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(This was asked due to the comments and downvotes on this Stackoverflow answer. I am not that good at maths, so was wondering if I had made any basic mistakes)

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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    $\begingroup$ That 'proof' looks perfectly fine to me. $\endgroup$ – Noldorin Jul 23 '10 at 8:56
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    $\begingroup$ 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. $\endgroup$ – Matt E Aug 1 '10 at 4:48
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    $\begingroup$ 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". $\endgroup$ – Matt E Aug 1 '10 at 4:51
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    $\begingroup$ 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 $0/0$. $\endgroup$ – Ryan Budney Sep 29 '10 at 15:02
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    $\begingroup$ It's true. Trying to search for "0/0" gives you a search for "00", so who can blame him/her? $\endgroup$ – Matt Gregory Mar 27 '11 at 12:14
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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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    $\begingroup$ 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. $\endgroup$ – bryn Jul 23 '10 at 11:24
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    $\begingroup$ @bryn, they're not axioms. They are, however, immensely useful properties that a value of 0/0 must have to be of any use. $\endgroup$ – badp Jul 23 '10 at 11:34
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    $\begingroup$ @Anixx Why would $\frac{x}{x}=1$ be true only if $x\neq -x$? $\endgroup$ – 5xum Dec 17 '14 at 7:56
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    $\begingroup$ @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. $\endgroup$ – 5xum Dec 17 '14 at 8:24
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    $\begingroup$ @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. $\endgroup$ – badp Dec 17 '14 at 9:52
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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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    $\begingroup$ So we can divide by zero in the trivial ring with $1=0$ :-) $\endgroup$ – Paŭlo Ebermann Jun 27 '11 at 0:54
  • $\begingroup$ So why one cannot define $0/0=0$? Is having multiplicative inverse of 0 needed for defining this? $\endgroup$ – Anixx Dec 17 '14 at 15:36
  • $\begingroup$ 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. $\endgroup$ – rgrig Dec 17 '14 at 21:58
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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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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 $x$ go to 0)

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

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

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    $\begingroup$ On the other hand, surely you'll agree that 0/0 is undefined in any (nonzero) ring even when you can't take limits! $\endgroup$ – Noah Snyder Jul 23 '10 at 16:03
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    $\begingroup$ but since +∞ is not necessarily commutative is really why its undefined? $\endgroup$ – Talvi Watia Sep 30 '10 at 0:01
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    $\begingroup$ @Talvi: commutativity is a property of operators (e.g. addition, multiplication), not of objects like ∞. $\endgroup$ – J. M. is a poor mathematician Oct 2 '10 at 1:48
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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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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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    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Jul 30 '10 at 22:44
  • $\begingroup$ 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. $\endgroup$ – ShreevatsaR Jul 30 '10 at 22:46
  • $\begingroup$ I think this answer doesn't really answer the question. It's not useless because from it, I can infer another answer that does answer the question but is very similar to Martigan's answer. This answer only asserts that a rational multiplicative inverse of 0 doesn't exist unlike their answer which asserts that a real multiplicative inverse of 0 doesn't exist. Since an answer that's so similar to another answer to the same question doesn't need to exist, I'm not sure this answer needs to exist either. $\endgroup$ – Timothy Nov 30 '17 at 5:14
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Everyone knows 0/0 is nullity.

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

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    $\begingroup$ 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. $\endgroup$ – Isaac Jul 31 '10 at 17:54
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    $\begingroup$ This should be CW. $\endgroup$ – Pedro Tamaroff May 4 '12 at 22:59
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    $\begingroup$ Even if the post is useless, this answer is just as useless as the post. That's why I downvoted it. An observation that doesn't answer the question doesn't warrant an answer. My answer assumes that the question is meaningful and answers that meaning. Also, everyone knows 0/0 is undefined doesn't prove 0/0 is undefined. You didn't give a reason why you think it's undefined. $\endgroup$ – Timothy Nov 30 '17 at 5:22

protected by Qiaochu Yuan Jun 27 '11 at 0:06

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