Re: ******* TRY THESE SCI.MATH **********

From: |-|erc (H_at_r.c)
Date: 01/29/05 

Date: Sat, 29 Jan 2005 15:17:18 +1000

"The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote in
> In sci.logic, |-|erc
> <H@r.c>
> wrote
> on Sat, 29 Jan 2005 11:34:01 +1000
> <3607ijF4pm1q0U1@individual.net>:
> > "The Ghost In The Machine" <ewill@sirius.athghost7038suus.net> wrote
> >> >> >> Here's a modified blackboard.
> >> >> >>
> >> >> >> **********
> >> >> >> * 2
> >> >> >> * 45
> >> >> >> * 3
> >> >> >> *
> >> >> >> * sum on the board = _50_
> >> >> >> *
> >> >> >> *********
> >> >> >
> >> >> >
> >> >> > What is the sum of numbers on the blackboard?
> >> >> >
> >> >> > Herc
> >> >> >
> >> >>
> >> >> 100.
> >> >>
> >> >> Did you want to introduce additional data? :-)
> >> >>
> >> >
> >> > So why did you put "sum on the board = 50"
> >> > Is sum on the board well defined?
> >>
> >> Is it?
> >>
> >> You asked for the sum of the numbers on the blackboard.
> >> 50 is one of the numbers.
> >>
> >> Ask the question correctly and you might get somewhere,
> >> especially since the board may have been partially
> >> erased.
> >>
> >> Is 1/3 part of {.3, .33, .333, ...} ?
> >>
> >
> > define in, define part of.
>
> If I'm not mistaken, "a is in B" is usually considered a
> primitive operation, but I'm not all that up on set theory.
> If one defines a set using a predicate, such as
> B = {b: P(b)}, where P : R -> boolean,
> then arguably the simplest method of figuring out whether
> a is in B is to evalute P(a).
>
> For this set P(b) is fairly easy to construct:
>
> (En)(isNatural(n) . b = (10^n - 1) /(3 * 10^n))
>
> For this sort of definition, one is better off transmuting
> this into something with an implicit quantifier and/or iterator:
>
> S_3 = {(10^n - 1) / (3 * 10^n): n in W}
>
> or one can approach it using a function:
>
> f_S3 : W -> S_3, f_S3(n) = (10^n - 1)/(3 * 10^n)
>
> This form is arguably the easiest to work with.
>
> 1/3 is in S_3 if and only if
>
> (En)(isNatural(n) . f(n) = 1/3)
>
> Or one can try to solve the equation (10^n - 1) / (3 * 10^n) for n.
>
> One naive method yields the following:
>
> f_S3(n) = (10^n - 1) / (3 * 10^n) = 1/3 - 1/(3 * 10^n) = 1/3 - 3 * 10^-n
>
> Equating 1/3 to f_S3(n) therefore requires
>
> 0 = -3 * 10^-n
>
> or n = +oo. Your favorite number! Unfortunately, +oo isn't
> a natural number, or even a real.
>
> Another method simply proves by induction that
>
> for all n in W, f_S3(n) = 1/3 - 1/(3 * 10^n) != 1/3.
>
> and this one's pretty simple.
>
> f_S3(0) = 0 = 1/3 - 1/3 != 1/3
>
> If f_S3(n) = 1/3 - 1/(3 * 10^n), then
>
> f_S3(n+1) = (10^(n+1) - 1) / (3 * 10^(n+1))
> = (10^(n+1) - 10 + 9) / (10 * (3 * 10^n))
> = (10^n - 1) / (3 * 10^n) + 3 / (10^(n+1))
> = f_S3(n) + 9/(3 * 10^(n+1))
> = 1/3 - 1/(3 * 10^n) + 9/(3 * 10^(n+1))
> = 1/3 - 10/(3 * 10^(n+1)) + 9/(3 * 10^(n+1))
> = 1/3 - 1/(3 * 10^(n+1))
>
> Therefore, f_S3(n) != 1/3, and 1/3 is not in S_3. But one
> can come up with elements that are arbitrairly close;
> if you pick an epsilon > 0, I can pick an N = ceil(-log10(3*epsilon)).
> If n > N, then 1/3 - f_S3(n) < epsilon.
>
> >
> > don't tell me to ask basic questions correctly you've made a total mess of.
> >
> > WHAT IS THE SUM OF NUMBERS ON THE BOARD?
> >
> > WHY HAVE YOU WRITTEN THIS FIGURE IS 50?
>
> I take it you're claiming that, because the halting problem
> requires itself to make itself unsolvable, that the
> problem is improperly specified?
>
> Let's explore this.
>
> After all, if one has a series of numbers on a
> hypothetical blackboard_0, contained within
> another blackboard which I'll call blackboard 1:
>
> +-------blackboard 1---------+
> | twenty-nine |
> | four |
> | |
> | +-----blackboard 0-----+ |
> | | | |
> | | 5 33 | |
> | | | |
> | | The sum of numbers | |
> | | on blackboard 0 | |
> | | is _50_ | |
> | | | |
> | | | |
> | | 12 | |
> | +----------------------+ |
> | |
> | seventeen-six |
> | |
> | The sum of numbers on |
> | blackboard 1 is |
> | _one hundred fourty-four_ |
> | |
> +----------------------------+
>
> one can ask with some puzzlement as
> to whether the proper answer to the question
>
> "What is the sum of numbers on blackboard 0?"
>
> is one of the following:
>
> [1] 50, because blackboard 1 contains no numerals,
> and the _50_ is an answer, not part of the
> numeric set.
>
> [2] 100, because _50_ is a perfectly valid numeral
> and therefore should be added.
>
> [3] 14 = 5+3+3+1+2
>
> [4] 19 = 5+3+3+1+2+5+0
>
