Question
zuhair
I have a question.
Imagine a bundle of stickes ||||........ , the first stick is at
position zero , the second at position 1, the third at position 2 , ad
infanitum. This bundle can be easily bijected to Peano's
natural numbers beginning from zero N={0,1,2,3,.........}, so it has
cardinality equal to
Aleph-0.
Now image a two dimensional stick bundle as below
.
.
.
|||||.......
|||||........
|||||........
This also can be bijected to N={0,1,2,3,} , this can be made easily in
a zigzag manner as below:
.
. .
16 . .
15 . . .
7 14 . .
68 13 . .
259 12 . .
13410 11......
So even this two dimensional bundle of stickes has cardinality equal to
Aleph-0 since it can
be bijected to N= {0,1,2,3,.......}
In a similar manner a three dimentional bundle can be imagined to be
bijected to N and thus
has also cardinality equal to Aleph-0.
In reality any n-dimensional bundle of stickes can be easily bijected
to N and thus has
Aleph-0 cardinality.
My question is what if we had an Aleph-0 dimensional bundle of stickes
, then can that
be bijected to N, and how can that be imagined.
Zuhair
zuhair
[Aleph-0]^n = Aleph-0 , n=1,2,3,4,........
But [Aleph-0] ^ [Aleph-0] = or > Aleph-0 ?
Zuhair
Arturo Magidin
You can stop using "bundles" and use numbers; in that case, what you
want is just the collection of all w-tuples
(a_0, a_1, a_2, ..., a_n, ...)
where the a_i are taking from a countable set.
This has cardinality (aleph_0)^{aleph_0}, which is the same as the
cardinality of the continuum; it is NOT bijectable with N.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Arturo Magidin
(aleph_0)^{aleph_0) = 2^{aleph_0} = c > aleph_0.
zuhair
Ok, I agree with you . But I have four questions
First:how can you demonstrate that this is the same as the cardinality
of continuum and how can you demonstrate that it is not bijectable to
N?
Second: is that subject has a relation to subject of "Selections and
the Multiplicative axiom"?
Third: Is (Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } > (Aleph-0) ^
(Aleph-0)
And what that can possibly represent, since (Aleph-0) ^ (Aleph-0) is
the cardinality of continuum
Then if (Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } > (Aleph-0) ^ (Aleph-0)
then what does
(Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } represent, could it represent
the cardinality of a MERGED
continuum.
Fourth: Is n ^ (Aleph-0 ) , n=2,3,4,5,........ is the same for all
values of n.
and Is n ^ (Aleph-0) = (Aleph-0) ^ (Aleph-0)
Since from what I knew 2^ Aleph-0 is the cardinality of continuum.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>
>Arturo Magidin wrote:
>> In article <1145788213.1...@v46g2000cwv.googlegroups.com>,
>> zuhair <zalj...@yahoo.com> wrote:
[.snip.]
>> >My question is what if we had an Aleph-0 dimensional bundle of stickes
>> >, then can that
>> >be bijected to N, and how can that be imagined.
>>
>> You can stop using "bundles" and use numbers; in that case, what you
>> want is just the collection of all w-tuples
>>
>> (a_0, a_1, a_2, ..., a_n, ...)
>>
>> where the a_i are taking from a countable set.
>>
>> This has cardinality (aleph_0)^{aleph_0}, which is the same as the
>> cardinality of the continuum; it is NOT bijectable with N.
>
>Ok, I agree with you . But I have four questions
>
>First:how can you demonstrate that this is the same as the cardinality
>of continuum and how can you demonstrate that it is not bijectable to
>N?
Once you know it has the same cardinality as the continuum, which is
the cardinality of P(N), the fact that it cannot be bijectable with N
is a simple corollary of Cantor's Theorem: no set is bijectable with
its power set.
As to how you know it has the cardinality of the continuum, it is
trivial that it has cardinality at least as large as 2^{omega},
since the set of all functions from omega to 2={0,1} is a subset of
omega^{omega}, the set of all functions from omega to itself. The
former has cardinality 2^{aleph_0}, the latter has cardinality
aleph_0^{aleph_0}. Showing that it has cardinality no more than
2^{aleph_0} is a bit trickier; I know I've done it in the past, but I
can't remember how right now. I'll check in my office tomorrow and get
back to you.
>Second: is that subject has a relation to subject of "Selections and
>the Multiplicative axiom"?
I have no idea what you mean by this.
>Third: Is (Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } > (Aleph-0) ^
>(Aleph-0)
Yes. Since aleph_0^{aleph_0} = c, then the left hand side is aleph_0^c
which is equal to 2^c (I believe it is essentially the same argument
as above); this is the cardinality of the set of all subsets of R; the
right hand side is merely c, the cardinality of R. Again, by Cantor's
Theorem, 2^c > c.
>And what that can possibly represent, since (Aleph-0) ^ (Aleph-0) is
>the cardinality of continuum
It represents the cardinality of the set of all subsets of the
continuum; which is the same as the cardinality of the set of all
functions from the continuum to itself.
>Then if (Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } > (Aleph-0) ^ (Aleph-0)
> then what does
>
>(Aleph-0) ^ { (Aleph-0) ^ (Aleph-0) } represent, could it represent
>the cardinality of a MERGED
>
>continuum.
I have no idea what a "MERGED continuum" ( or a 'merged continuum',
for that matter, leaving aside your shouting) is, so I have no idea
what you mean.
>Fourth: Is n ^ (Aleph-0 ) , n=2,3,4,5,........ is the same for all
>values of n.
Yes. Trivially, 2^{aleph_0} <= n^{aleph_0} <= (n+1)^{aleph_0} <=
(aleph_0)^{aleph_0} , and since both extremes are equal, all are
equal.
>and Is n ^ (Aleph-0) = (Aleph-0) ^ (Aleph-0)
Well, duh. Yes.
Arturo Magidin
Arturo Magidin <mag...@math.berkeley.edu> wrote:
[.aleph_0^{aleph_0}.]
>As to how you know it has the cardinality of the continuum, it is
>trivial that it has cardinality at least as large as 2^{omega},
>since the set of all functions from omega to 2={0,1} is a subset of
>omega^{omega}, the set of all functions from omega to itself. The
>former has cardinality 2^{aleph_0}, the latter has cardinality
>aleph_0^{aleph_0}. Showing that it has cardinality no more than
>2^{aleph_0} is a bit trickier; I know I've done it in the past, but I
>can't remember how right now. I'll check in my office tomorrow and get
>back to you.
Well, here is a way to show that aleph_0^{aleph_0} has cardinality no
more than 2^{aleph_0}. Let g:omega -> (omega x omega) be a bijection.
Given a map f:omega->omega, write
f(n) = b_{(0} + b_{n1}*2 + ... +b_{nk}*2^k
where b_{ni} =0 or 1, b_{nk}=1 (i.e., write it in binary); set
b_{nm}=0 for m>k. Then we can identify f with the element
(c_0, c_1, ..., c_r, ...) of 2^{omega} given by letting
c_i = b_{g(i)}.
This is an injection, since
f(n) = sum_{j=0}^{infty} c_{g^{-1}(n,j)}*2^j;
so this gives that the cardinality of omega^{omega} (which is the
cardinality of aleph_0^{aleph_0}) is at most 2^{aleph_0}. The other
inequality is trivial, as noted above, giving equality. If you use
"base n" instead of base 2 you get an injection from omega^{omega} to
n^{omega}, though this is not needed once you know that
2^{aleph_0}=aleph_0^{aleph_0}.
Dave Seaman
> As to how you know it has the cardinality of the continuum, it is
> trivial that it has cardinality at least as large as 2^{omega},
> since the set of all functions from omega to 2={0,1} is a subset of
> omega^{omega}, the set of all functions from omega to itself. The
> former has cardinality 2^{aleph_0}, the latter has cardinality
> aleph_0^{aleph_0}. Showing that it has cardinality no more than
> 2^{aleph_0} is a bit trickier; I know I've done it in the past, but I
> can't remember how right now. I'll check in my office tomorrow and get
> back to you.
