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.
ButΒ \(0\div 0\) 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 anyΒ nonzeroΒ 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 thatΒ \(0\div 0 = 1\) 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 thatΒ \(n\div n = 1\) 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 \(0\cdot1 = 0\cdot x\) , 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.Β 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 is NO number x that satisfies the equation 1/0 = x, or equivalently, 0*x = 1. In contrast, EVERY number x satisfies the equation 0/0 = x, or equivalently, 0*x = 0. But this 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 to think 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 is not enough information to determine the limit of 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 that if 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 limit does not exist: lim cos(1/x) x->0 By comparison, this equation has no 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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time = 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.
Yes, 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 is no way to determine from the speed and distance how long I have been running.
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My 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
Hi, 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.)
Hi 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?
When 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 unique answer.
Now, not all of what you are saying here is really about checking; sometimes you seem to be really trying to find the 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 what is the 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.
Thanks 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?
You 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.
Hello , In this sentence, if we make n equal to 1, the equation is correct. is it correct?
0/0 = 1
n * (0/0) = n * 1
(n*0)/0 = n
(0/0) = n
1 = n
Thanks.
Hi, Ramsy.
No, this is not correct. Because 0/0 is not defined, we can’t perform any operations on it, much less assume its value is 1.
Now, if you’re just experimenting with what would happen if we defined 0/0 as 1, observe that if n=2, you would end up proving that it is also equal to 1, which would be a contradiction. In other words, you are showing that your assumption would imply that all numbers are equal to 1! That is not an acceptable kind of mathematics, so it must be rejected.
Hello. This is in response to your communication with Jonathan Cender. I have had the same understanding that he has and I am trying to understand what you mean in your responses.
In regards to definitions, I am unclear on what you mean by this:
“When you apply a definition, you donβt need other rules (which are typically derived from the definition).”
As I imagine, the rule that a number divided by itself (or a number multiplied by its multiplicative inverse) equals one did not come about because some one figured that the same number in the numerator makes one. It came from a procedure. If I have 5 sets of 1/5, I have 5 parts that are all a fifth of a whole. Thus, when I put them together, I get 1. That is the same as adding five 1/5’s together or multiplying 5 to 1/5. But, obviously, that doesn’t work with a number whose denominator is 0.
To me, this seems like a case of putting the cart before the horse. That is, we have put the rule before what it took for people come to the rule. A pattern was found and it seems that we can’t accept that, at least in this context, it doesn’t work. Why is that such an issue?
Hi, Timothy.
As I said to Jonathan, this might work better as a submitted question than as a comment, so we could discuss it more conveniently.
I don’t think you are disagreeing with the conclusion that 0/0 is not defined, but just asking what my statements mean. Let us know If I’m wrong, and we can discuss that.
I think the main issue is that you are trying to approach this historically (though we don’t really know how these ideas were actually first conceived), and with a focus on processes, whereas we are approaching it as mathematics. In mathematics, we start with definitions, and reason logically from them. Processes are proved to be valid based on definitions. So we can’t start with processes such as “canceling” (and already known rules about when you can do it) as a way to explain the underlying mathematics.
(Often the logical relationship between ideas is quite different from their historical relationship. We are talking about a logical development from definitions, not about how someone first thought of it.)
So from our perspective, the “horse” that pulls everything is the definition, and the “cart” that follows is the rules and methods. We have to first determine whether 0/0 makes any sense (that is, show that the definition of division does not apply in this case), and then we know that you can’t cancel in this case.
I can’t force you to think as a mathematician; if your own understanding of how this works satisfies you, live with that. In fact, your thinking is probably more or less equivalent, as far as the reason “it doesn’t work”.
On the other hand, Jonathan’s initial question assumed that we were using cancelation in our explanation, and my answer was that we were not. He continued to assume we were canceling (illegally), and the whole point is that we are not. So this is a disagreement over a misunderstanding (which, in turn, I think is due to not seeing it from a mathematical perspective). Regardless of your perspective, division by 0 doesn’t work.
Greetings,
3/3=3^0=1
6/6=6^0=1
0/0=0^0=1
The math agreeth in the matching.