> And then there's the items on blackboard 1. If we
> take out blackboard 0, are the rest of the items:
>
> [A] 0, because there are no numerals?
>
> [B] 4+29+17+6 = 56?
>
> [C] 4+29+(17-6) = 44?
>
> [D] [B]+144?
>
> [E] [C]+144?
>
> [F] [D]+1 (look carefully; there's a 1 just after "blackboard" :-) )
>
> and so on. Or perhaps we should *not* take out
> blackboard 0, as it's part of blackboard 1.
> But never mind.
>
> We will now erase the blackboard and come back to the
> real coverage problem. For purposes of this argument I
> will make the following (rather silly) claim.
>
> The set T_10 = {k / 10^n: k in W, n in W, 0 <= k < 10^n}
> contains all reals in the interval [0,1).
>
> This claim is different from yours...but I don't see that much
> of a difference.
>
> The basis of this claim is fairly simple. If r is a real number
> in [0,1) then it has a decimal expansion. All prefixes of r are
> in T_10, by construction. (A prefix of length n of r
> is simply r's first n digits, not counting the initial "0.",
> and can be mathematically expressed as floor(r * 10^n) / 10^n.)
>
> Hence r is in ... no, wait, there is a problem. Is 1/3 in T_10?
> Lessee.
>
> If 1/3 is in T_10, then 1/3 = k / 10^n for some k and n in W.
> Therefore 3 * (1/3) = 3 * k / 10^n,
> 1 = 3 * k / 10^n, or 10^n = 3 * k.
>
> However, no positive power of 10 is a multiple of 3, and one can
> prove this by induction, as I shall do here:
>
> Obviously 10^0 = 1 is not a multiple of 3.
>
> If 10^n = 1 (mod 3), then 10^(n+1) = 10 = 10 - 9 = 1 (mod 3).
>
> Hence by induction 10^n = 1 (mod 3) for all n in W, and
> 1/3 is not in T_10.
>
> "But wait!" you might holler in agony. "You didn't use the
> Cantor Diagonalization Argument!"
>
> Certainly I didn't. However, it turns out I can.
> One can order T_10 in the following way, and mark
> out the diagonals (here, I've used []). The
> precise specification of this ordering I'll leave
> to the interested reader; it gets a bit involved.
>
> T_10 = {
> 0.[0]00000....
> 0.1[0]0000....
> 0.20[0]000....
> 0.300[0]00....
> 0.4000[0]0....
> 0.50000[0]....
> ...
> 0.010000....
> 0.110000....
> 0.210000....
> ...
> 0.020000....
> 0.120000....
> ...
> 0.990000....
> 0.001000....
> 0.101000....
> ...
> }
>
> 1/3 makes a perfectly good Cantor-type antidiagonal number for T_10,
> in this ordering. Admittedly, this doesn't prove all that much,
> beyond illustrating some flaws in your set inclusion logic.
> Besides, we knew that already.
>
> "But hold on!", you could wail in anguish, "You've not defined
> equals yet!"
>
> True. However, one can use some fairly conventional definitions
> here; one can either define equals in Q by extending equals in J:
>
> a/b = c/d iff ad = bc
>
> or by using the quantifier and infinite expansions:
>
> r = s iff (An)(isNatural(n) . r[n] = s[n])
>
> where r[n] is the n'th digit in r's expansion. (There are
> some issues regarding 0.999... and 1.000... which aren't
> all that easily tackled here -- but it's not usually that
> much of a problem for the Cantor diagonal argument, and
> it's not a big difficulty here, either.)
>
> I'll leave it to the interested reader to continue going
> down the steps back to Peano; the idea of '=' over the
> integers is very unambiguous.
>
> Now, since all of 1/3's digits are in fact '3', and
> the diagonal number above has all zeroes, for any
> number in T_10 in the given ordering, there's at least
> one digit that is unequal -- the n'th one. All the
> rest of them might very well be equal (unlikely anyway)
> but one digit being unequal is generally enough, absent
> the 0.999... = 1.000... issue, which doesn't apply here.
>
> "But wait a minute!" you may yell in excruciating pain.
> "All prefixes of the generated number are in the list!
> It's got to be in there!"
>
> Really?
>
> Let's revisit 1/3 and S_3 = {.3, .33, .333, ... }. All prefixes
> of 1/3 are of the form .333...3 . Therefore all prefixes
> of 1/3 are in S_3. However, 1/3 is not.
>
> At this point you might let out a primal scream.
> It might make you feel better. :-)
>
> I'll (re)tackle the halting problem another time.
>
> >
> >
> >
> > "Ralph Hartley" <hartley@aic.nrl.navy.mil> wrote in
> >> The Ghost In The Machine wrote: <snip>
> >>
> >> I'm afraid I didn't understand this at all.
> >>
> >> What are you trying to prove?
> >>
> >> Ralph Hartley
> >
> >
> > Time to wake up and drool to the accuphase you incoherent babbler
>
> What we have here is a failure to communicate.