One way is to use the laws of exponents.
aleph_0 ^ aleph_0 <= (2^aleph_0) ^ aleph_0
= 2 ^ (aleph_0 * aleph_0)
= 2 ^ aleph_0
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Arturo Magidin
Dave Seaman <dse...@no.such.host> wrote:
>On Mon, 24 Apr 2006 02:20:03 +0000 (UTC), Arturo Magidin wrote:
>> As to how you know it has the cardinality of the continuum, it is
>> trivial that it has cardinality at least as large as 2^{omega},
>> since the set of all functions from omega to 2={0,1} is a subset of
>> omega^{omega}, the set of all functions from omega to itself. The
>> former has cardinality 2^{aleph_0}, the latter has cardinality
>> aleph_0^{aleph_0}. Showing that it has cardinality no more than
>> 2^{aleph_0} is a bit trickier; I know I've done it in the past, but I
>> can't remember how right now. I'll check in my office tomorrow and get
>> back to you.
>
>One way is to use the laws of exponents.
>
> aleph_0 ^ aleph_0 <= (2^aleph_0) ^ aleph_0
> = 2 ^ (aleph_0 * aleph_0)
> = 2 ^ aleph_0
Yes, of course. <smacks forehead> For some reason I thought it should
be some sort of double inclusion.
This can also be used to show that aleph_0^{aleph_0^{aleph_0}} =
2^c, since
aleph_0^{aleph_0^{aleph_0}} = aleph_0^{c} <= (2^c)^c = 2^{c*c} = 2^c,
and the other inequality is easy.
zuhair
Let us try to imagine things.
I don't know if the following line of imagination is acceptable.
Let r by any real number
r can always be represented as a sequence of numerals. Let us work with
the binary numerals
so r = b_0 b_1 b_2 b_3 .................................
were b_i is either 0 or 1 or " . " (the decimal dot )
and only one of the b_i (s) above is the decimal dot.
now the set of all r is the set of all real numbers R. the cardinality
of which is c.
But 2^c is the set of all combinations of R.
what a combination of R could possibly represent. Can we say it
represents a point in space.
This is a strange concept to me imagine the combination between the two
real numbers
1.0101010........... and 1.110110110.............. ( Binary)
This would be the number 1.0101010.........1.110110110........... but
what is that number
is it a complex number??? a point in two dimensional space??? or simply
a Vector.
I think it would be a vector of numbers.
Zuhair
Arturo Magidin
[.snip.]
>
>Let us try to imagine things.
Us, kimosabe?
>I don't know if the following line of imagination is acceptable.
>
>Let r by any real number
>
>r can always be represented as a sequence of numerals. Let us work with
>the binary numerals
>
>so r = b_0 b_1 b_2 b_3 .................................
>
>were b_i is either 0 or 1 or " . " (the decimal dot )
>
>and only one of the b_i (s) above is the decimal dot.
The expression need not be unique (you need to choose whether to use a
trail of zeros or a trail of 1s when one or the other is possible.
>now the set of all r is the set of all real numbers R. the cardinality
>of which is c.
>
>But 2^c is the set of all combinations of R.
The CARDINALITY of the POWER SET of R is 2^c, where c is the
cardinality of R. This is also the cardinality of the set of all
functions from R to R. I have no idea what you mean by "combinations
of R", and in any case, "2^c" is a cardinal and as such not "equal" to
a set of things having something to do with R. Do keep straight the
difference between "being" something and "having the same cardinality"
as something.
>what a combination of R could possibly represent.
Depends entirely on just what the hell a "combination of R" could
possibly be.
>Can we say it
>represents a point in space.
We can say lots of things. Most of the things we can say, however, are
either incoherent or false.
The cardinality of the set of all points in R^3 is equal to the
cardinality of all points in R, i.e., it is equal to c. In fact, R^n
has cardinality c for every positive integer n.
>This is a strange concept to me imagine the combination between the two
>real numbers
Such is the case with most incoherent things.
>
>1.0101010........... and 1.110110110.............. ( Binary)
>
>This would be the number 1.0101010.........1.110110110........... but
>what is that number
Simple: it is not a number. You have now landed solidly in
"incoherence".
>is it a complex number??? a point in two dimensional space??? or simply
>a Vector.
None of the above.
2^c is the cardinality of the set of all functions from R to R. It is
also the cardinality of the power set of R, which is the set of all
subsets of R.
zuhair
Perhaps I didn't clarify my idea, so it seems incoherent, let me
present it again.
any real number can be represented as a specific sequence of numerals.
using the decimal numeral system for example it can be said that any
real number r can be represented by a string of numerals d_1 d_2 d_3
......... , were d_i = 0 or 1 or 2 or 3 or.... 9 or decimal dot, with
at most one of the d_i being a decimal dot. For example number 33 ,
here we have d_1 = 3 and d_2 =3 and d_3 = .
d_ j = 0 were j>3 .
Now the set of all possible permutations of d_1 d_2 d_3 .............
will be the set R.
and this has the cardinality of c.
Now 2^c is the cardinality of the power set of R.
The power set of R is the set of all combinationas that can come from
R.
Since any member of R is represented by d_1 d_2 d_3...............
Then the power set of R will have numerals that consists from a
combination of this representation .
For example 33.000...............1.00000000
now what that numeral represent for example. Does it represent a number
or what a vector ???
In order to describe this number we have 33. followed by Omega of zeros
which is followed by 1. which is followed by Omega of zeros. Now what
that sequence of numerals represent.
That is my question.
Zuhair
Dave Seaman
> The power set of R is the set of all combinationas that can come from
> R.
No. The power set of R is the set of all subsets of R.
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
[.snip.]
>any real number can be represented as a specific sequence of numerals.
>using the decimal numeral system for example it can be said that any
>real number r can be represented by a string of numerals d_1 d_2 d_3
>......... , were d_i = 0 or 1 or 2 or 3 or.... 9 or decimal dot, with
>at most one of the d_i being a decimal dot. For example number 33 ,
>here we have d_1 = 3 and d_2 =3 and d_3 = .
>d_ j = 0 were j>3 .
You once again forgot to exclude the possibility of non-unique
representation (despite the fact that I mentioned it explicitly). Is
there any point in my replying to your posts? Do you read the
responses?
You must also either exclude the possibility that there is some k>0
such that d_i=9 for all i>k, or else the possibility that there is
some k>0 such that d_i=0 for all i>k. Otherwise, your claim about
"specific sequence of numerals" is false. For example, the real number
1 can be represented either as 1.000000..... or as
0.999999999999999999....
>Now the set of all possible permutations of d_1 d_2 d_3 .............
>will be the set R.
No, it is not. For example, if the number you started with was "1",
written as 1.000000... then the set of all possible permutations of
these symbols does NOT yield the set of all real numbers. Numbers
whose decimal expansion have exactly k occurrences of the digit 7 can
only be obtained as permutations of each other, and no permutation of
it will yield a number whose decimal expansion has exactly (k+1)
occurrences of the digit 7 (or an infinite number of them).
>and this has the cardinality of c.
>
>Now 2^c is the cardinality of the power set of R.
>
>The power set of R is the set of all combinationas that can come from
>R.
No. The power set of R is the set of all SUBSETS OF R. Not the set of
all "combinations of R", a term which you continue to use without
saying just what it means.
>Since any member of R is represented by d_1 d_2 d_3...............
>
>Then the power set of R will have numerals that consists from a
>combination of this representation .
No. No. No. No. No.
Is five times enough, or do you need me to repeat it even more times?
>For example 33.000...............1.00000000
>
>now what that numeral represent for example. Does it represent a number
>or what a vector ???
It represents nonsense. Plain and simple.
The power set of R is not the set of all rearrangements of a
sequence. It is the set of all SUBSETS of R; the set of all sets, each
of which has all of its elements as real numbers.
>In order to describe this number we have 33. followed by Omega of zeros
>which is followed by 1. which is followed by Omega of zeros. Now what
>that sequence of numerals represent.
It represents nonsense.
>That is my question.
