Back in January, I discussed the issue of division by zero. There is a special case of that that causes even more trouble, in every field from arithmetic to calculus: zero divided by zero. I’ll look at several typical questions that we answered at different levels.

## Conflicting rules for division?

Let’s start here:

Zero Laws and L'Hopital's Rule Hi, I was just wondering - if you have 0/0 (zero divided by zero), which law takes precedence - a) zero divided by any number is zero, or b) any number divided by zero is undefined, or c) any number divided by itself is one? Thanks.

Sheryl knew three facts:

- \(0\div n = 0\) for any number
*n* - \(n\div 0\) is undefined for any number
*n* - \(n\div n = 1\) for any number
*n*.

But \(0\div 0\) fits all three rules, so what happens when math “rules” do battle? Is the result 0, or undefined, or 1? The answer turns out to be “undefined”, but there’s a lot to be considered. The first issue is to clarify those three rules.

Doctor Sonya answered:

Zero is a tricky and subtle beast - it does not conform to the usual laws of algebra as we know them. You are right that zero divided by any number (except zero itself) is zero. Put more mathematically: 0/n = 0 for all non-zero numbers n. You get into the tricky realms when you try to divide by zero itself. It's not true that a number divided by 0 is always undefined. It depends on the problem. I'm going to give you an example from calculus where the number 0/0 is defined. If you haven't had calculus yet, just let this sit in the back of your head, and refer to it again later.

Note the clarification: each of the three rules has been proved only under certain conditions:

- \(0\div n = 0\) for any
*nonzero*number n - \(n\div 0\) is undefined for any
*nonzero*number n - \(n\div n = 1\) for any
*nonzero*number n.

She then went into a little demonstration with limits in calculus; I’ll be getting into that later. (She isn’t really saying that 0/0 itself ever has a value, but that something that looks like it may, in the form of a “limit”.) For the moment, let’s stick to arithmetic, as she continues:

Here's another bit of weirdness with 0. Let's say that 0/0 followed that old algebraic rule that anything divided by itself is 1. Then you can do the following proof: We're given that: 0/0 = 1 Now multiply both sides by any number n. n * (0/0) = n * 1 Simplify both sides: (n*0)/0 = n (0/0) = n Again, use the assumption that 0/0 = 1: 1 = n So we just proved that all other numbers n are equal to 1! So 0/0 can't be equal to 1.

The assumption that \(0\div 0 = 1\) leads to a contradiction. And in fact, assuming *any* value leads to the same contradiction — unless you are willing to allow *all* numbers to be equal! For that reason, we have to say that it is simply undefined, so we can change that second rule back:

- \(n\div 0\) is undefined for
*any*number n

## Indeterminate

Here’s another question at the same level:

Defining 0/0 I'm in the tenth grade and I recently posed a question to my Algebra teacher on defining 0/0. Based on our own rules of math, I argued my teacher into agreeing that 0/0 must be defined as 1 simply because, even though zero is undefined, 0 = 0. And our math laws say that anything divided by itself equals 1. So my question is, based on these simple laws, shouldn't 0/0 = 1?

Jon is sticking with the rule that \(n\div n = 1\) for any number *n*, without exception. (It’s not clear what he means by saying that zero is undefined.) Doctor Shawn referred to our FAQ on division by zero, explaining that defining it would lead to contradictions; Jon came back with an attempt to avoid contradictions be declaring that zero *times* anything other than 1 is undefined. I responded,

It seems to me that you are trying to resolve an impossible problem by destroying one of the fundamental facts of math, that zero times anything is zero. Why would you want to do away with that, just so you can define 0/0? The fact is, we KNOW that 0*1 = 0 AND 0*2 = 0, so your reasoning shows equally well that 0/0 = 1 and that 0/0 = 2. When we find that an operation can produce two different results (and in fact, it could be anything at all), then we just accept that it is not defined. Any definition would cause inconsistencies in the rest of math, such as implying that 1 = 2, as you said. Rather than deny that 0*2 = 0 (which would mess everything else up anyway),we just call 0/0 indeterminate.

