What is wrong with the following argument (if you don't involve ring theory)?

Proposition 1: $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}$ is not equal to $0$

$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction

Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$

Therefore, $\frac{0}{0} = 0$.


Update (2015-12-01) after your answers:

Proposition 2: $\frac{0}{0}$ is not a real number

Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:

Suppose that $\frac{0}{0}= x$, where $x$ is a real number.

Then, either $x = 0$ or $x$ is not equal to $0$.

1) Suppose $x = 0$, that is $\frac{0}{0} = 0$

Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $


Therefore, it is not the case that $x = 0$.

2) Suppose that $x$ is not equal to $0$.

$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction

Therefore, it is not the case that $x$ is a real number that is not equal to $0$.

Therefore, $\frac{0}{0}$ is not a real number.


Update (2015-12-02)

If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.

Proposition 3: $\frac{0}{0}$ is not a real number

Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.

$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$

$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$


Update (2015-12-07):

How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?

Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$. Then, $\frac{0}{0} = 0$

Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.

$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$


Suggested definition of division of real numbers:

If $b \ne 0$, then

$\frac{a}{b}=c$ iff $a=bc$

If $a=0$ and $b=0$, then


If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.

A somewhat more minimalistic version:

Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.

Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.

$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$

$\therefore \frac{0}{0}=0$


  • 12
    $\begingroup$ Ever look at $\frac{\sin x}{x}$ as $x\to0$? That would be a limit that looks like $\frac{0}{0}$ yet equals one. $\endgroup$
    – JB King
    Commented Dec 1, 2015 at 15:36
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    $\begingroup$ @JBKing: Limits are not the same as evaluations. The former fails to address this question. $\endgroup$ Commented Dec 1, 2015 at 18:42
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    $\begingroup$ Lennart I understand not knowing basic TeX but you could have tried to copypaste the syntax in my edit to the first part before in posting the second... $\endgroup$ Commented Dec 1, 2015 at 21:04
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    $\begingroup$ 'Suppose "fish" is not equal to 42. That means "fish" is equal to some x for which x =/= 42. But there is no such x, therefore "fish" is equal to 42.' $\endgroup$ Commented Dec 2, 2015 at 4:48
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    $\begingroup$ 0/0 + 1/1 = (0*1)/(0*1) + (1*0)/1*0) Why do you think you can multiply both numerator and denominator of a ratio by 0 and the ratio's value will preserve? $\endgroup$
    – Cthulhu
    Commented Dec 2, 2015 at 10:24

16 Answers 16


The error is in the very first implication. If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$. There is no reason to believe that we can do this division and get a number.

Therefore you have a proof that $0/0$ cannot equal any nonzero $x$. Combine this with a proof that $0/0$ cannot equal zero, and you have proved that $0/0$ is not a number.

  • 20
    $\begingroup$ This. Merely saying something doesn't make it true or exist (or denial make it false either), and writing something down does't make it well-formed. $\endgroup$ Commented Dec 2, 2015 at 23:54
  • $\begingroup$ There is no proof above that 0/0 cannot equal to 0. Or more precisely, there is a proof but it is wrong, which has been indicated. $\endgroup$
    – Anixx
    Commented Dec 7, 2015 at 22:27
  • $\begingroup$ Waiting for a general theorem of non-numbers $\endgroup$
    – devinbost
    Commented Aug 8, 2023 at 10:25

Your proof assumes that $0/0$ is a number since your argument involves arithmetic operations.

Now, if we assume the usual rules of arithmetic for the rationals, why can't we do this: $$ \frac00+\frac11=\frac{0\times1+1\times0}{1\times0}=\frac00. $$ Then $$ 1=\frac11=0. $$

