| Topic: |
Science > Physics |
| User: |
"HERC777" |
| Date: |
13 May 2005 04:54:39 PM |
| Object: |
A simple undiagonalisable list - ILLUSTRATED |
FIXED FONT
0.1-----1......1..........1.........1
0.|2....|..2......2......2......2....
0.|.3...|.3...3.......3......3.......
0.|..4..|...4....4.......4.....4.....
0.|...5.|..5........5.......5........
0.|....6|......6.......6..........6..
0.7-----7.....7...7..........7.......
0........8--8------8.........8.......
0........|9.....9..|.........9.......
0........|.1---1...|......1.....1....
0........|.|2..|...|2.........2......
0........|.|.3.|...|..3..........3...
0........|.|..4|...|.......4..4......
0........|.5---5...|...5....5........
0........|......6..|....6.......6....
0........|.......7.|......7......7...
0........|........8|...8..........8..
0........9---------9........9..9.....
4DP <=>
where w = the natural numbers
and D = the decimal digits {0..9},
AnEcAd1d2[ new & cew & d1eD & d2eD & (d1=/=d2) & (c=/=d) ]
-> L[n,n]=d1
& L[c,c]=d2
& L[n,c]=d1
& L[c,n]=d2
This is an illustration of what I call the Quad-Digit-Property, or 4DP.
It's just a typical diagonal where each diagonal digit is repeated
numerous times in that real and is part of numerous 'squares' with
corresponding diagonal digits. In fact there must be atleast 9 squares
for each diagonal digit, one for each of {0..9}.
~
WHAT IS HE TALKING ABOUT UNDIAGONALISABLE LIST?
I CAN STILL MAKE A DIAGONAL TO EVERY LIST HE MAKES!!!
Let's play a game called Cantors list. I tell you the diagonal of my
list and you tell me a real that's not on the my list. Want to play?
DIAG = 0.135791357913579...
What's your real?
YOU : 0.246802468024680..
OK, you win, that's not on my list. But this time, I'll tell you what
class of sets I'm using. All the members are fractions of ninths.
My list was actually
0.1111111..
0.3333333..
0.5555555..
0.7777777..
0.9999999..
0.1111111..
hence the diagonal was 0.1357913579...
Lets try again using only the ninths fractions.
DIAG = 0.12345671234567...
What's your real?
Rather than EXPLAIN EVERYTHING, using only a single DIAG->ANTIDIAG
function, what reals can you come up with that are not on my list?
Herc
.
|
|
| User: "Randy Poe" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:01:31 PM |
|
|
HERC777 wrote:
WHAT IS HE TALKING ABOUT UNDIAGONALISABLE LIST?
I CAN STILL MAKE A DIAGONAL TO EVERY LIST HE MAKES!!!
Let's play a game called Cantors list. I tell you the diagonal of my
list and you tell me a real that's not on the my list. Want to play?
DIAG = 0.135791357913579...
What's your real?
YOU : 0.246802468024680..
Why do you insist on thinking there's only one choice
of what you call "anti-diagonal"?
What's wrong with this:
0.203684706895286...
or this:
0.4444444.....
or this:
0.626262626262...
or this:
0.00000000....
or any of the other uncountably many numbers which
differ in the n-th position from the n-th number of
your list, for all n?
OK, you win, that's not on my list. But this time, I'll tell you what
class of sets I'm using. All the members are fractions of ninths.
My list was actually
0.1111111..
0.3333333..
0.5555555..
0.7777777..
0.9999999..
0.1111111..
hence the diagonal was 0.1357913579...
Lets try again using only the ninths fractions.
DIAG = 0.12345671234567...
No it's not. If the 4th number in your list is 0.7777...
then the 4th digit of the diagonal is not 4.
- Randy
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:15:58 PM |
|
|
for the purposes of illustration
using only a single DIAG->ANTIDIAG
function, what reals can you come up with that are not on my list?
This is for a new list (but of the same type, repeating digits only).
DIAG = 0.12345671234567...
Anyone ?
Herc
.
|
|
|
| User: "Patricia Shanahan" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:40:22 PM |
|
|
HERC777 wrote:
for the purposes of illustration
using only a single DIAG->ANTIDIAG function, what
reals can you come up with that are not on my list?
This is for a new list (but of the same type, repeating
digits only).
DIAG = 0.12345671234567...
Anyone ?
Herc
I'm not clear about what proposition you are trying
to prove. Here are a couple of alternatives:
1. For every x in R, there exists a function f:N->R such
that x=f(n) for some n in N.
2. There exists a function f:N->R such that, for every x in
R, x=f(n) for some n in N.
If you are trying to prove one of these, please indicate
which. If you are trying to prove something else, please
state clearly what it is.
Thanks,
Patricia
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:56:33 PM |
|
|
I am proving that, given our current knowledge,
~proof(~[2])
Herc
.
|
|
|
|
|
| User: "Proginoskes" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:47:27 PM |
|
|
HERC777 wrote:
for the purposes of illustration
using only a single DIAG->ANTIDIAG
function, what reals can you come up with that
are not on my list?
This is for a new list (but of the same type,
repeating digits only).
DIAG = 0.12345671234567...
Anyone ?
