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February 4th, 2019, 05:46 AM   #81
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Quote:
 Originally Posted by skipjack How does the list reach a non-terminating number, given that adding finite-length numbers (for example, adding 1 to a finite-length number) always results in a finite-length number?
1) The list of natural numbers doesn't end, ie, all sequences of digits are in it.

2) Assumption: The mirror image of the digits in any real number in (0,1) exists.

3) Finite: ends. Infinite: doesn't end, endless.

How is it, that by taking one step at a time, I can embark on an endless journey?

February 4th, 2019, 07:06 AM   #82
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Quote:
 Originally Posted by zylo 1) The list of natural numbers doesn't end, ie, all sequences of digits are in it.
Consider the list of all finite strings of 1s. This list is endless (*), but, by definition, doesn't contain any strings of infinite length.

[(*) If it ended, it would have an element of maximal length, say the string consisting of n 1s, where n is some (finite) natural number. But then the string consisting of n+1 1s would be a longer string, and is still in the list as n+1 is also a (finite) natural number. This is a contradiction.]

What this example shows: the endlessness of a list of strings does not imply that the list contains strings of infinite length.

February 4th, 2019, 08:48 AM   #83
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Quote:
 Originally Posted by zylo 1) The list of natural numbers doesn't end, ie, all sequences of digits are in it.
The list of all regular $n$-gons of unit side doesn't include a circle or a straight line.

February 4th, 2019, 09:33 AM   #84
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Quote:
 Originally Posted by zylo 1) The list of natural numbers doesn't end, ie, all sequences of digits are in it.
(1) You have neither constructed such a list nor stated its existence as an assumption.

Quote:
 Originally Posted by zylo 2) Assumption: The mirror image of the digits in any real number in (0,1) exists.
(2) You haven't defined "mirror image" of a real number in (0, 1).

Quote:
 Originally Posted by zylo 3) How is it, that by taking one step at a time, I can embark on an endless journey?
(3) From your starting point of 1, your steps each involve adding 1, which is well-defined only for adding 1 to a finite-length number, and always gives a finite-length number. You can't reach 545454545... (which isn't finite-length) by adding 1, so you don't produce "all sequences of digits".

 February 4th, 2019, 10:05 AM #85 Senior Member   Joined: Dec 2015 From: somewhere Posts: 640 Thanks: 91 This is simply not defined yet. Which one is larger 222222... or 999999... (the two numbers are natural), but $\displaystyle \frac{2222...}{9999...}=2/9=0,\overline{ 2}$. Thanks from topsquark Last edited by skipjack; February 4th, 2019 at 10:35 AM.
 February 4th, 2019, 10:43 AM #86 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2222 The second number has greater magnitude, but the lengths of the decimal representations are equal. Thanks from topsquark
February 4th, 2019, 10:46 AM   #87
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Quote:
 Originally Posted by idontknow This is simply not defined yet. Which one is larger 222222... or 999999... (the two numbers are natural), but $\displaystyle \frac{2222...}{9999...}=2/9=0,\overline{ 2}$.
Actually that is $\displaystyle \lim_{n \to \infty} \left ( \dfrac{2}{9} \right )^n$. Of course this number does not exist except for zylo so we can't simply say that the limit goes to 0 because the numerator and denominator do not exist.

aka The fraction does not exist except in zylo's system of numbers.

-Dan

February 4th, 2019, 10:49 AM   #88
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Quote:
 Originally Posted by topsquark Actually that is $\displaystyle \lim_{n \to \infty} \left ( \dfrac{2}{9} \right )^n$.
I doubt you really meant that.

February 4th, 2019, 12:55 PM   #89
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Quote:
 Originally Posted by skipjack I doubt you really meant that.
What I was talking about is
Ha ha ha, ummmmm...

Ah well. Thanks for the catch.

-Dan

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