We add negative numbers and zero to natural sequence to make it closed under subtraction, the same thing happens with division (rational numbers) and root of -1 (complex numbers).

Why this trick isn't performed with division by zero?

J. M. ain't a mathematician
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    because it doesn't work as nicely. – Dustan Levenstein Mar 27 '12 at 19:53
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    Did you look over [this previous question](http://math.stackexchange.com/q/87781/742) about defining division by zero? – Arturo Magidin Mar 27 '12 at 19:56
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    You want the distributive law to hold unrestrictely and then it follows that $0\cdot x=0$ for all $x$. Thus it is impossible to construct the reciprocal of $0$, i.e. to divide by $0$. – Andrea Mori Mar 27 '12 at 19:59
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    To continue @AndreaMori s remark: If $j\cdot 0 = 1$ say, and you want the distributive law to hold, then $1 = j \cdot 0 = j \cdot (0 + 0) = j \cdot 0 + j \cdot 0 = 2$. – martini Mar 27 '12 at 20:02
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    As I mentioned in that previous question: you are going to lose things by doing this extension. The question is whether what you gain makes up for what you lose. When going from the integers to the rationals, for example, you lose the fact that any nonempty set of positives has a smallest element. When going from $\mathbb{R}$ to $\mathbb{C}$, you lose the fact that you have an ordered field (but you gain the Fundamental Theorem of Algebra). If you add a "division by zero", you are going to lose a *lot* of the basic algebraic properties you are familiar with; do you gain enough to make it up? – Arturo Magidin Mar 27 '12 at 20:04
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    There is an algebraic structure in which division by zero is always defined. Is is called a [wheel](http://en.wikipedia.org/wiki/Wheel_theory). Note that every commutative ring can be extended to a wheel. (We lose some algebraic properties, though.) – Dejan Govc Mar 27 '12 at 20:08
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    @DejanGovc Oh my God, this is so cool. How have I not learned of this before. I don't think I've ever read a mathematical paper with this much excitement. Thank you. – 000 Mar 28 '12 at 00:14
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    @user22144: glad to help =) – Dejan Govc Mar 28 '12 at 00:22
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    @izhak Here is an attempt to extend the Naturals - http://math.stackexchange.com/questions/1043958/does-these-arithmetic-rules-work-they-extend-the-number-system-by-a-zero-not-ba – Jonathan Cender Dec 01 '14 at 10:33
  • @DejanGovc That is a bit misleading. The "division" defined in wheels is not really what most people mean when they refer to division. In this structure, "divsion" just refers to some kind of abstract, general involutive homomorphism satisfying certain distributive laws. It does not, however, refer to an operation that serves to invert multiplication, which is actually what most mathematicians mean when they use the word "division." Wheel theory makes use of an idiosyncratic definition of the word. So it is not accurate to say that division by $0,$ in the sense that we want, is well-defined. – Angel May 02 '22 at 19:39

7 Answers7


You can add division by zero to the rational numbers if you're careful. Let's say that a "number" is a pair of integers written in the form $a\over b$. Normally, we would also say that $b\not=0$, but today we'll omit that. Let's call numbers of the form $a\over 0$ warped. Numbers that aren't warped are straight.

We usually like to say that ${a\over b} = {c\over d}$ if $ad=bc$, but today we'll restrict that and say it holds only if neither $b$ nor $d$ is 0. Otherwise we'll get that ${1\over 0} = {2\over 0} = {-17\over 0}$, which isn't as interesting as it might be. But even with the restriction, we still have ${1\over 2}={2\over 4}$, so the straight numbers still behave as we expect. In particular, we still have the regular integers: the integer $m$ appears as the straight number $m\over 1$.

Addition is defined as usual: ${a\over b} + {c\over d} = {ad+bc\over bd}$. So is multiplication: ${a\over b} \cdot {c\over d} = {ac\over bd}$. Note that any sum or product that includes a warped number has a warped result, and any sum or product that includes $0\over 0$ has a the result $0\over 0$. The warped numbers are like a hole that you can fall into but you can't climb out of, and $0\over 0$ is a deeper hole inside the first hole.

