# Division by $0$

I thought it was elementary to me, but I started to do some exercises and came up some definitions I have sort of difficulty to distinguish. In parentheses are my questions.

1. $\dfrac {x}{0}$ is Impossible ( If it's impossible it can't have neither infinite solutions or even one. Nevertheless, both $1.$ and $2.$ are divided by zero, but only $2.$ has infinite solutions so as $1.$ has none solution, how and why ?)

2. $\dfrac {0}{0}$ is Undefined and has infinite solutions. (How come one be Undefined and yet has infinite solutions ?)

3. $\dfrac {0}{x}$ and $x \ne 0$, it's okay for me, no problem, but if someone else wants to add something about it, feel free to do it.

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On 3, $0/x$, you need to say explicitly $x \ne 0$, to distinguish it from 2. – Henry Mar 11 '11 at 20:43
The following text was added in a suggested edit: "The 787 videos listed in the auto generated You-Tube channel: Division by zero is a good indication of the importance and common misunderstanding of this question. In my opinion, it is the obligation of this mathematical community to provide a greater variety of explanations to interested students at all levels." If this information is useful, it should be added as a comment, not by editing posts by other users. – Martin Sleziak Nov 8 '12 at 11:41
Yes, this makes sense! Clearly a question with +19 that has an answer of +50 hasn't received enough attention! – Asaf Karagila Nov 8 '12 at 13:33
It is undefined because it has infinite solutions. – Parth Kohli Feb 2 '13 at 15:42

The first question you need to ask is: What does "$a/b$" mean?

The answer is: "$a/b$ is the unique solution to the equation $bz = a$." (I'm using $z$ as the unknown, since you are using $x$ for other things).

(3) is perfectly fine: $0/x$, with $x\neq 0$, is the solution to $xz = 0$; the unique solution is $z=0$, so $0/x = z$. The reason it's unique is because $x\neq 0$, so the only way for the product to be $0$ is if $z$ is $0$.

In (1), by "impossible" we mean that the equation that defines it has no solutions: for something to be equal to $x/0$, with $x\neq 0$, we would need $0z = x$. But $0z=0$ for any $z$, so there are no solutions to the equation. Since there are no solutions to the equation, there is no such thing as "$x/0$". So $x/0$ does not represent any number.

In (2), the situation is a bit trickier; in terms of the defining equation, the problem here is that the equation $0z=0$ has any value of $z$ as a solution (that's what the "infinite solutions" means). Since the expression $a/b$ means "the unique solution to $bx= a$, then when $a=b=0$, you don't have a unique answer, so there is no "unique solution".

Generally speaking, we simply do not define "division by $0$". The issue is that, once you get to calculus, you are going to find situations where you have two variable quantities, $a$ and $b$, and you are considering $a/b$; and as $a$ and $b$ changes, you want to know what happens to $a/b$. In those situations, if $a$ is approaching $x$ and $b$ is approaching $y\neq 0$, then $a/b$ will approach $x/y$, no problem. If $a$ approaches $x\neq 0$, and $b$ approaches $0$, then $a/b$ does not approach anything (the "limits does not exist"). But if both $a$ and $b$ approach $0$, then you don't know what happens to $a/b$; it can exist, not exist, or approach pretty much any number. We say this kind of limit is "indeterminate". So there is a reason for separating out cases (1) and (2): very soon you will see an important qualitative difference between the first kind of "does not exist" and the second kind.

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Here's an example of how #2 is undefined. What makes it infinite is the fact that this can be repeated for any number. – KronoS May 2 '13 at 16:14
I like this answer for the first part, but I don't like the part about calculus.. 1) $\infty$ is not anything and 2) stated like that, it's as you never know what happens to $a / b$.. you just need more informations about the exact nature of the variables – Ant Apr 4 '14 at 17:31
In your last paragraph you need to add that when you have these a,b "approaching something" that they are continuous at this point. – Matthew Levy Dec 23 '14 at 17:07
I have borrowed your starting paragraph for my answer :-) – arivero Aug 16 '15 at 20:11

The key is to realize what a fraction $\frac ab$ really represents: $\frac ab$ is the number with the property that $\frac ab \cdot b = a$.