  • 3
    $\begingroup$ Thanks, this is exactly what I was looking for. Then one could construct an elementary proof of the proposition: There is no real number x such that 0/0 = x. $\endgroup$
    – Lennart
    Commented Dec 1, 2015 at 15:44
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    $\begingroup$ Best elementary proof by far. $\endgroup$
    – Insane
    Commented Dec 1, 2015 at 16:07
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    $\begingroup$ This proof presupposes that $\frac{1}{1}=\frac{1}{1}\frac{0}{0}$, which is needed to add the fractions. But this isn't the case, unless $\frac{0}{0}=1$ (and OP proved that it isn't). i.e., this argument does not prove that $\frac{0}{0}\neq 0$. $\endgroup$
    – vadim123
    Commented Dec 1, 2015 at 16:39
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    $\begingroup$ @vadim123: that depends on how you define what $\mathbb Q$ is, and how you define the addition of rationals. I personally define $$\frac ab+\frac cd$$ to be $$\frac{ad+bc}{bd}.$$ $\endgroup$ Commented Dec 1, 2015 at 19:01
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    $\begingroup$ This is why these sorts of insanities show up when trying to reconcile theories of physics. 0/0 is like a black hole - it is simultaneously enormous and yet occupies no space; attempt to use it like a number and it devours everything, which is exactly what numbers aren't supposed to do $\endgroup$
    – J...
    Commented Dec 4, 2015 at 11:47

In order to prove anything whatsoever about $\frac{0}{0}$, we need a definition of $\frac{a}{b}$ where $a$ and $b$ are, say, integers. The correct definition is $ \frac{a}{b} = ab^{-1}$, where the definition of $b^{-1}$ is the number (which is unique, when it exists, according to some basic properties of numbers) such that $bb^{-1} = 1$. But then one can see (again, assuming some basic properties of numbers) that $0^{-1}$ does not exist, so $\frac{0}{0}$ isn't even defined.

  • 1
    $\begingroup$ The mere existence of $0^{-1}$ is contradictory by ring theory arguments (what you refer to as "some basic properties of numbers", correct me if I'm wrong), but that's apparently not what the OP is talking about. $\endgroup$
    – Adar Hefer
    Commented Dec 2, 2015 at 20:51
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    $\begingroup$ @AdarHefer, more precisely, the existence of $0^{-1}$ implies that $1=0$, and hence that the ring is terminal. i.e. $0_R$ is invertible iff $R$ is the terminal ring. $\endgroup$ Commented Dec 3, 2015 at 13:24
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    $\begingroup$ I would say that any argument involving addition and multiplication invokes "ring theory." As others pointed out, the issue is that $\frac{0}{0}$ is not even a number, and I was trying to make the point that it's not a number by definition, rather than for some other reason. A lot of mathematics can be demystified by understanding that everything goes back to definitions. This is a difficult point to get across e.g. to calculus students, because, well, calculus has at its core profound definitions. $\endgroup$ Commented Dec 3, 2015 at 15:50

You can define $\frac{0}{0}$ to be anything you want; it can be $0$, or $1$, or $\pi$. But here's the catch: we want the choice of value for $\frac{0}{0}$ to be compatible with the usual laws of arithmetic.

From this standpoint, what you've really proved is that if $\frac{0}{0}$ is anything except $0$, then the law $2\frac{0}{0} = \frac{2 \cdot 0}{0}$ cannot hold. Therefore, the law: $a\frac{b}{c} = \frac{ab}{c}$ cannot hold, either. This suggests that simply defining $\frac{0}{0}=0$ might actually be a good idea. Unfortunately, this breaks another law of arithmetic, namely $\frac{a}{a} = 1.$

Since we cannot have the best of both worlds, perhaps it is best to simply leave $\frac{0}{0}$ undefined.

Actually, I think the best solution is to define division not of real numbers, but of affine subsets of $\mathbb{R}$. These are: the singleton subsets of $\mathbb{R}$, the empty subset, and $\mathbb{R}$ itself. So by passing to the affine subsets, we've effectively adjoined two new "points", namely $\emptyset$ and $\mathbb{R}$.

Now define that given affine subsets $Y$ and $X$ of $\mathbb{R}$, we have:

$$\frac{Y}{X} = \{r \in \mathbb{R} \mid Y \supseteq rX\}$$

You can check that the result of dividing one affine subset by another will itself always be affine.

Under these conventions, we have:

$$\frac{0}{0} = \mathbb{R}, \qquad \frac{1}{0} = \emptyset$$

This justifies the intuition that trying to divide $0$ by $0$ is somehow different from trying to divide a non-zero number by $0$.

