- #1

- 1,291

- 0

Is it 0 or undefined? I thought this was interesting, it seems a paradox in its own.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Nano-Passion
- Start date

- #1

- 1,291

- 0

Is it 0 or undefined? I thought this was interesting, it seems a paradox in its own.

- #2

- 148

- 12

Intuitively, dividing zero by zero makes no sense because you are asking 0=0*x for what number x? Well, x could be any real number and it would satisfy that equation. In other words, the expression 0/0 isn't defined to be a particular number, whereas when we define division as a/b for real numbers a and nonzero real numbers b, we mean a/b=x where x is the unique solution to the equation x*b=a. The expression a/b is supposed to spit out a single real number. In our 0/0 case, it sort of gives us literally every real number as an output, which means it is useless if we are trying to describe a specific number with it.

You might want to hide your post before the mathematicians see this and die of shock and mad rage! lol

- #3

DaveC426913

Gold Member

- 19,293

- 2,777

Thank you nucl, for reminding me of the proof that 0/0 has a good reason for being undefined.

- #4

- 1,291

- 0

Intuitively, dividing zero by zero makes no sense because you are asking 0=0*x for what number x? Well, x could be any real number and it would satisfy that equation. In other words, the expression 0/0 isn't defined to be a particular number, whereas when we define division as a/b for real numbers a and nonzero real numbers b, we mean a/b=x where x is the unique solution to the equation x*b=a. The expression a/b is supposed to spit out a single real number. In our 0/0 case, it sort of gives us literally every real number as an output, which means it is useless if we are trying to describe a specific number with it.

You might want to hide your post before the mathematicians see this and die of shock and mad rage! lol

That was a beautiful explanation nucl, I never thought about division just being the inverse of multiplication (duh!). Your explanation makes complete sense.

And why would the mathematicians see this and die of shock and mad rage? loll

- #5

- 148

- 12

That's just how they are! :P

- #6

- 22,089

- 3,296

Also see the FAQ on this topic: https://www.physicsforums.com/showthread.php?t=530207 [Broken]

Last edited by a moderator:

- #7

- 1,768

- 126

But, still, 0/0 wouldn't have a good interpretation because that would correspond to the function 0/x, which is zero everywhere. I guess you could send 0 to 0, so that the function is continuous. So, you could define 0/0 to be zero. But it would be very confusing and bad notation that wouldn't accomplish anything, since there's no need to describe the constant function equal to 0 by such a convoluted means. And again, you would need to be careful to point out that it's a mapping, rather than taking an inverse, but that's a moot point. Better not to discuss it at all than to cause all this confusion. So, yes, 0/0 is undefined.

- #8

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

- #9

mathman

Science Advisor

- 7,924

- 467

There is a distinction between (1) lim f/g, where f -> 0 and g -> 0 and (2) 0/0. Case (1) is undetermined, case (2) is undefined.

- #10

- 688

- 1

[itex]\frac{lim}{x -> 0}[/itex][itex]\frac{x}{x}[/itex]

and apply l'Hôpital's rule to obtain:

[itex]\frac{1}{1}[/itex] = 1

- #11

- 22,089

- 3,296

[itex]\frac{lim}{x -> 0}[/itex][itex]\frac{x}{x}[/itex]

and apply l'Hôpital's rule to obtain:

[itex]\frac{1}{1}[/itex] = 1

That are limits. Limits have nothing to do with our discussion here. Whether 0/0 is defined or not is independent from whether the limit is defined or not.

- #12

- 1,291

- 0

Also see the FAQ on this topic: https://www.physicsforums.com/showthread.php?t=530207 [Broken]

That was interesting, thank you. You said that [itex] \frac{1}{0}[/itex] isn't ∞ because as you approach 0 in a rational function then it can either be positive or negative infinity. So then would you be able to state:

[itex]\frac{0}{0}= -∞ < x < +∞ [/itex], where x exists anywhere on the extended real number line. Then the probability of x being a particular value on the one of the real numbers would be [itex]\frac{1}{∞}[/itex] would be undefined. Therefore, that might imply that [itex]\frac{1}{0}[/itex] is undefined.