0.2345678234567823456782345678...
That real number's not on your list.
If you reshuffle the reals on your list (k/9, where k is an integer and
where each k occurs an infinite amount of times), you can come up with
this diagonal (or any other, by using an appropriate subset), but that
doesn't change the fact that the actual number still isn't ON YOUR
(allegedly complete) LIST, and that's what Cantor's proof sets out to
do.
And isn't shuffling the items on the list like that game where you say
an integer out loud, then I say an integer out loud, and whoever says
the larger integer wins?
If you're not familiar with the subtleties of it, I'll give you the
courtesy of moving first. What is your integer? (Reply to this post
instead of saying it out loud, so that we can have an undisputable
record.)
--- Christopher Heckman
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 06:03:24 PM |
|
|
Step right up young man, the rules of the game are...
RULE 1 Only use the diag function anti(digit) =3D digit+1
(if digit>9 then digit=3D0)
[Herc]
The list only contains repeating digits
DIAG =3D 0.12345671234567...
Anyone ?
[Proginoskes]
0=2E2345678234567823456782345678=AD...
That real number's not on your list.
Right! And now for the major prize, while still abiding by the rules,
can you tell me another?
(hint : what is the list?)
Herc
.
|
|
|
| User: "Proginoskes" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 12:56:51 AM |
|
|
HERC777 wrote:
Step right up young man, the rules of the game are...
RULE 1 Only use the diag function anti(digit) =3D digit+1
(if digit>9 then digit=3D0)
[Herc]
The list only contains repeating digits
DIAG =3D 0.12345671234567...
Anyone ?
[Proginoskes]
0.2345678234567823456782345678=AD...
That real number's not on your list.
Right! And now for the major prize, while still abiding
by the rules, can you tell me another?
(hint : what is the list?)
By swapping the first two items on the list, you get:
1: 0.2222...
2: 0.1111...
3: 0.3333...
4: 0.4444...
..=2E.
So using the antidiagonal rule on this new list,
0=2E32456782345678923456789..., which is another real number which is not
on your new list, and since the new list contains exactly the same
number.
By suitably rearranging the items on the list, you can use the
antidiagonal process to get any antidiagonal that has an infinite
number of 2's, 3's, 4's, 5's, 6's, 7's, 8's and 9's on it.
By knowing what your list is, I can do even better:
THEOREM: If r is a real number with at least two distinct digits in its
base-10 expansion, then r is not on the list:
1: 0.1111...
2: 0.2222...
3: 0.3333...
4: 0.4444...
..=2E.
8: 0.8888...
9: 0.9999...
10: 0.1111...
11: 0.2222...
..=2E.
(BONUS QUESTION: What is the google-plex-th (10^(10^100)th) item on the
list?)
Looks like he's bolted again! :)
No, I got done with work (entering grades for my classes,
specifically), did some shopping, had supper, got home, watched
"NUMB3RS", and logged back on line. Have you been online the whole
time, then?
And why don't you want to play my game where we name numbers and
whoever names the bigger number wins?
Don't be shy, say you there! Walk Right Up Walk RIght Up
see if you can tell me TWO antidiagonals given the diagonal
0.12345671234567... using only the one antidiag function!
You can compose the antidiagonal funciton twice. (You haven't stated
whether this is one of your rules or not, though.) This would give you
0=2E34567893456789... Composing it three times gives
0=2E45678904567890... So I can get 9 antidiagonals out of one diagonal.
Why are you worried about how many antidiagonals there are which are
not on a given list, anyway? For Cantor's proof, you only need one to
show the list is incomplete.
--- Christopher Heckman
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 01:11:48 AM |
|
|
So using the antidiagonal rule on this new list
0.32456782345678923456789...
there's no 9 at the end but partial credit since you showed working.
what other classes of sets have the property that only given the
diagonal you can rearrange the digits of the diagonal in order to form
an antidiagonal?
Herc
.
|
|
|
| User: "Mike Terry" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 05:40:29 PM |
|
|
"HERC777" <herc777@hotmail.com> wrote in message
news:1116051108.402818.326940@z14g2000cwz.googlegroups.com...
So using the antidiagonal rule on this new list
0.32456782345678923456789...
there's no 9 at the end but partial credit since you showed working.
what other classes of sets have the property that only given the
diagonal you can rearrange the digits of the diagonal in order to form
an antidiagonal?
Herc
Herc, I've given a full answer to this question elsewhere in this thread, so
now you can just go there and continue with the next stage of you master
plan... :-)
Mike.
.
|
|
|
|
| User: "Proginoskes" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 01:33:21 AM |
|
|
HERC777 wrote:
So using the antidiagonal rule on this new list
0.32456782345678923456789...
there's no 9 at the end but partial credit since you
showed working.
what other classes of sets
Are "classes of sets" supposed to be the same as "lists"?
have the property that only given the diagonal you can
rearrange the digits of the diagonal in order to form
an antidiagonal?
If you have a list where each item is k/9, where k is an integer
between 0 and 9, it should work there, too. If by "rearrange" you mean
"change the order of", then there are a lot more, and it becomes a
tricky problem.
I'll have to think about it a bit.