Full disclosure: although passing to the affine subsets works algebraically, I'm not entirely sure how to put a topology or uniform structure or metric on the collection of affine subsets of $\mathbb{R}$. Until we can figure out how to do this, my proposed solution probably isn't that useful.

  • $\begingroup$ Your proposed solution is completely unworkable! For instance, $3/2$ is no longer equal to $1.5$; according to your definition, it equals $\{1.5\}$. And $(1/2)/2 = \{\{0.25\}\}$. Have you really thought this through? $\endgroup$
    – TonyK
    Commented Dec 2, 2015 at 19:35
  • 1
    $\begingroup$ @TonyK, I've edited. $\endgroup$ Commented Dec 3, 2015 at 1:11
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    $\begingroup$ The best part of this answer is the start: you can (or “one could”) define 0/0 how you like, but the problem is how to do it usefully – and of course we want a definition which plenty of people will accept. $\endgroup$
    – PJTraill
    Commented Dec 4, 2015 at 0:04

In my experience, when people have a problem with $\frac 0 0$ it generally means they have not developed a clear idea of the way mathematics forms models.

The character of mathematical systems

At first, every expression a pupil meets can be evaluated to a number they understand, but fairly soon $\frac 0 0$ and $\sqrt -1 $ turn up, yet they cling to the feeling that “if you can write it, it must mean something”. The point has to be made that we define what expressions mean in such a way as to help us solve problems, and sometimes these definitions just can’t be made to cover all cases without implying weird behaviour, and so we agree to consider some cases undefined. This makes maths sound a rather arbitrary process, but the constraint of solving problems actually makes it about the least contingent human activity there is. Of course, having made some definitions, we experiment with their implications, which gives rise to new problems, so pure mathematics feeds off itself as well as off applications. We discover that we can create things in an inevitable way.

Algorithmic v. solution seeking operations

We can distinguish algorithmic operations such as addition, multiplication and raising to powers, from solution-seeking operations such as subtraction, division and the taking of roots. Operations in the first class are defined by some terminating algorithm (such as repeated increment, addition or multiplication), often in the form of a recursive definition, which we show terminates. Operations in the second class are defined as solutions to equations, i.e. they are required to yield numbers which, when combined using previously defined operations, yield a certain result.

More generally, we may have conditions which are not written as equations, and we may be operating on other entities than ‘numbers’ (whatever they may be!1).

New numbers to make insoluble equations soluble

Sometimes we find that our condition can be satisfied by members of the set defined so far. For example, given numbers $a$ and $b$, we want to solve $ a + n =b $ for $n$, and call that $ b - a $; then, since $ 1 + 2 = 3 $, we say $ 3 - 1 = 2 $. At other times, we find that there is no such solution, e.g. there is no natural (counting) number $n$ such that $ 3 + n = 2 $ — this is when we ask ourselves if we can define $ 2 - 3 $. Thus we move from natural to integers.

Given a type of equation we are trying to solve, we hope we can define the new ‘numbers’ in a useful way:

  • They should include the old numbers, or something very like them.
  • We want to be able to apply the existing operations to them.
  • We want them to obey they same rules that the old numbers did.

As goblin’s answer said,

You can define $\frac 0 0$ to be anything you want; it can be 0, or 1, or π. But here's the catch: we want the choice of value for $ \frac 0 0 $ to be compatible with the usual laws of arithmetic.

Actually, though you can define it as you want, it may make you rather unpopular if you insist on something really quirky, so perhaps it is better to say “one could” etc.

Two ways of formalising a new number system

Typically we find that obeying the existing rules means that several expressions, such as $ 2 - 3 $ and $ 3 - 4 $, should have the same value. One way we deal with that is to define equivalence classes of pairs of numbers (if the operation is binary) that should yield the same value; we then show that we can define the various operations on the pairs so that they respect the classes and work as in the original situation.

Another way to deal with it is to introduce one or two new symbols and a syntax for combining them: we might then define integers as formal expressions of the form $+n$ or $-n$, and show that these formal expressions behave as desired. Similarly, introducing $i$ lets us define complex numbers.