But that is a very weak and inconclusive argument, I'm speaking gibberish haha.

Anyways, if you had to describe Peano axiom [of multiplication] in one or two sentences, what would it be?

Last edited by a moderator:

- #13

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

[itex]\frac{lim}{x -> 0}[/itex][itex]\frac{x}{x}[/itex]

and apply l'Hôpital's rule to obtain:

[itex]\frac{1}{1}[/itex] = 1

Or what about looking at

[itex]\frac{\lim}{x\to 0}\frac{2x}{x}[/itex] which has limit 2.

Or

[itex]\frac{\lim}{x\to 0}\frac{ax}{x}[/itex] which has limit a, for any number a.

- #14

jgens

Gold Member

- 1,581

- 50

That was interesting, thank you. You said that [itex] \frac{1}{0}[/itex] isn't ∞ because as you approach 0 in a rational function then it can either be positive or negative infinity.

Not quite. In the real numbers or the extended real numbers, the expression 1/0 is left undefined. In the real number system, this is done since

However, in the projective real number system, we define 1/0 = ∞. In the projective reals, there is only one infinite element and this element does not have sign. This is what makes this definition work.

The point is that the answer to some of these questions depends entirely on the context. In some number systems, 1/0 is undefined while in others it has a perfectly reasonable definition.

So then would you be able to state:

[itex]\frac{0}{0}= -∞ < x < +∞ [/itex], where x exists anywhere on the extended real number line.

So what you are suggesting here is that we define 0/0 as a collection of numbers. There is nothing inherently wrong with this, but there is also no real motivation to do so either. In my opinion, there are (aesthetic) reasons not to define 0/0 in this manner. In particular,

- If we adopt the convention that 0/0 =
**R**, then expressions like 1/2 are numbers while 0/0 is a set. There is nothing wrong with this, but it is inconvenient that some ways of stringing together numbers give numbers while other ways give sets. - While expressions like 1/2 can be interpreted as 2
^{-1}, we are forced to interpret 0/0 as an expression in it's entirety. In particular, things like 0 * 0^{-1}still make no sense, since the distributivity axiom for rings guarantees that 0 is not a unit in any ring.

Then the probability of x being a particular value on the one of the real numbers would be [itex]\frac{1}{∞}[/itex] would be undefined. Therefore, that might imply that [itex]\frac{1}{0}[/itex] is undefined.

If you want 0/0 to denote a value of

As a slightly unrelated note on probability, consider the following problem: If you select an integer at random from

- #15

- 1,291

- 0

As a slightly unrelated note on probability, consider the following problem: If you select an integer at random fromZ, what is the probability that the integer you chose is 0? It turns out the probability is zero. Therefore, there are events with probability 0 that can still occur. Likewise, there are events with probability 1 that do not occur. These are just some neat things that happen when you consider probability on infinite sample spaces.

Okay I agree with your post. Something with probability one does not have to occur. But how can something with probability 0 occur? It could be an infinitesimal and not occur that much I agree, [itex]lim_{Δx\ to 0}[/itex] isn't necessarily 0. But 0 is a bit of a different number.

What do you think? I'm still new to math so I might be wrong..

Last edited:

- #16

jgens

Gold Member

- 1,581

- 50

Okay I agree with your post. Something with probability one does not have to occur. But how can something with probability 0 occur? It could be an infinitesimal and not occur that much I agree, [itex]lim_{Δx\ to 0}[/itex] isn't necessarily 0. But 0 is a bit of a different number.

In the real number system, there are no infinitesimal elements. The same is true in the extended reals and projective reals as well. In fact, most mathematicians rarely (if ever) do any work that uses formal infinitesimals. There are number systems that have infinitesimal elements (like the hyperreal numbers), but most of these have roots in model theory and are fairly difficult to define formally. If you are interested, non-standard analysis is the subject that deals with the calculus of these infinitesimal numbers, but non-standard analysis is far from one of the more active areas of research in analysis.

