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

From: The Ghost In The Machine (ewill_at_sirius.athghost7038suus.net)
Date: 01/29/05 

Date: Sat, 29 Jan 2005 11:00:11 GMT

In sci.logic, |-|erc
<H@r.c>
 wrote
on Sat, 29 Jan 2005 15:17:18 +1000
<360kloF4rqefuU1@individual.net>:
> "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

What are the sums of numbers on these boards?

   +-----blackboard 1-----+
   | |
   | 5 33 |
   | |
   | The sum of numbers |
   | on blackboard 1 |
   | is _50_ |
   | |
   | 12 |
   +----------------------+

   +-----blackboard 2-----+
   | |
   | 5 33 |
   | |
   | The sum of numbers |
   | on blackboard 2 |
   | is _11_ |
   | |
   | 12 |
   +----------------------+

   +-----blackboard 3-----+
   | |
   | 5 33 |
   | |
   | The sum of numbers |
   | on blackboard 3 |
   | is _N_ |
   | |
   | 12 |
   +----------------------+

   +-----blackboard 4-----+
   | |
   | 5 33 |
   | |
   | This problem is |
   | hopelessly |
   | muddled. |
   | |
   | 12 teechur-> xx |
   +----------------------+

> 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 ................>

Is 1/3 equivalent to <0.31, 0.331, 0.3331, 0.33331, ... > as well?
How about <0.2, 0.29, 0.329, 0.3329, ... >? Perhaps
<1/5, 2/8, 3/11, 4/14, 5/17, 6/20, 7/23, ...> ?

I'm not entirely sure what you're trying to do here but it
looks like a variant of Cauchy sequences. One can of course
set up an equivalence of the reals with Cauchy sequences,
but if one associates 1/3 with a Cauchy sequence (a set),
1/3 need not be *in* the set, and in fact provably is not
in your case.

>
> 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?

S setMinus r =df
    T: (Ax)(Ay)( (x in S) . (y in T) => abs(x - y) = r)

if I'm reading
<NnmHc.85021$sj4.44690@news-server.bigpond.net.au>
correctly (found in Google Groups).

In short, S setMinus r = that set T such that every element
in T is r away from every element in S.

This is clearly impossible as written (take the two sets
X = {1,2,3,4,5, ...} and Y = {0.5, 1.5, 2.5,...} and try
to work out whether each pair of elements (x,y) can be
0.5 apart); I must assume it's more along the lines of
the simpler definition

S setMinus r =df {s - r: (As)(s in S)}

which means that for each element in the result there's an element
in S it's r away from.

Which one did you mean?

With that modification, {.3, .33, .333, ...} setMinus 1/3
yields {-1/30, -1/300, -1/3000, -1/30000, ...},
unless you really wanted

S setMinus r =df
    t: (Ae)(e > 0 => (Es)(d > 0 . s in S . abs(s - r - t) < e))

which would mean that {.3, .33, .333, ... } setMinus 1/3 = 0.

>
> Herc
> 

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

Relevant Pages

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