Whenever I asked my grade school and college teachers why $\frac00$ is undefined, they would show me a graph of $\frac1x$ and point out the vertical asymptote at $0$, noting that "Since it approaches $-\infty$ from the left and $\infty$ on the right, it could be $-\infty$, $\infty$, or anything in between. So, we say it's undefined."

My question is, then, why isn't it $\Bbb R$, or even $\{\Bbb{R,C}\}$?

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    $\begingroup$ You just explained why it's undefined. $\frac00$ is equal to the set of all numbers since anything times the denominator would equal to the numerator, and thus it's not a number. $\endgroup$
    – Shahar
    Aug 26 '14 at 15:29
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    $\begingroup$ @Shahar the question is why is it not the practice to say $\frac 00 = \Bbb R$ rather that $\frac 00$ "is undefined". $\endgroup$ Aug 26 '14 at 15:29
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    $\begingroup$ I don't see what the graph of $\frac 1x$ has to do with evaluating $\frac 00$. Perhaps the graph could be used to discuss $\frac 10$, which can't rightly be said to be "everything in between" $\endgroup$ Aug 26 '14 at 15:32
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    $\begingroup$ What would it help to say it's equal to $\mathbb R$? What problems would be easier to solve or have more elegant solutions? What theorems would be simplified? $\endgroup$
    – Jack M
    Aug 26 '14 at 15:33
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    $\begingroup$ I think you should explain why should it be $\Bbb{R}$ before asking why shouldn't it be so $\endgroup$ Aug 26 '14 at 15:34

Whenever questions like this one are asked, the answer should be: "In mathematics you can make up any rules you like, but what is it that you want to achieve?"

Division is defined the way it is not because your teacher says so, or because his teacher says so, but because its rules are chosen so that things work out in the best way possible. Many, many people thought about division and they agree that for many reasons it is better to leave $0/0$ undefined. However, if someone comes up with better rules, we'll use them. And if someone has different needs for division than everyone else, they have the freedom to use different rules (although they'll have to be pretty clever to think of better division than the one we use).

What do we want from division? Well, it should be useful in the sense that we can apply it in everyday real-world problems. It should also have good mathematical properties because then we will be able to do clever things with division. For instance, if division obeys the algebraic law $a \cdot (b / a) = b$ then we will be able to cancel out division and multiplication, which can potentially save us a lot of work.

You ask what's wrong with defining $0/0 = \mathbb{R}$. First of all, notice that you are suggesting that the quotient $0/0$ not be a number but a set of numbers. This will get you into trouble because I will ask annoying questions. For example, which of the following answers is supposed to be correct in your scheme of things:

  1. Is $(0/0)^2 = \mathbb{R}^2 = \{x^2 \mid x \in \mathbb{R}\} = \{y \in \mathbb{R} \mid y \geq 0\}$ correct?
  2. Is $(0/0)^2 = (0/0) \cdot (0/0) = \mathbb{R} \cdot \mathbb{R} = \{x \cdot y \mid x \in \mathbb{R} \land y \in \mathbb{R}\} = \mathbb{R}$ correct?
  3. Is $0/0$ a solution of the equation $2 x = 1 + x$ since $2 \cdot (0/0) = 2 \cdot \mathbb{R} = \mathbb{R} = 1 + \mathbb{R} = 1 + (0/0)$?

You may be able to suggest answers to these questions, but people will come up with many more. At the end of the day you will have to develop a coherent theory of computation where we can add, subtract, multiply and divide one set of reals by another set of reals. You will have to think carefully about which algebraic laws are still valid, and which are not. What is the payoff and is the result any better than leaving $0/0$ undefined?

You have the complete freedom to propose something new, but you do not have the freedom to make a proposal which you have not thought about carefully. This is why your teacher is shooting your suggestion down with a silly argument involving the graph of $1/x$: he doesn't want to think about all the things you're breaking, you should do that. If you think nothing breaks, write a research paper. If things break, well, you made a try.

Now let me describe a mathematical theory in which $0/0$ is equal to $\mathbb{R}$. It is called domain theory and it studies spaces, called domains, which are used to give mathematical meaning to datatypes in programming languages. One of such domain is the interval domain. Its elements are closed intervals $[a,b]$ together with the set $\mathbb{R}$ which we think of as an "infinite interval". We think of the interval $[a,b]$ as an approximate real number, i.e., a number whose value is known to be somewhere in the interval $[a,b]$. The interval $\mathbb{R}$ represents "completely unknown value".

