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April 28th, 2017, 08:32 AM   #21
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Quote:
Originally Posted by zylo View Post
101010......... makes no sense if there is a largest natural number.
There is no largest natural number, but every natural number is finite. Your example of 101010.......... (continuing without end) isn't a natural number.
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April 28th, 2017, 09:39 AM   #22
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The reason that 101010.......... (continuing without end) isn't a number is that
$$\sum_{n=0}^\infty a_n 2^n \quad a_n \in \{0,1\}$$
doesn't converge unless $a_n = 0$ for all $n \gt N \in \mathbb N$.

Last edited by v8archie; April 28th, 2017 at 09:48 AM.
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April 28th, 2017, 10:10 AM   #23
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Quote:
Originally Posted by v8archie View Post
The reason that 101010.......... (continuing without end) isn't a number is that
$$\sum_{n=0}^\infty a_n 2^n \quad a_n \in \{0,1\}$$
doesn't converge unless $a_n = 0$ for all $n \gt N \in \mathbb N$.
Convergence has nothing to do with it if we are discussing the set of infinite binary sequences. If we add the binary point, that has already been covered.

Let me count the infinite binary sequences for you.
Map them as follows:

1000000...... -> 1
0100000...... -> 2
1100000...... -> 3
0010000...... -> 4
.
.
.
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April 28th, 2017, 10:34 AM   #24
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Of course convergence has something to do with it!

So what is the value of your natural number represented by the sequence $$a_0a_1a_2\ldots \quad a_n \in \{0,1\}$$ when the series $$\sum_{n=0}^\infty a_n2^n$$ does not converge? This is exactly the definition of your supposed map, so without an answer your argument collapses.

Since natural numbers have no binary point, your attempted obfuscation is clearly irrelevant.

Last edited by v8archie; April 28th, 2017 at 10:57 AM.
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April 28th, 2017, 11:23 AM   #25
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Quote:
Originally Posted by zylo View Post
Let me count the infinite binary sequences for you.
Map them as follows:

1000000...... -> 1
0100000...... -> 2
1100000...... -> 3
0010000...... -> 4
.
.
.
Every one of those sequences contains a finite number of 1's and an infinite number of 0's. What about all the sequences containing an infinite number of 1's and an infinite number of 0's?

You seem to be an uncountably many binary sequences short of a full listing...

The binary sequences can be put into bijection with $P( \,\mathbb{N}) \,$ quite easily. If that mapping is 1-to-1 from $\mathbb{N}$ onto the set of infinite binary sequences, then we'd also have a mapping that is 1-to-1 onto $P( \,\mathbb{N}) \,$. Since you don't like Cantor's diagonal argument, consider his Theorem:

Let $f : \mathbb{N} \longrightarrow P( \,\mathbb{N}) \,$

Let $A = \{ n : n \notin f( \,n) \, \}$

If $f$ is surjective, then there exists a $k \in \mathbb{N}$ such that $f( \,k) \, = A$:

$k \in A \rightarrow k \notin f( \,k) \, = A$
$k \notin A \rightarrow k \in f( \,k) \, = A$

No $k$ can exist such that $f( \,k) \, = A$, so the assumption that $f$ was surjective was faulty.
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April 28th, 2017, 11:42 AM   #26
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I note
$\displaystyle p=a{_n}2{^n}+......+a{_1}2{^1}+a{_0}2{^0}$
$\displaystyle =a{_0}2{^0}+a{_1}2{^1}......+a{_n}2{^n}$
for all n.

So the count of infinite binary sequences
1000000...... -> 1
0100000...... -> 2
1100000...... -> 3
0010000...... -> 4
.
.
can be interpreted in either direction (map or actual number). The "last" number in the sequence is 1111111.............., which is the inverse of 000000000.
Note any natural number can be expressed as a unique infinite binary sequence and vice versa.

If the binary sequence wasn't infinite, the count of natural numbers would stop at some number.

Cantor's set of infinite binary sequences is simply the set of natural numbers in binary notation.
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April 28th, 2017, 11:42 AM   #27
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Quote:
Originally Posted by AplanisTophet View Post
You seem to be an uncountably many binary sequences short of a full listing...
In more ways than one.
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April 28th, 2017, 11:52 AM   #28
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Quote:
Originally Posted by zylo View Post
So the count of infinite binary sequences
1000000...... -> 1
0100000...... -> 2
1100000...... -> 3
0010000...... -> 4
.
.
can be interpreted in either direction (map or actual number).
And so you have to answer the question about convergence instead of pretending that you know what you are talking about.

As usual, you don't understand the difference between the concepts "arbitrarily large" and "infinite".

Last edited by v8archie; April 28th, 2017 at 11:56 AM.
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April 28th, 2017, 12:01 PM   #29
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Quote:
Originally Posted by zylo View Post
The "last" number in the sequence is 1111111..............
Since every natural number has a successor, what is the successor to 1111111..............? Or do you claim that there is now a greatest natural number? And that the natural numbers are not closed under addition or multiplication?
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April 28th, 2017, 01:05 PM   #30
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Quote:
Originally Posted by AplanisTophet View Post
You seem to be an uncountably many binary sequences short of a full listing...
LOL. Was that joke intended? As in a few cards short of a full deck?
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