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

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Discussion Overview

The discussion revolves around the proof of the statement that the sum of an infinite series of zeros, denoted as 0 + 0 + ... + 0, equals zero. Participants explore various approaches to rigorously establish this result, including the use of geometric series, limits, and induction, while also addressing the implications of infinite sums and the definitions involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose using geometric series to argue that since each term is zero, the sum must also be zero, but they question the rigor of this approach.
  • Others suggest that defining the infinite sum is crucial, noting that all partial sums are zero and that the limit of these sums as n approaches infinity is also zero.
  • A participant introduces the idea of using induction to prove the result, expressing satisfaction with this method.
  • Concerns are raised about the ambiguity of the notation used for infinite sums, with some participants emphasizing the need for clarity regarding the index set and summation rules.
  • There is a discussion about the implications of different types of infinity, such as aleph-null versus aleph-one, and how they relate to the summation of zeros.
  • Some participants challenge the assumptions made in earlier posts, questioning the validity of the claims without rigorous definitions.

Areas of Agreement / Disagreement

Participants express differing views on the rigor and definitions necessary for the proof. While some agree on certain approaches, there is no consensus on a definitive proof or the assumptions that must be clarified.

Contextual Notes

Limitations include the need for precise definitions of infinite sums and the implications of different summation rules. The discussion highlights the complexity of summation in various mathematical contexts and the assumptions that underlie the use of the term "zero."

  • #31
With the same logic ##1+ \cdots + 1\,=\, 1## because ##\{1\} \cup \cdots \cup \{1\} = \{1\}##?
That approach brings more problems than it is supposed to solve, especially as it is trivial to show the limit directly with the definitions (as done in posts 27 and 29).
 
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  • #32
mfb said:
With the same logic 1+⋯+1=11+ \cdots + 1\,=\, 1 because {1}∪⋯∪{1}={1}\{1\} \cup \cdots \cup \{1\} = \{1\}?

No beacuse defining the successor function for sets ##S(A)=A\cup\{A\}## you have that ##1=S(\emptyset)=\emptyset \cup \{\emptyset\}=\{\emptyset,\{\emptyset\}\}## so ##1+1 \not = 1## but ##\{\{\emptyset,\{\emptyset\}\},\{\{\emptyset,\{\emptyset\}\}\}\}=2##.
Ssnow
 
  • #33
That is different from what you suggested before, and it is still wrong. If you identify addition with the union of sets then x+x=x for all x because ##s \cup s=s\quad \forall s##.

If you identify the addition with ##f(s,t)=\{s,\{t\}\}## then 1+1 works but nothing else works any more.
 
  • #34
mfb said:
If you identify the addition with f(s,t)={s,{t}}f(s,t)=\{s,\{t\}\} then 1+1 works but nothing else works any more.

Yes, I was thinking of "classes" rather than "sets" ...
 

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