ST Alpha - "Here's what bugs me about the .9bar = 1 argument. Would .8bar also equal 1?" (Reading a Post)

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In response to "Yeah. You ask them if 0.99999999999999.....=1 and they say "that's right, homeboy" -- nm" by prayformojo

Here's what bugs me about the .9bar = 1 argument. Would .8bar also equal 1?

Posted by TWuG Dec 12 '12, 09:14

.7bar? Etc?

Because if you go out to infinity then that .whatever bar is indistinguishable from 1 for all practical purposes.

So the .x bar notation is just a flawed way of writing 1.

Responses:

no. -- nm - znufrii Dec 12, 09:40
0.8bar = 0.9, no bar. -- nm - Beryllium Dec 12, 09:20 8
- Eh, this is not right. Even if it was, it would have to be 0.888888(acceptable level of precision)88889 -- nm - Beryllium Dec 12, 09:33 2
  - That's the thing: "Acceptable level of precision" - TWuG Dec 12, 09:36 1
    - But, because we know that 0.333bar is 1/3, we could bounce it back to fractional format and get the answer of 1 -- nm - Beryllium Dec 12, 09:58
- *shrug* Maybe. But if one infinity can differ from another infinity then infinity means nothing. -- nm - TWuG Dec 12, 09:23 4
  - No. -- nm - TFox Dec 12, 09:48
  - 5/6 = 0.83bar ... which is night indistinguishable (at sufficient levels of repetition) from 0.84 -- nm - Beryllium Dec 12, 09:26 2
    - night=nigh -- nm - Beryllium Dec 12, 09:26 1
      - No, this is a bad example because it's not a 9 repeating. There's the kink. -- nm - Beryllium Dec 12, 09:28

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Here's what bugs me about the .9bar = 1 argument. Would .8bar also equal 1?

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