Therefore, it is often best not to resort with reasoning using infinitesimals. Without using their formal properties, it is easy for your intuition to deceive you. It turns out most people have terrible intuition when it comes to infinitesimals.

Now, it is important to note that [itex]lim_{h \to 0} h = 0[/itex]; that is, the value of the limit

Finally, keeping what I've said above in mind, something with probability 0 can occur in just the same manner as something with probability 1 not occurring.

- #17

- 1,291

- 0

In the real number system, there are no infinitesimal elements. The same is true in the extended reals and projective reals as well. In fact, most mathematicians rarely (if ever) do any work that uses formal infinitesimals. There are number systems that have infinitesimal elements (like the hyperreal numbers), but most of these have roots in model theory and are fairly difficult to define formally. If you are interested, non-standard analysis is the subject that deals with the calculus of these infinitesimal numbers, but non-standard analysis is far from one of the more active areas of research in analysis.

Therefore, it is often best not to resort with reasoning using infinitesimals. Without using their formal properties, it is easy for your intuition to deceive you. It turns out most people have terrible intuition when it comes to infinitesimals.

Now, it is important to note that [itex]lim_{h \to 0} h = 0[/itex]; that is, the value of the limitis0. The limit is not infinitesimally close to 0, but actually is 0. This is an extremely important point to understand.

Finally, keeping what I've said above in mind, something with probability 0 can occur in just the same manner as something with probability 1 not occurring.

Hey, thanks for your patience. You haven't really argued on how something with probability 0 can occur. I'm not completely convinced at the moment. I'll try to put it in words for the sake of argument; to me 0 is absolutely nothing, so for absolutely nothing to happen is a paradox. Can you throw in a bit of mathematics, I'm interested.

- #18

- 22,089

- 3,296

Hey, thanks for your patience. You haven't really argued on how something with probability 0 can occur. I'm not completely convinced at the moment. I'll try to put it in words for the sake of argument; to me 0 is absolutely nothing, so for absolutely nothing to happen is a paradox. Can you throw in a bit of mathematics, I'm interested.

Hmm, probability 0 is indeed a silly concept. Most people think of probability as throwing dice, and indeed: throwing a 5.5 with a dice has probability 0 and thus never happens. But it is important not to generalize this situations. There are some probability 0 situations which can happen.

As an example: choosing an arbitrary number in the interval [0,1]. It is clear that all numbers have the same probability p of being chosen. However, saying that a number has probability p>0 is wrong, since [itex]\sum_{x\in [0,1]}{p}\neq 1[/itex]. So we NEED to choose p=0. So choosing probability 0 for this is actually quite unfortunate and caused by a limitation of mathematics.

However, there is another way of seeing this. Probability can be seen as some "average" value. For example, if I throw dices n times (with n big), then I can count how many times I throw 6. Let [itex]a_n[/itex] be the number of 6's I throw. Then it is true that

[tex]\frac{a_n}{n}\rightarrow \frac{1}{6}[/tex]

So a probability is actually better seen as some kind of average.

Now it becomes easier to deal with probability 0. Saying that an event has probability 0 is now actually a limiting average. So let [itex]a_n[/itex] be the number of times that the event holds, then we have

[tex]\frac{a_n}{n}\rightarrow 0[/tex]

It becomes obvious now that the event CAN become true. For example, if the event happens 1 or 2 times, then we the probability is indeed 0. It can even happen an infinite number of times.

Probability 0 should not be seen as a impossibility, rather it should be seen as "if I take a large number of experiments, then the event will become more and more unlikely". This is what probability 0 means.

- #19

jgens

Gold Member

- 1,581

- 50

- If you choose an integer at random from
**Z**, what is the probability that the integer chosen is 0? - If you choose an integer at random from
**Z**, what is the probability that the integer lies between -N and N? - If you choose a real number at random from
**R**, what is the probability that the real number chosen is rational (or algebraic)?

