How far can you move the bar?

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 01/27/05 

Date: Thu, 27 Jan 2005 15:01:44 GMT

Let N, W, J, floor(), Q, and R have their conventional definitions.

Let T_10 = {k/10^n: k,n in W, n >= 0, 0 <= k < 10^n}.

Let L : N -> T_10 be a certain countable list of reals in [0,1).
These reals are constructed in the following manner:

L(0) = 0.000000...
L(1) = 0.300000...
L(2) = 0.330000...
L(3) = 0.333000...
L(4) = 0.333300...
L(5) = 0.333330...
L(6) = 0.333333...
...

L(n) = (10^n - 1) / (3 * 10^n) for any integer n.

Let the notation L(n,m) : N x W -> {0,1,2,3,4,5,6,7,8,9}
be the value floor(10^m * L(n)) - 10*floor(10^(m-1) * L(n))
for any integers m, n, m > 0, m >= 0. Define L(n,0) = 0.

More intuitively, L(n,m) = the m'th digit to the right of
the decimal point of L(n).

Let P(r,m) : [0,1) x N -> T_10 be floor(r * 10^m) / 10^m.

Let D(L, n) : (: N->T_10) x N -> {0,1,2,3,4,5,6,7,8,9}
    be defined as follows:
    if(L(n,n) = 3) then D(L,n) = 0
    else D(L,n) = 3

and let D(L) = sum(n in N) (D(L,n) * 10^(-n)).

Note that we do not assume D(L) is in T_10; it is, however, in R.

We can compute D(L) fairly readily, especially if we
notice that L(n,n) = 0 for all n in N. Therefore
D(L,n) is always 3 and D(L) = 0.333... .

The "bar" in this case refers to a certain integer b.
For b=2 this can be diagrammed:

L(0) = 0.00|0000...
L(1) = 0.30|0000...
L(2) = 0.33|0000...
L(3) = 0.33|3000...
L(4) = 0.33|3300...
L(5) = 0.33|3330...
L(6) = 0.33|3333...
...
D(L) = 0.33|3333...

and can be formally defined as

max({b: (Ea) (a in W . P(D(L),b) = P(L(a), b))})

It turns out that there is no such b;
P(D(L),b) = P(L(b), b), and therefore a = b.
Therefore any b in N will work; the bar can be set
anywhere in N.

So ... is D(L) in the list?

Well...

If we assume D(L) = L(n), then we note that
L(n,n) = 0 whereas D(L,n) = 3. Therefore no
such n can exist. 

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
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