So, in the first case if $x \ne 0$, there is no number $\frac x0$ with the property that $\frac x0 \cdot 0 = x$ since anything times zero is zero. So $\frac x0$ is undefined. In the second, any number $y$ has the property that $y \cdot 0 = 0$, so $\frac 00$ could represent any number $y$ according to the above characterization of a fraction, so $\frac 00$ is said to be indeterminate. Finally, if $x \ne 0$ then $0$ has the property that $0 \cdot x = 0$, so $\frac 0x$ is the number $0$.

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One should not use "indeterminate" here. That applies only to functions, not numbers. – Bill Dubuque Mar 11 '11 at 20:56
@BillDubuque I am currently learning elementary arithmetic. In mostly all contexts it is said that $\dfrac 00$ is indeterminate. – user103816 May 25 '14 at 9:48

Definition of Division

For every real number a and every nonzero real number b, the quotient a$\div$b,

or $\dfrac{a}{b}$, is defined by:

$$a\div b=a \cdot \frac{1}{b}.$$

Dividing by zero would mean multiplying by the reciprocal of 0.

But 0 has no reciprocal (because 0 times any number is 0, not 1.)

Therefore, division by 0 has no meaning in the set of real numbers.

Multiplicative property of 0

Prove:

If $a$ is any real number, then $a\cdot 0 = 0$ and $0\cdot a = 0$.

Proof:

Statement _________________Reason

1. $0 = 0 + 0$ ______________1. Identity property of addition

2. $a\cdot0 = a(0 + 0)$ __________2. Multiplication property of equality

3. $a\cdot0 = a\cdot0 + a\cdot0$ ________3. Distributive property of mult. with respect to add.

4. But $a\cdot0 = a\cdot0 + 0$ _______4. Identity property of addition

5. $\therefore$ $a\cdot0 + a\cdot0 = a\cdot0 + 0$ ____5. Transitive property of equality

6. $a\cdot0 = 0$ _______________6. Subtraction property of equality

7. $0\cdot a = 0$ _______________7. Commutative property of multiplication

Therefore, 0 times any number is 0, not 1.

(Source: Algebra: Structure and Method Book 1)

The two cases presented in an older edition of Book 2 of the above source are:

1. Dividing a nonzero number by zero, violates the multiplicative property of zero and therefore the properties of the real numbers upon which it is proven, as shown above.

2. Dividing zero by zero, which does not violate the multiplicative property of zero, but multiplication by zero is an operation that results in zero for every real number.

If $\dfrac{a}{0}$ = c, then $a = 0\cdot c$. But $0\cdot c = 0$. Hence, if $a$ is not equal to $0$, no value of $c$ can make the statement $a = 0\cdot c$ true, while if $a = 0$, every value of $c$ will make the statement true.

Thus, $\dfrac{a}{0}$ either has no value or is indefinite in value.

This separation into two cases, one of which results in no value satisfying the multiplicative property of zero and the other resulting in an indefinite value satisfying it, gives the impression that $\dfrac 0{0}$ is allowed.

The following argument starts with the equation: $$a \cdot \frac{1}{a} = 1$$ and notes that $a\neq 0$ because $0$ times any number is $0$.

Thus, the product of $0$ and no real number equals $1$.

This further reinforces the idea that $0$ has no number that when multiplied by it equals $1$.

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Assuming 0 has no reciprocal is begging the question. – Austin Mohr Sep 6 '11 at 16:42
@AustinMohr I am not "assuming" 0 has no reciprocal, as shown by the proof that 0 times any number is 0, and the definition that reciprocals are two numbers whose product is 1. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Nov 17 '12 at 8:31
@skullpatrol I just happened by this answer again, and I see you've expanded it very nicely. – Austin Mohr Jun 24 '13 at 1:42

Just to give another point of view (purely algebraic one):

Suppose you have a ring (a mathematical structure with addition and multiplication satisfying a bare minimum of laws you need to call them that), and in it there is a number $x$ such that $x \cdot 0 = 1$. Then $1 = x \cdot 0 = x \cdot (1 - 1) = x - x = 0$, which means that for any $y$ in this ring $y = y \cdot 1 = y \cdot 0 = y (1 - 1) = y - y = 0$, so this ring has just one element. Such ring is called trivial, and it is clearly not an extension of the ring of integer numbers.