This is perhaps the simplest such contradiction. We can’t make an arbitrary decision that one proof is valid and others are not, when there is no difference in the logic.

I used here the word “indeterminate”. At the level of arithmetic, this means what I just explained: that more than one value can be derived, so we can’t determine a single value, and so leave it undefined. It has a more nuanced meaning in calculus, as I next explained:

But you might be interested in some deeper facts about that indeterminate 0/0. When you get into calculus, you will learn that there are many problems that seem to lead to this "value," but you can often approach the problem differently and find which value "0/0" has in the specific case. You'll find that, in fact, it can have any value, not only 1. If we defined 0/0 to have one particular value, then all that work would become wrong!

The idea is that we can have an expression (function) in the form of a quotient, and as we move the variable toward a specific value, the numerator and denominator both approach zero, so that it looks as if the quotient was approaching 0/0. If that expression had one specific value, that would have to be the answer — but it turns out that the answer could be absolutely anything, depending on the particular functions involved. So we can’t define 0/0 without risking wrong answers.

## But what about limits?

Some students learn a little about the ideas I just described, called limits, and get a wrong impression. Here is a question that arose that way:

0 Divided by 0 What is the answer to 0 divided by 0? I think it is undefined because of this equation: 0/0 = x/1 X can equal any number and still satisfy that equation by the cross multiplication method. But I am starting to doubt myself because I was talking to a couple of friends of mine about it and they said that it is due to L'Hopital's Rule. I have looked it up on the Internet but all the explanations are in calculus terms, which are like Greek to me. Please help me figure this out.

Peter is saying that cross-multiplication in his equation yields \(0\times 1 = 0\times x\) , and any value of x will make that equation true, so that 0/0 can have any value. This is another way to describe what I said above. But his friends know enough about calculus to be dangerous! L’Hopital’s rule is a technique for finding limits, and it gives specific values, rather than leaving an undefined value. Doctor Mark told him what to say to his friends:

Well, your friends are wrong. L'Hopital's rule does *not* tell you what 0/0 is, because 0/0 is what is called an "indeterminate" quantity, which is to say that its value depends on what the situation is. To convince your friends of this, ask them the following question: "Find the limit of (ax)/x as a approaches 0 by using L'Hopital's rule." They will get "a" (trust me!). But if you just put x = 0 in this expression, you get 0/0. So, according to L'Hopital, 0/0 is equal to a.

Here we have a limit of a quotient, and whatever value \(a\) has, that will be the value L’Hopital’s rule gives for the limit. (Actually, we don’t need any fancy rules; the value of (ax)/x for any value of \(x\) is \(a\), and the limit of that has to be \(a\).)

Did you notice that I didn't say what "a" was? That's because it doesn't matter. You can pick a equal to anything you want. For instance, you could pick a = 1. Then you would get 0/0 = 1 Or pick a = - 3.14159. Then: 0/0 = - 3.14159. So as you can see, 0/0 can be anything you want it to be. On the other hand, in a particular problem, 0/0 might turn out to be something very precise (and that's where you really do need calculus to understand it!).

In calculus, the definition of an **indeterminate form** is a *form* (such as our quotient “approaching 0″/”approaching 0”) for which a limit can take more than one value, depending on how the parts of the expression approach their limits.

Then he backed up Peter’s arithmetic-level reasoning:

I think your argument for why 0/0 is undefined is a really good one. However, I have another way of understanding why 0/0 doesn't make sense, and it goes like this. One way of understanding the fraction a/b is to think of it as the answer to the following question: "If I had a dollars, and b friends, and I distributed those a dollars equally amongst my b friends, then how much money would each of my friends get?" The answer is that they would each get a/b dollars. You can see that this works for fractions like 6/3, or 5/10, and so on. But try it for 0/3. If you have 0 dollars, and 3 friends, and you distribute those 0 dollars (you're feeling generous...) equally amongst each of them, how much would each of your 3 friends get? Clearly, they would each get 0 dollars! Now try it for 3/0. If you have 3 dollars and 0 friends, and you....but how can you distribute any amount of money amongst friends who don't exist? So the question of what 3/0 means makes no sense! Now here's the kicker: What if you have 0 dollars and 0 friends? If you distribute those 0 dollars equally amongst your 0 friends, how much does each of those (nonexistent) friends get? Do you see that this question makes no sense either? In particular, if 0/0 = 1, then that would mean that each of your nonexistent friends got 1 dollar! How could that be? Where would that dollar have come from? Stand your ground, Peter...you're right, and they are wrong, and if they don't believe you, tell them to write to me, and I will set them straight.

