1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Proof that 0 + 0 +...+ 0 +... = 0

  1. Nov 21, 2017 #1
    Is there any rigorous way of proving this?

    I tried using geometric series of ever diminishing ratio and noticing that 0 is always less than each term of the series, then 0 + 0 +...+ 0 +... must be always less than ## \frac{1 } {1-r} - 1 ##. (*)

    Eventually, as r goes to 0 so does ## \frac{1 } {1-r} - 1 ##, so by (*) argument 0 + 0 +...+ 0 +... = 0.

    But I don't know if my proof is rigorous enough...

    I haven't found any proof of this by googling it so any help is appreciated.
     
  2. jcsd
  3. Nov 21, 2017 #2
    Certainly the sum on N zeroes as N approaches infinity (in this case, Aleph Null) is zero.
    I'm not sure that more of a proof is needed. But if you insist, try this:
    A = 0 + 0 + 0 + ...
    2A = 2x0 + 2x0 + 2x0 + ... = 0 + 0 + 0 + ...
    Therefore 2A = A; A = 0.
     
  4. Nov 21, 2017 #3
    If there's anything not rigorous it.s the statement of the result. Try to state 0+0+0 + .... + 0 = 0 as a limit and without using an ellipsis.
     
  5. Nov 21, 2017 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    First you need a rigorous definition what that means, because an infinite sum cannot end.
    If you mean 0+0+0+..., then it is sufficient to note that all partial sums are 0, and ##\lim_{n \to \infty} 0 = 0## directly follows from the definition of limits for sequences.
     
  6. Nov 21, 2017 #5
    Where in my post did I assert this?

    willem2 said that 0+0+0 + .... + 0 = 0 , but never did I make such a preposterous claim!!!
     
  7. Nov 21, 2017 #6
    Imagine an N-dimensional sphere of radius 1. The path along the diagonal for which all coordinates are positive hits the sphere when all coordinates are 1/sqrt(N).
    Now take the limit N to infinity. The diagonal hits the sphere where 0+0+0+... = 1 ;)
     
  8. Nov 21, 2017 #7

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Ah you didn’t have a zero at the end. The zero in the middle is still odd.
     
  9. Nov 21, 2017 #8

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    Induction?
     
  10. Nov 21, 2017 #9
    Induction indeed!
    I had thought of this before and since we concur I am happy now.
     
  11. Nov 21, 2017 #10

    fresh_42

    Staff: Mentor

    As long as you don't say, what ##0+0+\ldots +0+\ldots ## means, as long is every answer meaningless, because they all make assumptions on what it might mean. It is your turn to deliver. It is a summation, that's clear, so we have ##\sum_\mathcal{I} 0##. The dots further suggest that ##|\mathcal{I}| = \infty##. What has to be defined is: which infinity, and which summation rules. I assume the result will always be zero, although I'm not that sure there aren't any strange summation rules according to which it is not, but you asked for a rigorous proof, so both questions have to be answered in advance.
     
  12. Nov 21, 2017 #11
    Thanks for the input, but are you implying that if infinity is aleph-1, result may not be 0?

    I am very curious to learn now...
     
  13. Nov 21, 2017 #12

    fresh_42

    Staff: Mentor

    I'm only implying that it cannot be said for sure as long as the notation allows ambiguities. If we write ##0+0+\ldots ## we normally consider a countable index set for summation, but sums can sometimes consist of possibly uncountable many summands. E.g. if we consider the vevtor space of continuous functions, then it has (as every vector space, assuming the validity of the axiom of choice) a basis ##\{f_\iota\,\vert \,\iota \in \mathcal{I}\}## which is uncountable in this case. So if we write ##f= \sum_{\iota \in \mathcal{I}} c_\iota f_\iota ## we have therefore to add a the condition, that almost all (= all up to finitely many) ##c_\iota## are equal to ##0##. This is just meant as an example, that such sums aren't a priori finite or countable. Therefore the dots have to be explained, because otherwise it is just a convention to assume countably many. But this is only a convention and not put in stone.

    The summation itself is also only defined as a binary operation: ##(a,b) \mapsto a+b##. Already a third term makes an assumption: Is it ##a+b+c=(a+b)+c## or is it ##a+b+c=a+(b+c)##. Of course it doesn't matter if we speak about, e.g. real numbers, because addition is associative for them, i.e. the order doesn't matter. However, not all mathematical structures are associative and we cannot assume it without mention. So to imply ##0 \in \mathbb{N}## or ##0 \in \mathbb{Z}## or ##0 \in \mathbb{Q}## or ##0 \in \mathbb{R}## or ##0 \in \mathbb{C}## is another hidden assumption, or if you like, a convention, as the ##"+"## operation is usually only used for associative and commutative binary operations. Nevertheless, it is an assumption or a convention.