If you can divide by zero, then $1/0$ is defined and it has to satisfy the property above. Thus, division by zero is possible only in the trivial ring.

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What if I don't have a ring? What if I have a wheel? – JMCF125 Jan 18 '14 at 14:57
@JMCF125 then you roll – Alexei Averchenko Sep 7 '15 at 9:12

You will pretty much never see "0/0" or "x/0" in a situation where somebody expects you to actually interpret it as having a value. Other than simple mistakes, the times it comes up are usually of the following sort:

To solve the equation $ax=b$ for $x$, consider the expression "$b/a$".

1. If it is of the form "$0/0$", your equation has infinitely many solutions
2. If it is of the form "$x/0$" for nonzero x, your equation has no solutions
3. Otherwise, the division makes sense and the result is the unique solution

You might hear the word "form" mentioned to refer to the fact that we're talking about a formula, and not the number that might result from evaluating the formula. (especially in the context of limits)

It's important to note the difference, because the number system you use was created so that "$0/0$" and "$x/0$" are not allowed -- they are meaningless when viewed as arithmetic expressions. (the same is true for "$x/y$" if you do not already know that $y \neq 0$) There are lots of very good reasons why arithmetic was created in this way, and the other answers mention some of them.

I feel I should mention that there are other forms of arithmetic that behave differently. The projective numbers are often useful, and 1/0 makes sense in them (but $0/0$ does not). Wheel theory is somewhat more esoteric, but it provides an example where even $0/0$ makes sense. (and also clearly demonstrates the difficulties in accommodating its existence)

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Let me add this quick bit in addition to what has already been said (assuming you know a little set theory). Take the following semi-random example.

Suppose you have a function $f$ with $f(5) = 14$ and $f(6) = 11$. Going backwards, what do you get? Assuming $f$ is only defined for $5$ and $6$, $f^{-1}(14) = 5$ and $f^{-1}(11) = 6$, right? Now consider a different function, which like $f$ is only defined at two points... let's call it $g$ with $g(5) = 8$ and $g(6) = 8$. Now, what is $g^{-1}(8)$? Well, it wants to be both $5$ and $6$ if it's really the inverse, but defining it either way causes a contradiction so it must be undefined.

So what's the point? In algebra many operations (like multiplying by a nonzero number) are like $f$ and and many (multiplying by zero, for example) are like $g$. Here, draw the picture with arrows... with $f$ you know where the arrows came from, with $g$ two arrows point to the same place so you don't know where you 'came from'.

Now back to multiplication. Imagine the arrows if you multiply by $\frac{1}{2}$ versus if you multiply by $0$. You can see that essentially the same thing is going on, but with infinitely many numbers instead of just a few.

BTW: There is one partial 'fix' if you know a little set theory: (new notation here, NOTE THE CURLY 'SET' BRACKETS): $f^{-1}(\{8\}) = \{5,6\}$. Another example: if $h(x) = x^2$ then $h^{-1}(\{4,25\}) = \{2,5\}$. And back to were we started from, inverting division by zero: if you have $s(x) = 0$ then $s^{-1}(\{0\}) = (-\infty,\infty)$.

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If your domain consists of a two-element set, or a set with more than two elements (where any infinite set gets classified as having more than two elements) then Arturo's answer applies.

However, if your domain consists of {0}, then things work out a bit differently, as Bill Dubuque pointed out in a comment elsewhere.

For a system with domain of {0}, you'll still define a/b as the unique solution to (bz)=a. Now, (3) still works out as perfectly fine. x/0 means the unique solution to (0z)=x. Well, x can only equal 0 in this domain, so this means the unique solution to (0z)=0. z can also only equal 0 here, so the equation does have a unique solution of 0. 0/0 means the unique solution to (0z)=0. Well, z can only equal 0 here, and if z=0, the equation does hold, so (0/0)=0 here also.

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what does domain mean here? – Srivatsan Sep 2 '11 at 17:29
The domain consists of the set of elements under consideration. – Doug Spoonwood Sep 2 '11 at 17:37

Because dividing by zero destroys everything.

Reuben Hersh in "What is Mathematics, Really?" gives the following explanation (quotes). It articulates the point that we invent rules, as we go along, in order to account for extended powers of calculation. However, not unexpectedly, this often comes at a price. And each time we have to decide whether the cost is worth bearing or not.