Now, as Chris Eagle indicated, something must go wrong, but it's not as bad as it might seem at first. Addition and multiplication are still commutative and associative. You can't actually prove that $0=1$. Let's go through Chris Eagle's proof and see what goes wrong. Chris Eagle starts by writing $1/0 = x$ and then multiplying both sides by 0. 0 in our system is $0\over 1$, so we get ${1\over 0}{0\over 1} = x\cdot 0$, then ${0\over 0} = x\cdot 0$. Right away the proof fails, because it wants to have 1 on the left-hand side, but we have $0\over 0$ instead, which is different.

So what does go wrong? Not every number has a reciprocal. The reciprocal of $x$ is a number $y$ such that $xy = 1$. Warped numbers do not have reciprocals. You might want the reciprocal of $2\over 0$ to be $0\over 2$, but ${2\over 0}\cdot{0\over 2} = {0\over 0}$, not ${1\over 1}$. So any time you want to take the reciprocal of a number, you have to prove first that it's not warped.

Similarly, warped numbers do not have negatives. There is no number $x$ with ${1\over 0}+x = 0$. Usually $x-y$ is defined to be $x + (-y)$, and that no longer works, so if we want subtraction we have to find something else. We can work around that easily by defining ${a\over b} - {c\over d} = {ad-bc\over bd}$. But then we lose the property that $x - y + y = x$, which only holds for straight numbers. Similarly, we can define division, but if you want to simplify $xy÷y$ to $x$ you'll have to prove first that $y$ is straight.

What else goes wrong? We said we want ${a\over b} = {ka\over kb}$ when $a\over b$ is straight and $k\not=0$; for example we want ${1\over 2}={10\over 20}$. We would also like ${a\over b}+{c\over d} = {ka\over kb} + {c\over d}$ under the same conditions. If $c\over d$ is straight, this is fine, but if $d=0$ then we get ${bc\over 0} = {kbc\over 0}$. Since $bc$ could be 1, and $k$ can be any nonzero integer, we would have ${p\over 0} = {q\over 0}$ for all nonzero $p$ and $q$. In other words, all our warped numbers are equal, except for $0\over 0$. We have a choice about whether to accept this. The alternative is to say the law that $a + c = b + c$ whenever $a = b$ applies only when $c$ is straight.

At this point you should start to see why nobody does this. Adding a value $c$ to both sides of an equation is an essential technique. If we throw out techniques as important as that, we won't be able to solve any problems. On the other hand if we keep the techniques and make all the warped numbers equal, then they don't really tell us anything about the answer except that we must have used a warped number somewhere along the way. You never get any useful results from arithmetic on warped numbers: ${a\over 0} + {b\over 0} = {0\over 0}$ for all $a$ and $b$. And once you're into the warp zone you can't get back out; the answer to any question involving warped numbers is a warped number itself. So if you want a useful result out, you must avoid using warped numbers in your calculations.

So let's say that any calculation that includes a warped number anywhere is "spoiled", because we're not going to get any useful answer out of it at the end. At best we'll get a warped answer, and we're most likely to get $0\over 0$, which tells us nothing. We might like some assurance that a particular calculation is not going to be spoiled. How can we gain that assurance? By making sure we never use warped numbers. How can we avoid warped numbers? Oh... by forbidding division by zero!

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    Would it be fair to say that warped numbers are actually like $\mathbb{C}$? Ignored heavily in particular audiences and given a particular field of attention (i.e. Complex Analysis)? (We're speaking hypothetically, of course! :)) – 000 Mar 27 '12 at 21:42
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    No, I don't think it would be fair to say that. $\mathbb C$ is enormously useful and you sacrifice nothing when you use it. The warped numbers are nearly worthless. – MJD Mar 27 '12 at 21:43
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    Well, yes, the comparison diverges in that respect. However, I saw it only based on the way that people treat $\mathbb{C}$ in particular audiences. More specifically, complex numbers are almost treated like terrible creatures in highschool. – 000 Mar 27 '12 at 21:45
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    This is an interesting construction, something like constructing the field of fractions but giving a different structure. Does it have a standard name? – Nate Eldredge Mar 28 '12 at 17:35
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    @000 they really shouldn't be considered like that though; complex numbers *extend* the reals and maintain most useful properties. *If they did screw up like this* the comparison would be fair, but then no one would have bothered defining $i$ in the first place. – Robert Mastragostino Jun 20 '13 at 01:35
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    It's fair to call $0/2$ the recripocal of $2/0$; but it wouldn't be fair to call it the *inverse* of $2/0$. –  Aug 22 '14 at 06:20
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    There may be some warping going on [here](http://math.stackexchange.com/questions/1548487/is-this-a-correct-good-way-to-think-interpret-differentials-for-the-beginning-ca/1590019#1590019), but I can't be sure... –  Dec 29 '15 at 00:29
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    @MJD How do you get -17? –  Dec 30 '15 at 22:40
  • Your answer sounds like you are describing a tough problem that needs to be solved, rather than conclusive proof that the problem doesn't exist. – Giffyguy Aug 12 '16 at 03:04
  • I like your approach and assessment however; I think of $\frac{0}{0}$ differently. – Francis Cugler Mar 20 '17 at 21:26
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    I nearly didn't read this question, but this answer has made me glad I did. Excellently done. – Stella Biderman Mar 30 '17 at 17:38
  • I was actually thinking of something like $a+bk$ where $a,b \in \mathbb{R}$ for some arbitrary $k = \frac{1}{0}$. – Nirvana Mar 17 '21 at 06:44