And the answer is: it represents nonsense. You are utterly confused:
the reals cannot be obtained as the set of all rearrangements of a
single real, and the power set of the reals is not the "set of all
combinations of reals".
zuhair
Arturo Magidin wrote:
> In article <1145951829.2...@e56g2000cwe.googlegroups.com>,
> zuhair <zalj...@yahoo.com> wrote:
>
> [.snip.]
>
> >any real number can be represented as a specific sequence of numerals.
> >using the decimal numeral system for example it can be said that any
> >real number r can be represented by a string of numerals d_1 d_2 d_3
> >......... , were d_i = 0 or 1 or 2 or 3 or.... 9 or decimal dot, with
> >at most one of the d_i being a decimal dot. For example number 33 ,
> >here we have d_1 = 3 and d_2 =3 and d_3 = .
> >d_ j = 0 were j>3 .
>
> You once again forgot to exclude the possibility of non-unique
> representation (despite the fact that I mentioned it explicitly). Is
> there any point in my replying to your posts? Do you read the
> responses?
Well you have the right, in order to exclude non-unique representation
I will add a condition to what I have said above that the d_i which is
the decimal point shouldn't be d_1.
Second you think that number 1 can be represented as 0.99999........ (
decimal) , well I think that this is only an approximate representation
that is not correct.
The only representation of 1 is 1.0000......... and to me
0.9999........... < 1.
>
> You must also either exclude the possibility that there is some k>0
> such that d_i=9 for all i>k, or else the possibility that there is
> some k>0 such that d_i=0 for all i>k. Otherwise, your claim about
> "specific sequence of numerals" is false. For example, the real number
> 1 can be represented either as 1.000000..... or as
> 0.999999999999999999....
wrong see above.
>
> >Now the set of all possible permutations of d_1 d_2 d_3 .............
> >will be the set R.
>
> No, it is not. For example, if the number you started with was "1",
> written as 1.000000... then the set of all possible permutations of
> these symbols does NOT yield the set of all real numbers. Numbers
> whose decimal expansion have exactly k occurrences of the digit 7 can
> only be obtained as permutations of each other, and no permutation of
> it will yield a number whose decimal expansion has exactly (k+1)
> occurrences of the digit 7 (or an infinite number of them).
All of that is gibberish, I didn't say that the set of all real numbers
comes from rearrangment of
the symboles used in representing a single real number.
>
> >and this has the cardinality of c.
> >
> >Now 2^c is the cardinality of the power set of R.
> >
> >The power set of R is the set of all combinationas that can come from
> >R.
>
> No. The power set of R is the set of all SUBSETS OF R. Not the set of
> all "combinations of R", a term which you continue to use without
> saying just what it means.
Their is no difference between these two terms , they are synonum.
>
> >Since any member of R is represented by d_1 d_2 d_3...............
> >
> >Then the power set of R will have numerals that consists from a
> >combination of this representation .
>
> No. No. No. No. No.
>
> Is five times enough, or do you need me to repeat it even more times?
Repeat what . what I meant by combinations is what you call all
subsets.
>
> >For example 33.000...............1.00000000
> >
> >now what that numeral represent for example. Does it represent a number
> >or what a vector ???
>
> It represents nonsense. Plain and simple.
ah I c , this is an easy answer , Thanks.
>
> The power set of R is not the set of all rearrangements of a
> sequence. It is the set of all SUBSETS of R; the set of all sets, each
> of which has all of its elements as real numbers.
I agree, but you read me wrong.
>
> >In order to describe this number we have 33. followed by Omega of zeros
> >which is followed by 1. which is followed by Omega of zeros. Now what
> >that sequence of numerals represent.
>
> It represents nonsense.
Easy answer.
>
>
> >That is my question.
>
> And the answer is: it represents nonsense. You are utterly confused:
> the reals cannot be obtained as the set of all rearrangements of a
> single real,
I never said that
and the power set of the reals is not the "set of all
> combinations of reals".
Yes they are. But you don't understand the world combinations.
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>
>Arturo Magidin wrote:
>> In article <1145951829.2...@e56g2000cwe.googlegroups.com>,
>> zuhair <zalj...@yahoo.com> wrote:
>>
>> [.snip.]
>>
>> >any real number can be represented as a specific sequence of numerals.
>> >using the decimal numeral system for example it can be said that any
>> >real number r can be represented by a string of numerals d_1 d_2 d_3
>> >......... , were d_i = 0 or 1 or 2 or 3 or.... 9 or decimal dot, with
>> >at most one of the d_i being a decimal dot. For example number 33 ,
>> >here we have d_1 = 3 and d_2 =3 and d_3 = .
>> >d_ j = 0 were j>3 .
>>
>> You once again forgot to exclude the possibility of non-unique
>> representation (despite the fact that I mentioned it explicitly). Is
>> there any point in my replying to your posts? Do you read the
>> responses?
>
>Well you have the right, in order to exclude non-unique representation
>I will add a condition to what I have said above that the d_i which is
>the decimal point shouldn't be d_1.
>
>Second you think that number 1 can be represented as 0.99999........ (
>decimal) , well I think that this is only an approximate representation
>that is not correct.
Then you are, quite simply, wrong. 0.9999... represents the infinite
series 9/10 + 9/10^2 + ... + 9/10^n + ... which converges to 1, so
0.9999... is not an "approximate representation that is not correct",
but an EXACT representation that is EXACTLY equal to 1. Period.
>The only representation of 1 is 1.0000......... and to me
>0.9999........... < 1.
Then you are wrong. If that expression represented a number strictly
smaller than 1, call it x, then (1+x)/2 would be a number strictly
smaller than 1, and strictly larger than x.
What, pray tell, is the decimal representation of this number?
>> You must also either exclude the possibility that there is some k>0
>> such that d_i=9 for all i>k, or else the possibility that there is
>> some k>0 such that d_i=0 for all i>k. Otherwise, your claim about
>> "specific sequence of numerals" is false. For example, the real number
>> 1 can be represented either as 1.000000..... or as
>> 0.999999999999999999....
>
>wrong see above.
Indeed, wrong. You are just wrong, completely, absolutely, and
totally.
>> >Now the set of all possible permutations of d_1 d_2 d_3 .............
>> >will be the set R.
>>
>> No, it is not. For example, if the number you started with was "1",
>> written as 1.000000... then the set of all possible permutations of
>> these symbols does NOT yield the set of all real numbers. Numbers
>> whose decimal expansion have exactly k occurrences of the digit 7 can
>> only be obtained as permutations of each other, and no permutation of
>> it will yield a number whose decimal expansion has exactly (k+1)
>> occurrences of the digit 7 (or an infinite number of them).
>
>All of that is gibberish, I didn't say that the set of all real numbers
>comes from rearrangment of
>the symboles used in representing a single real number.
Nonsense. You wrote: "the set of all possible permutations of
d_1d_2d_3... will be the set R". And previously, you defined
"d_1d_2d_3..." to be a specific string of numerals which represented a
specific real number. So you did, indeed, state that the set of all
real numbers comes from rearranging the symbols used in representing a
single real number. If that is not what you ->MEANT<- to say, well,
that's your problem for not saying what you mean and saying nonsense
instead. However, I strongly suspect that what you meant to say was
also nonsense, just a different kind of nonsense.
>> >and this has the cardinality of c.
>> >
>> >Now 2^c is the cardinality of the power set of R.
>> >
>> >The power set of R is the set of all combinationas that can come from
>> >R.
>>
>> No. The power set of R is the set of all SUBSETS OF R. Not the set of
>> all "combinations of R", a term which you continue to use without
>> saying just what it means.
>
>Their is no difference between these two terms , they are synonum.
No, they are not. "Combination" has a very specific meaning; "subset"
has a very specific meaning. They are not synonimous. Each subset
corresponds to a "combination without repetition of objects taken form
a set", but not every combination corresponds to a unique subset.
>> >Since any member of R is represented by d_1 d_2 d_3...............
>> >
>> >Then the power set of R will have numerals that consists from a
>> >combination of this representation .
>>
>> No. No. No. No. No.
>>
>> Is five times enough, or do you need me to repeat it even more times?