Note that the question "why do we not divide by zero?" is loaded with the psychology of the exception. Meaning, it seems as if have a perfectly nice rule for all integers to begin with, and all of a sudden we exclude zero from it. This prompts the question "why do we exclude zero?" as a reasonable one. Especially since that's what it comes down to - excluding just one number. But understanding the process of creation (of definition) is important in order to understand why a plausible (or even an excellent) question may be missing a point. It is not the case that we "have a perfectly nice rule for all integers to begin with". Addition and multiplication start only with the natural numbers (positive integers starting with $1$) and only later do we move on to the reverse processes of subtraction and division. Given multiplication, therefore, it is not that we exclude zero from this new process we call "division", but rather that we find no good reason to extend the process in order to include zero - other than a psychological sense of ease that something holds for all numbers which is fine but not if it destroys all arithmetic. "God made the Integers, all else is man's work" goes the saying. It shouldn't come as a surprise that there should be a few of these exceptions, to this work of rules. (It turns out we can extend division to the negative integers without a problem, and with great gains, thus enhancing the psychology of a special case for zero. It's only a special case, and one shouldn't expect otherwise.)

Back to math. Suppose it were handy to introduce division by zero. What would be the cost of this rule? Hersh explains.

1/0 Doesn't Work (0 into 1 Doesn't Go)

Division by $0$ is not allowed. Why not? If it's allowed to introduce a symbol $i$ and say it's the square root of $-1$, which doesn't have a square root, why not introduce some symbol, say $Q$, for $1/0$?

We introduce new numbers, whether negative, fractional, irrational, or complex, to preserve and extend our calculating power. We relax one rule, but preserve the others. After we bring in $i$, for example, we still add, subtract, multiply, and divide as before. I now show that there's no way to define $(1/0)\cdot0$ that preserves the rules of arithmetic.

One basic rule is, $0\cdot$ (any number) $=0$.

(Formula I) So $0\cdot(1/0) = 0$.

Another basic rule is $(x)\cdot(1/x) = 1$, provided $x$ isn't zero. (But if we want $1/0$ to be a number, this proviso becomes obsolete.)

(Formula II) So $0\cdot(1/0) = 1$

Putting Formulas I and II together, $1=0$.

Addition gives $2=0$, $3=0$, and so on, $n=0$ for every integer $n$. Since all numbers equal zero, all numbers equal each other. There's only one number — $0$.

The supposition that $1/0$ exists and satisfies the laws of arithmetic leads to collapse of the number system. Nothing is left, except - nothing.

In the case of the imaginary $i$, it turns out that not only is there nothing destroyed (other than our sense of familiarity with numbers), but that things fall into place (Fundamental Theorem of Algebra).

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This is an attempt at some intuition for someone who doesn't know a lot of math.

The equation $\frac{a}{b}=x$ is the same as the equation $a=b\cdot$ x. You can look at $a=b\cdot x$ as asking the question "if I want to build $a$ out of $b$'s, how many $b$'s do I need?" (the $x$ is how many pieces of $b$ you need to build $a$)

Example

$\frac{4}{0.1}=x$ is the same as $4=0.1\cdot x$.

How many $0.1$'s do you need to build 4 out of $0.1$'s?

Well you need $\frac{4}{0.1}$ or $40$ pieces. Indeed, $40\cdot 0.1 = 4$.

So what about dividing by 1?

Example

$\frac{7}{1}=x$ is the same as $7=1\cdot x$.

How many $1$'s do you need to build 7 out of $1$'s?

Obviously you'll need seven pieces of one, beause $1+1+1+1+1+1+1=7$

So what about dividing by zero?

Example

$\frac{3}{0}=x$ is the same as $3=0\cdot x$.

How many $0$'s do you need to build 3 out of $0$'s?

Well, how can you build a number out of zeros? No matter what you do with your zeros, you're never going to be able to build any number out of zeros. If you could build a number out of zeros, you should be able to write a number, say 9, as a sum of zeros, which would mean that $9=0+0+0+0+0+0+\dots$, which can never be true no matter how many zeros you use.

So it doesn't really make sense to divide something by zero.