I would also like to point out that the IEEE floating-point standard allows division by zero. It incorporates three nonstandard numbers, called infinity, negative infinity, and "NaN", which is short for "Not a Number". $a\over 0$ is infinity, negative infinity, or NaN according as whether $a$ is positive, negative, or zero, respectively.

Infinity represents an overflow condition. So you might add two very large, positive numbers and get a result of infinity. As a result, IEEE floating point numbers fail to obey many of the usual mathematical laws. For example, $(a+b)+c = a+(b+c)$ may fail: suppose $a = b = -c$ and all are large. If $a+b$ overflows, the left-hand side of the equality will be infinity, but the right-hand side will be the finite quantity $a$.

The behavior in some cases can be very subtle and counter-intuitive. But it is widely used in practice.

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    Maybe it would be a great addition to elucidate on, "_The behavior in some cases can be very subtle and counter-intuitive. But it is widely used in practice._" I find myself quite curious about why it is widely used if it is counter-intuitive. – 000 Mar 27 '12 at 21:33
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    It's widely-used because it's the only thing that can be feasibly constructed in hardware. If your hardware can store numbers between -32768 and +32767, and you add 20000 + 20000, it has to do something, and that something is to report an overflow condition. But overflow (and underflow as well) depends on the order in which the operations are performed, as my example shows. So the numbers do not always behave like mathematically nice numbers, and that is the counter-intuitive part. – MJD Mar 27 '12 at 21:41
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    [Rather apropos...](http://xkcd.com/571/) – J. M. ain't a mathematician Mar 28 '12 at 16:35
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    It is widely used, in that next to nobody really thinks about overflow or rounding errors. And all hell breaks loose when it turns up... – vonbrand May 22 '14 at 23:05
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    Really wish they had used 0/0 = 0 as well. That way 0/a would be 0 for all a, and we would no longer have to deal with division by zero errors with floating point arithmetic at all. It seems fairly intuitive--for example, with averages and percentages: averaging zero elements results in an average of 0 and when you attempt something 0 times you have a success and failure rate of 0%. – Muhd Aug 07 '15 at 19:05
  • @Muhd and 100%. :D – Hakanai Mar 06 '22 at 19:43
  • @Muhd That is only useful and accurate in very specific circumstances. It is not worth adopting as a universal convention. – Angel May 02 '22 at 19:45

Here's a proof that $1/0$ can't exist:
Suppose $1/0=x$.
Then $1=0 \cdot x$.
So $1=(0+0) \cdot x$.
So $1=0\cdot x + 0 \cdot x=1+1$.
So $0=1$. Contradiction.

Thus, if we're going to extend our number systems so that dividing by $0$ makes sense, we need to change things so one step in this proof doesn't work. Let's have a look at what properties the proof used.

First I assumed that if $a/b=c$, then $a=b \cdot c$. This is the defining property of division. If we give it up, the resulting thing won't deserve to be called division at all.

Then I assumed that $0=0+0$. Making this false would be a pretty radical redefinition of addition: none of the extensions you mention involve redefining addition for the numbers we have already.

Next I used that $(a+b)\cdot c=a \cdot c + b \cdot c$. This is a key relationship between addition and multiplication. You could give this one up, but the result will be pretty weird.

Next I used that, if $a=a+b$, then $0=b$. Again, you could give up this key property of addition, but you'll get a weird structure as a result.

