Re: Is this SET of computable numbers computable?

From: |-|erc (contactvia_at_wwwadamskingdom.com)
Date: 03/29/04

  • Next message: Henning Fernau: "3-Hitting Set" 
    Date: Mon, 29 Mar 2004 18:59:57 +1000

    "Rupert" <rupertmccallum@yahoo.com> wrote in
    > <snip>
    > > In this example, n = 3 qualifies
    > > 3 8 3 5 7
    > >
    > > 1 0.8357.. is indexed as the 1st computable number.
    > >
    >
    > But we don't know it's a computable number. It hasn't even been
    > adequately specified as a number yet. The 3rd Turing machine might
    > fail to halt on some of the other digits.

    Correct. But they have 2 properties :

    1 It contains all the computable numbers, (superset)
    2 All digits up to the diagonal are specified.

    We can't guarantee infinite decimals on every number, but we can guarantee
    all computable numbers are present and counted for.

    I see no problem with this. When infinite digits are required we can't select
    a random number, the algorithm is just white box inspected to see 'what' number
    it is, just like algorithms are used now.

    The jth effectively computable number is guaranteed to halt for the 1st j digits.
    They could be defined as *atleast* rational, depending on the halting nature
    of further digits. The majority would work for all digits, if the 1st 200 values
    of a function work most likely all values will.

    So : does the set CONTAIN all computable numbers and is it amenible to diagonalisation?

    Herc

  • Next message: Henning Fernau: "3-Hitting Set"

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