July 14th, 2017, 09:58 AM  #21 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
How do you evaluate the first sum? Also,what do you mean by "approaches"?

July 14th, 2017, 10:15 AM  #22  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
Quote:
$\displaystyle .0110....=0\times2^{1}+ 1\times2^{2} + 1\times2^{3} + 0\times2^{4}+ .......= 3/8$ In the above scheme, there are an infinite number of natural numbers with the same first n digits. How do you narrow it down (converge) to a unique natural number?: With an infinite number of digits.  
July 14th, 2017, 10:30 AM  #23  
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141  Quote:
But $\sum a_n2^n$ does not approach a natural number as n grows larger.  
July 14th, 2017, 10:31 AM  #24  
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141  Quote:
When you write $\sum a_n 2^{n}$, you do not make this assumption. So while 0.111111111... is a good real number, if you swap it like you do, it is not: ...111111111  
July 14th, 2017, 11:53 AM  #25  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
.11111111 corresponds to the natural number 111111111......, which is, by above scheme: 111111...=1x2^0 + 1x2^1 + 1x2^2 + ....... There are an infinite number of natural numbers each one (of necessity} uniquely designated by an infinite sequence of digits. If you only accept natural numbers designated by a finite number of digits then you only have a finite number of natural numbers. What is .......111111? I make no reference to such a scheme. I am using a scheme of infinite digits to designate any natural number. I believe you are confusing my scheme with the conventional (practical) scheme: 1000000 instead of 00000000000010000000 (roughly)  
July 14th, 2017, 12:37 PM  #26  
Senior Member Joined: Jun 2014 From: USA Posts: 364 Thanks: 26  Quote:
Again, you're banging your head on a wall with the same nonsense over and over and over again...  
July 14th, 2017, 01:22 PM  #27  
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619  Quote:
Your original question was "Is an Infinite Binary Sequence a Natural Number?" and that means you can't explain your position by asserting that an infinite binary sequence is a natural number. You would be answering your own question by effectively changing the accepted meaning of the term "natural number". You certainly can't justify doing that and then using your new definition to interpret the term "natural number" in articles published before your definition was in use.  
July 14th, 2017, 01:53 PM  #28 
Senior Member Joined: Oct 2009 Posts: 406 Thanks: 141  
July 17th, 2017, 08:22 AM  #29 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
In conventional placement, which may feel more comfortable, any real decimal number is given by $\displaystyle ........a_{3/}a_{2}a_{1}.b_{1}b_{2}b_{3}......$ with infinite digits implied, or $\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}.\sum_{n=1}^{\infty}b_{n}\times 10^{n1}$ 37.234 is really .....0037.23400....... with ..... conventionally dropped. $\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}$ is the accepted expression for a natural number: ..........1111111 for example is a unique natural number: a_{n}=1 in previous formula. Problem with that? Do you have a problem with .1111111111....? Both infinite sequences can be uniquely associated with a natural number. If you counter CDA says they are uncountable, see Is an Infinite Binary Sequence a Natural Number? 
July 17th, 2017, 09:19 AM  #30 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619  

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binary, blah, infinite, natural, number, sequence 
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