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June 29th, 2016, 07:42 AM   #1
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Natural Numbers and the Continuum

Reverse binary/decimal representation of the natural numbers:
a0x2^0 + a1x2^1 + a2x2^2 + a3x2^3 + ....., an=0,1
a0x10^0 + a1x10^1 + a2x10^2 + a3x10^3 + ....., an=0,1,.....,9
ALL n.
01523 <-> 32510 (reversed <-> standard)

From the definition it follows there is a 1:1 correspondence between the natural numbers and the real numbers (continuum) by adding or removing a "decimal" point.
015 <-> .015 (0x10^0 + 1x10^1 + 5x10^2 <-> 0x10^-1 + 1x10^-2 + 5x10^-3)
1101 <-> .1101
00101.. <-> .00101..
33333..... <-> .33333

Finite and repeat natural numbers correspond to the rational numbers, and non-repeating natural numbers correspond to irrational numbers. Thus there is a 1:1 correspondence between the natural numbers and the continuum.

In decimal notation, corresponding to each point in [0,1) is an integer point on the line. The representation of all the points on the line by natural numbers is obvious.


i) The reals are countable.

ii) Cantor's Diagonal Argument is wrong because any countably infinite binary sequence corresponds to a natural number, by above.
The Argument fails because:
a) The countably infinite sequence of countably infinite binary digits is not square (2^n sequences per n digits, ALL n), or
b) The constructed sequence whose nth digit is the complement of the nth digit of the nth sequence in the list can only be at the end of the list, and so doesn't exist because the list has no end.

[Note by moderator: a natural number is finite (the summation you give doesn't converge unless it contains only a finite number of non-zero terms), so 33333... isn't a natural number. Also, if the "constructed sequence" cannot be in the list used, that confirms that an all-inclusive list is impossible, so Cantor's "not enumerable" conclusion is correct.]

Last edited by skipjack; June 29th, 2016 at 09:10 PM.
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June 29th, 2016, 10:23 AM   #2
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Originally Posted by zylo View Post
From the definition it follows there is a 1:1 correspondence between the natural numbers and the real numbers (continuum)
Ye gods, you are stupid.

Just saying "for all $n$" like some magic mantra doesn't make what you say any less incorrect. Instead of repeating your incoherent nonsense over and over again, try learning what that phrase actually means.
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Last edited by skipjack; June 29th, 2016 at 09:12 PM.
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June 29th, 2016, 05:57 PM   #3
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v8archie, please address other members respectfully, regardless of how impertinent their ideas may be.

As this thread is redundant I am closing it.
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