--- Christopher Heckman
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 02:19:47 AM |
|
|
Are "classes of sets" supposed to be the same as "lists"?
I think it has to be a property of sets, because if you gave the list
the property with a particular ordering, then shuffling the diagonal
might knock it out from having that property, so you might only get one
digit swap.
i.e. the property should hold for all permutations of elements of the
list.
Herc
.
|
|
|
|
|
|
| User: "Virgil" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 02:39:06 PM |
|
|
In article <1116050210.997292.199740@f14g2000cwb.googlegroups.com>,
"Proginoskes" <proginoskes@email.msn.com> wrote:
HERC777 wrote:
Step right up young man, the rules of the game are...
RULE 1 Only use the diag function anti(digit) = digit+1
(if digit>9 then digit=0)
[Herc]
The list only contains repeating digits
DIAG = 0.12345671234567...
Anyone ?
[Proginoskes]
0.2345678234567823456782345678...
That real number's not on your list.
Right! And now for the major prize, while still abiding
by the rules, can you tell me another?
(hint : what is the list?)
By swapping the first two items on the list, you get:
1: 0.2222...
2: 0.1111...
3: 0.3333...
4: 0.4444...
...
So using the antidiagonal rule on this new list,
0.32456782345678923456789..., which is another real number which is not
on your new list, and since the new list contains exactly the same
number.
By suitably rearranging the items on the list, you can use the
antidiagonal process to get any antidiagonal that has an infinite
number of 2's, 3's, 4's, 5's, 6's, 7's, 8's and 9's on it.
By knowing what your list is, I can do even better:
THEOREM: If r is a real number with at least two distinct digits in its
base-10 expansion, then r is not on the list:
1: 0.1111...
2: 0.2222...
3: 0.3333...
4: 0.4444...
...
8: 0.8888...
9: 0.9999...
10: 0.1111...
11: 0.2222...
...
(BONUS QUESTION: What is the google-plex-th (10^(10^100)th) item on the
list?)
Looks like he's bolted again! :)
No, I got done with work (entering grades for my classes,
specifically), did some shopping, had supper, got home, watched
"NUMB3RS", and logged back on line. Have you been online the whole
time, then?
And why don't you want to play my game where we name numbers and
whoever names the bigger number wins?
Don't be shy, say you there! Walk Right Up Walk RIght Up
see if you can tell me TWO antidiagonals given the diagonal
0.12345671234567... using only the one antidiag function!
You can compose the antidiagonal funciton twice. (You haven't stated
whether this is one of your rules or not, though.) This would give you
0.34567893456789... Composing it three times gives
0.45678904567890... So I can get 9 antidiagonals out of one diagonal.
Why are you worried about how many antidiagonals there are which are
not on a given list, anyway? For Cantor's proof, you only need one to
show the list is incomplete.
--- Christopher Heckman
In fact if g is any bijection, g:N ->N and f:N -> R is any listing of
reals, then the "antidiagonal" for the composition of functions fg: N ->
R, produces another anti-diagonal, though not all such fg antidiagonals
need be distinct.
.
|
|
|
| User: "The Ghost In The Machine" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 06:00:03 PM |
|
|
In sci.logic, Virgil
<ITSnetNOTcom#virgil@COMCAST.com>
wrote
on Sat, 14 May 2005 13:39:06 -0600
<ITSnetNOTcom#virgil-C1E68A.13390614052005@comcast.dca.giganews.com>:
In article <1116050210.997292.199740@f14g2000cwb.googlegroups.com>,
"Proginoskes" <proginoskes@email.msn.com> wrote:
HERC777 wrote:
Step right up young man, the rules of the game are...
RULE 1 Only use the diag function anti(digit) = digit+1
(if digit>9 then digit=0)
[Herc]
The list only contains repeating digits
DIAG = 0.12345671234567...
Anyone ?
[Proginoskes]
0.2345678234567823456782345678...
That real number's not on your list.
Right! And now for the major prize, while still abiding
by the rules, can you tell me another?
(hint : what is the list?)
By swapping the first two items on the list, you get:
1: 0.2222...
2: 0.1111...
3: 0.3333...
4: 0.4444...
...
So using the antidiagonal rule on this new list,
0.32456782345678923456789..., which is another real number which is not
on your new list, and since the new list contains exactly the same
number.
By suitably rearranging the items on the list, you can use the
antidiagonal process to get any antidiagonal that has an infinite
number of 2's, 3's, 4's, 5's, 6's, 7's, 8's and 9's on it.
By knowing what your list is, I can do even better:
THEOREM: If r is a real number with at least two distinct digits in its
base-10 expansion, then r is not on the list:
1: 0.1111...
2: 0.2222...
3: 0.3333...
4: 0.4444...
...
8: 0.8888...
9: 0.9999...
10: 0.1111...
11: 0.2222...
...
(BONUS QUESTION: What is the google-plex-th (10^(10^100)th) item on the
list?)
Looks like he's bolted again! :)
No, I got done with work (entering grades for my classes,
specifically), did some shopping, had supper, got home, watched
"NUMB3RS", and logged back on line. Have you been online the whole
time, then?
And why don't you want to play my game where we name numbers and
whoever names the bigger number wins?
