Fun with 0 / 0

Calligar · 2016-01-01 15:32:48

One of the things I thought was pretty interesting was something my friend showed me some time back. Very basic and easy, but amazingly I had never thought about it before. 0 / 0 = undefined, why? Well, that's been answered here before and there are LOTS of reasons for that already. But when you start thinking anything * 0, you get 0.

So 1 * 0 = 0. 2 * 0 = 0. 3.9425 * 0 = 0. (pi) * 0 = 0. So then if you divide by 0, you'd think you'd get the reverse. If you divided by 0, you could get any real number, no? My friend (I think jokingly) brought up 0 / 0 = (a variable). So for example, 0 / 0 = v, because v * 0 = 0. And while that may not be correct, I actually thought that was pretty cool and wanted to share it with other people. Any thoughts?

Relentless · 2016-01-01 21:12:25

Yes! It is a neat implication, one of many when working with 0. Your thinking is that the solution to 0/0 is v in v*0=0; but then v=0/0 can be literally anything. One of the fundamental issues with dividing by zero is that if you allow it, then just as you get 1*0=0, 3.9425*0=0, you also get 1 = 3.9425

Personally, I think rather than saying that x/0 is equal to some undefined variable v, it is best to say that it is a question that does not make sense. When you think about 20 divided by 5, you are thinking about how many times you subtract 5 from 20 to get 0. So how many times do you subtract 0 from 1 to get 0? Well, you never will. How many times do you subtract 0 from 0 to get 0? As many as you like. They are really nonsense questions from that perspective.

Last edited by Relentless (2016-01-01 21:23:39)

zetafunc · 2016-01-01 22:55:31

There are frameworks in which division by zero are defined. For instance, the extended complex plane (also called the Riemann sphere), given by adopts arithmetic conventions in which addition, multiplication and division by 0 or infinity are defined (although certain expressions such as ∞ - ∞ or ∞/∞ are left undefined). We can then identify the Riemann sphere with the unit 2-sphere using an isomorphism, which is called stereographic projection -- one of the foundations of spherical geometry.

Relentless wrote:

Personally, I think rather than saying that x/0 is equal to some undefined variable v, it is best to say that it is a question that does not make sense. When you think about 20 divided by 5, you are thinking about how many times you subtract 5 from 20 to get 0. So how many times do you subtract 0 from 1 to get 0? Well, you never will. How many times do you subtract 0 from 0 to get 0? As many as you like. They are really nonsense questions from that perspective.

I think we have to be careful when we decide whether or not a question makes sense: many results in mathematics don't necessarily have to coincide with our natural human intuition. For instance, if we assume the axiom of choice, then we end up with the Banach-Tarski paradox, which challenges our geometric intuition. It also doesn't make intuitive sense to consider the factorial of a non-integer, nor the Riemann zeta function with complex argument -- but they can be made sense of via a process called analytic continuation, the latter of which gives rise to the Riemann hypothesis, perhaps the most famous unsolved problem in mathematics.

Last edited by zetafunc (2016-01-02 00:46:00)

Calligar · 2016-01-02 07:35:59

Relentless wrote:

One of the fundamental issues with dividing by zero is that if you allow it, then just as you get 1*0=0, 3.9425*0=0, you also get 1 = 3.9425

Interesting point (it took me a little bit to realize what you were saying). It is at least one of the issues calling it v. But again, I never listed it saying it was correct, I just thought that such a simple thing like that was worth mentioning out of interest.

Relentless wrote:

Personally, I think rather than saying that x/0 is equal to some undefined variable v, it is best to say that it is a question that does not make sense. When you think about 20 divided by 5, you are thinking about how many times you subtract 5 from 20 to get 0. So how many times do you subtract 0 from 1 to get 0? Well, you never will. How many times do you subtract 0 from 0 to get 0? As many as you like. They are really nonsense questions from that perspective.

