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Does 0+0+0+0+0+0+0+ =1?

  1. Jul 12, 2013 #1
    1.)does an infinite number of zeroes summed up equal a finite value?

    2.)does dx=0?

    3.)does 2xdx=0?

    4.)the probability of picking 100 in the range of all natural numbers is zero?

    how can the probability be zero? shouldn't the probability of this be dN? ie.. an infinitesimally small chance? i just don't see how this can be zero.. say you pick a natural number Ni, and you say that this number has a zero probability of being picked, and then you pick any number in the range of natural numbers. One thing for certain is that you WILL pick a number.. so then you end up with Nii, but someone i know says that all the natural numbers have a probability of zero of being picked. but we know we just picked one... so if the probability was truly zero.. why and how did i just pick Nii?

    how can the answer to number 3 be yes? that would imply that every differential equation e.g: (x+y)dx(x-y)dy=0 would just be zero because (x+y)dx that just equals zero.... (x-y)dy that just equals zero... no work to be done...
    no reason for naming the damn things... homogeneous odes.... exact equations....

    and if 2xdx=0 you're summing up a bunch of zeros how can this EVER reach a finite value other than zero regardless of how many times you're summing it up?
  2. jcsd
  3. Jul 12, 2013 #2
    Define "summing an infinite number of zeroes". Are you talking about a series with terms all 0? Then we have that ##0 + 0 + 0 + ... = 0##. Did you have something different in mind?

    Please define what you mean with ##dx##. In the standard intepretation of differential forms, this is false.

    Define "probability". This is a very subtle point. There is no uniform probability distribution on the natural numbers. So when talking about probability of picking a certain number, you need to be very specific on what distribution you're using.
    Last edited: Jul 12, 2013
  4. Jul 12, 2013 #3


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    I cannot agree with my honorable colleague. Clearly ##0 + 0 + 0 + ... = 2##.
  5. Jul 12, 2013 #4


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    (Just kidding. Obviously ##0 + 0 + 0 + ... = 0##. How could it be anything else?)
  6. Jul 12, 2013 #5
    Oh God. What did I write :cry: I corrected it.
  7. Jul 12, 2013 #6


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    It's not even a full moon yet.
  8. Jul 12, 2013 #7
    1) "dx" has no value if you aren't working in the hyperreals. The entire epsilon-delta definitions for limits were constructed to avoid working with infinitesimals as numbers, and for good reason; as it causes unnecessary ambiguity. So no, [itex]dx \neq 0[/itex].

    However, infinitesimals are negligible quantities as they stand, that is why they are called "infinitesimals"; this is emphasized in the hyperreal theory with the standard part function. It is also emphasized in the sum for the Riemann integral.

    2) Look above. An infinitesimal isn't a number you can work with, but it does behave like a real number under some circumstances, for example in the chain rule where you can cancel infinitesimals. However, as I said above, something like [itex]2x\,dx[/itex] is not a number.

    3) Obviously [itex]0+0+0... = 100[/itex]. No, just kidding. If you are interpreting it as the sum [itex]\displaystyle \sum 0[/itex], then obviously it is 0. However, there are other interpretations of this statement. Take the sequence [itex]\displaystyle a_n = \sum^{n}_{k=1} \frac{1}{n} = 1[/itex]. This is a constant sequence, but it can also be interpreted as [itex]0+0+0...[/itex] as [itex]n\to\infty[/itex] from a certain angle of view.

    As another example, the Riemann sum for an integral can also be interpreted as an infinite sum of zeroes; because it is the sum of the areas of infinitely many intervals with widths infinitely small. As the width goes to zero, this is also an interpretation of the sum you wrote. You must be more precise.

    4) The probability distribution function is important here, but applying the usual distribution function, yes; this is correct. However, this function is not so well defined with infinite sets. The reason these probabilities "add up to 1" is because of the sum sequence I gave in 3). There are other probability distribution functions; but the usual "there is an equal chance for each event to happen" distribution leads to this absurd-ish result.

