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 July 14th, 2017, 09:58 AM #21 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 How do you evaluate the first sum? Also,what do you mean by "approaches"? July 14th, 2017, 10:15 AM   #22
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Quote:
 Originally Posted by skipjack How do you evaluate the first sum? Also,what do you mean by "approaches"?
Quote:
 Originally Posted by zylo The infinite binary (or decimal) sequence $\displaystyle a_{1}a_{2}a_{3}.......$ defines a unique natural number or unique binary fraction, depending on interpretation: Natural Number: $\displaystyle \sum a_{n}2^{n}$ which approaches a unique natural number as n -> infnity. Binary Fraction: $\displaystyle \sum a_{n}2^{-n}$ which approaches a unique fraction in [0,1) as n -> infinity. Note the unique association between natural number and binary fraction.
$\displaystyle 0110.....=0\times2^{0}+ 1\times2^{1} + 1\times2^{2} + 0\times2^{3}+ .......= 6$
$\displaystyle .0110....=0\times2^{-1}+ 1\times2^{-2} + 1\times2^{-3} + 0\times2^{-4}+ .......= 3/8$

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
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Quote:
 Originally Posted by zylo The infinite binary (or decimal) sequence $\displaystyle a_{1}a_{2}a_{3}.......$ defines a unique natural number or unique binary fraction, depending on interpretation: Natural Number: $\displaystyle \sum a_{n}2^{n}$ which approaches a unique natural number as n -> infnity. Binary Fraction: $\displaystyle \sum a_{n}2^{-n}$ which approaches a unique fraction in [0,1) as n -> infinity. Note the unique association between natural number and binary fraction.

But $\sum a_n2^n$ does not approach a natural number as n grows larger. July 14th, 2017, 10:31 AM   #24
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 Originally Posted by zylo $\displaystyle 0110.....=0\times2^{0}+ 1\times2^{1} + 1\times2^{2} + 0\times2^{3}+ .......= 6$ $\displaystyle .0110....=0\times2^{-1}+ 1\times2^{-2} + 1\times2^{-3} + 0\times2^{-4}+ .......= 3/8$ 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.
Yes, sure but there is a crucial difference. When you write $\sum a_n 2^n$ you assume that only finitely many of the $a_n$ are nonzero.
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
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Quote:
 Originally Posted by Micrm@ss Yes, sure but there is a crucial difference. When you write $\sum a_n 2^n$ you assume that only finitely many of the $a_n$ are nonzero. 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
Not at all.

.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
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 Originally Posted by zylo If you only accept natural numbers designated by a finite number of digits then you only have a finite number of natural numbers.
Nope. There are infinitely many natural numbers with a finite number of digits.

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
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 Originally Posted by zylo .11111111 corresponds to the natural number 111111111......, which is, by above scheme: 111111...=1x2^0 + 1x2^1 + 1x2^2 + .......
You haven't previously justified referring to 111111.... as a natural number. You were using the equation 111111...=1x2^0 + 1x2^1 + 1x2^2 + .... to define 111111...., but the right-hand side is a sum with an infinite number of non-zero terms, and you haven't explained what that means or why it should be called a natural number. You suggested the sum approaches a natural number, but haven't explained what "approaches" means in this context.

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
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 Originally Posted by zylo Not at all. .11111111 corresponds to the natural number 111111111......, which is, by above scheme: 111111...=1x2^0 + 1x2^1 + 1x2^2 + .......
Can you prove this is a natural number? July 17th, 2017, 08:22 AM #29 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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 $\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}.\sum_{n=1}^{\infty}b_{n}\times 10^{n-1}$ 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
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Quote:
 Originally Posted by zylo $\displaystyle \sum_{n=1}^{\infty}a_{n}\times 10^{n}$ is the accepted expression for a natural number
No, it isn't. You can't express 1 in that form, and 1011011101111011111.... isn't associated with a specific natural number in any commonly accepted way. Tags binary, blah, infinite, natural, number, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zylo Topology 25 June 4th, 2016 11:55 PM zylo Topology 8 April 8th, 2016 08:19 PM Zman15 Elementary Math 10 March 28th, 2015 06:42 AM Shen Elementary Math 2 June 5th, 2014 07:50 AM durky Abstract Algebra 1 March 15th, 2012 11:28 AM

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