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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?

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because it doesn't work as nicely. –  Dustan Levenstein Mar 27 '12 at 19:53
Did you look over this previous question about defining division by zero? –  Arturo Magidin Mar 27 '12 at 19:56
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
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
There is an algebraic structure in which division by zero is always defined. Is is called a wheel. 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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4 Answers

up vote 30 down vote accepted

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
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
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
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
@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 1:35
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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.

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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
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
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
@N. S. good point –  Chris Eagle Mar 27 '12 at 21:20
No, $0+x=x$ is the defining property of $0$. –  Chris Eagle Mar 28 '12 at 10:08
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I would also like to point out that the IEEE floating-point specification 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
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
Rather apropos... –  J. M. Mar 28 '12 at 16:35
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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: On Hypotheses Underlying Mathematics, explores your question with this in mind. A link to it may be found on my user page. 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.
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