Don't be shy, say you there! Walk Right Up Walk RIght Up
see if you can tell me TWO antidiagonals given the diagonal
0.12345671234567... using only the one antidiag function!
You can compose the antidiagonal funciton twice. (You haven't stated
whether this is one of your rules or not, though.) This would give you
0.34567893456789... Composing it three times gives
0.45678904567890... So I can get 9 antidiagonals out of one diagonal.
Why are you worried about how many antidiagonals there are which are
not on a given list, anyway? For Cantor's proof, you only need one to
show the list is incomplete.
--- Christopher Heckman
In fact if g is any bijection, g:N ->N and f:N -> R is any listing of
reals, then the "antidiagonal" for the composition of functions fg: N ->
R, produces another anti-diagonal, though not all such fg antidiagonals
need be distinct.
Maybe not, but it's very easy to produce a countable number of
such diagonals. Let the list be, just for giggles:
1: 0.400000000
2: 0.030000000
3: 0.003000000
4: 0.000300000
....
and one possible antidiagonal is of course .3444... . However, one can
put that 3 anywhere in the antidiagonal if one reorders the list
(though the reasoning is slightly silly, as the only digits here are
0, 3, and 4); none of these numbers are anywhere in the *set*, though
if the set is properly dense one can get as close as one desires.
One such set is my (?) test set T_b = {k/b^n: k,b in N, k < b^n}.
This set is dense over [0,1) and has all of HERC's desired properties
as far as I can tell, but it does not cover the rationals
(e.g., 1/(b+1) is not in the set, and if b > 2 1/(b-1) isn't, either),
let alone the reals.
Of course, once one has generated a countable number of diagonals,
it's easy enough to diagonalize *that*. Here's one generator;
if one has a list L: N -> [0,1), then one can generate a family
of diagonals (another list, basically) g_n(L), for any nonnegative
integer n:
g_n(L)(i) = 3 if L(i)(i) != 3 and floor(n/2^i) is odd
g_n(L)(i) = 4 if L(i)(i) == 3 and floor(n/2^i) is odd
g_n(L)(i) = 5 if L(i)(i) != 5 and floor(n/2^i) is even
g_n(L)(i) = 6 if L(i)(i) == 5 and floor(n/2^i) is even
and then one can generate *another* number h:
h(i) = 3 if g_i(L)(i) != 3 and L(i)(i) != 3
h(i) = 4 if g_i(L)(i) == 3 and L(i)(i) != 4
h(i) = 5 if g_i(L)(i) == 3 and L(i)(i) == 4
h is clearly not one of the g_n's, but it also is a diagonal of L.
This process can be repeated indefinitely, though one might have
to diagonalize bigger "chunks" of the number (take N digits at
a time, rathre than 1) as one goes on.
There's an awful lot of diagonal numbers for a list... :-)
--
#191,
It's still legal to go .sigless.
.
|
|
|
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 06:53:35 PM |
|
|
Looks like he's bolted again! :)
Don't be shy, say you there! Walk Right Up Walk RIght Up see if you
can tell me TWO antidiagonals given the diagonal
0.12345671234567... using only the one antidiag function!
anyone?
Herc
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 11:26:45 PM |
|
|
You wonder why I start new threads when everyone ignores the points I
make.
USING THE FUNCTION NEWDIGIT = OLDDIGIT+1 mod10
CAN YOU CALCULATE A MISSING REAL FROM THIS LIST?
0.333333333...
0.444444444...
0.555555555...
0.333333333...
0.444444444...
0.555555555...
0.333333333...
....
YES THEY GET THIS FAR
0.456456456...
RIGHT!
GIVEN THE DIAGONAL IS 0.123123123
(and I'm still only using recurring digit reals)
CAN YOU CALCULATE A MISSING REAL ?
YES THATS EASY
0.234234234...
THIS ONE NOBODY CAN DO?
GIVEN THE DIAGONAL IS 0.123123123
(and I'm still only using recurring digit reals)
CAN YOU CALCULATE ****ANOTHER**** MISSING REAL ?
HINT : WHAT IS THE LIST?
THE DIAGONAL IS 0.123123123...
THE NUMBERS ARE ALL RECURRING DIGITS.
THE LIST IS
0.1111111..
0.2222223..
0.3333333..
0.1111111..
0.2222222..
0.3333333..
...
USING THE FUNCTION NEWDIGIT = OLDDIGIT+1 mod10
CAN YOU CALCULATE A MISSING REAL FROM THIS LIST?
0.23423423.. IS ONE WHATS ANOTHER?
* REARANG|E THE FUCKING LIST *
0.111..
0.222..
0.333..
...
=
0.333..
0.222..
0.111..
....
MISSING REAL = 0.432432432...
THAT FUCKING HARD. 10 minutes blackboard demo to 15 year olds takes 2
weeks around here.
Herc
.
|
|
|
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 05:35:07 PM |
|
|
HERC777 wrote:
for the purposes of illustration
using only a single DIAG->ANTIDIAG
function, what reals can you come up with that are not on my list?
This is for a new list (but of the same type, repeating digits only).
