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Thanks in advance.

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12 Answers 12

up vote 106 down vote accepted



Given that answer, let's discuss your points out of order:





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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 at 20:11



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@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.



Multiplicative property of 0


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


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:


  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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@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. –  Rigor Nov 17 '12 at 8:31




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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







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Because dividing by zero destroys everything.




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











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So what about dividing by 1?





So what about dividing by zero?





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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protected by Michael Albanese Jul 7 '14 at 6:27


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