What you’re saying here is that you can apply the reasoning used to prove that a/a = 1, namely that a^1/a^1 = a^{1-1} = a^0 = 1, to the case where a = 0. The problem is that this only shows that 1 is a possible value of 0/0. The reason 0/0 is undefined is that that value is not unique, which was the main point of this post; so showing a different way to obtain 1 doesn’t undermine what we have said.
As we said, “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!” We showed that it can equally be shown that 0/0 can equal any other number. In particular, using your reasoning, you could say that 0/0 = (5*0)/0 = 5*0^1/0^1 = 5*0^0 = 5*1 = 5.
But what’s happening more specifically in your work is that the rules you are applying, that a^m/a^n = a^{m-n} and that a^0 = 1, like other rules we mentioned, assume that a is non-zero. Most importantly, it is not universally true that 0^0 = 1; see our FAQ on that subject.
Please explan why 0/0=0 is wrong? zero times any number equals zero, So why is it that any number divied by zero is not zero? 100×0=0 100/0=0. I must be Ignorant to some rule that says 0/0 can not be zero. From my understanding the issue is with the rule that n/n=1 which 0/0 proves that rule is not correct for 0 but is correct for the rest of the numbers. So why is it no one has made the rule n/0=0 they did that for nx0=0? Thank you for taking the time to explain this to me so my ignorance can solved.
Hi, Anthony.
I have explained this several times. Quoting myself from a comment above, “Finding that 0/0 = 0 is not a contradiction, because the whole point is that it could be any number (and therefore canβt be defined to be any one of them).”
The problem is not that 0/0 can’t be 0, but that it can be anything with equal justification; defining it as any one value leads to contradictions (particularly with the calculus sense of indeterminacy, where it would mess up the concept of limits).
But when you suggest n/0 = 0, that breaks many rules; it makes no sense. (See Why Canβt You Divide by Zero?) And the fact that n*0 = 0 is not a made-up rule; that can be proved from other rules.
\( \frac{0}{0} \) is indeterminate. If to think about graphical representation of this relation this can be a numeric line, something similar to \( f(x) = \frac{x}{0}, x = 0 \), in this case the solutions will be all the values of line (0, y).
\( \frac{10}{0} \) is undefined, meaning it does not have solutions at all.
More than that, I do consider number 0 parity undefined, because when checking number 0 parity, dividing \( \frac{0}{2} \), even if the remainder is 0, because the quotient is 0, that means that number 0 parity is undefined.
Here some logical contradictions arise when considering 0 even number, because we check number parity dividing absolute value number by smallest even absolute value number, if 0 is even, it is automatically smallest even absolute value number, therefore we must divide by 0 instead of 2. Contradiction.
Hi, Vitalie.
The graph you suggest isn’t really valid (since it wouldn’t really be a function), but it does hint at the correct distinction between indeterminate and undefined.
But your comments about parity (even/odd) make no sense to me; you seem to be starting from a wrong definition. A number is even when it can be obtained by multiplying an integer by 2; there is no need to talk about division, much less to reject a quotient of 0. If someone has said that we have to divide by the smallest absolute value of an even number, they must have had only positive integers in view; that statement is easily corrected by saying that an even number is a multiple of the smallest even positive integer. (But why would you use the word “even” in its own definition?) Do you have a source for your definition?
**Definition:**
> Numbers parity is counting number of units or object if they are odd or even.
> Zero has no units or objects, therefore number zero parity is undefined.
**Demonstration:**
3: πππ
2: ππ
1: π
0: zero apples, therefore zero parity is undefined
$$ f(x) = \frac{0}{x} = (x,0) \cap (0,y) $$
But your “definition” is not the real definition (I asked for a source!); it is not even a definition at all, since it uses the word “even” without defining it. All you are doing is stating your own idea about what “even” should mean.
When we use words, we use the definitions others agree on. And that definition is well known:
Wikipedia:
MathWorld:
There is a reason for preferring a definition that includes zero; it makes many rules (such as that the sum of two even numbers is even) consistently true, and it eliminates what, by your “definition”, would be a gap in the list. In general, mathematicians prefer such inclusive definitions, rather than arbitrarily excluding special cases as you want to do.
