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## Main Question or Discussion Point

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

[itex] \aleph_0 [/itex] # of digits?

[itex] \aleph_0 [/itex] # of digits?

- Thread starter cragar
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- #1

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

[itex] \aleph_0 [/itex] # of digits?

[itex] \aleph_0 [/itex] # of digits?

- #2

pwsnafu

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Every natural number has a finite number of digits.

- #3

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Adding to the above (which is correct), the set of infinite digit strings is uncountable.

- #4

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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

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Each

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.

The

- #6

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You can't. This doesn't change the fact that every natural number has a finite number of digits.I mean you couldn't put a bound on the number of digits.

- #7

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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).

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- #9

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[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

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No. The contradiction does not verify that every natural number has a base 10 representation with only finitely many digits.Could I use this as a proof by contradiction to verify it?

- #11

Bacle2

Science Advisor

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If I understood you correctly, you want to compose all strings of finite length

[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?

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

Ʃ

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

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ya thats what i am kinda saying

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