These are two important ways to build a formal ‘working model’ of a number system, but there are many others, such as Dedekind cuts, von Neumann’s construction of $\mathbb N$ as ‘free sets’ and Conway’s construction of Surreal numbers.

The error in the question

The problem with $\frac 0 0$ in this question arises partly from failing to recognise that the formal expressions we introduce have only those properties we (consistently) define them to have and anything that follows from those. A bigger problem, though, is not realising that the formal expressions do not automatically mean anything, until we have shown that they reproduce the old numbers as a subset. Once we realise that they do not automatically mean anything, it is easier to accept that our definition may exclude expressions like $\frac 0 0$ (or call them ‘undefined’) in order to ensure desirable behaviour of the new number system.

Another error in the question as first formulated was to assume that the new expressions behave like the numbers they were constructed from, rather than considering that as something to be investigated.

The solution set approach

Goblin’s answer, (after making the essential point that we decide how to define expressions) went on to suggest an alternative definition of division in terms of the solution set of $a \times n = b$. This suggests you consider $ \frac 0 0 $ as $ \mathbb R $ (why not $ \mathbb Q $ or $ \mathbb C $ … or even $ \mathbb Z $ or $ \mathbb N $ ?), but it does not help you deal with $n^2 = -1$, which needs the approach described above.


1 As Dedekind asked, Was sind und was sollen die Zahlen? “What are numbers and what should they be?” or just possibly “… what is the point of them”!

  • $\begingroup$ Good answer. $\;$ $\endgroup$ Commented Dec 4, 2015 at 14:55
  • $\begingroup$ I'm not good at math at all but I could understand this answer. +1 :) $\endgroup$ Commented Dec 5, 2015 at 22:54
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    $\begingroup$ Regarding the last paragraph before the Notes: I like to think of $0_\mathbb{R}$ as different from $0_\mathbb{Q}$, which is, in turn, different from $0_\mathbb{C}$. So $\frac{0_\mathbb{R}}{0_\mathbb{R}} = \mathbb{R}$, but $\frac{0_\mathbb{Q}}{0_\mathbb{Q}} = \mathbb{Q}.$ $\endgroup$ Commented May 5, 2016 at 10:40

How do you define a fraction?

The definition I know is that for any number $a \neq 0$ we say the symbol $\frac{1}{a}$ satisfies $a \cdot \frac{1}{a} = 1$. One can show that this is well defined for any $a \neq 0$ and unique. Further, the symbol $\frac{b}{a}$ for any numbers $b,a$ with $a\neq 0$ is defined as $\frac{b}{a}:=b \cdot \frac{1}{a}$.

If $\frac{1}{0}$ would exists, then clearly by definition we would have $$\frac{0}{0}=0 \cdot \frac{1}{0} = 1.$$ However, the symbol $\frac{1}{a}$ is not definied for $a=0$, because there exists no number $b$ with $0 \cdot b = 1$, so there is no value to which we can assign the symbol $\frac{1}{0}$.

Example of the above definition:

We can compute $\frac{1}{2}$, because we know

$$ 2 \cdot \frac{1}{2} = \frac{1}{2} +\frac{1}{2} =1$$ and since

$$ 2 \cdot 0.5 = 0.5 + 0.5 = 1 $$ it follows that $$\frac{1}{2} = 0.5$$ holds.

  • 1
    $\begingroup$ I feel like this answer would tie together better if your example said "2x1/2 =1" and "2x0.5 = 1". I mean it's obvious they're equal, but I did a double take for a sec, purely based on the different format. $\endgroup$
    – Jeff
    Commented Dec 2, 2015 at 3:21
  • $\begingroup$ @Jeff thanks your right. I edited my answer. $\endgroup$
    – Adam
    Commented Dec 2, 2015 at 9:44

If $\frac00 = 0$ then $0\times0=0$ would work as a proof. The problem is that also $3\times0=0$ and $7\times0$ and so on would all work.

So $\frac00=0,\space\frac00=1,\space\frac00=3,\space\frac00=$ just about anything.

There are too many possible results, we want such operations to return defined results (one and only one result), otherwise the operation itself is not actually defined. In fact, dividing any number by $0$ is 'undefined', 'illegal'.

