# Explanation of method for showing that $\frac{0}{0}$ is undefined

Ignoring limits, I would like to know if this is a valid explanation for why $\frac00$ is undefined:

$x = \frac00$
$x \cdot 0 = 0$

Hence There are an infinite number of values for $x$ as anything multiplied by $0$ is $0$.

However, it seems to have got comments, with two general themes.

Once is that you lose the values of $x$ by multiplying by $0$.

The other is that the last line is:

$x \cdot 0 = \frac00 \cdot 0$

as it involves a division by $0$.

Is there any merit to either argument? More to the point, are there any major flaws in my explanation and is there a better way of showing why $\frac00$ is undefined?

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That 'proof' looks perfectly fine to me. – Noldorin Jul 23 '10 at 8:56
Dear Jacob, Your explanation is pretty good, but I would make the following slight change at the end: rather than have the conclusion be that there are an infinite number of x solving the equation, interpret it as follows: there is no well-determined x that solves the equation (because any x will do!). Thus we cannot find a well-determined value for 0/0. – Matt E Aug 1 '10 at 4:48
Dear 97832123, I think you could be more generous to the OP, and interpret the question as asking "Is this a correct explanation as to why mathematicians leave 0/0 undefined"? The answer to the question is then essentially "yes". – Matt E Aug 1 '10 at 4:51
The key part of your argument Jacob is that $0*x=0$ for all $x$. It can be useful to identify exactly which underlying algebraic fact causes this -- it's a general phenomenon in rings. Moreover, it follows from the distributivity law. So one way to say why we don't define $0/0$ is that it would force us to give up on distributivity. While $0/0$ has no compelling definition, distributivity is a compelling idea. So that's why we don't bother trying to define $0/0$. – Ryan Budney Sep 29 '10 at 15:02
It's true. Trying to search for "0/0" gives you a search for "00", so who can blame him/her? – Matt Gregory Mar 27 '11 at 12:14

For "all" x,

$$\frac0x = 0 \overset{?}{\implies} \frac00 = 0$$

For "all" x,

$$\frac x x = 1 \overset{?}{\implies} \frac00 = 1$$

Moreover, if one could say $\frac00 = k, \forall k$, we could then say $2 = 3$ — just divide both sides by 0 and get $k = k$, which is patently true.

Since there is no reasonable value $\frac00$ can have, $\frac00$ must be undefined.

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Neither 0/x = 0 nor x/x = 1 would be considered axioms though -- they're just observations that can be proved for any x other than zero, so I don't think it makes sense to pit them off against each other. – bryn Jul 23 '10 at 11:24
@bryn, they're not axioms. They are, however, immensely useful properties that a value of 0/0 must have to be of any use. – badp Jul 23 '10 at 11:34
@Anixx Why would $\frac{x}{x}=1$ be true only if $x\neq -x$? – 5xum Dec 17 '14 at 7:56
@Anixx I don't think you're quite getting the idea that badb is trying to present. He is saying that for all $x\neq 0$, we know that $\frac 0x=0$, so any sensible definition of $\frac 00$ will preserve this property, thus setting $\frac 00$ to $0$. On the other hand, for all $x\neq 0$, we know that $\frac xx=1$, so any sensible definition of $\frac 00$ will set $\frac00$ to $1$. The conclusion is that $\frac00$ has no sensible definition. – 5xum Dec 17 '14 at 8:24
@Anixx It's a common mathematical process: since the property is true for all $x > 0$ and for all $x < 0$ and $x \in \mathbb R$, then the property is also extended by continuity for $x = 0$. Of course doing it here does not lead to a correct result: that's the whole point of the answer. – badp Dec 17 '14 at 9:52

The existence of a (multiplicative) inverse for the (additive) zero is inconsistent with the other field axioms. As you noticed, the crux is that anything multiplied by zero gives zero. Let us then establish this fact first.

0 is identity for addition: $0+0=0$
multiply by some x: $(0+0)\cdot x=0\cdot x$
distributivity: $0\cdot x + 0\cdot x = 0\cdot x$
add the additive inverse of $0\cdot x$ to both sides: $(0\cdot x + 0\cdot x) + (-(0\cdot x)) = 0\cdot x + (-(0\cdot x))$
associativity of addition: $0\cdot x + (0\cdot x + (-(0\cdot x))) = 0\cdot x + (-(0\cdot x))$
definition of "additive inverse": $0\cdot x + 0 = 0$
zero is additive identity: $0\cdot x = 0$

Let's assume now that there is a multiplicative inverse of 0, denoted by Z.