DIAG = 0.12345671234567...
function Y = ANTIDIAG(DIAG)
Let X = decimal expansion of pi.
for n=1,2,3,.....
let x_n = n-th digit of X.
let d_n = n-th digit of DIAG.
if (x_n == d_n)
set y_n = (d_n + 1) mod 10
else
set y_n = x_n
endif
end
return Y.
------------------
For that matter, I don't need pi or the if test, I
was just being cute, so I'd generate a non-repeating
decimal from your repeating one.
Here's a simpler version:
function Y = ANTIDIAG(DIAG)
for n=1,2,3,.....
let d_n = n-th digit of DIAG.
set y_n = (d_n + 1) mod 10
end
return Y.
----------------------------
I guarantee that neither Y will be on your list.
- Randy
.
|
|
|
| User: "Poker Joker" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 07:43:55 PM |
|
|
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:1116023706.980292.39930@g44g2000cwa.googlegroups.com...
function Y = ANTIDIAG(DIAG)
Let X = decimal expansion of pi.
for n=1,2,3,.....
let x_n = n-th digit of X.
let d_n = n-th digit of DIAG.
if (x_n == d_n)
set y_n = (d_n + 1) mod 10
else
set y_n = x_n
endif
end
return Y.
------------------
For that matter, I don't need pi or the if test, I
was just being cute, so I'd generate a non-repeating
decimal from your repeating one.
Here's a simpler version:
function Y = ANTIDIAG(DIAG)
for n=1,2,3,.....
let d_n = n-th digit of DIAG.
set y_n = (d_n + 1) mod 10
end
return Y.
Are those Y's uncomputable?
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 08:17:21 PM |
|
|
=> Here's a simpler version:
=>
=> function Y = ANTIDIAG(DIAG)
=> for n=1,2,3,.....
=> let d_n = n-th digit of DIAG.
=> set y_n = (d_n + 1) mod 10
=> end
=>
=> return Y.
Are those Y's uncomputable?
I didn't notice that, the return is outside of the functions scope.
Its still computable using lazy parameter evaluation, it just gives the
required digits as they are needed by whatever function calls them.
defun y (stream) (cons (add (head stream) 1) (y (tail stream)))
first ( 5 (y "123456789" )) = "23456"
Herc
.
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 08:17:51 PM |
|
|
Poker Joker wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:1116023706.980292.39930@g44g2000cwa.googlegroups.com...
function Y = ANTIDIAG(DIAG)
for n=1,2,3,.....
let d_n = n-th digit of DIAG.
set y_n = (d_n + 1) mod 10
end
return Y.
Are those Y's uncomputable?
I don't think so.
Certainly the second one isn't, since it's a repeating
decimal with exactly the same period as Herc's, and I
already know in advance what each digit is. I'm just
replacing 1->2, 2->3, ... 8->9, 9->0 on Herc's diagonal.
And the offset can be anything, it needn't be 1. Nor is
that the only simple substitution rule. I could replace
pairs of numbers by other pairs for instance. I could
use mod 8 arithmetic. There are literally infinitely
many possibilities.
- Randy
.
|
|
|
| User: "Poker Joker" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 08:42:23 PM |
|
|
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:1116033471.531347.167690@g49g2000cwa.googlegroups.com...
Poker Joker wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:1116023706.980292.39930@g44g2000cwa.googlegroups.com...
function Y = ANTIDIAG(DIAG)
for n=1,2,3,.....
let d_n = n-th digit of DIAG.
set y_n = (d_n + 1) mod 10
end
return Y.
Are those Y's uncomputable?
I don't think so.
Certainly the second one isn't, since it's a repeating
decimal with exactly the same period as Herc's, and I
already know in advance what each digit is. I'm just
replacing 1->2, 2->3, ... 8->9, 9->0 on Herc's diagonal.
And the offset can be anything, it needn't be 1. Nor is
that the only simple substitution rule. I could replace
pairs of numbers by other pairs for instance. I could
use mod 8 arithmetic. There are literally infinitely
many possibilities.
- Randy
How does your algorithm return a 'Y' value if the list is made up
of uncomputable reals?
.
|
|
|
| User: "Randy Poe" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 10:08:34 PM |
|
|
Poker Joker wrote:
How does your algorithm return a 'Y' value if the list is made up
of uncomputable reals?
It doesn't. I thought you were referring to Herc's
"undiagonalizable" list of repeating decimals. The
algorithm was a response to his demand that somebody
produce an algorithm which would generate a number not
on the list.
In general, for Cantor's proof all you need to do is convince
yourself that a number exists which is different from
number n in position n, for all n. I fail to see why
anybody would think this is impossible in general.
- Randy
.
|
|
|
|
|
| User: "Virgil" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 08:26:14 PM |
|
|
In article <1116033471.531347.167690@g49g2000cwa.googlegroups.com>,
"Randy Poe" <poespam-trap@yahoo.com> wrote:
Poker Joker wrote:
"Randy Poe" <poespam-trap@yahoo.com> wrote in message
news:1116023706.980292.39930@g44g2000cwa.googlegroups.com...
function Y = ANTIDIAG(DIAG)
for n=1,2,3,.....
let d_n = n-th digit of DIAG.
set y_n = (d_n + 1) mod 10
end
return Y.
Are those Y's uncomputable?