>
>Repeat what .
That you are, and continue to, speak nonsense.
> what I meant by combinations is what you call all
>subsets.
Still nonsense. A subset of R consists of a collection of elements of
R. This is not a rearrangement of a presentation of a number, nor does
it corresopnd to such a rearrangement, nor does it correspond to a
concatenation of two such representations. You are speaking nonsense,
over and over and over and over again.
Here is a subset of R:
{pi^r | r a positive real number}.
Pray tell, what are the "numerals that consist from a combination of
this representation" (->what<- representation?) that represents this
subset and no other?
>> The power set of R is not the set of all rearrangements of a
>> sequence. It is the set of all SUBSETS of R; the set of all sets, each
>> of which has all of its elements as real numbers.
>
>I agree, but you read me wrong.
No, you speak wrong. So far, you either speak falsehoods or nonsense.
>> >In order to describe this number we have 33. followed by Omega of zeros
>> >which is followed by 1. which is followed by Omega of zeros. Now what
>> >that sequence of numerals represent.
>>
>> It represents nonsense.
>
>Easy answer.
Sometimes the easy answer is correct. What you wrote above is
nonsense, and your assertions about representations of real numbers
are alternating between nonsense and falsehoods.
>> >That is my question.
>>
>> And the answer is: it represents nonsense. You are utterly confused:
>> the reals cannot be obtained as the set of all rearrangements of a
>> single real,
>
>I never said that
You did say that. Exactly and explicitly. You perhaps did not MEAN
that, but you did a rather poor job of saying what you meant in that
case. What you said is exactly what I wrote above.
>and the power set of the reals is not the "set of all
>> combinations of reals".
>
>Yes they are. But you don't understand the world combinations.
I understand it perfectly well. You, on the other hand, don't even
know what a real is, apparently.
zuhair
Dave Seaman wrote:
> On 25 Apr 2006 00:57:09 -0700, zuhair wrote:
>
> > The power set of R is the set of all combinationas that can come from
> > R.
>
> No. The power set of R is the set of all subsets of R.
They are exactly the same.
Zuhair
Virgil
> Second you think that number 1 can be represented as 0.99999........ (
> decimal) , well I think that this is only an approximate representation
> that is not correct.
> The only representation of 1 is 1.0000......... and to me
> 0.9999........... < 1.
If 0.9999........... < 1 then x = 1 - 0.9999... > 0
So how much larger than zero is this x?
Is it large enough so that x/2 is smaller?
zuhair
[x/n] + [x/(n^2)] + [x/(n^3)] +.............+ [ x/(n^k)] = [x/(n-1)] -
[ x/{ ( n^k) ( n-1 ) } ]
For n=2,3,4,5,.............
and K= 1,2,3,4,5,............
now 0.99999............ = 9/ 10 + 9 / 10^2 + 9 / 10^3 +
....................= 9/9 - 9/[ (10^Omega) *9} = 1 - [1/(10^Omega)]
Now [1/(10^Omega)] > zero
because if [1/(10^Omega)] = 0 then 0* 10^Omega = 1 which is
impossible.
Then 0.9999......... < 1
Zuhair
zuhair
1- 0.9999 = 1/( 10^ Omega) =x
of coarse x/2 = x.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>
>Virgil wrote:
>> In article <1145991892.1...@i39g2000cwa.googlegroups.com>,
>> "zuhair" <zalj...@yahoo.com> wrote:
>>
>>
>> > Second you think that number 1 can be represented as 0.99999........ (
>> > decimal) , well I think that this is only an approximate representation
>> > that is not correct.
>> > The only representation of 1 is 1.0000......... and to me
>> > 0.9999........... < 1.
>>
>> If 0.9999........... < 1 then x = 1 - 0.9999... > 0
>>
>> So how much larger than zero is this x?
>>
>> Is it large enough so that x/2 is smaller?
>
>[x/n] + [x/(n^2)] + [x/(n^3)] +.............+ [ x/(n^k)] = [x/(n-1)] -
>[ x/{ ( n^k) ( n-1 ) } ]
>
>For n=2,3,4,5,.............
>and K= 1,2,3,4,5,............
>
>now 0.99999............ = 9/ 10 + 9 / 10^2 + 9 / 10^3 +
>....................= 9/9 - 9/[ (10^Omega) *9} = 1 - [1/(10^Omega)]
10^omega is not a number; it is an ordinal, but it is not a
number. (10^omega)*9 is an ordinal, but not a number.
Division of ordinals is undefined. So 9/[ (10^omega)*9] is nonsense.
Subtraction of ordinals is undefined; so even if 9/[(10^omega)*9] were
some meaningful ordinal, "9/9 - 9/[ (10^omega)*9]" would be nonsense.
Even if 9/9 - 9/[ (10^omega)*9] made sense as an ordinal, the left
hand side of your "equation" is a real number, and the right hand side
is an ordinal. So the expression you have would STILL be nonsense.
>Now [1/(10^Omega)] > zero
That is nonsense.
I notice, by the by, that all your nonsensical gyrations failed to
answer his question.
Arturo Magidin
If both 0.999... and 1 are real numbers, then their difference, your
"x", is a real number. Hence so is x/2.
So x/2 = x is an equation of real numbers. Subtracting x/2 from both
sides we get x/2 = 0. Multiplying through by 2 we get x=0.
Therefore, x=0. Since x = 1-0.9999999...., that means that
0 = 1 - (0.99999....)
which means that 0.999.... = 1,
as was originally claimed and as you denied.
Now, if it is NOT the case that both 0.999.... and 1 are real numbers,
then it would clearly be the case that it is the former which is not a
real number (unless you now want to claim that 1 is not a real
number?). But in that case, your original assertion concerning how
real numbers are bijectable with sequences of numerals would be a
false statement, since certain sequences of numerals would not be real
numbers.
Either way, you said something false in at least one place.
Virgil
Does not answer the question of whether x/2 is smaller than x.
and is a false summation formula anyway.
The infinite sum 1 + a + a^2 + a^3 + ... is undefined for |a| >=1 and
converges to the value 1/(1-a) when |a| < 1.
so that (9/10)*(1 + 1/10 + 1/100 + ...) = (9/10)(1/(1 - 1/10)) = 1
Oscar Lanzi III
cardinality of the reals.
Why? Well, the cardinality is aleph_0^aleph_0. As aleph_0 > 2,
aleph_0^aleph_0 >= 2^aleph_0. But we also have aleph_0 < 2^aleph_0,
hence aleph_0^aleph_0 <= 2^(aleph_0*aleph_0). Now aleph_0*aleph_0 =
aleph_0. So, aleph_0^aleph_0 <= 2^aleph_0 and aleph_0^aleph_0 >=
2^aleph_0, both of which can be simultaneously satisfied only if
equality holds. QED.
--OL
zuhair
Arturo Magidin wrote:
> In article <1145996666....@j33g2000cwa.googlegroups.com>,
> zuhair <zalj...@yahoo.com> wrote:
> >
> >Virgil wrote:
> >> In article <1145991892.1...@i39g2000cwa.googlegroups.com>,
> >> "zuhair" <zalj...@yahoo.com> wrote:
> >>
> >>
> >> > Second you think that number 1 can be represented as 0.99999........ (
> >> > decimal) , well I think that this is only an approximate representation
> >> > that is not correct.
> >> > The only representation of 1 is 1.0000......... and to me
> >> > 0.9999........... < 1.
> >>
> >> If 0.9999........... < 1 then x = 1 - 0.9999... > 0
> >>
> >> So how much larger than zero is this x?
> >>
> >> Is it large enough so that x/2 is smaller?
> >
> >[x/n] + [x/(n^2)] + [x/(n^3)] +.............+ [ x/(n^k)] = [x/(n-1)] -
> >[ x/{ ( n^k) ( n-1 ) } ]
> >
> >For n=2,3,4,5,.............
> >and K= 1,2,3,4,5,............