Illustrations like this always have some loopholes; but the fact is that no matter how much you say you gave to each nonexistent friend, no one can prove you wrong! You could say that each of them got $1, so that you spent a total of 0 times $1 = $0; or that each of them got a million dollars.

## Any number? Or a detour sign?

People keep writing to ask about 0/0, many thinking they have a resolution to this “problem”. The next one led to a long discussion I will only dip into; but it turns out that in a sense he was essentially right.

The Indeterminate Nature of 0/0 I have been fiddling around with dividing 0 by 0, and have come up with an interesting theory. I have not yet seen any proof that this theory cannot work, so as of now I think it is valid. It goes as follows: My theory is that: 0/0 = any number To start, use the definition of division equation where: a/b = c because c*b = a In the situation of 0/0 = c, then c*0=0 for any real value of c (I am not familiar with how to work with infinity or non-real numbers so I have left them out.) ... I hope my writing has made sense. If you can think of a way to tear down this theory, I am anxious to hear it. Please keep in mind that I am not arguing 1/0, because that is still undefined, I am only arguing for 0/0.

Doctor Ian first answered, taking “any number” to mean any *particular* number, and showing the contradiction. After Rob wrote back, Doctor Rick replied, focusing on his misunderstanding of the word “indeterminate”:

I don't know exactly what your theory means, and neither did Dr. Ian. You say, "0/0 = any number." The natural way to interpret this phrase is, "You can choose any number x, arbitrarily, and the equation 0/0 = x is true." Dr. Ian read it this way, I believe, and from this he derived that 1 = 2. This causes big problems for math, as he said. Your response is that this isn't what you meant. You say that "the value of 0/0 is dependent on the situation."The word I would use for this is "indeterminate."Here is what we mean by "indeterminate." The value of 1/0 is called "undefined" because there isNO numberx that satisfies the equation 1/0 = x, or equivalently, 0*x = 1. In contrast,EVERY numberx satisfies the equation 0/0 = x, or equivalently, 0*x = 0. Butthis does not mean that you could substitute any number arbitrarily for 0/0. It depends on the context - on the details of the problem in which you encountered 0/0. When you run into a problem whose solution appears to be 0/0, there may be a solution, but you'll have to back up and try to find it by another approach. When you find it, you could say that 0/0 "equals" that value, but only for that particular problem - just as you say.

It appears that Rob wants 0/0 to stand for every number at the same time. But if we take that seriously, it will cause as much trouble as taking 0/0 to have any one value. Any expression we write has to have only one value, or it becomes practically meaningless.

After a comment on limits, he says,

Thus, a good way tothink of the "indeterminate" 0/0 is as a "detour sign". It says in effect, "The road ahead is washed out. The town beyond it is still there, but you'll have to find another way to reach it." To bring this around to the word that you don't like, it says, "The solution can NOT be DETERMINED by this method. There may be a solution, but you must determine it some other way."

After more useful discussions of Rob’s points, he closes with this:

Have I made sense yet? I think the key concept you have been missing is this problem-specific nature of 0/0. The difference between "any number" and "indeterminate" is that "indeterminate" means "it MAY be any number, but you can't tell just by looking at 0/0 itself."

We could say that 0/0 is like a pronoun, such as “it”. It can stand for any object at all; but you don’t know what that object is except from the context, and it doesn’t stand for every object in the universe all at the same time. Outside of a context, it has no meaning at all.

