July 13th, 2017, 07:32 AM  #1 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,431 Thanks: 105  Is an Infinite Binary Sequence a Natural Number?
Is an infinite binary sequence, interpreted as coefficients of 2$\displaystyle ^{n}$, a natural number? $\displaystyle a_{0}a_{1}a_{2}.....\equiv a_{0}\times 2^{0}+a^{1}\times 2^{1} + a_{2}\times 2^{2}+....., \\ \text{where } a_n \in \{0,1\} \\ \text{Example:} \\ 00100....=0\times2^{0}+0\times2^{1}+1\times 2^{2}+0\times2^{3}......=4$ Last edited by skipjack; July 14th, 2017 at 01:29 AM. Reason: for clarity 
July 13th, 2017, 08:08 AM  #2 
Senior Member Joined: Oct 2009 Posts: 439 Thanks: 147 
It definitely is if only finitely many of the $a_i$ are nonzero. For an infinite sequence, you'll need to look at the socalled 2adics, which make sense of them, but quite surprising things can happen for those. For example, in the 2adic number system, the following makes sense: Let $x=1 + 2 + 2^2 + 2^3 + ...$ Thus $x1 = 2(1+2+2^2 + 2^3 + ...) = 2x$ Thus $x=  1$. Hence the $2$adic number $1+2+2^2 + 2^3 + ...$ is none other than the integer $1$. 
July 13th, 2017, 08:55 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra 
I really can't believe you are asking this. First, because it's been explained to you many times before; and second, because you haven't flat out stated that it is. Micrm@ss is correct. An infinite binary sequence having infinitely many nonzero digits can't be a natural number because in that case your sum $$ a_{0}a_{1}a_{2}\ldots \equiv a_{0}\times 2^{0}+a^{1}\times 2^{1} + a_{2}\times 2^{2}+\ldots$$ doesn't converge. 
July 13th, 2017, 09:47 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,431 Thanks: 105 
Convergence has nothing to do with it. Natural numbers don't converge. What does the sequence of natural numbers converge to? If you accept the existence of an infinite sequence of digits in a binary fraction, do they disappear when you remove the binary point? As for micrm@ss, x is an infinite number so the calculation is meaningless. Infinity1=Infinity. 
July 13th, 2017, 09:52 AM  #5 
Senior Member Joined: Oct 2009 Posts: 439 Thanks: 147  
July 13th, 2017, 01:06 PM  #6  
Senior Member Joined: Aug 2012 Posts: 1,998 Thanks: 569 
Zylo meets Micrm@ss. This is going to be good. I'm getting out the popcorn. Quote:
The above equatlon represents the same fact on a computer with infinitely many bits to represent a number. Last edited by Maschke; July 13th, 2017 at 01:17 PM.  
July 13th, 2017, 01:21 PM  #7  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra  Quote:
Convergence has everything to do with it. "Removing the decimal point" is not a mathematical operation.  
July 13th, 2017, 01:29 PM  #8 
Senior Member Joined: Aug 2012 Posts: 1,998 Thanks: 569  Calling Pedantry Man!! I run into a phone booth and  wait, I haven't seen a phone booth in years. Where do superheroes change clothes these days? Given a real number $x \in (0,1)$, it has a binary representation $x_1 x_2 x_3 \dots$ with $x_i \in \{0, 1\}$. As usual we ignore the pesky dualrepresentation issue. Now we define a map from $\mathbb R \to 2^{\mathbb N}$, the space of all binary sequences or bitstrings, by $x_i \mapsto x_i$. Wait, that was way too easy. Of course "removing the decimal point" is a perfectly welldefined mathematical operation that maps each binary representation to its corresponding bitstring. It's not injective because of the dual representation problem. Last edited by Maschke; July 13th, 2017 at 01:33 PM. 
July 13th, 2017, 02:12 PM  #9  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra  Quote:
You have defined an operation in considerably more detail, in particular describing the range. Zylo's effort didn't which, in the context of a discussion about natural numbers, is distinctly problematic.  
July 13th, 2017, 02:17 PM  #10  
Global Moderator Joined: Dec 2006 Posts: 19,502 Thanks: 1739  Quote:
However, defining an arbitrary infinite binary sequence as being equal to a sum of just $n$+1 terms doesn't make sense, as it leaves $n$ undefined. Edit: if you intended an infinite number of terms in the sum, refer to this post. Last edited by skipjack; July 14th, 2017 at 01:54 AM.  

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binary, blah, infinite, natural, number, sequence 
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