> >
> >now 0.99999............ = 9/ 10 + 9 / 10^2 + 9 / 10^3 +
> >....................= 9/9 - 9/[ (10^Omega) *9} = 1 - [1/(10^Omega)]
>
> 10^omega is not a number; it is an ordinal, but it is not a
> number. (10^omega)*9 is an ordinal, but not a number.
No you are wrong. Omega is a Transfinite Ordinal number. review
Cantor.
Also number 9 is an ordinal number.
Can you tell me what is the difference between real number and ordinal
number.
>
> Division of ordinals is undefined. So 9/[ (10^omega)*9] is nonsense.
>
> Subtraction of ordinals is undefined; so even if 9/[(10^omega)*9] were
> some meaningful ordinal, "9/9 - 9/[ (10^omega)*9]" would be nonsense.
>
> Even if 9/9 - 9/[ (10^omega)*9] made sense as an ordinal, the left
> hand side of your "equation" is a real number, and the right hand side
> is an ordinal. So the expression you have would STILL be nonsense.
Your reply is nonsensical. All numbers 9 and 10 and Omega can be
ordinal numbers.
>
>
> >Now [1/(10^Omega)] > zero
>
> That is nonsense.
Oh yes it makes full sense.
>
> I notice, by the by, that all your nonsensical gyrations failed to
> answer his question.
Wrong.
zuhair
Arturo Magidin wrote:
> In article <1145996821.9...@g10g2000cwb.googlegroups.com>,
> zuhair <zalj...@yahoo.com> wrote:
> >
> >Virgil wrote:
> >> In article <1145991892.1...@i39g2000cwa.googlegroups.com>,
> >> "zuhair" <zalj...@yahoo.com> wrote:
> >>
> >>
> >> > Second you think that number 1 can be represented as 0.99999........ (
> >> > decimal) , well I think that this is only an approximate representation
> >> > that is not correct.
> >> > The only representation of 1 is 1.0000......... and to me
> >> > 0.9999........... < 1.
> >>
> >> If 0.9999........... < 1 then x = 1 - 0.9999... > 0
> >>
> >> So how much larger than zero is this x?
> >>
> >> Is it large enough so that x/2 is smaller?
> >
> >1- 0.9999 = 1/( 10^ Omega) =x
> >
> >of coarse x/2 = x.
>
> If both 0.999... and 1 are real numbers, then their difference, your
> "x", is a real number. Hence so is x/2.
Define real number? I think what you mean by real number is what
Bertrand Russell calls Inductive numbers, or simply speaking numbers
that follow the inductive properties.
>
> So x/2 = x is an equation of real numbers. Subtracting x/2 from both
> sides we get x/2 = 0. Multiplying through by 2 we get x=0.
Wrong, first of all I what you mean by real numbers, numbers that
follow inductive rules of summation subtraction multiplication
etc...........
Then of coarse it is obvious that 0.99999......... is not a real
number in that sense .
now 1- 0.9999........= x , here x is of coarse not a real number.
and accordingly x/2=x is not an equation of real numbers.
now subtracting x/2 from both sides will yeild x/2 - x/2 = x - x/2
since x/2=x
then this leads to x - x = x/2
and x-x = x
in reality subtraction of x - x is something undetermined.
The whole of your reasoning is not applicable here.
>
> Therefore, x=0. Since x = 1-0.9999999...., that means that
>
> 0 = 1 - (0.99999....)
>
> which means that 0.999.... = 1,
>
> as was originally claimed and as you denied.
>
>
> Now, if it is NOT the case that both 0.999.... and 1 are real numbers,
> then it would clearly be the case that it is the former which is not a
> real number (unless you now want to claim that 1 is not a real
> number?). But in that case, your original assertion concerning how
> real numbers are bijectable with sequences of numerals would be a
> false statement, since certain sequences of numerals would not be real
> numbers.
Yes I agree with that , here you are making a reasonable statement. I
agree I was wrong
I don't like that category called Real. I believe that the category
Ordinal is better.
So I will substitute what I said about the real numbers into ordinal
zuhair
x/2 = x , it is clear of coarse x/2 is not smaller than x , it is equal
to x.
>
> and is a false summation formula anyway. It is not false.
>
> The infinite sum 1 + a + a^2 + a^3 + ... is undefined for |a| >=1 and
> converges to the value 1/(1-a) when |a| < 1.
This is wrong. The summation formula I have provided is the true one.
> so that (9/10)*(1 + 1/10 + 1/100 + ...) = (9/10)(1/(1 - 1/10)) = 1
Wrong.
Zuhair
Tony Orlow
It's Lil'un, Silly. And, sure, there's a Lil'un/2, but we're assuming log2
(Big'un) bits, of course, to be consistent, so it's not included. Luv Ya,
Virgil! Rock on, Dog! :)
--
Smiles,
Tony
Tony Orlow
Hi Zuhair, how are you? Having fun getting answers and abuse? Enjoy! This is
what life's about. :)
Omega? Oh, Zuhair, why? Why, oh why, oh why, oh Zuhair, must you? (sigh)
Well, I suppose it's a good refutation of Virgil's statement, since it fits
with standard theory that omega=omega+1 -> 1/(10^omega)= 1/(10^omega+1).
Anyway, to address your questions, the power set of the continuum is the set of
all subsets, that is, the set of all unique sets of real numbers. So, your
combinations of these reals should be as sets, each pair being separated by
commas, not combined into a single string by concatentation. In fact, forget
the digital representation. That's a different kind of infinity.
As an alternative image, consider c to be the 1 dimensional continuum, the real
line. Then c^n would be a continuous space of n dimensions. This is still
considered to be the same cardinality as c, as aleph_0^n is aleph_0 for finite
n. You could consider c^aleph_0 to be a countably infinite-dimensional space,
but this is equal to (2^aleph_0)^aleph_0 = 2^(aleph_0^2)=2^aleph_0=c. So, this
hyperdimensional space is considered to contain no more points than the 1D
line. In order to get to the next higher cardinality, you would have to have at
least 2^c, which can be considered the size of that set of all possible sets of
reals. If you want a geometric image of this, consider the number of vertices
on a c-dimensional hypercube, in an uncountably infinite-dimensional space.
That's 2^c. I'm sure that clarified your image, didn't it, Zuhair? heh.
To all the standard bearers in the group, please correct me if I said anything
incorrect above, besides the opinions expressed of course. :)
--
Smiles,
Tony
Arturo Magidin
Which is what I said. In this context, "number" refers to "real
number". Duh.
>Also number 9 is an ordinal number.
>
>Can you tell me what is the difference between real number and ordinal
>number.
Can you?
An ordinal is a well-ordered set such that for each a in A,
a = {x in A : x<a}.
A real number is many things, depending on your construction: it can
be a Dedkind cut of rationals, or an equivalence class of Cauchy
sequences of rationals, etc. In any case, it is not "a well-ordered
set such that for each a in A, a = {x in A : x < a}.
>> Division of ordinals is undefined. So 9/[ (10^omega)*9] is nonsense.
>>
>> Subtraction of ordinals is undefined; so even if 9/[(10^omega)*9] were
>> some meaningful ordinal, "9/9 - 9/[ (10^omega)*9]" would be nonsense.
>>
>> Even if 9/9 - 9/[ (10^omega)*9] made sense as an ordinal, the left
>> hand side of your "equation" is a real number, and the right hand side
>> is an ordinal. So the expression you have would STILL be nonsense.
>
>Your reply is nonsensical. All numbers 9 and 10 and Omega can be
>ordinal numbers.
And how, pray tell, Oh Great And Glorious One, do you define quotients
and subtraction of ordinals, which is exactly what I said above?
Duh.
>> I notice, by the by, that all your nonsensical gyrations failed to
>> answer his question.
>
>Wrong.
No. The post failed to contain an answer to the question. Its entire
contents were nonsense. And you are quite clearly either a crank or a
troll. Bye.
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> Virgil wrote:
> > In article <1145996666....@j33g2000cwa.googlegroups.com>,
> > "zuhair" <zalj...@yahoo.com> wrote:
> >
> > > Virgil wrote:
> > > > In article <1145991892.1...@i39g2000cwa.googlegroups.com>,
> > > > "zuhair" <zalj...@yahoo.com> wrote:
> > > >
> > > >
> > > > > Second you think that number 1 can be represented as 0.99999........
