0 divided by 0

  • #1
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[tex]\frac{0}{0}[/tex]

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

Answers and Replies

  • #2
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Well, assuming you are working with the field of real numbers, 0 doesn't have an inverse multiplicative element. There is no real number x such that 0*x=1. So the expression 0/0 which really means 0 times the multiplicative inverse of 0 has no meaning, since the multiplicative inverse of 0 isn't in the field of reals and doesn't actually exist. Thus 0/0 is undefined.

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
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Thank you nucl, for reminding me of the proof that 0/0 has a good reason for being undefined.
 
  • #4
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Well, assuming you are working with the field of real numbers, 0 doesn't have an inverse multiplicative element. There is no real number x such that 0*x=1. So the expression 0/0 which really means 0 times the multiplicative inverse of 0 has no meaning, since the multiplicative inverse of 0 isn't in the field of reals and doesn't actually exist. Thus 0/0 is undefined.

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
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That's just how they are! :P
 
  • #6
micromass
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Also see the FAQ on this topic: https://www.physicsforums.com/showthread.php?t=530207 [Broken]
 
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  • #7
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Actually, in complex analysis, it's common to say that the function 1/z maps 0 to infinity when infinity is considered as the point at infinity on the Riemann sphere. So, in that sense, you could say 1/0 = infinity (when it's done naively by calculus students, this is wrong because they don't have a mapping in mind). You could restrict this to get 1/x when x is a real variable. The trick here is that you have to identify negative infinity with positive infinity. This isn't to say that 0 has an inverse. It is just that it is now included in the domain of the function 1/x and the point at infinity is added to the range.

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
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We might note that many text refer to "0/0" as "undetermined" rather than "undefined" because if you have a limit of a fraction where the numerator and denominator both go to 0, the actual limit itself can exist and can be anything.
 
  • #9
mathman
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We might note that many text refer to "0/0" as "undetermined" rather than "undefined" because if you have a limit of a fraction where the numerator and denominator both go to 0, the actual limit itself can exist and can be anything.

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
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What about looking at:

[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
micromass
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What about looking at:

[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
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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?
 
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  • #13
HallsofIvy
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What about looking at:

[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
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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 R contains no infinite elements. In the extended real numbers, this is done for precisely the reason you stated above; in particular, the left-hand limit of 1/x as x → 0 is -∞ while the right-hand limit is +∞.

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.
So, while there is technically no issue with defining 0/0 = R, I still think there is sufficient reason not to. Also, I do not think you gain any utility from defining 0/0 = R, so why do it in the first place?

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 R then you need to choose a value when you define it. Otherwise, when we write 0/0, it could literally mean any real number; there would be no way of actually picking out which value of 0/0 we want. This way of defining 0/0 is problematic.

As a slightly unrelated note on probability, consider the following problem: If you select an integer at random from Z, 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.
 
  • #15
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As a slightly unrelated note on probability, consider the following problem: If you select an integer at random from Z, 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..
 
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  • #16
jgens
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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 is 0. 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.
 
  • #17
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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 is 0. 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
micromass
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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
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Here are some simple examples:
  • 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)?
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.

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 from Z), the probability is 0.

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 is 0, but rather is something else, what do you propose that something else should be?
 
  • #20
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[tex]\frac{0}{0}[/tex]

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 , 0x10100000000000000000 = 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.
 
  • #21
jgens
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For example we can say that 1/0 = ∞ because 0x1 = 0 , 0x10100000000000000000 = 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 R, we conventionally leave the expression 0/0 undefined. So saying "0/0 = x, where x can be any number" is nonsense. However, saying that every real x satisfies 0 * x = 0 is motivation for leaving 0/0 undefined, is perfectly reasonable.

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.
 
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  • #22
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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 , 0x10100000000000000000 = 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 :approve:, thanks for sharing.

Here are some simple examples:
  • 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)?
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.

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 from Z), the probability is 0.

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 is 0, 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
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.
 
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  • #24
Pythagorean
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I take indeterminate to be consistent with "any value".
 
  • #25
pwsnafu
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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, nobody in this thread claimed the equation 0x = 0 was undefined! It simply has an infinite number of solutions, so it does not have a unique solution. That's the whole point: we can't use that equation to define division if the solution is not unique.

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.
 
  • #26
chiro
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One needs to be careful with this kind of thing.

If we are putting this in the context of evaluating a limit where we get an indeterminate form (like 0/0), then in cases the limit might be able to have a determinate value that makes sense. In this vein, if the limit is in the context of a limit of some sort that deals with a functional representation, then this needs to be considered.
 
  • #27
0/0
one of many ways of writing all the no.s in one go.
 
