Question about natural numbers.

  • Thread starter cragar
  • Start date
  • #1
2,544
2

Main Question or Discussion Point

Are there an [itex] \aleph_0 [/itex] # of natural numbers with an
[itex] \aleph_0 [/itex] # of digits?
 

Answers and Replies

  • #2
pwsnafu
Science Advisor
1,080
85
Every natural number has a finite number of digits.
 
  • #3
806
23
Adding to the above (which is correct), the set of infinite digit strings is uncountable.
 
  • #4
2,544
2
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.
 
  • #5
pwsnafu
Science Advisor
1,080
85
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.
 
  • #6
806
23
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.
 
  • #7
441
0
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).
 
  • #8
296
0
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.
 
  • #9
2,544
2
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?
 
  • #10
jgens
Gold Member
1,580
49
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.
 
  • #11
Bacle2
Science Advisor
1,089
10
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.
 
  • #12
2,544
2
ya thats what i am kinda saying
 

Related Threads on Question about natural numbers.

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
1K
  • Last Post
Replies
2
Views
745
  • Last Post
Replies
17
Views
733
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Top