I don't think so.
Certainly the second one isn't, since it's a repeating
decimal with exactly the same period as Herc's, and I
already know in advance what each digit is. I'm just
replacing 1->2, 2->3, ... 8->9, 9->0 on Herc's diagonal.
And the offset can be anything, it needn't be 1. Nor is
that the only simple substitution rule. I could replace
pairs of numbers by other pairs for instance. I could
use mod 8 arithmetic. There are literally infinitely
many possibilities.
- Randy
Uncountably many!!!
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 08:37:27 PM |
|
|
I still need the puzzle I gave to Proginoskes answered. If I post the
solution you will all be confused, someone else play the list debunker
and answer the quiz.
Given
1/ I am using only reals of the form, 0.000000, 0.111111., 0.22222,..
0.999999
2/ The only digit altering function allowed is CHANGE(digit) = digit +
1 mod 10
3/ The diagonal is 0.123456712345671234567...
Q1/ What obvious real is missing?
Q2/ What other reals are missing?
(hint: reverse engineer what the list is)
Herc
.
|
|
|
| User: "Mike Terry" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 09:31:12 PM |
|
|
"HERC777" <herc777@hotmail.com> wrote in message
news:1116034647.329946.158880@g14g2000cwa.googlegroups.com...
I still need the puzzle I gave to Proginoskes answered. If I post the
solution you will all be confused, someone else play the list debunker
and answer the quiz.
Given
1/ I am using only reals of the form, 0.000000, 0.111111., 0.22222,..
0.999999
2/ The only digit altering function allowed is CHANGE(digit) = digit +
1 mod 10
3/ The diagonal is 0.123456712345671234567...
Q1/ What obvious real is missing?
I guess you want us to say 0.234567823456782345678... ?
Q2/ What other reals are missing?
(hint: reverse engineer what the list is)
Every real other than 1/9, 2/9, 3/9, 4/9, 5/9, 6/9, 7/9.
(Your list consists of only these seven numbers repeated.)
Now what do you think this shows?
Regards,
Mike.
Herc
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
13 May 2005 10:25:40 PM |
|
|
I orignally had.. name another missing real but using the same
antidiag function.
your answer was any real other than 1/9, 2.9...
that includes 0.50000000 as missing from the list.
that does not use the same antidiag function on the diagonal
0.12345671234567.. as in Q1.
therefore that is wrong for breaking rule 2.
use the hint, given the class of reals is stated and the diag is
0.12345671234567... what is the list?
Herc
.
|
|
|
| User: "Mike Terry" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 05:35:19 PM |
|
|
"HERC777" <herc777@hotmail.com> wrote in message
news:1116041140.841822.249100@f14g2000cwb.googlegroups.com...
I orignally had.. name another missing real but using the same
antidiag function.
your answer was any real other than 1/9, 2.9...
that includes 0.50000000 as missing from the list.
that does not use the same antidiag function on the diagonal
0.12345671234567.. as in Q1.
therefore that is wrong for breaking rule 2.
Well I wondered what "rule 2" was supposed to mean - "question 2" just asked
what other reals were missing, which I answered! Even now it's not clear to
me what you're asking, so I'll clarify what I think you mean, and you can
agre or not. Then I'll have another go at giving the answer you want...
-----------------------
You have a list L, with a diagonal of 0.12345671234567.., and given the
numbers you have told us you are using, we know exactly what your list must
be.
You want us to find all the possible AntiDiagonals of ... what?... all the
different permutations applied to your list? (This is what I think you
mean.)
To be clear, a permutation of a list (x[i]) is a new list (x[p(i)]) where p
is a permutation of N = {1,2,3,4}.
(And in case you're not familiar with "permutation", a permutation of N is
just a bijection from N to N. And a bijection is a mapping that is 1-1 and
onto. So you could think of a permutation of a set as specifying a
"rearrangement" of a set - every element in the set goes to a new element in
the set, and no two elements go to the same new element, and every element
is "gone to" by some element in the original set. In the case of lists,
every POSITION in the old list has gone to a new position in the new list,
no two old list positions have moved to the same new list position, and
every new list position has come from some old list position.
Sorry for boring you if you already know what a permutation is! ;-))
-----------------------
Is the above what you're asking?
If so, here is a working out of your answer...
First, your list contains infinitely many of each of the elements of the set
S = {0.1111.., 0.2222.., 0.3333.., 0.4444.., 0.5555.., 0.6666.., 0.7777..}
A list L is a permutation of your list iff it consists only of elements of S
and in addition contains infinitely many of each element of S. For the
diagonal this implies it will contain only the digits {1,2,3,4,5,6,7} and
moreover will contain infinitely many of each digit.
So the answer to your question is that the missing numbers obtained by
AntiDiag are precisely the real numbers in [0,1] contain only the digits
{2,3,4,5,6,7,8} and moreover will contain infinitely many of each digit.