If you post 300 lines in 10 minutes you have a failure to have relevant thoughts.
does anyone understand this babble?

what is the sum of numbers on the board? THATS THE QUESTION
your answer, in numerical form was a 10,000 digit number, WRONG!

you can't draw a picture of a board on the board and exclude those numbers
from the real board.

1/3 itself is equivalent to <0.3, 0.33, 0.333 ................>

That means if you map the infinite set to the number line 1/3 is drawn.
I have no idea why none of you can cope with simple facts like this,
remember set-minus, set at a time reasoning for the simple 3GL sci.mathers?

Herc

Relevant Pages

  • Re: ******* TRY THESE SCI.MATH **********
    ... > another blackboard which I'll call blackboard 1: ... > contains all reals in the interval [0,1). ... > Cantor Diagonalization Argument!" ... > "All prefixes of the generated number are in the list! ...
    (sci.logic)
  • Re: ******* TRY THESE SCI.MATH **********
    ... > another blackboard which I'll call blackboard 1: ... > contains all reals in the interval [0,1). ... > Cantor Diagonalization Argument!" ... > "All prefixes of the generated number are in the list! ...
    (sci.math)
  • Re: ******* TRY THESE SCI.MATH **********
    ... >> And then there's the items on blackboard 1. ... > what is the sum of numbers on the board? ... set up an equivalence of the reals with Cauchy sequences, ... S setMinus r = that set T such that every element ...
    (sci.math)
  • Re: ******* TRY THESE SCI.MATH **********
    ... >> And then there's the items on blackboard 1. ... > what is the sum of numbers on the board? ... set up an equivalence of the reals with Cauchy sequences, ... S setMinus r = that set T such that every element ...
    (sci.logic)
  • Re: ******* TRY THESE SCI.MATH **********
    ... >> And then there's the items on blackboard 1. ... > what is the sum of numbers on the board? ... set up an equivalence of the reals with Cauchy sequences, ... S setMinus r = that set T such that every element ...
    (comp.theory)