Finally I used that $1$ does not equal $0$. Again, an "extension" which messes with the numbers we have already like that is a pretty odd extension.

In short, while you can extend your number system to allow division by $0$, doing so requires breaking far more fundamental rules of logic and arithmetic than needed in constructing the rationals or the complexes. As a result, the structures you get are not very pleasant and not all that useful.

Chris Eagle
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    Why don't you stop your proof after $1=0\cdot x$ therefore $1=0$? – Gerenuk Mar 27 '12 at 20:17
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    Because I want to explain exactly why $0 \cdot x=0$ is something we want to keep, since to my mind it's less obviously important and less fundamental than the other properties I mention. – Chris Eagle Mar 27 '12 at 20:19
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    You can stop at $1=0\cdot x + 0 \cdot x$ because you already know that $0 \cdot x=1$... – N. S. Mar 27 '12 at 21:19
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    @N. S. good point – Chris Eagle Mar 27 '12 at 21:20
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    Isn't $0\cdot x=0$ the definition of zero, so whenever you use $0$ you exactly mean this property? – Gerenuk Mar 28 '12 at 08:05
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    No, $0+x=x$ is the defining property of $0$. – Chris Eagle Mar 28 '12 at 10:08
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    Of course, if you to define 1/0 it would probably be +infinity, in which case 0*x is no longer a valid operation when x is +/- infinity, rendering your proof invalid. So, all this means is that if we allow division by zero, we will need to handle infinities properly as well. Breaking logic is not required, but one will need to be careful with certain cases of arithmetic, just as you do now with division by zero. – Muhd Aug 07 '15 at 18:58
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    Obviously, you would also have to change the definition of division. – Muhd Aug 07 '15 at 19:07
  • @Muhd If you are changing the definition of division, then the operation you are working with is, strictly speaking, no longer division, but some other operation. And there does not exist a consistent way of doing arithmetic with $-\infty$ and $\infty.$ – Angel May 02 '22 at 19:46

The "trick" of extending the number system you ask about is addressed by Patrick Suppes in his Introduction to Logic, Chapter 8, Sections 5 and 7, titled respectively "The Problem of Division by Zero" and "Five Approaches to Division by Zero." (These sections begin on pages 181 and 184 of the linked PDF, not the pages listed in the table of contents.) Your trick is the fourth of the five approaches listed in Section 8.7. As Suppes notes, this approach is the "solution which is probably most consonant with ordinary mathematical practice."

One example of approach four is the Riemann Sphere or extended complex plane. Some discussion of it in relation to division by 0 may be found in the entry for Division by Zero at MathWorld–A Wolfram Web Resource.

There are, however, contexts in which division by zero can be considered as defined. For example, division by zero $z/0$ for $z \in \mathbb{C}^* \neq 0$ in the extended complex plane $\mathbb{C}-^*$ is defined to be a quantity known as complex infinity.

What is gained? Roger Penrose stresses two advantages in his books The Road to Reality, The Emperor's New Mind, and Shadows of the Mind:

  1. Constructing a map without a hole
  2. Modeling subatomic phenomena

Can approach four be carried a step further? I've done some work in an effort to show that yes, it can. The extra step is to extend the number system in the course of division by a new number zero. My paper, Replacing 0: A NonEuclidean Arithmetic, explores your question with this in mind. As the subtitle suggests, the example of nonEuclidean geometries is followed by altering the axioms of standard arithmetic to allow a new number zero.

  1. The different number of nothing is first defined in terms of division and then subtraction (instead of the usual way of first defining it in terms of subtraction).
  2. The new number zero has been designed to replace the number 0 in such a way that division by it results in quotients that are not real or complex as allowed by the 4th approach.
  3. Quotients are unique, unlike with complex infinity in $\mathbb{C}-^*$.
  4. When dividends are limited to the reals, a real plane can be constructed.
  5. The notation used turns out to match a notation Penrose uses in The Road to Reality for $n$-real-dimensional space so it is easy to see that spaces of higher dimension can be constructed.
Jonathan Cender
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The argument that "it's 'better' for algebraic structures to have such-and-such properties and they don't work if you can divide by zero" is sort of missing the point.