Examples:
0.444455888236666744445588823666674444558882366667... is an AntiDiag
0.3434343434343434... is NOT an AntiDiag (but is missing)
(e.g. does not contain infinitely many 2s)
0.2222222222... is NOT an AntiDiag (and is not missing)
(e.g. does not contain infinitely many 5s)
0.23456788888888888888888888... (only 8s repeating) is NOT an AntiDiag
(e.g. does not contain infinitely many 5s)
0.1234567823456782345678... is NOT an AntiDiag
(contains a 1 digit)
0.5000000000000.. is NOT an AntiDiag (as you pointed out)
(contains a 0 digit)
Of course, to be an AntiDiag, numbers don't need to be periodic like my
first example to be an AntiDiag - just as long as each of the digits
2,3,4,5,6,7,8 appear infinitely often, and no other digits appear.
use the hint, given the class of reals is stated and the diag is
0.12345671234567... what is the list?
See above - now the same question again:
WHAT DO YOU THINK ALL THIS PROVES?
Regards,
Mike.
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
14 May 2005 08:42:48 PM |
|
|
Good answer, and you have stumbled onto the constraint for a reversible
infinite shuffle that was bothering you in the other thread.
The point is I didn't ask...
WHAT ARE THE ANTIDIAGONALS OF PERMUTATIONS OF THE LIST?
I was only IMPLYING that.
And all I told you was the diagonal, and the class of set I was using.
Corrolory : for certains classes of sets, the diagonal can be shuffled.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Any other classes of set this works for?
;
;
;
;
;
;
;
;
I really think this opens a tin of worms about the process of
diagonalisation, Cantors missing real is not only
0.1347654567098762255654567654...
Its 0. [0|1|2|3|4|5|6|7|8|9] [0|1|2|3|4|5|6|7|8|9]
[0|1|2|3|4|5|6|7|8|9]...
That's what I define as a NoNumber!
Herc
.
|
|
|
| User: "Mike Terry" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
15 May 2005 08:20:52 AM |
|
|
"HERC777" <herc777@hotmail.com> wrote in message
news:1116121368.840827.283720@g43g2000cwa.googlegroups.com...
Good answer, and you have stumbled onto the constraint for a reversible
infinite shuffle that was bothering you in the other thread.
I haven't stumbled on anything, as I can see quite clearly where I'm going
thank you! :)
The point is I didn't ask...
WHAT ARE THE ANTIDIAGONALS OF PERMUTATIONS OF THE LIST?
I was only IMPLYING that.
And all I told you was the diagonal, and the class of set I was using.
Corrolory : for certains classes of sets, the diagonal can be shuffled.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
What exactly is this a corrolory of? And what exactly does it mean?
If by "shuffled" you just mean permuted (I recommend you stop saying
"shuffled" and stick with "permuted" as the latter has a clear mathematical
meaning), then any list L can be permuted to get a new diagonal, but in
general the diagonal is not necessarily a permutation of the original
diagonal. Sometimes it is, but so what?
Also there are some lists L, such that if the diagonal is D, then by
suitably permuting L it is possible to generate a list with any diagonal
that is a permutation of D. Again, so what?
(A not very interesting example of such an L:
L = (0.11111.., 0.11111.., 0.11111.., 0.11111.., ...)
D = 0.11111...
So any permutation of D is also D, and the original list L (which is a
trivial permutation of L) has the right diagonal).
None of this presents any contradiction for Cantor's proof.
Any other classes of set this works for?
;
;
;
;
;
;
;
;
I really think this opens a tin of worms about the process of
diagonalisation,
Really? How so?
Cantors missing real is not only
0.1347654567098762255654567654...
Missing real number from what? You're not making sense I'm afraid...
Its 0. [0|1|2|3|4|5|6|7|8|9] [0|1|2|3|4|5|6|7|8|9]
[0|1|2|3|4|5|6|7|8|9]...
For any list L you come up with, this will clearly not be so, and you have
consistently failed to prove any of this. Each time you come up with a new
list permutation construction it turns out to be faulty in some way and
never produces the result you want.
In particular, nothing you have done so far proves your last statement - if
you think it does, then present a mathematical proof again from scratch and
I'll point out where you're going wrong...
Regards,
Mike.
That's what I define as a NoNumber!
Herc
.
|
|
|
| User: "HERC777" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
15 May 2005 09:04:49 AM |
|
|
In particular, nothing you have done so far proves your last statement
- if
you think it does, then present a mathematical proof again from
scratch and
I'll point out where you're going wrong...
Regards,
Mike.
No you won't MIke, you'll make a plebian futile crybaby argument and
the religious sci.math drones will cheer you on to your satisfaction
that you've preserved the faith. Give me a week to take some of you up
to speed on elementary logic before I waste bandwidth with a proof.
FIXED FONT
0=2E1-----1......1..........1...=AD......1
0=2E|2....|..2......2......2....=AD..2....
0=2E|.3...|.3...3.......3......3=AD.......
0=2E|..4..|...4....4.......4....=AD.4.....
0=2E|...5.|..5........5.......5.=AD.......
0=2E|....6|......6.......6......=AD....6..
0=2E7-----7.....7...7..........7=AD.......
0=2E.......8--8------8.........8=AD.......
0=2E.......|9.....9..|.........9=AD.......
0=2E.......|.1---1...|......1...=AD..1....
0=2E.......|.|2..|...|2.........=AD2......
0=2E.......|.|.3.|...|..3.......=AD...3...
0=2E.......|.|..4|...|.......4..=AD4......