> > > > > (
> > > > > decimal) , well I think that this is only an approximate
> > > > > representation
> > > > > that is not correct.
> > > > > The only representation of 1 is 1.0000......... and to me
> > > > > 0.9999........... < 1.
> > > >
> > > > If 0.9999........... < 1 then x = 1 - 0.9999... > 0
> > > >
> > > > So how much larger than zero is this x?
> > > >
> > > > Is it large enough so that x/2 is smaller?
> x/2 = x , it is clear of coarse x/2 is not smaller than x , it is equal
> to x.
In that case
x - x/2 = x/2 = 0
so
2*x/2 = x = 0
But then 0.999... = 1 after all, and Zuhair is WRONG!
> >
> > The infinite sum 1 + a + a^2 + a^3 + ... is undefined for |a| >=1 and
> > converges to the value 1/(1-a) when |a| < 1.
> > so that (9/10)*(1 + 1/10 + 1/100 + ...) = (9/10)(1/(1 - 1/10)) = 1
>
> Wrong.
Zuhair fails mathematics.
Tony Orlow
Hi Virgil. Is that a new cardigan? It looks nice. Can I borrow it at the next
Nonlinear Informatics and Beany Baby Convention? No? You fail charm school. :)
Zuhair is wrong, and you are right, of course, in the standard world. Since
reals are distinguished by having finite differences, there is no standard real
difference between 1.000... and 0.999... But, you don't have to be nasty about
it. WRONG! BAD! NO! BAD DOG! NO! Shall I smack you with a newspaper? That
wouldn't be nice. Here's a Milk Bone. Good Boy! ;)
It's certainly true, in standard math, that those values are equal. Any finite
number of digits, and you have a finite difference, but with an infinite number
of 9's and 0's, the difference becomes infinitesimal, and the values become
indistinguishable, on the FINITE level. However, any specific infinity of
digits gives a specific infinitesimal difference, so in nonstandard theories
one may consider these infinitesimally distinguishable hyperreals. This is
where relatively infinite levels of quantity may be continuous where the
infinitesimal level may be considered discrete. Archimedes would agree, with a
little convincing, no doubt. Have a nice afternoon!
--
Smiles,
Tony
zuhair
Hi Tony, How are you.
What you said is pritty convinicing. Now I know how standard
mathematicians look at the number which is represented by the numeral
0.99999....... , they think of it as a finite number, and most of the
time when they say real number they mean a number that obeys inductive
laws, or simply speaking a finite number.
The way I interpret numeral 0.9999........ is that it represents a sort
of an infinite number
that approaches 1.000000000...... but never reaching it. and the
difference between them
is what you call an infentismal difference and sure it is not a finite
difference.
The difference is 1/ ( 10^ Omega). this is because the ordinal number
of the repeatition of numeral 9 in 0.9999.... is Omega, or in simple
words the numeral 0.9999...... contains Omega of nines.
However I can think of another even bigger number than 0.9999...... and
yet smaller than 1.0000... , and that would be 0.9999...... 9 here we
have Omega +1 nines on the right of the decimal point and so 1 -
0.999.....9 = 1/[10^(Omega+1)] . which has a smaller
ordinal value than 1/( 10 ^ Omega ).
Virgil asked if the x/2 is smaller than x , according to the
transfinite laws division of a transfinite number by a finite number
reaveals the same transfinite number.
So 1/ 10^Omega cannot be halfed, but yet there exists numbers that are
larger than
1-x and yet smaller than 1, these like 1/ [10^(Omega+1)] etc.
Virgil said that the summation formula that I have presented earlier
and i will repeat it below is a false summation formula, does any body
agree with him:
[1/n] + [1/(n^2)] + [1/(n^3)] +......+[1/(n^k)] = [1/(n-1)] -
[1/{(n-1)(n^k) }]
n= 2,3,4,5,............
k= 1,2,3,4,............
Of coarse as k grow bigger, [1/{(n-1)(n^k) }] becomes smaller, and
thus the left side of the equation will coverge to 1/( n-1) , but it
will never reach it, because as I said before
the term [1/{(n-1)(n^k) }] cannot equal zero at finite or infinite
level because zero summed by itself finitelly or infinitely will always
equal zero.
Accordingly my proove that 0.9999......... < 1 is more logical.
But Tony as you said if standard mathematics looks at 0.999........ as
representing a finite number , then there is no real( finite) number
other than 1 which it could better symbolize.
Zuhair
zuhair
Ok, Tony , this is very nice, it seems you are reading my ideas.
Zuhair
zuhair
Tony Orlow wrote:
> zuhair said:
> >
> > Virgil wrote:
> > > In article <1145991892.1...@i39g2000cwa.googlegroups.com>,
> > > "zuhair" <zalj...@yahoo.com> wrote:
> > >
> > >
> > > > Second you think that number 1 can be represented as 0.99999........ (
> > > > decimal) , well I think that this is only an approximate representation
> > > > that is not correct.
> > > > The only representation of 1 is 1.0000......... and to me
> > > > 0.9999........... < 1.
> > >
> > > If 0.9999........... < 1 then x = 1 - 0.9999... > 0
> > >
> > > So how much larger than zero is this x?
> > >
> > > Is it large enough so that x/2 is smaller?
> >
> > 1- 0.9999 = 1/( 10^ Omega) =x
> >
> > of coarse x/2 = x.
> >
> > Zuhair
> >
> >
>
> Hi Zuhair, how are you? Having fun getting answers and abuse? Enjoy! This is
> what life's about. :)
>
> Omega? Oh, Zuhair, why? Why, oh why, oh why, oh Zuhair, must you? (sigh)
>
> Well, I suppose it's a good refutation of Virgil's statement, since it fits
> with standard theory that omega=omega+1 -> 1/(10^omega)= 1/(10^omega+1).
I think this is not precise.
In standard theory Omega +1 > Omega, but 1+ Omega = Omega, so for
your statements to be true you should replace Omega +1 by 1+ Omega.
I refuted Virgils statements because in standard theories Omega/n were
n is finite
is always Omega itself.
Smiles
Zuhair
zuhair
I confess I have scarce mathematical knowledge, and I don't insist on
my beleives, I am only questioning to know the truth.
>From what I know from Bertrand Russell's books Number x is the class
of all similar classes containing x members ( the definition seems
cyclic but it is not ). similar classes means the existence of one to
one relation between their members ( bijection).
So number two is the class of all doubles and number three is the class
of all triples. Review " Introduction to mathematical philosophy".
An Ordinal number is the number terms in a series, it is not the
series. It is the class of all similar series.( similar has the same
meaning above).
A cardinal number is the number of members in a set, ie the class of
all similar sets, and it is not affected by order.
What is a real number ? I don't have an exact definition , but from
what is writtin in this forum it seems to be a number which possess
inductive properties, the kind of a number Bertrand Russells refers to
as inductive number or finite number, perhaps?.
I heard that division is not defined for ordinals , but I don't know
why? Division of transfinite cardinals by finites is defined, why for
transfinite ordinal numbers is not defined I don't really know.
However addition multiplication and even exponentiation is defined for
ordinals.
you say that 1/ (2^ Omega ) is non sense, perhaps but I don't know why
it is so. why division and subtraction is not defined for ordinals?
zuhair
Even if these are not defined , it shouldn't preclude us from arriving
at some conclusion using general concepts.
you see 2^Omega is defined.
now the question is what is 1/( 2^Omega) , from the general rule x/y=z
means z: yz=x
Now 1/(2^Omega) means a number if multiplied by 2^Omega will result in
1.
Now what is that number exactly? it is not defined.
But whatever is that number , yet it cannot be zero.
why?
because if 1/(2^Omega) = 0 this means that 0* (2^Omega) = 1 , which is
impossible.
If 1/(2^Omega)< 0 , then this means that a positive number multiplied
by a negative number is yielding a positive number which is impossible.
So we are left with 1/(2^Omega) >0 which is fairly logical.
