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    Default Cardinality of the number of digits in an irrational number

    I know that there's an infinite number of digits in each irrational number, but how infinite? Aleph-null? Aleph-one? Higher? It seems to me like it has to be aleph null because you can put the digits into bijection with the whole numbers (in pi, 3 is the first digit, 1 is the second digit, 4 is the third digit, etc.) And yet it also seems to me that if it were aleph null then you woukd be able to count all the real numbers (out of order) in omega^2 (=still aleph null) steps by just reflecting the counting numbers over the decimal point

    Spoiler
    Show

    1.) 0.0
    2.) 0.1
    3.) 0.2
    4.) 0.3
    5.) 0.4
    6.) 0.5
    7.) 0.6
    8.) 0.7
    9.) 0.8
    10.) 0.9
    11.) 0.01
    12) 0.11
    .
    .
    .
    omega.) 1.0
    omega+1) 1.1
    .
    .
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    2Omega.) 2.0
    .
    .
    .


    But we know that the cardinality of the reals is strictly above aleph null so it seems that if that's the case then mustn't the number of digits in at least some irrational numbers be strictly above aleph null as well?
    Last edited by Bohandas; 2020-10-27 at 10:52 PM.
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    Orc in the Playground
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    Default Re: Cardinality of the number of digits in an irrational number

    As you said in the opener, there are countably many digits.

    As for the spoiler: for which ordinal do you get 1/3 = 0.3333....? (Hint: you never will; the only numbers you will reach will be of the form p/(10^i) for integers p, i - and 1/3 cannot be expressed in such a way)
    Last edited by uncool; 2020-10-28 at 01:01 AM.

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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by uncool View Post
    As you said in the opener, there are countably many digits.

    As for the spoiler: for which ordinal do you get 1/3 = 0.3333....? (Hint: you never will; the only numbers you will reach will be of the form p/(10^i) for integers p, i - and 1/3 cannot be expressed in such a way)
    Didn't you just express 1/3 as a decimal though in your example?
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    Default Re: Cardinality of the number of digits in an irrational number

    Ok, this is a bit tricky, but working with infinities is always a bit weird.

    First things first: number of digits is countable, so for irrational numbers it is aleph 0.

    Now to how real (or just irrational ones - doesn't matter) numbers are uncountable: there is a very simple proof that a set of all infinite sequences has to be uncountable, which boils down to the fact that regardless of how you create a countable list of all sequences, there will always be a simple method to build a sequence that will never be on the list.

    This I think has some relation to the more general case of set of all subsets of an infinite set having larger cardinality than the original set.
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    Default Re: Cardinality of the number of digits in an irrational number

    The mirroring trick doesn't get you all reals, or even all rationals.

    The whole point of being countably infinite is that you can show an ordering that arrives at any number in the set in a finite number of steps. That ordering takes an infinite number of steps to get to 1.
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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by Bohandas View Post
    Didn't you just express 1/3 as a decimal though in your example?
    Yes, but it still doesn't appear in your list. Every individual number in your list has a finite number of digits.

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    Orc in the Playground
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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by Bohandas View Post
    Didn't you just express 1/3 as a decimal though in your example?
    How are you defining "decimal"?

    If you mean "expressible through the standard decimal system, possibly using limits", then yes - but your list doesn't cover limits. If you mean "numbers of the form p/(10^i) for integers p, i", as I said in my hint, then no.

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    Troll in the Playground
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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by warty goblin View Post
    The mirroring trick doesn't get you all reals, or even all rationals.

    The whole point of being countably infinite is that you can show an ordering that arrives at any number in the set in a finite number of steps. That ordering takes an infinite number of steps to get to 1.
    The mirroring trick is not meant to construct real numbers. It is just a proof by contradiction: if reals are countable, you should be able to make a complete infinite list of all infinite sequences over a finite set (for example 0 to 9 numbers in a decimal system). Going from that premise it is easy to show that for infinite sequences there is always at least one element of the set of all sequences that cannot be on this list. This is more than enough to prove that reals are not countable.

    As for the construction of real numbers, I specifically did not want to go there as it is more involved and harder to understand for people without the relevant knowledge. Showing that reals are uncountable from construction is also less clear than the simple proof by contradiction.
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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by Radar View Post
    The mirroring trick is not meant to construct real numbers. It is just a proof by contradiction: if reals are countable, you should be able to make a complete infinite list of all infinite sequences over a finite set (for example 0 to 9 numbers in a decimal system). Going from that premise it is easy to show that for infinite sequences there is always at least one element of the set of all sequences that cannot be on this list. This is more than enough to prove that reals are not countable.

    As for the construction of real numbers, I specifically did not want to go there as it is more involved and harder to understand for people without the relevant knowledge. Showing that reals are uncountable from construction is also less clear than the simple proof by contradiction.
    I was referring to the OP's mirroring numbers over the decimal, not your argument. Sorry if that was unclear.
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    And he lay in his blood on the highway, with the bunch of lace at his throat.


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    Troll in the Playground
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    Default Re: Cardinality of the number of digits in an irrational number

    Quote Originally Posted by warty goblin View Post
    I was referring to the OP's mirroring numbers over the decimal, not your argument. Sorry if that was unclear.
    Oh, ok. No problem.
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    Default Re: Cardinality of the number of digits in an irrational number

    First, Bohandas's list of numbers does not include *any* number with aleph-null digits, which includes all repeating decimals like 1/3, as well as all irrational numbers. Every number on the list has a finite number of digits after the decimal point.

    Yes, the number of digits in irrational numbers between zero and one is aleph-null. [You can choose to include all rational numbers as well, by counting an infinite string of zeroes.]

    Therefore the number of different numbers between 0 and 1 is 10^aleph-null, which is uncountable.
    Last edited by Jay R; 2020-11-24 at 03:26 PM.

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