We can define arithmetical operations on intervals to get the so called interval arithmetic. The rules of interval arithmetic are different than those of usual arithmetic. Among other things we define $[a,b] / [c,d] = \mathbb{R}$ when $c \leq 0 \leq d$ because that's better than leaving division by zero undefined. In other words, the interval domain gives a coherent theory of computation with intervals. One of the intervals is $\mathbb{R}$ and it represents the "undefined" value, commonly known as NaN in floating-point arithmetic.

In conclusion: you are right, we can define $0/0 = \mathbb{R}$. However, we cannot do this in isolation from the rest of artithmetic. Instead we need to develop a whole new theory of computation. This theory is useful in applications that deal with approximate values, but not in traditional college aglebra.

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    $\begingroup$ I wish I could +1 this 10 more times! $\endgroup$ Aug 27 '14 at 2:23

For the same reason that $\lim_{x\to \infty} \sin(x)$ is not defined, rather than $[-1,1]$.

The Epsilon-Delta definition of the limit requires that the limit is an element of the set, and not a set itself.


The most basic reason is that it is more useful to define division as a single-valued function. We can think of division as a function whose inputs are numbers in $\mathbb{R} \times \left( \mathbb{R} \setminus \{0\} \right)$ -- the first factor corresponding to the numerator and the second factor corresponding to the denominator -- and whose output is a number in $\mathbb{R}$.

We cannot extend the definition to include $0$ as a denominator because either the function has no value (e.g. in the case of $1/0$) or is multi-valued in the case of $0/0$, using the definition of $a/b$ as the value $c$ such that $a = bc$. However, provided we require that $b \neq 0$, then such a definition does determine a single-valued function.


The basic idea of the quotient $a/b$ is that $b\cdot(a/b)=a$ for all possible $a$ and $b$. (This is true for real and complex numbers, but also more generally.) Note that this structure makes sense only if the quotient of any two numbers is a number. If it is some other object, the basic algebraic properties of division cease to hold and you need to define all algebraic operations for these new objects. And if you start with numbers, you typically want to end up with numbers. Leaving $0/0$ (and $1/0$) undefined seems to be the simplest way of dealing with the issues stemming from the fact that you can never uniquely solve $0x=a$ for $x$.

  • $\begingroup$ so $0\times\Bbb R \neq 0$? $\endgroup$ Aug 26 '14 at 15:49
  • $\begingroup$ @Supuhstar, depends on what you mean. It can make sense to say $0\times\mathbb R=0$ in some cases. But if we define $0/0=\mathbb R$, we need to be able to define all operations, which does not look promising. What are $\mathbb R/\mathbb R$, $1/\mathbb R$, $\mathbb R+1$ and $\mathbb R\times\mathbb R$? Losing the structure of a field is not helpful. $\endgroup$ Aug 26 '14 at 16:05
  • $\begingroup$ I'm a little rusty on discrete math, but wouldn't those be "a set of all values in ℝ divided by themselves" (infinite $1$s), "a set of all values in ℝ, each divided into 1", "a set of all values in ℝ, each added to 1", and "the dot-product of the sets ℝ and ℝ", respectively? W|A illustrates: po.st/TP9rwP po.st/WtYos4 po.st/bAzaeT po.st/DSNi0e $\endgroup$ Aug 26 '14 at 17:28
  • $\begingroup$ @Supuhstar, the point is that if you assign those natural sounding values to the expressions, then you no longer have results like $a+b=a+c\iff b=c$. If $a$ happens to be $\mathbb R$, this is not true. And having special cases when this fails makes life (or algebra) unnecessarily difficult. $\endgroup$ Aug 26 '14 at 17:34

There is no theoretical reason why you can't do that. But the practical reason is that it would create more problems than it solves.


Because the operations of multiplication and addition are defined for elements of $\mathbb R$, and their inverses are defined for elements of $\mathbb R$, except that zero has no multiplicative inverse. These properties are inherited from the definitions of groups, rings and fields and their basic properties, which form the essential backbone of the arithmetic operations as they are commonly defined and used.

As it happens, in a wider setting, it is possible to define the multiplication of sets in various ways, including the multiplication of "Ideals" - as a field $\mathbb R$ has just two ideals - the whole of the real numbers, and the zero ideal. This is a context in which we can make sense of your comment. But this is algebra well beyond grade school level, and involves some careful definitions of the objects and operations concerned.


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