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 January 13th, 2016, 06:12 PM #21 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 You explicitly list various finite decimals, but those are immaterial, as it's undisputed that they're countable. Their countability doesn't imply that the infinite decimals are countable. You're effectively asserting that such an implication exists, but it doesn't, and a mere assertion doesn't constitute a proof. You could provide 3, 3.1, 3.141, 3.1415, 3.14159, . . . $\pi$, and I would be happy with that. However, that way of dealing with irrational numbers uses a subset of the rationals, and there are uncountably many subsets of the rationals, so you don't have a countable number of countable numbers with that approach.
January 13th, 2016, 06:41 PM   #22
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 Originally Posted by zylo My proof deals with infinite decimals:... n=1-> infinity
I know what an infinite decimal is.

Your problem is that every $n$ is finite. You appear not to understand that the notation $\sum \limits_{n=1}^\infty a_n$ (which is what you are effectively doing: $a_n = 10^n$) includes only finite values of $n$ because there is no infinite natural number.

February 29th, 2016, 03:19 AM   #23
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 Originally Posted by zylo I'm not listing the infinite decimals, I'm counting them
You counted terminating decimals. You gave the counts for 1-place, 2-place and n-place decimals, and never counted (or mentioned within your argument) any specific non-terminating decimal.

February 29th, 2016, 06:24 AM   #24
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Quote:
 Originally Posted by zylo I'm not listing the infinite decimals, I'm counting them
There is no difference between listing and counting. They are the same operation.

February 29th, 2016, 08:46 AM   #25
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Quote:
 Originally Posted by zylo v8archie, 1) Every real is a cut is a decimal. The decimals can be counted. -> The reals can be counted.
I don't think you know what the word "countable" means.

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