My Math Forum Cantors Diagonal Argument for n digits.

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 July 13th, 2017, 10:12 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 Cantors Diagonal Argument for n digits. The number of binary sequences for n digits is always greater than n, for all n. Ex, n=2 10 01 11 00 11=00 is in the list. 00 01 10 11 01=10 is in the list.
 July 13th, 2017, 10:40 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 680 I am not clear why you have posted this. Do you have a question? And I don't understand what "11= 00" or 01 = 10" could mean. It is straight forward to prove that the set of all binary numeral with up to n digits (counting leading 0s) is $2^n$. It is also easy to prove that $2^n> n$ for all positive integers n.
July 13th, 2017, 11:47 AM   #3
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Quote:
 Originally Posted by zylo The number of binary sequences for n digits is always greater than n, for all n. Ex, n=2 10 01 11 00 11=00 is in the list. 00 01 10 11 01=10 is in the list.
What, exactly, are you trying to say?

I hope we're not back to the nonsense of refuting Cantor's "diagonal argument"! It has
been proven by a legitimate mathematical method and holds true. Whether or not one
chooses to believe the concepts that it espouses is an entirely different matter.

 July 13th, 2017, 01:25 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2265 Math Focus: Mainly analysis and algebra More to the point Cantor's diagonal argument doesn't talk about finite sequences (those that cease after $n$ digits), it talks about infinite sequences (those that never cease).
 July 13th, 2017, 03:19 PM #5 Senior Member   Joined: Oct 2009 Posts: 141 Thanks: 59 This is interesting. As the OP shows, you can indeed extend Cantor's argument to show that $2^n > n$ for finite numbers $n$. Never thought of that, thanks!
 July 13th, 2017, 03:55 PM #6 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,570 Thanks: 613 Math Focus: Wibbly wobbly timey-wimey stuff. Again? -Dan Thanks from greg1313
July 13th, 2017, 04:01 PM   #7
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Quote:
 Originally Posted by topsquark Again?
Not for long, if at all.

July 13th, 2017, 06:12 PM   #8
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Quote:
 Originally Posted by zylo The number of binary sequences for n digits is always greater than n, for all n.

...couldn't help myself.

 July 13th, 2017, 06:46 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,939 Thanks: 2265 Math Focus: Mainly analysis and algebra Or, indeed any $n \not \in \mathbb N$.
July 14th, 2017, 10:14 AM   #10
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Quote:
 Originally Posted by zylo The number of binary sequences for n digits is always greater than n, for all n.
Cantor used non-terminating binary sequences, not binary sequences n digits long.

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