr 13 '12 at 2:42
  • $\begingroup$ @AlexBecker And I suppose just writing 0 without a sign does not denote that it is positive, right? $\endgroup$ – Camilo Martin Apr 13 '12 at 2:47
  • $\begingroup$ @CamiloMartin Correct, $0$ is neither positive nor negative. $\endgroup$ – Alex Becker Apr 13 '12 at 2:51
  • $\begingroup$ @Camilo Remember the limit is not about the "destination" but rather about the "way". Quite a deep thought for a mathematical abstraction, right? $\endgroup$ – Pedro Tamaroff Apr 13 '12 at 3:16

It matters if you compactify the real line with $2$ infinities, as you implicity did. If you use the real line Alexandroff Compactification (which seems like what you were naively assuming when you said "Aren't both numbers equal by all measurable accounts?"), you can actually define $1/0=\infty$ in order to make this function continuous in the whole line (Notice you also put $1/\infty=0$). But you may lose some generality in arithmetic properties by doing that, and have to define things wisely in order to keep the maximum generality possible (or simply don't define those things at all).

  • $\begingroup$ I'll write myself off as "too uninformed to have asked the question in the first place". $\endgroup$ – Camilo Martin Jan 24 '15 at 3:58

You have to be careful here; you're poking at some advanced mathematics. Before delving into the discussion at hand, concerning your "infinity in [its] own dimension" remark, have a look at this. Now, onto why $\frac{1}{0}$ is (for the most part) problematic. We shall begin with a brief discussion of the concept of number and division.

We may only consider the concept of a non-negative integer, as the rest follows :) While there are a number of good, and technical, definitions of this type of mathematical object, it will suffice for us to interpret it as a set of sets. So, for example we may say $0 = \varnothing = \{\}, 1 = \{0\} = \{\{\}\},$ $2 = \{0, 1\} = \{\{\},\{\{\}\}\},$ and so on (Reference: Advanced Calculus, Shlomo Sternberg). Now, we may conveniently define $\mathbb{Z}^*$ (the set of non-negative integers) to be the set containing all objects describable in the form "$\{\{\}_1,\{\{\}\}_2,..., \{...\{\}...\}_{n - 1}\}$", where $n \in \mathbb{Z}^+$ and the indices have only been shown for convenience. Notice, that this disallows the number "$\infty$", since "$\infty$" is defined as (for our purposes) the object such that no number is greater than or equal to it. Indeed, this is problematic given our current framework, hence the link at the beginning. Anywho, we can now also define the binary operators "$+$", "$-$", "$\div$", and "$\times$". I'll leave the details to you, and only focus on "$\div$".

$$ \mathbf{\tag{1} Def:}\ \forall x, y \in \mathbb{Z}, \ {x} \div {y} = \frac{x}{y} = r \in \mathbb{Z}^* : (r \times y = x ) \iff (x \div y = r : y \text{ can be subtracted } r \text{ times from } x) \text{ (i.e., } x - r \times y = 0). $$

Certainly, $x \geq y$ and $y \neq 0$ or $x < y$ and $x,\ y \neq 0$ is a necessary condition for $x \div y$ to be defined. As a corollary to what we have covered so far, $\forall x \in \mathbb{Z}^*,\ x \div 0$ is undefined, for there is simply no integer, $r$, such that $r \times 0 = x$ if $x > 0$; if $x = 0$, $x \div 0$ is still undefined, though, for a subtly different reason which I'll allow you the pleasure of working out for yourself.

The fact that "$\lim_{x \rightarrow \infty} x^{-1} = \pm \infty$" is merely a subtlety of the real number system, the notion of limit, and mathematicians rejecting the idea of defining infinity with coarse machinery.

  • $\begingroup$ As much as this explanation is very well laid out and certainly interesting, I did not understand the formula and the subsequent statement, but that's my fault. Other than that, the final paragraph clears up that my doubt stems from the notion of limit. (I should probably have stated I'm a layperson, but +1 nonetheless for a compendious answer.) $\endgroup$ – Camilo Martin Apr 13 '12 at 14:29

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