Unfortunately dividing any other number by 0 creates a lot more issues. I've worked with it before with my friend. 1 / 0 can not equal v like 0 / 0 can. That point I made only works with 0 / 0. As you pointed out, if you divide 1 / 0, you can not reach a number. There isn't a number in existence (that I'm aware of) that can answer that question. Some may argue the answer is ∞. But then comes an issue with using ∞ as a number, which it isn't (or at least not exactly in that way). Plus there are issues to that. If say 1 / 0 = ∞, what does 2 / 0 = ? Does it also equal ∞, does it equal 2∞? How do you even begin to make sense of it? You can make sense of it, but I don't know anything in mathematics that will answer it except for maybe...1 / 0 = undefined (or something of that nature). Probably already realized all this, but at least interesting to point out.

zetafunc · 2016-01-02 09:04:37

Calligar wrote:

So for example, 0 / 0 = v, because v * 0 = 0.

Why? If such an operation were allowed, then we'd be able to construct arguments such as:

"1*0 = 0
2*0 = 0

dividing through by zero yields

1 = 0/0 = 2

Therefore, 1 = 2."

0/0 is often called an 'indeterminate form' since one can obtain numerous different answers depending on how it is approached.

Calligar · 2016-01-02 09:38:21

That's what Relentless also pointed out...

Relentless wrote:

One of the fundamental issues with dividing by zero is that if you allow it, then just as you get 1*0=0, 3.9425*0=0, you also get 1 = 3.9425

It is an issue with how you are looking at it. If I were to try to argue the case, I'd say it's a bit flawed. 1 ≠ 0/0, nor does 2. v = 0/0, and it will remain undefined unless there's something else that will allow it to be defined. In other words, v may = 1, v may = 2, but unless it is defined, you can't simply say it is 2. However, if v = 0/0, and for some reason v were defined as 3 in this case. Then only 3 = 0/0, and not 2. It would have to be specific to that 0/0. However, please don't try to look at it this way. Anything divided by 0 is undefined or the like, and for reason. I'm not trying to argue that 0 / 0 should equal v. I only was pointing out something interesting that helps you look at 0 / 0 that I found interesting (which doesn't mean it is correct).

Last edited by Calligar (2016-01-02 09:40:09)

zetafunc · 2016-01-02 10:45:48

Calligar wrote:

1 ≠ 0/0, nor does 2. v = 0/0, and it will remain undefined unless there's something else that will allow it to be defined. In other words, v may = 1, v may = 2, but unless it is defined, you can't simply say it is 2. However, if v = 0/0, and for some reason v were defined as 3 in this case. Then only 3 = 0/0, and not 2. It would have to be specific to that 0/0.

I'm finding it rather difficult to make sense of this. It seems as though you're merely replacing "0/0" with the symbol "v". In particular, you say that:

"...unless there's something else that will allow it to be defined."

"...for some reason v were defined as 3 in this case."

What would this "something else" be that would allow the undefined quantity 0/0 to be well-defined? You will need to exhibit your own framework in which 0/0 is well-defined. Moreover, if you make the choice to assign 0/0 with a value of 3 (which, in doing so, you implicitly assume that 0/0 is not undefined), then immediately you're faced with the same contradiction as before: it would not "be specific to that 0/0" (which is a statement I also can't make sense of).

Calligar · 2016-01-02 13:09:21

Okay, just for fun. Let's say that what my friend said is true (which I'm not saying it is). In that example, v (or any variable), and not just any number would have to be equal to 0 / 0. What you are trying to do is get to the answer 0 / 0. As it stands, if you were to multiply 0 by any number, you would get 0. Therefore, using that, 0 / 0 does not make 1, nor 2, nor 3. It remains an undefined variable unless you were using it in some defined manner.

Now how can the variable possible be defined. So I'll make up a problem to try to give an example. So the problem would go something like this: 0÷0+b = a-1. If b is 3, show that 0(a) = 0(4).
0÷0+b = a-1
0÷0+b+1 = a
0(0÷0+b+1) = 0(a)
0[(0÷0)+(+b+1)] = 0(a)
0(0÷0)+0(b+1) = 0(a)
0+0(b+1) = 0(a)
*0(b+1) = 0(a)

filling in 3 for b....
0(3+1) = 0(a)
0(4) = 0(a)
0 = 0

Now after doing all that, my point arises with the specific variable being defined, while 0÷0 still existing. 0÷0 still however remains undefined in this way. That's the first part of what I was talking about...

