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:)

]]>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∞.

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

]]>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 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.

]]>(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?

SoHere, 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.

Calligar wrote:

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

Calligar wrote:

I don't think that's necessarily the issue: we've defined The issue is, with this, you can

notdefine "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.

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).