Now, this issue doesn’t really belong in this post; we came close to the topic in Kids Ask About Even and Odd Numbers, though that doesn’t get into zero or negative numbers. At the end there are links to answers including Is Zero Even?, which I will be making a new post about soon.
By the way, your final “equation” makes no sense.
As I have said to others, if you have further questions, please do so through the Ask a Question link at the top of the page, not through comments. That is how we handle extended discussions, especially when they are off-topic as this is.
Dear Dave, what to say?
Mathematics some time has misleading preconceptions and sophism.
Let me show one of them.
$$ \frac{1}{10} = 0, 1 \rightarrow 10 \cdot 0 + 1 = 1 $$
$$ \frac{10}{1} = 10, 0 \rightarrow 1 \cdot 10 + 0 = 10 $$
$$ β΄ \frac{10}{0} = 0, 10 \rightarrow 0 \cdot 0 + 10 = 10 $$
Bibliograpy:
https://www.etymonline.com/word/imply
https://www.etymonline.com/word/sophism
If you are saying this is an example of a sophism, you are right (though I don’t know where the preconception is).
First, you are using a faulty notation for integer division with remainder (Euclidean division). I know of no formal notation, but in early school grades you would write, say, \(1\div10=0\text{ rem }1\), or the like.
Second, in your third line, the quotient of zero could be replaced by any integer, since 0 times anything is 0, so in terms of integer division, division by zero could almost be described as indeterminate – except, …
Third, the remainder is defined as having to be strictly less than the divisor, so a remainder of 10 on division by 0 is improper. Therefore, this actually just shows another reason we can’t define division by 0 (even in terms of integer division).
I assume this was the point you wanted to make.
Fourth, your implication arrows go the wrong direction. Integer division is defined by the multiplications you put on the right, not the other way around.
As for the words, I hope you understand that etymology is not the same as meaning; the history of a word does not tell us what it actually means today. I don’t know why you chose to show this.
I think I found confirmation of my thought in Octave program, which is exclusively Mathematical Tool.
Here it is.
me@amadeus:~$ octave -q
octave:1> mod(10, 0)
ans = 10
octave:2>
me@amadeus:~$
Therefore, using general formula seems to be correct:
$$ \frac{a}{b} = (q, r) \rightarrow b \cdot q + r = a $$
$$ βΈ« \frac{10}{0} = (0, 10) \rightarrow 0 \cdot 0 + 10 = 10 $$
You asked for some sources.
Here it is one interesting, that touch the theme of modulo with zero, that in programming languages are not supported.
Function Reference: mod
https://octave.sourceforge.io/octave/function/mod.html
As I understand it, your claim is that 10 divided by 0 is 0, because that is equivalent to saying that the remainder of that division is 10. And you are justifying that claim using the mathematical software Octave, which gives that answer.
But not only are you missing the fact that software programs do not define mathematics, you have not read your reference carefully. What it says is,
What this means is that the usual definition of remainder does not apply in this case, because x/0 is undefined, so they arbitrarily choose to say that the modulus in this case is equal to the dividend.
In fact, they point out that the related function rem uses “opposite conventions”, namely
We can’t use such conventions for special cases in mathematics, because they would break consistency.
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0/0=1 THINK ABOUT IT HOW MANY TIMES CAN 0 GO INTO 0 1 SO IT IS NOT IMPOSIBLE AND IM ONLY 10 YEARS OLD
Hi, Leland.
You could say that 0 goes into 0 one time, because \(0=0\times1\). But, as explained in the post, you could just as well say that 0 goes into 0 two times, since \(0=0\times2\). In fact, you could claim that 0/0 has any value at all for exactly the same reason.
It’s not because 0/0 is impossible, but because it is “too possible” (in the sense that too many answers work), that causes us to say it can’t be defined. We simply can’t choose one answer to be correct. (And once you learn algebra, you’ll understand what we said here, that if we did make one choice, it would mess up all of mathematics.