Now subtraction is not defined so 1 - ( 1/2^Omega) is non sense,
this is not the complete story.
We know that 1/ (2^ Omega) >0 and that is what we need.
Also we know that 1/(2^ Omega) should be a very small number yet
positive. so the subtraction
should yeild a number that is smaller than one, that is all what we
need.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
[.snip.]
>I confess I have scarce mathematical knowledge, and I don't insist on
>my beleives, I am only questioning to know the truth.
>
>>From what I know from Bertrand Russell's books Number x is the class
>of all similar classes containing x members ( the definition seems
>cyclic but it is not ). similar classes means the existence of one to
>one relation between their members ( bijection).
Try reading Halmos's "Naive Set Theory" instead.
A binary relation <= on a set S is called a partial order if and only
if it satisfies the following properties:
(i) Reflexivity: a<= a for all a in S.
(ii) Antisymmetry: if a<=b and b<=a, then a=b, for all a,b in S.
(iii) Transitivity: for all a,b,c in S, if a<=b and b<=c, then a<=c.
A partial order is a total order if for all a,b in S, either a<=b or
b<=a.
A total order on S is a well order on S if and only if for every nonempty
subset X of S, there exists x in X such that x<=y for all y in X
(i.e., every nonempty subset has a least element).
An ordinal is a well-ordered set S such that for all a in S, the
subset {x in S : x<a } is equal to the element a.
Some examples of ordinals: the empty set, ordered by set inclusion, is an
ordinal. This is usually called 0.
The set whose only element is the empty set is an ordinal; {0} is
usually called "1".
The set {emptyset, {emptyset}} is an ordinal, ordered by set
inclusion. This is {0,1}, and is usually called "2".
The set {0,1,2} is an ordinal (ordered by inclusion), and usually
called "3".
Continuing in this manner you obtain all the natural numbers: if you
have already defined all the natural number 0 through n, then "n+1" is
the ordinal {0, 1, 2, 3,..., n}, ordered by set inclusion.
In general, if a is an ordinal, ordered by set inclusion, then the set
a U {a}, which contains all elments of a, plus a itself, is also an
ordinal, denoted a+1. Such an ordinal is called a "successor ordinal".
omega is the union of the ordinals 0, 1, 2, 3, 4, ..., n, ...
And so on.
A cardinal is an ordinal which cannot be bijected with any ordinal
strictly smaller than itself.
>So number two is the class of all doubles and number three is the class
>of all triples. Review " Introduction to mathematical philosophy".
Please read some MATH, not some philosophy. These definitions don't
really work inside set theory, as Russell himself demonstrated. You
need to go to class theory, and even there they are a bit dodgy.
>An Ordinal number is the number terms in a series, it is not the
>series. It is the class of all similar series.( similar has the same
>meaning above).
>
>A cardinal number is the number of members in a set, ie the class of
>all similar sets, and it is not affected by order.
Both of these notions are at odds with the standard
definitions. Ordinals are specific sets, as are cardinals; they are
not "quantities" or "number of elements" or any such animal. No wonder
you sound like a loony.
>What is a real number ? I don't have an exact definition , but from
>what is writtin in this forum it seems to be a number which possess
>inductive properties, the kind of a number Bertrand Russells refers to
>as inductive number or finite number, perhaps?.
No. None of these things.
An ordinal is finite if and only if it cannot be bijected with a
proper subset of itself. The "natural numbers" are the finite
ordinals.
One can define an addition and a multiplication of natural numbers
inductively. For any natural number n, we define
n + 0 = n
n + (a+1) = (n+a) + 1
for every natural number a.
Then one can define multiplication also inductively:
n * 0 = 0
n * (a+1) = (n*a) + n
for every natural number n.
Once you have the natural numbers, you can define the integers. One
way is to consider the collection of all pairs (n,m) of natural numbers;
define an equivalence relation so that (a,b) ~ (n,m) if and only if
a+m = b+n. The integers are the equivalence classes under this
relation.
One can show that every equivalence class contains either a pair of
the form (n,0) or a pair of the form (0,n), with n a natural
number. If n=0, we call this equivalence class "0". Otherwise, we
denote the class of (0,n) also by "n", and the class of (n,0) by
"-n".
(Intuitively, the pair (a,b) represents the solution to the equation
"a+x=b"; define addition and multiplication accordingly)
It is then an easy exercise to show that the map that sends the
natural number n to the class of (n,0) respects addition and
multiplication and order, so that we can view the natural numbers as
contained in the integers, and this abuse of notation is justified.
Once we have the integers, you can define the rationals. Consider the
set of all pairs (a,b) of INTEGERS, with b nonzero. Define the
equivalence relation (a,b) ~ (x,y) if and only if ay=bx. (Intuitively,
the pair (a,b) represents the solution to the equation ax=b).
A "rational number" is an equivalence class under this. We denote the
equivalence class of (a,b) by "a/b"; each equivalence class contains
one and only one pair (a,b) such that gcd(a,b)=1, and then "a/b" is
the "expression in least terms". Define addition by (a,b) + (x,y) =
(ay+bx,xy), and multiplication by (a,b)*(x,y)=(ax,by).
It is then also an easy exercise to show that the rationals contain a
copy of the integers, namely the integer n corresponds to the rational
n/1 (or to the class of (n,1)).
Once you have the rationals, the reals can be constructed in any
number of ways.
A "Dedekind Cut" of the rationals is a partition of the rationals into
two sets, (A,B), such that:
(i) A U B is all the rationals.
(ii) A/\B is empty.
(iii) for each a in A and b in B, a<b.
(iv) A and B are each nonempty.
These partitions come in three flavors: either A has a largest element
(in which case B has no smallest element); or B has a smallest
element (in which case A has no largest element); or A has no largest
and B has no smallest element.
For example, the partition A = {x : x <=0}, B = {x : x>0} is of the
first type.
The partition A = {x : x<1}, B = {x: x>=1} is of the second kind.
The partition that has B = { x: x>0 and x^2 >= 2}, and has A =Q-B
is of the third kind.
The real numbers can be defined in terms of equivalence classes of
these Dedekind cuts. Intuitively, in either case 1 or case 2, the cut
represents a rational; in case 3, it represents an irrational.
Read "Continuity and irrational number" by Richard Dedekind; in
"Essays on the Theory of Numbers" by Richard Dedekind, translated by
Wooster Woodruff Beman. Dover Publications, Inc.
Alternatively, one can define a "distance" of rational numbers by the
usual means: the distance between a/b and c/d is |ad-bc|/bd, where
|ad-bc| is the absolute value of ad-bc.
A sequence of rationals is function from the natural numbers to the
rationals.
We say a sequence (a_0, a_1,...) (usually denoted {a_i}) converges to
the rational number Q if and only for every N>0 there exists M>0 such
that if n>M, then |a_n-Q|<1/N.
We say a sequence (a_0, a_1, ...) is a "Cauchy sequence" if and only
if for every N>0 there exists M>0 such that if n,m>M, then
|a_n-a_m|<1/N.
It is easy to verify that if a sequence converges to some rational,
then it is Cauchy, though the converse does not hold.
We can define an equivalence relation among sequences by saying that
the sequence {a_i} and the sequence {b_i} are "equivalent" if and only
if the sequence {a_i-b_i} is a Cauchy sequence.
It is an easy exercise to show that if {a_i} converges to q and {a_i}
is equivalent to {b_i}, then {b_i} also converges to q. And if {a_i}
is Cauchy and {b_i} is equivalent to {a_i}, then {b_i} is also Cauchy.
The "real numbers" can be defined to be the set of all equivalence
classes of Cauchy sequences of rationals. Those that converge
correspond to the rationals; those that do not converge to a rational
correspond to the irrationals.