Calligar wrote:

1 ≠ 0/0, nor does 2. v = 0/0, and it will remain undefined unless there's something else that will allow it to be defined. In other words, v may = 1, v may = 2, but unless it is defined, you can't simply say it is 2.

The second part...

However, if v = 0/0, and for some reason v were defined as 3 in this case. Then only 3 = 0/0, and not 2. It would have to be specific to that 0/0.

Making this very simple, if 0/0 = v, and v = 3, then 0/0 = 3. 0/0 = v, but it also = 4, or 5, or 6. Since v is now defined, v is just 1 of infinite things 0/0 may be. Yet if you use an undefined variable, like in the example I gave where "a" never gets defined (but "v" does). Even in step 2 where 0÷0+b+1 = "a", that's as much of a definition as you're going to get because you still can't determine what 0÷0 is, making a an even different undefined variable showing how this all may seem simple, but it can very quickly get more complicated. So I repeat again, there's reason this remains undefined. Comparing it to a variable is something interesting, but if you actually want to go deeper in understanding it, it will get more and more difficult as you advance.

I would also like to add that I made a bit of a mistake, saying "only 3 = 0/0, and not 2", like what I was saying above, is it's specific to the variable "v". If 3 is defined as "v", then v ≠ 2, and 0/0 = um...let's say "w" which remains undefined, unless you define that as well, then you must use another variable as long as it remains undefined.

Anyway, let me make my case clear now, 0/0 ≠ v. It was an interesting comparison, comparing the answer to a variable, but it just isn't as simple as that. I am still interested in what people have to comment on it because I still find the comparison really interesting as it reminds me of something a kid would do. But I want to make sure everyone understands I don't actually think this is how it works...

* *edit* I changed "v" to "b" because I noticed I put "v" by mistake....

Last edited by Calligar (2016-01-05 18:03:01)

Relentless · 2016-01-02 15:20:29

I understood that I'm not sure if you can say that 0/0 could be anything, honestly, unless you create a framework for it. But in elementary arithmetic, although no answer is satisfactory, I think some answers make more sense than others.

In particular, if you graph a few functions, you will get answers for what functions approach when various things get arbitrarily close to 0, and they will not be 3.9425. I stated in the thread "What is 0^0?" what these limits are.
For example, you mentioned that the answer to 1/0 could be infinity. But if you graph 1/x, you will find that it could be infinity ... or it could be minus infinity, depending on which direction you approach x=0 from. The same holds true for any nonzero real number in the numerator.

As for 0/0, if you graph x/x, it is always 1. If you graph 0/x, it is always 0 (although x has to be positive). So, the answer to 0/0 could be either 1 or 0, or plus or minus infinity, in a sense. But it doesn't work unless there is one consistent answer.

Last edited by Relentless (2016-01-02 15:24:27)

zetafunc · 2016-01-04 05:02:18

Calligar wrote:

Okay, just for fun. Let's say that what my friend said is true (which I'm not saying it is). In that example, v (or any variable), and not just any number would have to be equal to 0 / 0. What you are trying to do is get to the answer 0 / 0. As it stands, if you were to multiply 0 by any number, you would get 0. Therefore, using that, 0 / 0 does not make 1, nor 2, nor 3. It remains an undefined variable unless you were using it in some defined manner.

In other words, you start by assuming that 0/0 is undefined? I'm still having trouble understanding precisely what you mean: are we simply replacing "0/0" with "v", here? If not, can you give a more precise definition of v?