When dividing a non-zero integer by zero, we have the same problem. only, in that case, we simply have no-results.

ie: $\frac60 =$ number which multiplied by $0$ equals $6$.

We can say $\frac60=\infty$ only to cut short on a theorem, stating that the limit of $\frac mn$ with m integer and $n$ going to $0$ is infinite. It is obviously a notation-artifact. Otherwise we should say that $\infty\times0$ equals $6$, and $7$, and $8$, and $9$...

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    $\begingroup$ I would say that this presupposes that 0/0 = 1 (or the ring definition of division). If 0/0 = 0, then, for example, 3*0 = 0 <=> (3*0)/0 = 0/0 <=> 3*(0/0) = 0/0 <=> 3*0 = 0 <=> 0 = 0. It leads to nowhere, but not to 0/0 = 3. $\endgroup$
    – Lennart
    Commented Dec 1, 2015 at 18:15

The first fault is in the very statement. You assume that there is a $0/0$ in the first place, you have to have that before you even assume that it's not equal to zero.

Then even if there were a $0/0$ you're assuming that there is a $2{0\over0}$ and then that it is equal to ${2\cdot 0\over0}={0\over0}$.

Finally you assume that if $2x=x$ you must have that $x=0$, even if you've got this far there's no reason to think that $x$ is a number and must obey the rules for numbers.

Your updated proposition is true, but your proof has some issues.

In the first case you use the identity ${a\over b}+{c\over d} = {ad+bc\over bd}$ which is only shown to be true if $bd\ne0$.

In the second case you use the identity $a{b\over c} = {ab\over c}$ which is only shown to be true if $c\ne 0$.

That is you use twice identities where the prerequisite is not fulfilled.


The multiplicative semigroup of a field is not a group, but only an inverse semigroup. In the context of semigroups, we call $y$ an inverse element of $x$, if $x=xyx$ and $y=yxy$. An inverse semigroup is a semigroup where each element has an unique inverse element, such that an inverse operation $^{-1}$ can be defined via $x=xyx \land y=yxy \Leftrightarrow y=x^{-1}$. The unique multiplicative inverse element of $0$ in this context is $0$, so $\frac{0}{0}=0\cdot 0^{-1}=0\cdot 0=0$. But note that we also have $\frac{1}{0}=0$.

Note however that the rules for dealing with fractions get more complicated, because the sum of two symbolic fractions is not necessarily a pure symbolic fraction: $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}+0_d\frac{a}{b}+0_b\frac{c}{d}$. Here $0_x:=1-\frac{x}{x}$ is idempotent, because $\frac{x}{x}$ is idempotent, i.e. $0_x^2=1-2\frac{x}{x}+\frac{x^2}{x^2}=1-\frac{x}{x}=0_x$. The details of this standard representation are are worked out in a paper by Bergstra et al.

Inverse semigroups are the appropriate generalization of groups if we look at partial bijections instead of total bijections. In the context of rings, a ring whose multiplicative semigroup is an inverse semigroup is called a strongly regular ring. The category of strongly regular rings is related to the category of skew fields in a similar way as the integers are related to the prime numbers (and the same applies to the relation of the category of regular commutative rings to the category of fields). The name meadow has also been proposed for such rings. I have collected some scattered informations on those subjects in a blog post.

If we add the universal formula $a = 0 \lor (ab = ac \Rightarrow b = c)$ characterizing integral domains to the equations characterizing regular commutative rings, then we end up with a finite set of universal formulas characterizing the category of fields, as I learned from a paper by Hiroakira Ono. From this we can learn that the class of fields is closed under isomorphism, substructures and ultraproducts, see theorem 2.20 in chapter V "Connections with Model Theory" of Burris and Sankappanavar's A Course in Universal Algebra.

Now goblin decided to give an answer with a slightly different spin, but goblin has played with meadows himself in the past. He probably noticed that inversion defined like this is not a continuous function, and so kept on searching. I decided to mention this, not in order to avoid partial functions, but in order to point out that partial functions (and especially partial bijections) do have nice properties on their own, which are different from the properties of total functions.

