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I just finished a discrete mathematics course and I was wondering if I could prove that the paradox here is wrong using simple logic without using advanced mathematics such as limits and sequences e.g. consider the equation $x+2=x+4$ , it obviously has no solution in $\mathbb{R}$ . The proof is as follows assume it does have a solution $x \in \mathbb{R} \implies 2=4$ (contradiction) $\implies x \notin \mathbb{R}$ as required.

So I did the same for $\sum_{i=0}^{\infty} 2^i$, that is assume that $\sum_{i=0}^{\infty} 2^i=x \in \mathbb{R} \implies$ (the rest of the proof goes as here) $\implies x=-1$(contradiction)$\implies x \notin \mathbb{R}$ as required.

Does this proof count as valid argument and if not were did I go wrong ?

Edit: It seems there have been a misunderstanding and it is my fault for not being clear . My goal was determine where does my argument fail not that the proof here is not convincing . The main points where my argument fails as pointed out are :

  1. I need the notion of limits and sequences to define $\sum_{i=1}^{\infty} 2^i$ .
  2. $x=-1$ is indeed not a contradiction as I assumed.
abc1455
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    Yes, in 2-adics. – Oscar Lanzi Jul 17 '20 at 19:41
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    I don't understand the claimed contradiction. Why can't $x=-1$? (I'll note that there are a couple of possible arguments, but you haven't given any of them...) – Xander Henderson Jul 17 '20 at 19:48
  • There are no valid proofs of paradoxes. Otherwise they wouldn't be paradoxes. – fleablood Jul 17 '20 at 19:54
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    @abc1455 That is certainly one argument. I also have to wonder: if you want an argument without limits or sequences, then what does the notation $\sum_{j=0}^{\infty} 2^j$ even mean? – Xander Henderson Jul 17 '20 at 19:54
  • @fleablood As I read the question, the asker wants to know if their argument successfully proves that the paradox is false (i.e. that $\sum 2^j = -1$ is a false statement). They aren't trying to prove it, but rather to disprove it. The title is a bit misleading... – Xander Henderson Jul 17 '20 at 19:56
  • Related: https://math.stackexchange.com/questions/37327/infty-1-paradox – Ottavio Jul 17 '20 at 19:59
  • @XanderHenderson : How should I incorporate these facts that you pointed out into the argument ? – abc1455 Jul 17 '20 at 20:00
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    How do you define $\sum_{i=0}^\infty 2^i$ if not as the limit of the sequence of partial sums? You can't really prove a statement about the limit of a sequence without using limits or sequences. – Vercassivelaunos Jul 17 '20 at 20:03
  • I'll be honest: I don't see how the answers provided in the thread to which you have linked in your question are insufficient. I am going to vote to close this question as a duplicate of that question. If you explain how the other question fails to answer your question, please edit your question to clarify the problem. – Xander Henderson Jul 17 '20 at 20:05
  • I personally think that your proof is correct because x is indeed not a real number. It is a divergent series. –  Jul 17 '20 at 20:06
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    Exactly as @Vercassivelaunos said - if you want a rigorous proof - you must use limits, because that's the definition of an infinite sum, and there is no way around it. If you want a non-rigorous, intuitive proof - really just saying "we are adding more and more positive terms that are getting hugely bigger, so the series diverges to ${+\infty}$" is more than enough. – Riemann'sPointyNose Jul 17 '20 at 20:11
  • @Vercassivelaunos : I wanted to know where the proof breaks down as the same argument is used to calculate $\sum_{i=0}^{\infty} 2^{-i}$ – abc1455 Jul 17 '20 at 20:15
  • $x = -1$ is not a contradiction. The contradiction is that for any $N \in \mathbb N$ then if $n > N$ then $|-1 - (\sum_{k=0}^n 2^k|= |(2^{n+1}-1)-(-1)|=|2^{n+1}|>2^{N+1} \ge 2$. So there is no $N$ for which $n > N$ would result i $|-1-(\sum_{k=0}^n2^k| < 1$. A contradiction that for every $\epsilon > 0$ then is an $N\in \mathbb N$ so that $n > N$ will always mean $|x - \sum_{k=0}^n 2^k| < \epsilon$. ... that was an unstated and ingnored but essential assumption in the statement $\sum_{k=0}^\infty 2^k = x \in \mathbb R$. – fleablood Jul 17 '20 at 20:18
  • @abc1455 As fleablood has pointed out - there is an unstated assumption that the limit actually exists. We can do that for the series of ${2^{-i}}$ because the limit exists. But otherwise all the steps become invalid. Also, ${x=-1}$ in itself does not explain the contradiction, as pointed out by Xander Henderson. – Riemann'sPointyNose Jul 17 '20 at 22:09
  • @XanderHenderson : sorry for the misunderstanding . It is my fault for not being clear enough , my intention was to know if my argument fails and if it fails where does it fail . As you pointed out at the beginning $x=-1$ is not a contradiction , I just wasn't able to realize it back then . – abc1455 Jul 18 '20 at 00:04
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    Well, it could be a contradiction. We claim $1 < 1+2 < 1+2+4 < ..... < x$ then $x>0$ so $x=-1$ is contradiction. But we have to claim that as as we are always adding positive values so $x > 1+2 + 4 + ..... + 2^n$ for all $n$ may or may not be acceptable. One this one hand "c'mon! It's just common sense!" is compelling... but is it valid? But if we go to that level is $2x = 2(1 + 2 + 4 + ...) = 2+4+8+... $ actually valid? And is $x=1+2+4+...$ so $x-1 = 2+4 +8+...$ valid? Eventually we have to address the elephant in the room: What does $1+2+4+8+...$ mean? – fleablood Jul 18 '20 at 01:00

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