Calligar wrote:

Now how can the variable possible be defined. So I'll make up a problem to try to give an example. So the problem would go something like this: 0÷0+b = a-1. If b is 3, show that 0(a) = 0(4).
0÷0+b = a-1
0÷0+b+1 = a
0(0÷0+b+1) = 0(a)
0[(0÷0)+(+b+1)] = 0(a)
0(0÷0)+0(b+1) = 0(a)
0+0(b+1) = 0(a)
0(v+1) = 0(a)
filling in 3 for b....
0(3+1) = 0(a)
0(4) = 0(a)
0 = 0
Now after doing all that, my point arises with the specific variable being defined, while 0÷0 still existing. 0÷0 still however remains undefined in this way. That's the first part of what I was talking about...

By starting with the expression 0÷0+b = a-1, aren't you assuming that 0/0 is defined to begin with? Moreover, isn't the identity 0*a = 0*4 true for all a, independent of your expression?

Calligar wrote:

Yet if you use an undefined variable, like in the example I gave where "a" never gets defined (but "v" does). Even in step 2 where 0÷0+b+1 = "a", that's as much of a definition as you're going to get because you still can't determine what 0÷0 is, making a an even different undefined variable showing how this all may seem simple, but it can very quickly get more complicated. So I repeat again, there's reason this remains undefined. Comparing it to a variable is something interesting, but if you actually want to go deeper in understanding it, it will get more and more difficult as you advance.

But why, then, are you allowed to manipulate an expression involving terms which are undefined?

There are other reasons why 0/0 is undefined: for instance, if 0 were its own multiplicative inverse, then the real numbers would then fail to be a field, since we would then end up with 0 = 1, which contradicts the crucial field axiom that 0 ≠ 1 (since the real numbers are a complete, ordered field). If you want 0/0 to be defined, then you'll have to make the assumption that 1 = 0.

A simple construction in which 0/0 can be defined is to consider the ring R = {0} -- called the "trivial ring" -- in which addition (+) and multiplication (*) are defined such that 0 + 0 = 0 and 0*0 = 0. Then 0 serves as its own multiplicative inverse, and in particular, 0/0 = 0, as desired. This ring is unique in the sense that it is the only ring consisting of one element (up to isomorphism).

Calligar · 2016-01-05 21:10:56

zetafunc wrote:

In other words, you start by assuming that 0/0 is undefined? I'm still having trouble understanding precisely what you mean: are we simply replacing "0/0" with "v", here? If not, can you give a more precise definition of v?

Here, let me make this easier; let's start off like this.
0×0 = 0, 1×0 = 0, 2×0 = 0, 3×0 = 0, etc.. Therefore, instead of listing the unlimited different answers, one just simply puts the variable "v". So it looks like this: v×0 = 0. Therefore, this reasoning looks at it in a way that if it were v×0=0, then 0÷0 = v. The issue is, with this, you can not define "v" or it won't work. If you were to say 0÷0 = 1, then 0÷0 ≠ 2. That is why it remains "v" and stays undefined. However, I personally don't see this as a solution either, and going deeper with this has its difficulties to say the least.

Another thing I wanted to clarify:

Relentless wrote:

For example, you mentioned that the answer to 1/0 could be infinity.

I apologize if I left that impression. I'll make try to make this clear now. I do not believe 1/0 = ∞. Not even close actually. Not only do you have an issue of using infinity as a number, but it just won't work, at least not that simply. Like I left in my earlier example, if 1/0 = ∞, then what is 2/0 = ?. Logically, one would start to conclude 2/0 = 2∞, however when dealing with infinity, that doesn't seem to make sense. Putting this in this simple form, when you divide 1 by 0, what number do you get? When you divide 2 by 0, what number do you get? Surely you don't get the same number for both problems, do you? Infinity is also not a number. And while it can at times be used similarly to one, it doesn't mean it is one. That's at least how I like to think about it (putting it simply).

zetafunc · 2016-01-05 21:50:11

Calligar wrote:

Here, let me make this easier; let's start off like this.
0×0 = 0, 1×0 = 0, 2×0 = 0, 3×0 = 0, etc.. Therefore, instead of listing the unlimited different answers, one just simply puts the variable "v". So it looks like this: v×0 = 0.

So v is some real number? Looks OK so far. (I assume you're working over , since you mentioned π.) Or is v a set, i.e. ?

Calligar wrote:

Therefore, this reasoning looks at it in a way that if it were v×0=0, then 0÷0 = v.