I will prove that the probability of the second statement is 0:

Let [(2N)

I should probably write this more formally and nicely, but it captures the point. So there's your example. If you don't think that the limit actually

- #20

- 785

- 15

Is it 0 or undefined? I thought this was interesting, it seems a paradox in its own.

Hii , Nano-Passion !!

This is a very interesting question.

0/0 is neither ∞ nor 0. It is what is "Indeterminate".

For example we can say that 1/0 = ∞ because 0x1 = 0 , 0x10

But in case of 0/0 , every equation is satisfied !

0/0 = x , where x can be any number. So this is kinda indeterminable.

Here is the best explanation of 0/0 by Doctor Math : http://mathforum.org/library/drmath/view/55722.html

Read it , it is very interesting.

- #21

jgens

Gold Member

- 1,581

- 50

For example we can say that 1/0 = ∞ because 0x1 = 0 , 0x10^{100000000000000000}= 0. So we assume that somehow at an undefined place that is ∞ 0 will become 1.

I disagree. Even in the projective number system where 1/0 = ∞, we still leave 0 * ∞ undefined, and for very good reason.

0/0 = x , where x can be any number. So this is kinda indeterminable.

I am being nitpicky, but I dislike this way of writing things. While it is true that 0 * x = 0 for any x in

Edit: Having read through the doctor math article, I do not quite agree with his/her argument. While the article provides good intuitive reasons behind why we should leave 0/0 undefined, the way he talks about limits is wrong. For example, while it is true that the limit [itex]\lim_{x \to 0} \frac{ax}{x} = a[/itex], this is very different than saying that the value of 0/0 is a. In particular, the former is correct while the latter is false. The article confuses the value of a function at 0 and the limit of the function at 0.

Last edited:

- #22

- 1,291

- 0

Hii , Nano-Passion !!

This is a very interesting question.

0/0 is neither ∞ nor 0. It is what is "Indeterminate".

For example we can say that 1/0 = ∞ because 0x1 = 0 , 0x10^{100000000000000000}= 0. So we assume that somehow at an undefined place that is ∞ 0 will become 1.

But in case of 0/0 , every equation is satisfied !

0/0 = x , where x can be any number. So this is kinda indeterminable.

Here is the best explanation of 0/0 by Doctor Math : http://mathforum.org/library/drmath/view/55722.html

Read it , it is very interesting.

Hey , thanks for sharing.

In each case, the probability in question is 0. The third statement is a little more complicated, but it has a nice proof once you have measure-theoretic concepts.

- If you choose an integer at random from
Z, what is the probability that the integer chosen is 0?- If you choose an integer at random from
Z, what is the probability that the integer lies between -N and N?- If you choose a real number at random from
R, what is the probability that the real number chosen is rational (or algebraic)?

I will prove that the probability of the second statement is 0:

Let [(2N)^{m}] = {-(2N)^{m}, ... , (2N)^{m}}. Then for a fixed m, the probability of choosing an integer between -N and N is (2N)^{1-m}. By letting m → ∞, we see that the probability goes to 0. In particular, in the limiting case (when we are choosing elements fromZ), the probabilityis0.

I should probably write this more formally and nicely, but it captures the point. So there's your example. If you don't think that the limit actuallyis0, but rather is something else, what do you propose that something else should be?

The third question is very interesting actually haha.

I agree with you, when you are dealing with an infinite amount of numbers then it would be 0. But what about a finite amount of numbers? Can a number occur with 0 probability, such that n is a finite number [in this case let us limit n to a world consisting only of 50 digits].

Hmm, probability 0 is indeed a silly concept. Most people think of probability as throwing dice, and indeed: throwing a 5.5 with a dice has probability 0 and thus never happens. But it is important not to generalize this situations. There are some probability 0 situations which can happen.

As an example: choosing an arbitrary number in the interval [0,1]. It is clear that all numbers have the same probability p of being chosen. However, saying that a number has probability p>0 is wrong, since [itex]\sum_{x\in [0,1]}{p}\neq 1[/itex]. So we NEED to choose p=0. So choosing probability 0 for this is actually quite unfortunate and caused by a limitation of mathematics.