It is important to keep in mind that this definition of $\frac{x}{0}$ leads to discontinuous operations, so it cannot be used to evaluate limits. And for meromorphic functions, using this definition would even destroy part of the inherent structure, but declaring division by zero as undefined will do so too.


Actually there are solid arguments why $0/0$ can be defined as $0$.

Here is a graphic of the function $f(x,y)=x/y$:

enter image description here

The function is odd against both $x$ and $y$ variables.

Along the axis $x=0$ it is constant zero.

Along the axis $y=0$ it is unsigned infinity, but its Cauchy principal value is constant zero.

Along the diagonal $y=x$ it is constant 1.

Along the diagonal $y=-x$ it is constant -1.

As all odd functions its behavior around zero does not have any preference to positive or negative values. As all odd analytic functions, this function in zero(if defined) can either be zero or unsigned infinity. But unsigned infinity is not a part of affinely extended real line or complex plane, which is usually used in analysis (it is part of projectively extended real line/complex plane). As such, the only variant that remains is $0$.

Note also that we can define a Cauchy principal value of a two-variable function in a point $(x_0,y_0)$ as a limit of the following integral:

$$\lim_{r\to 0}\frac 1{2\pi}\int_0^{2\pi}f(x_0+ r \sin (t),y_0+r \cos(t)) dt$$

This integrates the values of a function over a circle around a point. For analytic functions as long as the radius of the circle r comes to 0, the value of the integral comes to the value of the function in $(x_0,y_0)$. For our function, in $x_0=0, y_0=0$ the value of the integral is constant zero (because the function is odd) and does not depend on r. Consequently, the limit is also zero.

We can extend our analysis to complex numbers, but we again will get the same result, $0$.

  • 1
    $\begingroup$ I had the impression that there is a qualitative difference between the statements "the principal Cauchy value of $p$ is $q$" and "$p = q$"; specifically, that the first statement is weaker and does not imply the second statement. Have I misunderstood something? $\endgroup$
    – David K
    Commented Jan 6, 2016 at 19:57

Using the definition of a Ring, it is pretty simple to show that $$\forall a, 0\times a=a\times 0=0$$ Thus, If we were to define division (only in the case where $c=ab=ba$): $$\forall a, a=\frac{0}{0}$$ Which can be true if and only if your ring is the Zero ring $$\{0\}$$ Therefore, when $\frac{0}{0}$ exists, it is actually $0$ (but there aren't any other numbers in the ring)

  • $\begingroup$ The second statement does not follow from axioms of ring. Division is not defined on rings. $\endgroup$
    – Anixx
    Commented Dec 18, 2015 at 11:49
  • $\begingroup$ You are right, I corrected my answer $\endgroup$
    – Uri Goren
    Commented Dec 18, 2015 at 14:03
  • $\begingroup$ You can define division this way, but it is not the only possible way, actually. One can define 0/0=0 and have everything fine in a ring (not only empty ring). $\endgroup$
    – Anixx
    Commented Dec 19, 2015 at 11:24

The answer to this might be better understood in layman's terms. Going back to grade-school level word puzzles, $\frac{x}{0}$ means:

If you have $x$ items, and you distribute them equally among $0$ people, how many items does each have?

The answer is not a number—it is "I can't answer that" because it rests on an invalid assumption, exposed by noticing that each doesn't refer to any people. The correct response is "each who?" or "there are no people!"

Just because a math problem is expressed using normal numbers and operators, taking the form of other valid math problems to which an answer is possible, doesn't mean that it must have an answer nor that the answer is a number. Division by 0 really means "don't divide."

What's the result of a division where you don't divide? Whatever the answer is, it's $NaN$—not a number.

Using a logic analogy (since math is ultimately just a specialized branch of logic), if someone asks you a question that normally has a true/false answer, such as "Have you stopped beating your wife?" but the premise is wrong because you have never done so, then neither true nor false is a correct answer. It is either the case that the answer is outside the domain of Boolean, or there is no answer at all. One cannot do further Boolean or true-false logic on such a non-Boolean. Attempting to do so is a category mistake.

Similarly, doing any number math with the result of the requested operation $\frac{x}{0}$ is also a category mistake. Proceeding as though it must equal some number $n$ is to act like "What does the color blue taste like?" is a sensible question. But blue doesn't taste like anything, because it isn't a flavor.

