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July 13th, 2017, 07:32 AM   #1
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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
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July 13th, 2017, 08:08 AM   #2
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It definitely is if only finitely many of the $a_i$ are nonzero.
For an infinite sequence, you'll need to look at the so-called 2-adics, which make sense of them, but quite surprising things can happen for those.

For example, in the 2-adic number system, the following makes sense:

Let $x=1 + 2 + 2^2 + 2^3 + ...$
Thus $x-1 = 2(1+2+2^2 + 2^3 + ...) = 2x$
Thus $x= - 1$.

Hence the $2$-adic number $1+2+2^2 + 2^3 + ...$ is none other than the integer $-1$.
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July 13th, 2017, 08:55 AM   #3
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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 non-zero 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.
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July 13th, 2017, 09:47 AM   #4
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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. Infinity-1=Infinity.
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July 13th, 2017, 09:52 AM   #5
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Quote:
Originally Posted by zylo View Post
C
As for micrm@ss, x is an infinite number so the calculation is meaningless. Infinity-1=Infinity.
You should read up on the 2-adics.
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July 13th, 2017, 01:06 PM   #6
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Zylo meets Micrm@ss. This is going to be good. I'm getting out the popcorn.

Quote:
Originally Posted by Micrm@ss View Post

Hence the $2$-adic number $1+2+2^2 + 2^3 + ...$ is none other than the integer $-1$.
As I understand it, this fact links the $2$-adics with the two's-complement arithmetic commonly used in computer hardware. For example on an 8-bit computer, -1 = 11111111 in two's-complement.

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.
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July 13th, 2017, 01:21 PM   #7
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Quote:
Originally Posted by zylo View Post
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?
Is that what passes for a coherent argument in your world? You are conflating different sequences. Bit of consideration should see you right.

Convergence has everything to do with it.

"Removing the decimal point" is not a mathematical operation.
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July 13th, 2017, 01:29 PM   #8
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Quote:
Originally Posted by v8archie View Post

"Removing the decimal point" is not a mathematical operation.
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 dual-representation 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 well-defined 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.
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July 13th, 2017, 02:12 PM   #9
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Originally Posted by Maschke View Post
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?
Grow up.

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.
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July 13th, 2017, 02:17 PM   #10
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Quote:
Originally Posted by zylo View Post
Is an infinite binary sequence, interpreted as coefficients of 2$\displaystyle ^{n}$, a natural number?
Your example showed that you defined the sequence as having a value equal to a sum with last term $a_n$, though you presumably intended that term to be $a_n2^n$. Such a sum has exactly $n$+1 terms and its value is a natural number; the question of convergence doesn't arise.

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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