    It would be better to use a distribution function like [itex]P(n) = 2^{-n-1}[/itex]. However, this function would have "unnatural" results; because the probability to pick 0 is 1/2.
    Last edited: Jul 12, 2013
  9. Jul 12, 2013 #8
    Yes. ##\displaystyle \sum_{n=1}^{\infty}\left[0\right]=0##, which is finite.
    Only if ##x## is something called a closed form, which is a differential form that has an exterior derivative of 0.
    Let's suppose ##2\neq0## (unless, of course, jbunn knows something we don't. :tongue:). If ##x\neq0## and ##dx\neq0##, then ##2x \, dx \neq 0##.
    Yes. This is actually a question of measure theory and sigma algebras.
  10. Jul 12, 2013 #9

    3.) why would you say that 0+0+0+0+... = some number other than zero if that was not what you meant? and i don't understand what you're saying when you say sequence and then you give me a sum, then you say it equals 1 which.. i then assume you're talking about the sequence and not the series because the series diverges to infinity.. what was the sum for?..
    i don't know what angle to look at that from to end up with what you got..

    "approaches zero" does NOT mean "equals" zero.

    why is there being an equal probability for all natural numbers absurd? and before you were telling me that the probability was zero. there is no probability whatsoever for a natural number to be picked at random out of a range of all natural numbers. that's what "zero probability" means in this case.. you said that. do you still stand by that statement? or are you still saying that since all natural numbers have a zero probability of being chosen, therefore they are all equal?

    and forgive me but i don't understand what you guys are referring to in this context when you say distribution. why do we need a function to determine the probability of a natural number being picked in a range of all natural numbers? it's not like 1 is 600 times more probable of being picked than 63; it's RANDOM... so again, why do we need a distribution function to determine probability for this?
  11. Jul 12, 2013 #10
    The Force is strong in this one. I like him. :biggrin:

    Who says the distribution has to have different values for different numbers? Consider a uniform probability distribution.
  12. Jul 12, 2013 #11
    There is no uniform probability distribution on the natural numbers.
  13. Jul 12, 2013 #12
    I'm not suggesting there is. I'm attempting to provide intuition on a distribution in which "1 is not 600 times more likely than 63." :tongue:
    Last edited: Jul 12, 2013
  14. Jul 12, 2013 #13


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    What probability would you assign to each number? Let us call this probability ##p##. If ##p = 0## then the total probability is
    $$\sum_{n\in \mathbb{N}} p = \sum_{n\in \mathbb{N}} 0 = 0$$
    whereas if ##p > 0## then
    $$\sum_{n\in \mathbb{N}} p = \infty$$
    But a probability distribution must sum to ##1##, not ##0## or ##\infty##.

    By the way, this same fact (a countably infinite sum of a nonnegative constant is either ##0## or ##\infty##) is the key reason we are able to construct sets of real numbers that are not Lebesgue measurable.
  15. Jul 12, 2013 #14
    In past 8, you said yes to his question. This is false.

    And the notion of "closed form" has nothing to do with this thread. The very definition of ##dx## is that it is a function ##dx:\mathbb{R}\rightarrow \mathbb{R}^*## such that ##dx(p):\mathbb{R}\rightarrow \mathbb{R}## is a linear function such that ##dx(p)(h)=h##. So it is not equal to ##0##. More generally, we have that ##df(p)(h) = f^\prime(p) h##, so ##df(p) = f^\prime(p) dx##.
  16. Jul 12, 2013 #15
    The sets of real numbers you are discussing (I'm guessing the Vitali set?) are not what we are discussing here, correct? I thought all singletons in ##\mathbb{R}## have Lebesgue measure 0, which implies that the probability of their selection when considering all real numbers is 0.

    See above.

    This depends on how we define ##x##. ##dx## is the exterior derivative of some form ##x##. For example, if we set ##x=d\beta## (that is, if x is exact), then ##dx=0##.
    Last edited: Jul 12, 2013
  17. Jul 12, 2013 #16
    Again, there is no uniform probability distribution on the set of all real numbers. If you want to talk about the "probability of selecting a real number", then you need to specify a specific distribution.
  18. Jul 12, 2013 #17
    As far as I understand, my thinking applies to any given distribution on the real numbers, not just a (nonexistent :frown:) uniform one. Consider a Gaussian distribution, for example. The probability of picking a given number is still 0 because singletons have measure 0. The only time I can think this might not apply to an (existing) distribution might be some form of Dirac Delta distribution, but I don't have time to check right now. I'll be happy to discuss this with you privately when I get back from dinner, but I suggest we refocus to the OP's questions.
  19. Jul 12, 2013 #18
    Your thinking applies to distributions we call absolutely continuous. There are many other kinds of distributions on ##\mathbb{R}##.
  20. Jul 12, 2013 #19


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    "In past 8," is this Irish or autocorrect?
  21. Jul 12, 2013 #20
    An autocorrect function that is using an Irish dictionary. :frown:
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