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#19
Nov2811, 10:45 PM

PF Gold
P: 1,622

Here are some simple examples:
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)^{1m}. 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
Nov2911, 04:18 AM

P: 758

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. 


#21
Nov2911, 10:17 AM

PF Gold
P: 1,622

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. 


#22
Nov3011, 01:26 AM

P: 1,306

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]. 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
Jan3112, 04:01 AM

P: 75

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 "welldefined."
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. 


#25
Jan3112, 05:31 AM

Sci Advisor
P: 839

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
Jan3112, 05:33 AM

P: 4,579

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
Jan3112, 07:39 AM

P: 3

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


#28
Jan3112, 08:17 AM

Mentor
P: 18,346

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. 


#29
Mar712, 06:46 AM

P: 70

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. 


#30
Mar1212, 09:53 AM

P: 28

not paradox, but a prohibition, as it is mathematically undefined.



#31
Mar1212, 10:34 AM

P: 28

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
Mar1212, 10:53 AM

Mentor
P: 15,202

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 [i]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 contradictionfree. 


#33
Mar1212, 11:32 AM

P: 28

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
Mar1212, 12:06 PM

P: 28

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
Mar1212, 12:31 PM

Mentor
P: 15,202

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 nonzero 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. 


#36
Mar1212, 12:55 PM

P: 28




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