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Question about natural numbers.

 
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Aug24-12, 01:09 AM   #1
 

Question about natural numbers.


Are there an [itex] \aleph_0 [/itex] # of natural numbers with an
[itex] \aleph_0 [/itex] # of digits?
 
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Aug24-12, 02:01 AM   #2
 
Every natural number has a finite number of digits.
 
Aug24-12, 02:12 AM   #3
 
Adding to the above (which is correct), the set of infinite digit strings is uncountable.
 
Aug24-12, 02:19 AM   #4
 

Question about natural numbers.


ok I understand what you guys are saying but it still seems strange to me.
I feel like that is saying the natural numbers are not bounded but they have a finite number of digits. I mean you couldn't put a bound on the number of digits.
 
Aug24-12, 02:47 AM   #5
 
Quote by cragar View Post
ok I understand what you guys are saying but it still seems strange to me.
I feel like that is saying the natural numbers are not bounded but they have a finite number of digits. I mean you couldn't put a bound on the number of digits.
Each individual natural number has a finite number of digits.
The entire set is unbounded.
 
Aug24-12, 12:20 PM   #6
 
I mean you couldn't put a bound on the number of digits.
You can't. This doesn't change the fact that every natural number has a finite number of digits.
 
Aug24-12, 12:56 PM   #7
 
For any natural number you pick, I can pick one with more digits. For example, if you picked x I could pick 10x, or 100,000,000,000,000,000x.

However all three of those numbers have a finite number of digits.

As the natural numbers get larger and larger so do the number of digits.

Say you have f(x) = # of digits x has for all natural numbers.

Then it is certainly true that as x approaches infinity, so does f(x).
 
Aug24-12, 03:18 PM   #8
 
To make the above a bit more rigorous, the number of digits in a natural number [itex]n[/itex] is given by [itex]\lfloor \log_{10}(n) \rfloor[/itex] and this obviously goes to infinity.
 
Aug24-12, 08:15 PM   #9
 
I could see the problem with saying that there are natural numbers with an
[itex] \aleph_0 [/itex] of digits because then I would have 10 choices for each number in the slot and I would have [itex] 10^{\aleph_0} [/itex] numbers which would be uncountable and a contradiction because the set of naturals is countable. Could I use this as a proof by contradiction to verify it?
 
Aug24-12, 08:37 PM   #10
 
Quote by cragar View Post
Could I use this as a proof by contradiction to verify it?
No. The contradiction does not verify that every natural number has a base 10 representation with only finitely many digits.
 
Aug25-12, 02:06 PM   #11
 
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Quote by cragar View Post
I could see the problem with saying that there are natural numbers with an
[itex] \aleph_0 [/itex] of digits because then I would have 10 choices for each number in the slot and I would have [itex] 10^{\aleph_0} [/itex] numbers which would be uncountable and a contradiction because the set of naturals is countable. Could I use this as a proof by contradiction to verify it?
If I understood you correctly, you want to compose all strings of finite length

with terms in {0,1,..,9} . If you write those strings as

Ʃi=0Nai10i

and let N→∞ , then(a) problem is that your sum will diverge much of the time, so that

many of those strings are not natural numbers.
 
Aug25-12, 03:27 PM   #12
 
ya thats what i am kinda saying
 
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