A Formula for Generating Irrational Normal Numbers?

Discussion in 'General Math' started by Guest, Feb 21, 2006.

  1. Guest

    Guest Guest

    This may be a really stupid question, but is there some kind of formula by
    which you could generate an infinite number of irrational normal numbers?
     
    Guest, Feb 21, 2006
    #1
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  2. r = 0.123456789 10111213141516171819 202122... is normal and irrational.
    for all n in N, n + r is normal and irrational.
     
    William Elliot, Feb 22, 2006
    #2
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  3. Guest

    Aggie1956 Guest

    I will give this a try:

    The rational numbers (that is those numbers that can be placed in the form
    of a/b , where a is a natural number and b is a counting number.) can
    be enumerated (that is they can be counted off one by one. NOT counted but
    placed in an order to be counted off).
    Consider:
    0/1
    1/1 2/1 3/1 4/1 5/1 6/1 ...
    1/2 2/2 3/2 4/2 5/2 6/2 ...
    1/3 2/3 3/3 4/3 5/3 6/2 ...
    1/4 2/4 3/4 4/4 5/4 6/4 ...
    ...

    Counting off:
    0 , 1/1 , 2/1 , 1/2 , 1/3 , 2/2 , 3/1 , 4/1 , 3/2 , 2/3
    , 1/4 , 1/5 , 2/4 , 3/3 , 4/2 , 5/1 , 6/1 , 5/2 ...

    If you would continue counting off the rationals following the pattern there
    would be no rational number missed nor expected to be missed in the counting
    off.

    However, I don't think the irrationals can be enumerated. You on the other
    hand have asked for a formula to generate an infinite number of irrational
    normal (?) numbers.

    Consider:
    The product of two rational number produces a rational number. Such as
    (a/b)(c/d) = ac/bc
    The product of two irrationals can produce an irrational number or a
    rational number. Such as:
    (sqrt 2)(sqrt3) = sqrt 6
    (sqrt 3)(sqrt3) = 3
    The product of a rational number and an irrational number produces an
    irrational number.

    So an equation could such as :

    X = (a/b) ( sqrt5) where a is a natural number (0,1,2,3,...) and b
    is a counting number (1,2,3,4,...)
    and of course sqrt5 is an irrational number (but any irrational
    number would do).

    This so give you an infinite number of irrational numbers but not all of
    them.

    Hope this is of some help.
     
    Aggie1956, Feb 22, 2006
    #3
  4. Guest

    ...al Guest

    According to a paper titled "Cryptography Based on Transcendental
    Numbers" by J. Pieprzyk, H. Ghodosi, C. Charnes and R. Safavi-Naini
    there is a class of numbers that are more than irrational, they are
    transcendental and normal.

    The numbers are of the form a^b
    where a is a natural number not 1
    and b is a real quadratic irrational. (e.g. sqrt(2) or sqrt(3))

    Since there are infinitely many a's and b's, There are at least an
    infinite number of such numbers.

    As for generating the decimal expansion of these numbers
    (to some arbitrary length)
    a suggestion was made, by Robert Israel, a regular poster to sci.math,
    that one can use the biniomial series to generate the digits.
    This works but the program I wrote converges slowly as a and b become large.
    This may be due to my implementation.

    Another suggestion, mentioned in the paper from the top line, is to use
    something called the 'Splitting circle method' referred to by A.
    Schonage in a paper called
    "The fundamental theorem of algebra in terms of computational complexity".
    I have a copy of this paper however; it's application to decimal
    expansion is beyond me.
    Perhaps someone in this newsgroup can help.

    Or do as I do and fire up Mathematica or Maple, plug in the formula and
    generate 30k or 50k digits in a reasonable amount of time.

    ....al
     
    ...al, Feb 23, 2006
    #4
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