  • #28
micromass
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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.

Did you understand anything which has been said in this thread?? Mathematics is common sense and the things we do in mathematics are perfectly sensible.

Furthermore, you claim that -1/-1=1 and 1/1=1, which is correct.
How does it follow from that that (-1+1)/(-1+1)=1???? I don't see how you infer that.


0/0
one of many ways of writing all the no.s in one go.

No it's not.
 
  • #29
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0/0 means that you don't have enough information about the problem.

It's like trying to solve a system of 2 equations by using only one of the equations.

I will give you an example to understand what I mean.

Consider a iron tube with length L, the tube is filled so you can't see through it. Look at the attached picture:

- in the left view I'm showing you the pipe from the front, at an angle θ = 90°, so you can only see that its section is round
- in the right view I'm showing it to you from an angle θ = 45°

If you have never seen that pipe before can you tell me from the left view what is its length? If no then why?

Look at the way you can compute L:

L = a/cos(θ)

In the left view L = 0/0 this is why you can't say what is the length. However you know the trick and move a little to the right of the pipe, now you have more information and you can say that L = a / cos(45°).

Every time you get to 0/0 it means you should look for more information.
 

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  • #30
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not paradox, but a prohibition, as it is mathematically undefined.
 
  • #31
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When you take a divided by b, a must be something divided by something, example, 10 apples divided by 2 persons = 5 each!

If 0 apples divided by 2 persons, then 0 each!!

If 10 apples divided by 0 person, it doesn't make any sense because no one's there to share.

Combining the both, should it be 0, 1 (since X amount of apples divided by X person) or undefined??
 
  • #32
D H
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Combining the both, should it be 0, 1 (since X amount of apples divided by X person) or undefined??
None of the above.

1/0 is undefined using the standard definition of the real numbers. 0/0, on the other hand, is indeterminate.

Division is the inverse of multiplication. a/b=c means c is the unique number such that a=b*c. In the case of 1/0, there is no such number c∈ℝ such that 1=0*c. So 1/0 is "undefined".

In the case of 0/0, every number c∈ℝ satisfies 0=0*c. This makes it appear that one could assign any number whatsoever as the value of 0/0. This has two problems, both of them killer. One obvious problem is the lack of uniqueness. An even bigger problem is that assigning any one specific value opens the door to all kinds of contradictions. Mathematical systems must be contradiction-free.
 
  • #33
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Give it this way, it's like solving a system of equations using matrices. It can be:
a) insufficient data, i.e. lack of equations;
b) more than enough data, which makes some equations redundant;
c) sufficient data with more than one answer, i.e. line of solution, plane of solution, infinite solution etc.

I don't think it's insufficient data. When performing division, you need 2 data, a divide by b. Now, a is zero, b is also zero. Both data are available, it's just not possible to perform. Just an opinion anyway.
 
  • #34
28
0
None of the above.

1/0 is undefined using the standard definition of the real numbers. 0/0, on the other hand, is indeterminate.

Division is the inverse of multiplication. a/b=c means c is the unique number such that a=b*c. In the case of 1/0, there is no such number c∈ℝ such that 1=0*c. So 1/0 is "undefined".

In the case of 0/0, every number c∈ℝ satisfies 0=0*c. This makes it appear that one could assign any number whatsoever as the value of 0/0. This has two problems, both of them killer. One obvious problem is the lack of uniqueness. An even bigger problem is that assigning any one specific value opens the door to all kinds of contradictions. Mathematical systems must be contradiction-free.


very simple.
All theorem built from axioms, and mathematics axiom is what we often say "difficult' to prove, because its
the pillar of math, we have to start with something.

the Reals are defined that way,
hence one of the axiom define it so, if A is an element of R, B is an element of R,
A/B is an element of R, where B=/= 0.

which implies 0/x exist in the real, except when x=0. - (Z)

to prove statement Z, would require you to use the axiom, which obviously the axiom cant be proven
with examples, because it is the truthfulness of the axiom that makes the example valid.

You dont prove axioms, unless you are to bring in some philosophical argument, which
is not so relevant in the rigorous context.
 
  • #35
D H
Staff Emeritus
Science Advisor
Insights Author
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very simple.
I think you completely missed the point of my post, which was that the correct term for 0/0 is that it is indeterminate rather than undefined. 1/0 is undefined, but 0/0 is indeterminate.

0/0 cannot be given meaning, period. 1/0 can be given a meaning in various contexts. a/0 is ∞ on the projective real number line for all non-zero a. In complex analysis, a/0 (with a≠0) is sometimes treated as complex infinity, a number whose magnitude is greater than any real number but whose argument is indeterminate.
 

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