(*) $0\times Z=1$

From the last two relations, 1=0, which contradicts another field axiom (often forgotten), which is:

$1\ne0$

Therefore, you either accept that 0 has no inverse or you change at least one of the field axioms---you can't have both at the same time. In a sense, it is a matter of convention which axioms you choose. In practice, some sets of axioms lead to more useful consequences. For example, if you want 0 to have an inverse and drop the axiom saying that 1 does not equal 0, then the 'arithmetic' you end up doing won't be very interesting.

In short, you are right.

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So we can divide by zero in the trivial ring with $1=0$ :-) – Paŭlo Ebermann Jun 27 '11 at 0:54
So why one cannot define $0/0=0$? Is having multiplicative inverse of 0 needed for defining this? – Anixx Dec 17 '14 at 15:36
The notation $x/y$ means $x\times y^{-1}$, where $y^{-1}$ is the multiplicative inverse of $y$. If you change this convention, which would be rather drastic, then you'd have to provide an alternative. – rgrig Dec 17 '14 at 21:58

I think that ignoring limits is problematic.

If there was a limit of the function $f(x,y)=x/y$ for $x,y \to 0$ regardless of how the limit is performed, then one would define that value to be $f(0,0)$, even if everything else is strange. Since the limiting value depends on the way the limit is done, choosing a value for $f(0,0)$ is counterproductive as it gives a non-continuous function. Better to have a continuous function over a slightly smaller domain.

This also forces the point that if you do have a limit process that results in the evaluation of $f(0,0)$, you realize early on that you should examine the limit carefully rather than use $\lim g(x) = g(\lim x)$ (which is only true for continuous functions, of course.)

By the way, this might be too trivial, but I'll give an example of how the limiting value depends on the limit:

$\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$, (both $\sin(x)$ and $x$ goto 0)

$\displaystyle \lim_{x \to 0} \frac{\cos x - 1}{x} = 0$ (again, both num and denum goto 0, but num goes "faster")

$\displaystyle \lim_{x \to 0} \frac{\sqrt{x}}{x} =+\infty$.

PS, sorry about the latex notation, hopefully, we could have it working nicely like in MO...

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On the other hand, surely you'll agree that 0/0 is undefined in any (nonzero) ring even when you can't take limits! – Noah Snyder Jul 23 '10 at 16:03
but since +∞ is not necessarily commutative is really why its undefined? – Talvi Watia Sep 30 '10 at 0:01
@Talvi: commutativity is a property of operators (e.g. addition, multiplication), not of objects like ∞. – J. M. Oct 2 '10 at 1:48

Ok, here is one reason why dividing by $0$ has no meaning...

Dividing by $0$ would mean to multiply by the inverse of $0$.

However, the inverse of $x$ is the number $y$ for which $x \times y=1$.

Since $\forall x \in \mathbb{R}, 0 \times x=0$, $0$ does not have any inverse.

Hence, you can't divide by $0$.

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is that an $\emptyset$ empty set? – user103028 Nov 4 '14 at 12:00
@VitalieGhelbert What do you mean by "is that an empty set"? What is supposed to be the empty set? – Martigan Nov 4 '14 at 12:03
@VitalieGhelbert I KNOW what is an empty set... But your comment make no sense relatively to my answer... What is supposed to be the empty set in my answer??? – Martigan Nov 4 '14 at 12:07
@VitalieGhelbert You are confusing arithmetic and group theory. You can't divide by $0$. The problem is not that there is no answer, it is that there is no question! – Martigan Nov 4 '14 at 12:28
There are several fields, and in some fields, if you choose the right axioms, you can define things which are not defined under other theories. Under the common models of mathematics used by almost everyone by default, you can't divide by $0$. If you want to do "unusual" things, you should imperatively specify clearly under which theory you are acting, what are your axioms. – Martigan Nov 6 '14 at 10:01

EDIT:I botched my original answer, here's what I actually meant:

Fractions are equivalence classes of pairs of integers subject to (a,b)=(c,d) iff ad=bc, with addition and multiplication extended from {(a,1)} as a copy of the integers by multiplication defined component-wise in general.