0=2E.......|.5---5...|...5....5.=AD.......
0=2E.......|......6..|....6.....=AD..6....
0=2E.......|.......7.|......7...=AD...7...
0=2E.......|........8|...8......=AD....8..
0=2E.......9---------9........9.=AD.9.....
4DP <=3D>
where w =3D the natural numbers
and D =3D the decimal digits {0..9},
AnEcAd1d2[ new & cew & d1eD & d2eD & (d1=3D/=3Dd2) & (c=3D/=3Dd) ]
-> L[n,n]=3Dd1
& L[c,c]=3Dd2
& L[n,c]=3Dd1
& L[c,n]=3Dd2
blame the inconsistency on my logician George, but you get the idea!
Can (an infinite set of) numbers with this property have their diagonal
independantly permuted?
Herc
.
|
|
|
| User: "Mike Terry" |
|
| Title: Re: A simple undiagonalisable list - ILLUSTRATED |
15 May 2005 02:32:04 PM |
|
|
"HERC777" <herc777@hotmail.com> wrote in message
news:1116165889.565091.55740@g49g2000cwa.googlegroups.com...
FIXED FONT
0.1-----1......1..........1.........1
0.|2....|..2......2......2......2....
0.|.3...|.3...3.......3......3.......
0.|..4..|...4....4.......4.....4.....
0.|...5.|..5........5.......5........
0.|....6|......6.......6..........6..
0.7-----7.....7...7..........7.......
0........8--8------8.........8.......
0........|9.....9..|.........9.......
0........|.1---1...|......1.....1....
0........|.|2..|...|2.........2......
0........|.|.3.|...|..3..........3...
0........|.|..4|...|.......4..4......
0........|.5---5...|...5....5........
0........|......6..|....6.......6....
0........|.......7.|......7......7...
0........|........8|...8..........8..
0........9---------9........9..9.....
4DP <=>
where w = the natural numbers
and D = the decimal digits {0..9},
AnEcAd1d2[ new & cew & d1eD & d2eD & (d1=/=d2) & (c=/=d) ]
-> L[n,n]=d1
& L[c,c]=d2
& L[n,c]=d1
& L[c,n]=d2
Hmmm, is this an example of the sort of logic you'll be teaching us over the
next week? Well, good job nobody's paying you for the lessons! :)
Seriously - one of your problems is you need to express what you say more
precisely, otherwise it's difficult to know what to agree or disagree with.
For example, looking at what you've written above I would comment:
* what is 4DP? (it appears on one side of a double implication,
so that would make it a statement of some kind). But I think
maybe you are trying to *define* a property of (certain) lists?
* Where does d come into things?
* You have quantifiers AnEc... but it looks to me as though you have got
confused about the whole thing, e.g. put them in the wrong order or
something. Let's look at what you actually said...
You started An..., so in my head I think OK let's look at n=1.
AnEc... , so I know there is a c to go with n=1...
(You intend n and c as row/column numbers, right?)
AnEcAd1Ad2, so I know for this c, for ANY d1 and d2 the following
statement is to be true... In particular it is to be true for
d1=8 and d2=9
AnEcAd1Ad2 [...] Apart from not knowing what d is(!), the [] bit
seems to be satisfied, so the right hand side of the --> should
also be satisfied, in particular L[n,n] = L[1,1] = d1 = 8.
However, looking at your FIXED FONT example list, I can see
L[1,1] = 1, so no choice of c can satisfy this, so the whole
clause starting AnEc.... is false. I don't think this is
what you meant. (Obviously no list is ever going to have
such a property as you've written it!)
In short, you've made a bit of an effort at being precise (which is good)
but you've just got what wanted to say totally garbled!
Here is what I think you meant to say:
---------------------------
(N = set of natural numbers {1,2,3..}
D = set of digits {0,1,2...9})
Definition: A list L of real numbers in [0,1] is said to have the "4DP"
property if:
An e N, Ad e D, Em e N :
[ m >= n
& L[n,n]=L[n,m]
& L[m,m]=L[m,n]=d
]
---------------------------
In words, the list has the 4DP property if given any diagonal element in
position [n,n], and any digit d we want to "swap it for", there is a
diagonal element further down the list at [m,m] which will allow us to do
this simply by virtue of swapping elements [n] and [m] in the original list.
blame the inconsistency on my logician George, but you get the idea!
Can (an infinite set of) numbers with this property have their diagonal
independantly permuted?
Instead of asking all these questions, why don't you present a list with the
4DP property if you think one exists?
And if you do find one, remember that (my) 4DP only allows you to do one
swap operation. It's not clear whether the list after performing a swap
still always have the 4DP property. You could no doubt have a stronger 4DPE
(extended) property of lists, that would guarantee that after performing a
swap, the new list would always still have 4DPE. This would allow you to do
any finite number of swaps like this, but you'll never be able to extend
this to allow an infinite number of swaps, which is what you really want I
think... (After an infinite number of swaps, there is no guarantee that the
new list is a permutation of the old one - in fact first you would have to
be define exactly what you meant by the product of an infinite number of
swaps...)
Hope this helps,
Mike.
Herc
.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Related Articles |
|
|