For a similar discussion, see:

Division by Zero: Indeterminate or Undefined?

## Distinguishing the terms

I’ll close with this question, at the calculus level:

Undefined and Indeterminable ... at the Same Time? Are the terms "indeterminate form" and "undefined/no solution" mutually exclusive? In other words, can an expression be considered both an indeterminate form AND undefined? Of course it could be neither -- but is it only ever one or the other? I know it is possible for an expression to be ONLY undefined, as is 1/0. But can something like 0/0 or 0^0 be considered both undefined and indeterminate? Or are they just indeterminate?

Doctor Vogler replied:

The phrase "indeterminate form" is used in the context of limits, whereas "undefined" refers to evaluating functions, and "no solution" refers to solving equations or similar problems. Let's look at some examples from each of these different contexts.

These ideas can overlap, but they are typically answering different questions.

Consider the function f(x) = (x^2 - 4)/(x - 2) It is undefined at x = 2, even though it has a limit as x -> 2. Indeed, the limit has the indeterminate form 0/0 and therefore one way to evaluate the limit is using L'Hopital's Rule. (Another way is dividing out the common factor.)

This function simplifies to \(f(x) = x + 2\ (x \ne 2)\), so the limit is \(2 + 2 = 4\). So although the function itself is undefined there, and the *form* of the limit is 0/0 (indeterminate), the limit is not undefined.

But if you are taking a limit at a finite point (not a limit at infinity), then you can relate evaluating the limit as x -> c to evaluating the function at c. In this context, you can compare "undefined" to "indeterminate." I should warn you that the limit as x -> c does not depend on the value of the function at c. However, indeterminate forms apply when the function in question is some operation (like the quotient) of two *continuous* functions at c. So, for example, we say that 0/0 is an indeterminate form because if ... f(x) = g(x)/h(x) ... for continuous functions g and h (at the point c) and g(c) = h(c) = 0, this isnot enough information to determine the limitof f(x) as x -> c. But it is enough information if, for example, h(c) = g(c) = 1.

This is a case where we can say that the limit has an indeterminate form, and we can also try to evaluate the expression at the given value, and say that it can’t be evaluated — it is undefined.

Normally, it is true thatif your limit has an indeterminate form, then evaluating the function at that point will yield an undefined result, in the sense of the continuous functions referred to above. However, I can think of one exception to this rule: the last example you named! For several reasons, the expression 0^0 is generally understood to evaluate to 1, even though (in limits) this is an indeterminate form, because as x -> 0, 0^(x^2) -> 0 and x^0 -> 1. Of course, a function with a "removable discontinuity" is undefined at the point in question even if the limit does not have an indeterminate form. So in that sense, the two concepts are not really related.

How about Michael’s mention of “no solution”, and the similar phrase “does not exist”?

Finally, I would say that this limitdoes not exist: lim cos(1/x) x->0 By comparison, this equation hasno solution: x^2 + 1 = (x - 1)(x + 1)

These are entirely different contexts; “solution” applies only to equations (or, more generally, to “problems”).

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Peter Spasovtime = distance/speed , e.g. t = (5 m)/(2 m/s) = 2.5 s

if any object travels zero distance at zero speed, a fancy way of saying standing still, then t = (0 m)/(0 m/s), meaning time could be anything while the object is standing still.

Dave PetersonYes, that’s a good illustration of the fact that 0/0 is

indeterminate, in the sense that it could conceivably be answered by any number. If I run at 0 meters per second, how long does it take me to cover 0 miles? I’ll be at the zero meter mark at every possible time; there isno way to determinefrom the speed and distance how long I have been running.Pingback: Is {0} Closed Under Division? Thoughts, and Second Thoughts – The Math Doctors

Jonathan CenderMy question is about the following from the article above.

“0/0 = x, or equivalently, 0*x = 0”.