But you can't make the deduction . You haven't justified why you can perform this operation over the real numbers: this step requires an assumption that 0 has a multiplicative inverse on . It doesn't: if it did, then the reals would cease to be a field. If you want to construct a set of axioms under which this operation is defined, then you'll have to take 0 = 1 (and hence, inductively, 1 = 2 = 3 = ...). I've provided a suitable construction (the trivial ring R = {0}) in the previous post. If you want another justification, then you'll need to construct your own set of axioms for the real numbers under which the operations of addition and multiplication are well-defined and are consistent with 0 having a multiplicative inverse.

Calligar wrote:

The issue is, with this, you can not define "v" or it won't work. If you were to say 0÷0 = 1, then 0÷0 ≠ 2. That is why it remains "v" and stays undefined. However, I personally don't see this as a solution either, and going deeper with this has its difficulties to say the least.

I don't think that's necessarily the issue: we've defined v as being either a real number or . The issue is the implication above.

Last edited by zetafunc (2016-01-06 02:52:50)

zetafunc · 2016-01-06 01:53:42

After some thought, I'm convinced that what you are trying to do is define the quotient 0/0 to be a set, i.e. In that case, can you define a set of rules for which addition and multiplication of sets of real numbers by other sets of real numbers makes sense? For instance, which of these "multiplications" would you like to be true?

(These are not usually how products of sets are defined, by the way: the Cartesian product (the third one) is the norm.)

Whichever you choose (and there are probably other possibilities), you will need to give a detailed explanation of your theory. In particular, which theorems and algebraic laws are still valid? Which ones aren't? And more importantly, is there any reason why we'd do this instead of leaving 0/0 to be indeterminate?

There are arithmetic constructions out there which do similar things: for instance, the so-called "interval arithmetic". Wheel Theory is an algebra in which 0/0 is used, for which there is a more thorough treatment here. You've mentioned "going deeper" with this -- could you elaborate?

Last edited by zetafunc (2016-01-07 09:34:13)

Calligar · 2016-01-09 12:13:14

Okay, I just want to make sure you have caught this...

Calligar wrote:

Anyway, let me make my case clear now, 0/0 ≠ v. It was an interesting comparison, comparing the answer to a variable, but it just isn't as simple as that. I am still interested in what people have to comment on it because I still find the comparison really interesting as it reminds me of something a kid would do. But I want to make sure everyone understands I don't actually think this is how it works...

Basically, all this time, I wasn't trying to prove it, unless I was making arguments for the fun of it, which aren't entirely correct. I was merely trying to explain that there was an interesting comparison my friend made. The logic (and possibly one of the flaws too) is that when dealing with multiplication, division is the opposite. Therefore, whatever you multiply, you can turn around to divide. So 4×2 = 8 and 2×4=8. So, 8÷2 = 4 and 8÷4 = 2. Using that logic, comes the 0÷0, the problem is. Since v×0 = 0 and 0×v = 0, then 0÷0 = v and 0÷v = 0. We already know that 0 divided by anything = 0 (unlike anything divided by 0), therefore, there will be less argument against that. The problem is saying 0÷0, as you start running into more issues.

Now another one of the issues with what I did, which is pretty much wrong, was with this example:

Calligar wrote:

0÷0+b = a-1
0÷0+b+1 = a
0(0÷0+b+1) = 0(a)
0[(0÷0)+(+b+1)] = 0(a)
0(0÷0)+0(b+1) = 0(a)
0+0(b+1) = 0(a)
*0(b+1) = 0(a)
filling in 3 for b....
0(3+1) = 0(a)
0(4) = 0(a)
0 = 0

Let me do something quick that will make it seem wrong.
2 ≠ 3
0(2) ≠ 0(3)
0 ≠ 0

Talking to my friend recently about this, he will still argue it's correct, actually making the argument both 0s are not equal. In the example he gave, he would have taken it further and reversed the whole thing as well. He also would not have done the last thing I did where I said 0 = 0. So I think that's the end of me trying to make arguments for his case . Now here...

zetafunc wrote:

You've mentioned "going deeper" with this -- could you elaborate?

