Is there any way of treating $\frac{1}{0}$ without breaking maths? I tried just the variable $\lambda$, thinking that it would be easy to manipulate: $$\frac{2}{0} = 2\cdot\frac{1}{0} = 2\lambda$$ But I soon realised that you couldn't divide any numbers by it because $$\frac{2}{\lambda} = \frac{2}{0}^{-1}=\frac{0}{2}=0, \text{ implying that } 2=0\lambda, \text{ or 0.}$$

Is there any system/way of making this work without destroying maths and everything it stands for?

  • 3
    $\begingroup$ If you want to play this game, you should make a distinction between $1\cdot0$ and $2\cdot0$. $\endgroup$ – Yves Daoust Oct 25 '16 at 21:22
  • $\begingroup$ An answer to the question in this link math.stackexchange.com/questions/144715/… refers to "wheel theory" in which division is always defined. $\endgroup$ – Mark Bennet Oct 25 '16 at 21:28
  • $\begingroup$ you can take $\bar{R}:=\mathbb{R}\cup \left\{\infty\right\}$ and define $\frac{1}{0}\cdot r:=\infty$ for all $r\in\bar{R}$ and $\frac{1}{0}+ r:=\infty$. everything depends on what you want to have; if you want $\frac{1}{-0}=-\frac{1}{0}$ and a well defined addition, then you need $\infty=-\infty$, otherwise you would have to define $-\infty+\infty$ and would most probably be getting trouble with associativity then. in the end you will definitely loose properties. you will definitely loose the field property for instance. $\endgroup$ – Max Oct 25 '16 at 21:28
  • 1
    $\begingroup$ What does fair division have to do with the question? $\endgroup$ – Asaf Karagila Oct 25 '16 at 21:32
  • 1
    $\begingroup$ @SimpleArt in Riemann sphere there is only one $\infty$ with no sign the same like zero. en.wikipedia.org/wiki/Riemann_sphere $\endgroup$ – Pentapolis Oct 25 '16 at 22:09

Any rule you can come up with that has you dividing by $0$ will eventually lead to a contradiction. Since there is no satisfactory definition, we have to say that the operation is undefined.

About the best you can do.

If you have a function such as $y=\frac{x^2-1}{x-1}$ you can talk about the behavior of the the function in the neighborhood of $x=1$

When $x = 1$ you are dividing by zero and the function not defined.

But if you look at values of $x$ very near $1,$ you will see that the closer $x$ gets to $1,$ the closer $y$ gets to $2$

When $x$ is in the neighborhood of $1, y$ is in the neighborhood of $2.$

  • $\begingroup$ Aw man! Stealing my example limit. XD $\endgroup$ – Simply Beautiful Art Oct 25 '16 at 21:31

Even in symbolic form, there is no way to include division by zero $\frac{1}{0}$ into the reals, complex numbers, or any (non-zero) ring that does not produce a contradiction given the other ring axioms. It's true in a ring that $0 \cdot a = a \cdot 0 = 0$ for all $a \in R$, but apparently $0 \cdot \lambda = 1$, hence $1 = 0$.

  • $\begingroup$ I really don't understand what rings and axioms are - I haven't been taught about them. @basket $\endgroup$ – user366469 Oct 25 '16 at 21:55
  • $\begingroup$ @OllyBritton Axioms are the rules of math, like why $1+1=2?$ One such rule is a rule of multiplication, which basically goes along the lines $0\times a=0$. These are the "rules", and you aren't supposed to break them. $\endgroup$ – Simply Beautiful Art Oct 25 '16 at 22:33
  • $\begingroup$ Loosely speaking, rings are places where we can add, subtract, and multiply. They are defined in such generality that anything that meets our loosest notions of algebra is a ring, and is subject to arguments that follow from the definition. $\endgroup$ – basket Oct 25 '16 at 22:36

I would say no. By our usual conventions, division by zero is undefined (and even worse, anything like $\frac{0}{0}$ is called an indeterminate form.

Nice try though, it's good to experiment. You'd be breaking established rules in algebraic structure if you try this.


Nice try, mainly I meant nice try in disproving yourself, because when testing the waters, the ability to prove/disprove something based on how you decided to define things is a great way of determining if your definitions make any sense mathematically.

However, instead of dividing by zero, which happens to be impossible, a better problem would be to divide by something small. So small that it is almost $0$, but not quite there yet. For lack of better term, we'll call this magical almost $0$ the number $\epsilon$. Funny thing is we can divide by $\epsilon$, especially if the thing we are dividing by is almost $0$ too.

I want to look at the following function:


Particularly, I want to study it at $x=1$, though this results in division by $0$. So instead, I study the function at $x=1+\epsilon$ and $x=1-\epsilon$.


So I guess we can agree that it makes the most sense that $f(1\pm\epsilon)=2!$ This... idea. It is known as a limit, and it usually only works when we have:


In the event that we have $\frac10$, the numbers will get infinitely big. Try $f(x)=\frac1x$ around $x=0$ for example.

A good note is that we avoid any problems with this idea of division by zero since we aren't actually using $0$. We're just using very small numbers, where all basic math still holds. This way, we can't construct weird proofs of $2=0$.

  • $\begingroup$ @SimplyArt Instead of treating $\frac{1}{0}$, why not treat it as a vector? I know this wouldn't make sense mathematically, but it would work, surely. If $\frac{1}{0} = [x=1, y=0]$. This would mean that multiplication would work, addition would work and division would work (hopefully). $\endgroup$ – user366469 Oct 26 '16 at 17:58
  • $\begingroup$ When I say [x=1, y=0] I mean 1 at the top and 0 at the bottom. $\endgroup$ – user366469 Oct 26 '16 at 17:59
  • $\begingroup$ @OllyBritton That would imply things like $\frac10\ne\frac20$ and, I'm not entirely sure, but $\frac12\ne\frac24$. Specifically, you'd have to set up rules, or axioms, and rings, as basket mentions $\endgroup$ – Simply Beautiful Art Oct 26 '16 at 18:52