If we allow pairs of the form (a,0) for some integer a, we have with the multiplication axioms (a,0)=(b,0) for any a and b, and in particular all of them coincide with (0,1)(a,0)=(0,0) as multiplication is component-wise.

But (0,0)=(a,b) for every pair, so our construction gives us the trivial ring. Hence, to get a non-trivial structure, we must disallow pairs of the form (a,0), including (0,0) and that gives us the rationals.

Anyway, the above is the standard reason why 0/0 is undefined. You can actually define it, though to do that you must change some of the fundamental arithmetic properties. For a fun read, check out the following paper on the topic: http://www2.math.su.se/~jesper/research/wheels/wheels.pdf

Original answer: By definition (or by construction) a/b, where a and b are natural numbers, is defined to be the rational number which when multiplied by b gives you a. From this you have that 0/0=0.

Different fractions a/b and c/d are defined to be equal if ad=bc. This allows us to extend the arithmetic of the integers to an arithmetic of fractions. From this you have that 0/0=c/d for every rational number c/d since x0=0 for all x in any system where multiplication distributes over addition (proof: x0=x(0-0)=x0-x0=0).

Thus, if you allow 0/0 among your fractions, and you attempt to extend the arithmetic of the integers, all your fractions must be 0.

(note: the reason you cannot allow a/0 among your fractions is when a is not equal to 0 is because if you extend the arithmetic to where you have multiplication distributing over addition, you would have 0x=0 for all x and so the existence of a/0 is only consistent if a=0, which is the case above).

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Hmm, not really. Fractions $a/b$ are only defined for $b\neq0$ so there is no need to prove that there is no such thing as $0/0$. Moreover, you most certainly not have that $0/0=0$, because the left hand side is meaningless, and meaningless things cannot be equal to zero. – Mariano Suárez-Alvarez Jul 30 '10 at 22:44
Even accepting the definition in your first sentence, "the rational number which when multiplied by 0 gives you 0" can be absolutely any rational number at all, not necessarily 0. – ShreevatsaR Jul 30 '10 at 22:46

Everyone knows 0/0 is nullity.

(yes, this is a joke post, but that was a serious article)

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That article is one of the most annoying instances of media completely failing to understand the topic on which they were reporting. Non-math people kept passing that article along to the math faculty where I was working at the time it was published and we kept having to explain that it was somewhere between nonsense and a slightly-altered version of IEEE floating-point arithmetic. – Isaac Jul 31 '10 at 17:54
This should be CW. – Pedro Tamaroff May 4 '12 at 22:59

Suppose that division by $0$ is possible. Then consider the following equation,

$$x=0\implies1\cdot x=0\cdot x\implies \large{\large{\color{blue}{1=0}}}$$

This implies that the successor of $0$ is equal to it and that contradicts the Peano Axioms. The above was possible because we assumed that division by $0$ is possible.

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It's undefined because $0/0 = 0$ is against the identity $a/a = 1$. Or at least, that's how I understand it, although there are probably deeper explanations that involve the consistency of algebra.

You could also approach it in a simple manner from the direction that what you have is $0^1 \cdot 0^{-1}$ which would give you $0^0$ which is a very disturbing expression (and could be taken as either 0 or 1 depending on your definition of exponents).

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0⁰ = 1 if you want polynomials like x + 1 = x¹ + x⁰ to work for x = 0... :) – badp Jul 23 '10 at 9:41
Which makes 0/0 even more nonsensical then :) – workmad3 Jul 23 '10 at 11:14
Except that the 0^-1 is undefined (the original question didn't ask why 1/0 was undefined, but that seems acceptable), so there is no need for us to be disturbed. – bryn Jul 23 '10 at 11:18
As I pointed out though, 0/0 can be written out as $0^1 * 0^-1$ which relies on the undefined $0^-1$. The answer by @badp is much better explained than my answer though (and is pretty much what I was attempting to say :) ) – workmad3 Jul 23 '10 at 12:16

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