I may have given a brief example of taking it a little deeper, but to be honest, I am not really sure I want to go too much deeper into that (at least at this time), because that's beginning to go into things that don't exist (at least within my knowledge) and also complicated things that already exist, expanding on rules or changing other ones. And I'm not even the one that came up with it, making it more difficult to explain something my own friend did (which is frustrating especially when I make certain mistakes). I'd prefer to not use things like 0÷0 at all as other than just working on the math of it as I don't really have any use for that (so just going into the pure math for the fun of it I guess).

The reason I brought it up here is not to argue what this said is correct, just a curious example that may get you to think about it some more (especially those who aren't familiar with why 0÷0 is undetermined). I know a few years ago (I think I even brought it up on this forum), I honestly thought 0÷0 = only 0. Knowing that 1÷0 = undefined, but specifically was wondering why that was the case with 0÷0. I figured it out later, but when I talked to my friend about it, and he had this whole thing set up for it already that used pretty much basic algebra to try to show it. I thought it was a very curious example, wrong or not. Basically, I was curious about other peoples reactions to that, because I honestly thought it was pretty cool the first time I saw it (even if I don't really believe that to be the case).

Also, I'm not ignoring all the cases you gave zetafunc, I'm honestly still looking into those. The wheel theory I find pretty interesting. But I don't know how much further I can actually answer your questions to be honest, at least at this time. Also, I'm not familiar enough with some of the things your saying. For instance, I'm honestly not sure what you mean each time you're using ∈ nor am I the most familiar with set theory (which I have done very little work with; better just to say I don't know set theory), though I believe I understand the rest of it (could be mistaken).

Note: Also saying 0÷0 = v = ℝ probably is fine, though, I'm unsure if my friend would put it the same way as I'm not sure if he would restrict it to only real numbers or possible include even more.

bobbym · 2016-01-09 12:30:30

I'd prefer to not use things like 0÷0 at all as other than just working on the math of it as I don't really have any use for that

What people find useful is very time relevant, the Calligar of the future will certainly find uses for 0^0 = 1.

Calligar · 2016-01-09 12:46:28

0^0 = 1 is actually quite interesting. It seems to go against what one would think. Let's see, how does the one proof go again...? 1 = A^n/A^n = A^(n-n) = A^0...something like that...

Oh, rofl, I see! 0^0/0^0. I was just about to post and caught that .

Relentless · 2016-01-09 18:31:36

0^0 is equivalent to 0/0, as far as I know. Both of them, especially 0^0, have strong arguments for being 1. Unfortunately, they both have solid arguments for being 0 as well. Several people have mentioned recently that it is useful to assume 0^0 = 1 for the purpose of many formulas (in fact, I saw this just yesterday when deriving the Kelly formula). That is what bobbym is referring to.

Calligar · 2016-03-26 09:10:03

Oh yeah, I realize 0/0 is also used for that purpose with convenience. One of the things that actually gets me interested in these kinds of things is seeing how it may potentially be something other than what you think it is when you look more at it. So in arithmetic, you learn 0 / 0 = (undefined), but then you begin to think, wait a second, what does that even mean exactly? Why is it even undefined? You learn more math later on, as you move on to algebra, calculus, um...set theory, also things you just simply study on your own, etc., which may provide you with even more reasoning for why it is like that. But then, my friend tells me of something like this he concocts, and then, you just kind of laugh and say, this is seemingly quite simple and appears to make sense while also finding it quite clever (well I do anyway). Even if it isn't true and doesn't work, I find the whole thing very interesting and still wonder if it does work with more explanation. It still think on some level it could fit in, but there's so much I don't know, I'm not even able to say that with confidence.

thickhead · 2016-04-21 19:40:12

If somebody starts with zero asset and go on multiplying his assets he will never become rich. for that he has to add assets. Of course once he adds then he can go on multiplying.