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I usually argue as the other answerers have but recently I have started using a slightly different argument.

I argue that multiplying by $1/x$ undoes the action of multiplying by $x$. So for example $11*2=22$ and $22*(1/2)$ brings you back to $11$.

Now we have $0*11=0$ but you can't get back to $11$ by multiplying as zero times anything is zero.

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We can divide by $0$.

Let $X$ be the set of numbers we are debating on.

The multiplicative inverse of $x\in X$, namely $x^{-1}$ or $\frac{1}{x}$, is defined as

$x\cdot x^{-1} = id$, where $id$ does not change any element in $X$ when we multiply, for instance, $1 \in \mathbb{Q}$.

For any $y \in X$, $0 \cdot y=0$, so we require $id = 0$, so let $id = 0$.

Then, for any $x \in X,$ $x\cdot id = 0$.

We get $X = \{0\}$.

We can divide by $0$ on $\{0\}$, since $0\cdot 0 = 0$. But this is not an interesting case.

So, we usually assume $X\not =\{0\}$, so that we cannot divide by $0$.

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The first question you need to ask is: What does "a/b" mean?

The answer is: "a/b is a representative of a class of pairs of numbers $(a,b) \in Z \times Z$ such that two pairs $(a,b)$ and $(c,d)$ are equivalent if $ad=cb$."

This equivalence fails -is not transitive- if we include the pair $0/0$, because any pair should be equivalent to it, even two arbitrary pairs which are not equivalent between them. So we need to exclude the "indeterminate" pair, correcting the above definition to say instead $(a,b) \in Z \times Z - \{ (0,0) \}$.

All the elements of the form $0/a$ are equivalent. And similarly all the elements of the form $a/0$ are equivalent. Call them the classes $0/1$ and $1/0$

We can define multiplication operating in each $Z$ set, i.e. $(a,b).(c,d) \equiv (ac,bd)$. With this definition, the identity is the class represented by $(1,1)$.

We find that our multiplication is not defined for a product of elements in the classes $0/1$ and $1/0$. The product is precisely the excluded pair $(0,0)$. So if we want multiplication to be always well defined in our set of all the fractions, we need to exclude one of the two classes. Traditionally we exclude the class $1/0$, the class that the original post calls the "Impossible class".

Arriving here, we see that really the motivations to exclude the "undefined element" and the "impossible class" are very different: the former, being able to match any other fraction, completely destroys the transitivity of the equivalence relation, while the later simply is a nuisance to close the multiplication.

Both classes had the potential to do the role of a multiplicative zero: $$1/0 . (a,b) = (a,0) \propto 1/0$$ $$0/1 . (a,b) = (0,b) \propto 0/1$$ But now that we have declared the first one "impossible" it is clear that we now must kept the second one to be used as the traditional zero.

The last steps should be to define a sum operation such that $(a,b) + 0/1 = (a,b)$ and a mapping of the integer numbers into the fractions, $Z \longrightarrow Z \times Z - \{ (0,0) \}$, preserving the operations of sum and multiplication already defined in the integers, and mapping the zero and identity of the integers to the corresponding classes in the fractions. But as far as the OP questions goes, we do not need even to finish the construction :-)

EDIT it could be interesting to try to argue that, in the same way that with the usual definition of sum we have $(a,b) + 0/1 = (a,b)$, we should have that $(a,b)+ 1/0 = 1/0$. So we would be in accord with the usual meaning of "infinity" that we assign to this class. In fact it is so if we define $$(a,b) + (c,d) \equiv (ad+bc,bd)$$

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Among the rules for arithmetic with fractions are the formulas for adding and subtracting,

$${a\over b}\pm{c\over d}={ad\pm bc\over bd}$$

Now what if we allow division by $0$? Keeping in mind the properties $x+0=x$, $x\cdot0=0$, and $x-x=0$ for all numbers $x$, we find first that

$$0={1\over 0}-{1\over 0}={1\cdot0-0\cdot1\over0\cdot0}={0\over0}$$

and then that, for all $x$,

$$x={x\over1}={x\over1}+0={x\over1}+{0\over0}={x\cdot0+1\cdot0\over1\cdot0}={0\over0}=0$$

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## protected by Michael AlbaneseJul 7 '14 at 6:27

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