  • If you don't have a temperature, what is your temperature?
  • If you punch no one in the nose, how mad will they all be?
  • If you forgot to buy a coffee cup, does your empty cup have 0 ml caffeinated or 0 ml decaffeinated coffee in it?
  • Of all the times you've successfully committed suicide, which one hurt the most?
  • Were all the parents you never had blue-eyed or green-eyed?
  • How many ccs does a little white lie occupy?
  • $\begingroup$ That's silly. What people? You've completely missed the point. To the downvoters: comments please? This is an awesome answer! $\endgroup$
    – ErikE
    Commented Dec 6, 2015 at 20:01
  • $\begingroup$ It's silly to make such a comment. Future down voters may see my comment before downvoting so it had some utility. I understand stackexchange. $\endgroup$
    – ErikE
    Commented Dec 6, 2015 at 20:34
  • $\begingroup$ I think you're missing the point. It's silly to talk about nonexistent humans. You're having trouble breaking out of your familiar, but invalid, frame of reference. That is the exact point my answer makes. You glibly gave an answer to a nonsense question as if you were showing me up! That's why I referred to logic, not math, and tried to take it out of the realm of numbers via a word problem, to prompt seeing the answer through clearer thinking. $\endgroup$
    – ErikE
    Commented Dec 6, 2015 at 20:56
  • 1
    $\begingroup$ No assuming allowed. Math. Fake proof? Strikethrough. Real proof, with logic. $\endgroup$
    – ErikE
    Commented Dec 9, 2015 at 1:55
  • 1
    $\begingroup$ He doesn't want a fake proof. He wants to debunk a fake proof. $\endgroup$
    – ErikE
    Commented Dec 9, 2015 at 8:08

The problem is that you assume $\frac00$ is = some $x$ belong to the real number group.$\frac00$ is just not defined for you to assume it to be $=x$.

Basically its one of the seven indeterminate forms of mathematics.Look here for the complete list-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms

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    $\begingroup$ No, $\frac{0}{0}$ isn't defined in the way that it does not equal any $x$. It just isn't defined. $\endgroup$
    – Thomas
    Commented Dec 7, 2015 at 13:39

Try to cancel 0/0 with different numbers each time you will get different solutions for 0/0 . Suppose 1×0/1×0 you get solution 1. Suppose 2×0/0×1 you get solution 2 . Suppose n×0/0×1 you get solution n. ∵0/0 has infinitely many solutions 0/0 is not defined.


0/0 is indeterminate. Which means it has infinite number of values.

(0)(1) = 0     =>     0/0 = 1
(0)(2) = 0     =>     0/0 = 2
(0)(4) = 0     =>     0/0 = 3 .....

From above it can be shown that 0/0 has infinite number of values which make it meaningless. Hence the term.


Well, then $n \cdot 0=0\; \forall \mathbb{R} \implies \frac{0}{0}=n=\frac{1}{n}$ so, $\frac{0}{0}$ is pretty much any real number. This is definitely not the case. This is why it is undefined.

  • $\begingroup$ I don't accept this argument. If 0/0 = 0, then n*0 = 0 <=> (n*0)/0 = 0/0 <=> n*(0/0) = 0/0 <=> n*0 = 0. It leads nowhere. $\endgroup$
    – Lennart
    Commented Dec 1, 2015 at 17:54
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    $\begingroup$ I would say this definitely is the case, which why it fails to be (uniquely) well-defined. $\endgroup$ Commented Dec 1, 2015 at 18:41
  • $\begingroup$ That is another way to look at it. No matter how you look at it, it doesn't make sense. @Lennart $\endgroup$ Commented Dec 2, 2015 at 1:29
  • $\begingroup$ And where did I accept that $\frac{0}{0}=0$??? It's your argument that doesn't make sense to me. @Lennart $\endgroup$ Commented Dec 2, 2015 at 1:35
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    $\begingroup$ I don't know why you guys keep downvoting this answer. Maybe this is what they call 'reputation bias'. $\endgroup$ Commented Dec 3, 2015 at 3:35

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