Last edited by thickhead (2016-04-21 19:43:03)

thickhead · 2016-04-24 23:03:07

Last edited by thickhead (2016-04-24 23:08:13)

rileywkong · 2016-06-21 06:41:13

Calligar wrote:

As you pointed out, if you divide 1 / 0, you can not reach a number. There isn't a number in existence (that I'm aware of) that can answer that question. Some may argue the answer is ∞. But then comes an issue with using ∞ as a number, which it isn't (or at least not exactly in that way). Plus there are issues to that. If say 1 / 0 = ∞, what does 2 / 0 = ? Does it also equal ∞, does it equal 2∞? How do you even begin to make sense of it? You can make sense of it, but I don't know anything in mathematics that will answer it except for maybe...1 / 0 = undefined (or something of that nature). Probably already realized all this, but at least interesting to point out.

As far as I know, ∞ = 2∞. Think of it this way:

Say we have a hotel that has ∞ floors, with 1 room on each floor. Let's say that the hotel is full. Then, a party of ∞ comes along. How do they make space for them? By moving the person in floor #1 to floor #2 and fitting 1 person from the party in it. Same for the person in #2, #3, so on.

Or, in math language (aka Engrish), if we have a set that has ∞ terms, we can multiply each term by 2 to fit in an additional ∞ terms.

So, writing that out in mathematical terms, ∞ = 2∞.

Last edited by rileywkong (2016-06-21 07:04:04)

Stangerzv · 2016-06-21 19:51:54

If you have an equation of x=y, it is always passing through the coordinate {0,0}. Since,

it concludes that :)

thickhead · 2016-06-25 01:21:44

Well, if y=2x 0/0=2 . actually you can assign a value to 0/0 so that your function has no discontinuity. the value depends on the function.

Stangerzv · 2016-06-25 02:11:04

Basically 0/0 can be anything. Unless we tend to go beyond normal understanding of infinity, we can actually get 0/0=1. For example

let

Taking natural long on both sides yields

Then

If we could consider the "undefined" cancelling each other equal to zero.

Thus

We can argue like forever with this thing but this is how mathematics progresses Perhaps 0/0=An Apple:)

zetafunc · 2017-05-23 00:51:23

I will be making a video about this discussion within the next week or so.

Math Is Fun Forum

#1 2016-01-01 15:32:48

Fun with 0 / 0

#2 2016-01-01 21:12:25

Re: Fun with 0 / 0

#3 2016-01-01 22:55:31

Re: Fun with 0 / 0

#4 2016-01-02 07:35:59

Re: Fun with 0 / 0

#5 2016-01-02 09:04:37

Re: Fun with 0 / 0

#6 2016-01-02 09:38:21

Re: Fun with 0 / 0

#7 2016-01-02 10:45:48

Re: Fun with 0 / 0

#8 2016-01-02 13:09:21

Re: Fun with 0 / 0

#9 2016-01-02 15:20:29

Re: Fun with 0 / 0

#10 2016-01-04 05:02:18

Re: Fun with 0 / 0

#11 2016-01-05 21:10:56

Re: Fun with 0 / 0

#12 2016-01-05 21:50:11

Re: Fun with 0 / 0

#13 2016-01-06 01:53:42

Re: Fun with 0 / 0

#14 2016-01-09 12:13:14

Re: Fun with 0 / 0

#15 2016-01-09 12:30:30

Re: Fun with 0 / 0

#16 2016-01-09 12:46:28

Re: Fun with 0 / 0

#17 2016-01-09 18:31:36

Re: Fun with 0 / 0

#18 2016-03-26 09:10:03

Re: Fun with 0 / 0

#19 2016-04-21 19:40:12

Re: Fun with 0 / 0

#20 2016-04-24 23:03:07

Re: Fun with 0 / 0

#21 2016-06-21 06:41:13

Re: Fun with 0 / 0

#22 2016-06-21 19:51:54

Re: Fun with 0 / 0

#23 2016-06-25 01:21:44

Re: Fun with 0 / 0

#24 2016-06-25 02:11:04

Re: Fun with 0 / 0

#25 2017-05-23 00:51:23

Re: Fun with 0 / 0

Board footer