The integers and the rational numbers both have clear interpretations: the integers represent discrete positions on, or translations of, a number line (or translations of the natural numbers themselves, where the elements are allowed to "fall off" the end), while the (positive) rational numbers represent all possible commensurability relations between lengths. They happen to be constructed from the natural numbers in ZFC but they have an importance in their own right as well.

So, the real question to ask is, what are you trying to model with division by zero? If you divide a cake into $n$ parts without making overlapping cuts, you have to make $n - 1$ cuts. So if you want to divide by zero you have to say what it means to cut something -1 times. If you generalize the meaning of division to include quotienting a set by a group then this is still the case: there is no group with zero elements. This suggests considering semigroup actions, but in any case it is better to decide what the meaning of division (or reciprocal) is first, and then work out its properties afterwards, instead of blindly asserting some axioms and hoping they give you something interesting. Then you will understand why certain things are possible or not.

One place where division by zero has a clear meaning is in the projective line. There, taking the reciprocal means reflecting over the $y = x$ line, and if the horizontal line is zero, the its reciprocal is the vertical line, aka $\infty$.

Ibrahim Tencer
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I am not an expert but from reason and experience, I think that by introducing number lines of zero and infinity as an extension to real number line, undefined expressions like $\frac{0}{0},\frac{x}{0},0\times\infty,$etc. can be well defined.

The number line of zero is as follows: enter image description here

The number line of infinity is as follows: enter image description here

Note-: Both these number lines have negatives too. Separate real number lines exist at every point of infinity number line and separate infinitesimal number lines exist at every point of real number line.

Now if we want to compute $0\times\infty$, we should first define where $0$ stands in the zero number line and also where $\infty$ stands in the infinity number line. For example: $$(0\times3)\times(\infty\times5)=(\infty^{-1}\times3)\times(\infty\times5)=3\times5\times\infty^{-1}\times\infty=15$$

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  • This is an interesting idea, but this sounds more like [Dual_number](https://en.wikipedia.org/wiki/Dual_number), but with an added element $\delta=\frac{1}{\epsilon}$. In this case, you write $\delta$ as $\infty$ and $\epsilon$ as $\infty^{-1}$. It also is similar to hyperreal numbers. – Nirvana Feb 01 '21 at 08:18

Define $0k=1$ for some new element $k$. Certainly $k$ is not a real number, so let us extend $\mathbb{R}$ to a new set $\mathbb{E}$ which includes every $a+bk$ for all $a,b \in \mathbb{R}$. Note that, because of k's property, an existing real number like 60 would not be represented as $60+0k$, but rather as $59+0k$.

We now define addition and multiplication of these numbers as follows: $$(a_{1}+b_{1}k)+(a_{2}+b_{2}k) = (1+a_{1}+a_{2})+(b_{1}+b_{2})k\\ (a_{1}+b_{1}k)\times{(a_{2}+b_{2}k)} = \begin{cases} b_{2}\;\;\; & a_{1} = -1, b_1=0\\ b_{1} &a_{2}=-1,b_{2}=0 \\ (a_{1}+a_{2}+a_{1}a_{2})+(b_1+a_2b_1+b_2+a_1b_2+b_1b_2)k& \text{otherwise} \\ \end{cases}$$

An example of addition: $$2k+(1+k)\\ =(0+2k)+(1+k)\\ =(1+0+1)+(2+1)k\\ =2+3k$$ An example of multiplication: $$2k\times{5}\\ =(0+2k)\times{(4+0k)}\\ =(0+4+(0\times{4}))+(2+(4\times{2})+0+(0\times0)+(2\times0))k\\ = 4+10k $$ We also define subtraction and division: $$ x-y=x+(y\times{-1})\\ \text{rec}(a+bk)=\begin{cases} k\;\;\; & a = -1, b=0\\ \frac{1}{a+1} &a\neq{-1},b=0 \\ 0& \text{otherwise} \\ \end{cases}\\ x/y=x\times{\text{rec}(y)} $$ Properties $(\mathbb{E},+,-,\times)$ satisfies:

  1. $(x+y)+z = x+(y+z)$
  2. $x+y=y+x$
  3. $x+0=x$
  4. $x-x=0$
  5. $x\times{1}=x$
  6. $x\times{(y+z)} = x\times{y}+x\times{z}$

But this extension does not satisfy some other expected properties like $x/x=1$, $k\times{x} = kx$ for $x \in \mathbb{R}$, and $x\times{(y\times{z})} = (x\times{y})\times{z}$.

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