Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

First of all, I am aware of all the laws of algebra , so my question is about intuition .

We know that $\frac{x}{y}$ undoes the operation $x \times y$ . Shoudn't $(x/0)$ undo $x \times 0$ ? That means , if $x \times 0$ destroys x , maybe $\frac{x}{0}$ should restore it. A kind of math memory if you like.

To be more clear , I know it's impossible in our current number system as it is , so I am asking , How can we tweak the laws ?

Maybe something like this :

$x \times 0$ = 0 [x] (destroying)

$\frac{0 [x]}{0}$ = x [0] (restoring)

share|cite|improve this question
In a sense that’s exactly why division by $0$ makes no sense: since $x\cdot 0=0$ for all $x$, you’d have to have $\frac00=x$ for all $x$, which is absurd. – Brian M. Scott Jan 27 '13 at 21:47
Why the down votes? – Git Gud Jan 27 '13 at 21:48
@YACP It's a perfectly legitimate question at the right level. Are we imposing a minimal difficulty for a question to be asked here on MSE now? I hope not. – Git Gud Jan 27 '13 at 21:54
@GitGud: Probably because some people downvote questions for style regardless of the substance. I consider this petty and unhelpful, but such matters are entirely up to the individual. – Brian M. Scott Jan 27 '13 at 21:55
If you delete all the previous content of the question, the answers won't make sense since they've lost their context. If you have another question, post a new question. Deleting the question is unfair to those that answered your question. Without a very compelling reason, we won't delete questions with upvoted answers. – robjohn Feb 6 '13 at 19:21

No It should not. First of all, $x\cdot 0=0$, that would mean $\frac{x\cdot0}{0}\cdot0=\frac{0}{0}\cdot0=x$ for EVERY x.

We have this time and time again. There is no way of defining $\frac{0}{0}$ without adding additional information.

About this "additional information": Non-Standard analysis does exactly this. It basically "saves" the original values. Numbers are stored as sequences instead of values and therefore carry additional information that can be used to restore the original number. However, $\frac{1}{0}$ is not well-defined even in NSA, as there is no additional information given.

share|cite|improve this answer

You are absolutely right! Let $R$ be a set of numbers with addition and multiplication where division by $0$ is well-defined.

Suppose that we have unity elemenet $1 \in R$ (if not there is no point of talking about division, at least in my opinion). Then $0 = 0 \times (1/0) = 1$. Therefore, for any $x \in R$, we have $0 = 0 \times x = 1 \times x = x$. Thus $R = \{0\}$.

What we showed is that if division by $0$ is well-defined for a reasonable algebraic structure $R$ with $+, \times$, then such $R$ can contain only one element, which we usually call $0$.

Again, I don't think division by $0$ is impossible, but it forces our number system to become so uninteresting. You can also consider defining a new number system very carefully, but unless you have a good reason and intuition, not many people will be interested in studying it.

share|cite|improve this answer

One I was thinking about numbers as smooth functions on neighborhood of $0$. So $0$ could by written as $x$, $x^2$ .. etc.(but they are different zeroes). They "primary value" would be limit in zero.

Example: $1/0$ could be written as $1/x^2$. And $\frac{1}{x^2} x^2 = 1$.

There are problems as what would $1/x$ mean etc. Probably it would take a lot of time to make definitions precise to get a field. And I guess it wouldn't be very usefull theory but you could divide by zero in some sence.

share|cite|improve this answer
This is what Non-standard analysis basically does. Numbers are represented by monotonuous series. This easily defines a ring over the so-called hyperreal numbers $\mathbb{R}*$ – CBenni Jan 27 '13 at 23:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.