Is cancellation involved in “equivalently”? If so, how is this warranted? Cancellation meaning

0/0 = x

multiply 0/1 by both sides

0/0 * 0/1 = x * 0/1

then cancel to arrive at

0 = x * 0

Again, how is cancellation warranted? I am not aware of any normal arithmetic rule allowing cancellation of zero unlike other numbers. Following normal arithmetic rules gives

0/0 * 0/1 = x * 0/1

multiply numerators and denominators

(0*0)/(0*1) = (x*0) / 1

0/0 = 0

Dave PetersonHi, Jonathan.

Doctor Rick is applying the definition of division that Rob had given: “a/b = c [whenever] c*b = a”. That is, division is the inverse operation to multiplication. Replacing a, b, and c with 0, 0, and x respectively, we find that 0/0 = x is “equivalent” in this sense to x*0 = 0. Since this is true for any x, we can’t identify one number x that is the appropriate value of 0/0; it is indeterminate.

The whole point, of course, is that this division is

undefined, so normal rules won’t actually apply! The rules apply only to “numbers” that have a specified value. That’s why what you suggest doesn’t work. (You could, however, use the very fact that multiplication rules would fail if we defined 0/0 to have some value, to conclude that it does not; but that is harder for most people to follow.)Jonathan CenderHi Dave,

Thanks for your reply.

Request clarification on using cancellation specifically.

I understand that Doctor Rick is applying the definition of division. And I understand that the definition does not apply to 0. My question goes to how we check if numbers correctly satisfy the definition.

As I understand it, cancellation is normally involved in the checking process. For example, using cancellation normally, we can substitute 4,2 for a,b to see there is an integer c = 2 but for 3,4 we can check and see there is no integer equal to c. We do not just have to do it in our heads.

However, when checking if 0 is correct, how do we do that using normal rules? Since cancelling 0 is not a rule in normal arithmetic why should it be allowed in this particular instance? Or is there another way different than how I showed above?

Dave PetersonWhen you apply a definition, you don’t need other rules (which are typically derived from the definition).

If I claim that 12/4 = 3, I can apply the definition to check my claim by seeing whether it is true that 12 = 4*3. The latter is true, so the former is true.

If I claim that 12/5 = 3, I check whether it is true that 12 = 5*3. It is not, so my claim is wrong.

If I claim that 12/0 = 1, I check whether it is true that 12 = 0*1. It is not, so my claim is wrong. I have not used any rules of arithmetic to do that.

If I claim that 0/0 = 1, I check whether it is true that 0 = 0*1. Hmmm … It is true! But it would have been true if I had tried any other answer. This is what is unique about 0/0; the problem is not that I can’t find an answer, but that there are too many possibilities. There is no

uniqueanswer.Now, not all of what you are saying here is really about

checking; sometimes you seem to be really trying tofindthe answer, which is not the same thing. In my case of 12/5, having seen that 3 was not the right answer, I might now want to find whatisthe right answer, by writing 12 = 5*x and trying to find a number that works by looking through a multiplication table. I find none, and can convince myself that there is no such integer (and switch over to looking for a quotient and remainder). Even here, I am not using any rules or “canceling”. I really don’t understand what you mean by your claim that canceling is normally involved in checking.Jonathan CenderThanks for addressing “cancelling” directly.

Cancelling is involved solving for x in an equation or reducing a fraction.

So cancel the fives in the fraction 5*x/5 and the result is x. In other words divide the fives for the result 1*x which equals x.

This procedure can be substituted for looking up tables and trying out various numbers.

You mention that “[I] seem to be really trying to find the answer.” I am happy to address cancelling in those terms. Can we say “solving an equation”?

So for the equation 0/0 = n

usual step in solving an equation like this is to multiply the inverse of the denominator by both sides

(0/0)*(0/1) = n*(0/1)

by distributive property

(0*0)/(0*1) = (n*0)/1

Multiply by 0 since division by 0 is not defined

0/0 = 0

This gives a different result than plugging in numbers into the definition.

Why the difference? Is there a flaw in my arithmetic?

Dave PetersonYou are missing the point. There are several issues:

I could go into each of these, and more, in depth, but this is the sort of discussion that doesn’t belong in comments. If you have further questions, please do so through our Ask a Question site. That’s what we’re here for.