However, there is another way of seeing this. Probability can be seen as some "average" value. For example, if I throw dices n times (with n big), then I can count how many times I throw 6. Let [itex]a_n[/itex] be the number of 6's I throw. Then it is true that

[tex]\frac{a_n}{n}\rightarrow \frac{1}{6}[/tex]

So a probability is actually better seen as some kind of average.

Now it becomes easier to deal with probability 0. Saying that an event has probability 0 is now actually a limiting average. So let [itex]a_n[/itex] be the number of times that the event holds, then we have

[tex]\frac{a_n}{n}\rightarrow 0[/tex]

It becomes obvious now that the event CAN become true. For example, if the event happens 1 or 2 times, then we the probability is indeed 0. It can even happen an infinite number of times.

Probability 0 should not be seen as a impossibility, rather it should be seen as "if I take a large number of experiments, then the event will become more and more unlikely". This is what probability 0 means.

Hey, thanks for the reply.

To me probability rings to my neurons as a tendency to become a value over a period of time or over n times. But that is just my definition of course. If we take this definition in that context, then perhaps a probability of 0 would imply that it has 0 tendency to become any value over a period of time or n times. But then I guess this doesn't hold true in the mathematical context.

I wonder, if something has 0 probability in Quantum Mechanics, can it happen? I suppose it can, which would support your statement.

- #23

- 75

- 0

The seeming human inability to accept that the equation 0*x = 0 is satisfied by any value one could possibly imagine for x, to the point where one must suggest the algebraic manipulation of that formula via division to be "undefined," or "indeterminate" is just beyond (IMHO) "retarded." It is perfectly "well-defined."

X = "any value" one can possibly imagine.

Therefore, 0/0 = anything you may wish it to equal, including 1, which typically is the value one obtains when dividing some positive or negative quantity by itself.

-1/-1 = 1

1/1 = 1

But (-1 + 1)/(-1 + 1) is somehow not equal (at least) to 1?

Please...

A classic example of how mathematics can "straightjacket" common sense. An equation with infinite solutions is NOT undefined, nor is it "indeterminate" excepting to the mind that requires finite solutions to questions with infinite answers.

X = "any value" one can possibly imagine.

Therefore, 0/0 = anything you may wish it to equal, including 1, which typically is the value one obtains when dividing some positive or negative quantity by itself.

-1/-1 = 1

1/1 = 1

But (-1 + 1)/(-1 + 1) is somehow not equal (at least) to 1?

Please...

A classic example of how mathematics can "straightjacket" common sense. An equation with infinite solutions is NOT undefined, nor is it "indeterminate" excepting to the mind that requires finite solutions to questions with infinite answers.

Last edited:

- #24

Pythagorean

Gold Member

- 4,214

- 272

I take indeterminate to be consistent with "any value".

- #25

pwsnafu

Science Advisor

- 1,080

- 85

Please...

A classic example of how mathematics can "straightjacket" common sense. An equation with infinite solutions is NOT undefined, nor is it "indeterminate" excepting to the mind that requires finite solutions to questions with infinite answers.

I don't know if you are being sarcastic, or just trolling, or if you don't understand what is being said.

Division must satisfy the requirements of being a binary operation. That is, it must have two inputs and output. That means if I take 2 and 3, and evaluate 2 divided by 3, I need the same answer to occur. I am not allowed to say its equal to 4 on Tuesdays but equal to 5 on Fridays. Math does not work like that. We don't allow 0/0 to be "anything we wish". Logic must be consistent.

Secondly,

Thirdly, as to your assertion that

[itex]\frac{-1+1}{-1+1} = 1[/itex]

Multiply both sides by -1 to obtain

[itex](-1) \frac{-1+1}{-1+1} = (-1)(1)[/itex]

We use the -1 on the left to multiply the numerator

[itex]\frac{1+-1}{-1+1} = -1[/itex]

Rearrange the numerator

[itex]\frac{-1+1}{-1+1} = -1[/itex]

Maybe to you [itex]-1=1[/itex] is "common sense". It isn't to me.

Share: