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0 divided by 0 |
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| Nov26-11, 10:37 PM | #1 |
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0 divided by 0
[tex]\frac{0}{0}[/tex]
Is it 0 or undefined? I thought this was interesting, it seems a paradox in its own. |
| Nov27-11, 01:16 AM | #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 |
| Nov27-11, 01:23 AM | #3 |
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Thank you nucl, for reminding me of the proof that 0/0 has a good reason for being undefined.
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| Nov27-11, 01:48 AM | #4 |
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0 divided by 0That 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 |
| Nov27-11, 02:36 AM | #5 |
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That's just how they are! :P
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| Nov27-11, 04:16 AM | #6 |
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Also see the FAQ on this topic: http://www.physicsforums.com/showthread.php?t=530207
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| Nov27-11, 02:32 PM | #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. |
| Nov27-11, 03:03 PM | #8 |
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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.
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| Nov27-11, 03:23 PM | #9 |
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Recognitions:
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| Nov27-11, 04:43 PM | #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 |
| Nov27-11, 04:52 PM | #11 |
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| Nov28-11, 01:07 AM | #12 |
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[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? |
| Nov28-11, 12:45 PM | #13 |
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[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. |
| Nov28-11, 04:01 PM | #14 |
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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.
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. |
| Nov28-11, 09:42 PM | #15 |
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What do you think? I'm still new to math so I might be wrong.. |
| Nov28-11, 10:11 PM | #16 |
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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. |
| Nov28-11, 10:27 PM | #17 |
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