    Even in case it is a usual addition, there are concepts (outside usual calculus) which consider different summation processes, e.g. the Ramanujan summation. Of course this is nothing we have to bother here, as it is more of an analytical tool than a summation rule, but it shall demonstrate, that the summation isn't defined in itself. Again we have the convention here, that ##"+"## denotes the summation of real numbers if not stated otherwise, but you've asked for rigor, in which case it has to be said.

    That brings me to another point which I previously have forgotten to list. What is ##0\,##? We all followed the convention here, that it is the neutral element of addition, however, nobody has said so. E.g. ## x \cdot y := \dfrac{2xy-x-y}{xy-1}## defines a multiplication with neutral element ##0##. Because of this overall convention about the notation of ##0##, it would be abhorrent to use ##0## differently. Nevertheless, in a purely logical framework it had to be mentioned.

    Finally, to destroy any doubts here, those conventions make absolutely sense and there is usually no need to mention them. So don't request it. But in this internet context here, where nobody knows in advance, what people know, what they are currently dealing with, or - as here - to which extend the term rigor has been meant, things are far less clear in advance.
     
  14. Nov 21, 2017 #13
    I may have misstated. I was referring to this summation:
    $$ \sum_{n=1}^{\infty}{0} $$
    In that case, the number of "add"s is Aleph Null. It can't by Aleph-1 because the terms in the summation can be listed serially.
    $$ Note \space that \space this \space \space is \space not \space \aleph_0 \times 0 \space because \space \aleph_0 \space is \space not \space a \space number.$$
     
  15. Nov 21, 2017 #14
    True, but in your proof how are you equating A = 0 + 0 + 0 + ... = 0 + 0 + 0 + ... = 2 A in advance, without first establishing that a limit exists? This is just begging the question. If you haven't proven that a limit exists first this is like saying: 0 = 1 - 1 +1 -1 +.... = 1 -1 +1 -1 +1 -.... = 1
     
  16. Nov 21, 2017 #15
    Thank you for your answer. I hadn't imagined that such a simple question as this has so many implications and I am grateful that you took the time to write such a lengthy reply.
    Still, you haven't answer my question: Suppose we are having aleph-1 many uncountable summands now, so that ## \sum_{n=1}^{\infty}{0} ## does not apply, is there a way of proving that 0 + 0 + 0 + ... = 0, or is it the case that such a question is nonsense and no such proof exists at all?
     
  17. Nov 21, 2017 #16

    fresh_42

    Staff: Mentor

    I'm a bit rusty on that matter, but transfinite induction may be a possibility.
     
  18. Nov 21, 2017 #17

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    Hmmm I thought I posted an answer but now it is gone.

    Write the integers base ##-2##, the negative binary system. Every integer is expressed as sums of powers of ##-2##. One knows that ##0=1+ -1## since they are inverses of each other. In the negative binary system this is the addition ##1 + 11## since ##-1 = (-2)^1+(-2)^0##.

    Adding ##1## to ##11##, one gets in the ones column ##1 + 1 = 110## so write down the zero and carry the ##11##. Now in the ##-2##'s column one again has ##1+1## so write down another zero carry the ##11##. This goes on forever and generates an unending sum of zeros.

    If one thinks of the infinite addition of zeros as signifying all of these computations then the answer is it equals zero.
     
    Last edited: Nov 21, 2017
  19. Nov 21, 2017 #18

    OCR

    User Avatar

    ...And Bob's your uncle ! . :biggrin:
     
  20. Nov 22, 2017 #19
    Ok. How about this:
    I may have misstated. I was referring to this summation:
    $$ \sum_{n=1}^{\infty}{2^{-n}} = 1$$
    Multiplying both sides by zero:
    $$ \sum_{n=1}^{\infty}{0\times2^{-n}} = 0$$
    Thus:
    $$ \sum_{n=1}^{\infty}{0} = 0$$
     
  21. Nov 22, 2017 #20
    Nice proof and better than mine. I like it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Proof that 0 + 0 +...+ 0 +... = 0
  1. 0^0=0 method (Replies: 10)

  2. Proof 0=4 (Replies: 2)

  3. A^0=1 proof (Replies: 26)

Loading...