It is this latter definition that gives rise to the numerical
representation. A decimal expansion
N.d1d2d3....
with N an integer, di an integer between 0 and 9, is short hand for
the sequence
(N, N+d1/10, N+(d1/10)+(d2/100), ..., N + (d1/10) + (d2/100) + ... + (dn/10^n),...)
which can easily be verified is a Cauchy sequence; so the decimal
expansion represents the EQUIVALENCE CLASS of cauchy sequences
corresponding to this sequence. It is again a trivial exercise to
show, for example, that the Cauchy sequence represented by
1.000000... (which is the constant sequence (1,1,1,1,...)) and the
Cauchy sequence represented by 0.9999.... (which is the sequence
(9/10, 99/100, 999/1000, ....)) are equivalent Cauchy
sequence. Therefore, a fortiori, they represent the same "real
number".
>I heard that division is not defined for ordinals , but I don't know
>why? Division of transfinite cardinals by finites is defined,
No, it is not defined. At least, not in STANDARD cardinal theory. You
can define anything you want, naturally. But for example, the USUAL
meaning of division is that "a/b" represents the UNIQUE element c such
that b*c = a. This does not work for cardinals, except in the very
limited situation in which a and b are both finite, and b divides a
(in the usual sense of natural numbers).
Do read some set theory, man.
> why division and subtraction is not defined for ordinals?
Because division is defined in terms of the inverses for
multiplication, and subtraction is defined in terms of the additive
inverses. Neither of them exist for ordinals nor for cardinals.
--
Dave Seaman
> We can define an equivalence relation among sequences by saying that
> the sequence {a_i} and the sequence {b_i} are "equivalent" if and only
> if the sequence {a_i-b_i} is a Cauchy sequence.
But that would make any two Cauchy sequences equivalent. I think what
you mean is {a_i} and {b_i} are equivalent iff {a_i-b_i} converges to 0.
Arturo Magidin
Dave Seaman <dse...@no.such.host> wrote:
>On Thu, 27 Apr 2006 15:13:03 +0000 (UTC), Arturo Magidin wrote:
>
>> We can define an equivalence relation among sequences by saying that
>> the sequence {a_i} and the sequence {b_i} are "equivalent" if and only
>> if the sequence {a_i-b_i} is a Cauchy sequence.
>
>But that would make any two Cauchy sequences equivalent. I think what
>you mean is {a_i} and {b_i} are equivalent iff {a_i-b_i} converges to 0.
Quite right. Sorry about that. That's what happens when I try to
squeeze what is about two weeks of lectures into a single post.
zuhair
Tell me what is that trivial exercise.
Zuhair
>
>
>
> >I heard that division is not defined for ordinals , but I don't know
> >why? Division of transfinite cardinals by finites is defined,
>
> No, it is not defined. At least, not in STANDARD cardinal theory. You
> can define anything you want, naturally. But for example, the USUAL
> meaning of division is that "a/b" represents the UNIQUE element c such
> that b*c = a. This does not work for cardinals, except in the very
> limited situation in which a and b are both finite, and b divides a
> (in the usual sense of natural numbers).
>
> Do read some set theory, man.
>
>
> > why division and subtraction is not defined for ordinals?
>
> Because division is defined in terms of the inverses for
> multiplication, and subtraction is defined in terms of the additive
> inverses. Neither of them exist for ordinals nor for cardinals.
Thanks .
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
[.snip.]
>> Alternatively, one can define a "distance" of rational numbers by the
>> usual means: the distance between a/b and c/d is |ad-bc|/bd, where
>> |ad-bc| is the absolute value of ad-bc.
>>
>> A sequence of rationals is function from the natural numbers to the
>> rationals.
>>
>> We say a sequence (a_0, a_1,...) (usually denoted {a_i}) converges to
>> the rational number Q if and only for every N>0 there exists M>0 such
>> that if n>M, then |a_n-Q|<1/N.
>>
>> We say a sequence (a_0, a_1, ...) is a "Cauchy sequence" if and only
>> if for every N>0 there exists M>0 such that if n,m>M, then
>> |a_n-a_m|<1/N.
>>
>> It is easy to verify that if a sequence converges to some rational,
>> then it is Cauchy, though the converse does not hold.
>>
>> We can define an equivalence relation among sequences by saying that
>> the sequence {a_i} and the sequence {b_i} are "equivalent" if and only
>> if the sequence {a_i-b_i} is a Cauchy sequence.
Note the correction: this should read "the sequence {a_i-b_i}
converges to 0".
>> It is this latter definition that gives rise to the numerical
>> representation. A decimal expansion
>>
>> N.d1d2d3....
>>
>> with N an integer, di an integer between 0 and 9, is short hand for
>> the sequence
>>
>> (N, N+d1/10, N+(d1/10)+(d2/100), ..., N + (d1/10) + (d2/100) + ... + (dn/10^n),...)
>>
>> which can easily be verified is a Cauchy sequence; so the decimal
>> expansion represents the EQUIVALENCE CLASS of cauchy sequences
>> corresponding to this sequence. It is again a trivial exercise to
>> show, for example, that the Cauchy sequence represented by
>> 1.000000... (which is the constant sequence (1,1,1,1,...)) and the
>> Cauchy sequence represented by 0.9999.... (which is the sequence
>> (9/10, 99/100, 999/1000, ....)) are equivalent Cauchy
>> sequence. Therefore, a fortiori, they represent the same "real
>> number".
>
>Tell me what is that trivial exercise.
Can you be bothered to try anything that contradicts your
preconceptions or challenges your ignorance, or must everything be
done for you?
The first sequence is {a_i} with a_i = 1 for all i. The second
sequence is {b_j} with b_j = (10^j-1)/10^j for each j.
By definition, {a_i} is equivalent ot {b_j} if and only if
{a_i - b_i} converges to 0. First, let c_i = a_i - b_i. Then
c_i = 1 - [(10^j-1)/10^j] = [10^j - 10^j + 1]/10^j = 1/10^j.
Does the sequence {1/10^j} converge to 0? According to the DEFINITION,
the sequence {c_i} converges to 0 if and only if for every N>0 there
exists M>0 such that, for all n>M, |c_n - 0| < 1/N.
So, let N>0. Then there exists M such that 10^M > N. Therefore, for
all n>M, we have
|c_n-0| = |c_n| = c_n = 1/10^n < 1/10^M < 1/N.
Thus, for every N>0 there exists M>0 such that for all n>M,
|c_n-0|<1/N. This proves, BY DEFINITIION, that the sequence
{c_i}={a_i-b_i} converges to 0. BY DEFINITION, this means that the
sequence {a_i} and the sequence {b_i} are equivalent. This means, BY
DEFINITION, that the real number corresponding to the equivalence
class of the sequence {a_i} (which was the real number represented by
the decimal expansion 1.0000....) and the real number corresponding to
the equivalence class of the sequence {b_i} (which was the real number
represented by the decimal expansion 0.9999....) are the same real
number, since the two equivalence classes are the same.
zuhair
hmmm.............
Jee thanks.
I should study that subject.
Zuhair
Tony Orlow
Okay, symbolically, 0.999... is an infinite digital string, but the value isn;t
infinite. It's important, I believe, to distinguish between symbolic infinities
and quantitative ones. But okay....
>
> The difference is 1/ ( 10^ Omega). this is because the ordinal number
> of the repeatition of numeral 9 in 0.9999.... is Omega, or in simple
> words the numeral 0.9999...... contains Omega of nines.
Oh, you're causing me pain, Zuhair. It hurts so bad.... :)
You are going to run into trouble if you try to use omega or aleph_0, I
guarantee it. But, I'll agree that if you have some specific infinite number n
of 9's, you will have a difference of 1/10^n, which would be infinitesimal.
>
> However I can think of another even bigger number than 0.9999...... and
> yet smaller than 1.0000... , and that would be 0.9999...... 9 here we
> have Omega +1 nines on the right of the decimal point and so 1 -
> 0.999.....9 = 1/[10^(Omega+1)] . which has a smaller
> ordinal value than 1/( 10 ^ Omega ).
Right, for any number of digits n, you can always add another, and get 1/10 the
previous difference.
>
> Virgil asked if the x/2 is smaller than x , according to the
> transfinite laws division of a transfinite number by a finite number
> reaveals the same transfinite number.
OW!! Oh, that hurt......oh the pain.....(wince)
Yes, those are the rules for transfinite numbers, in the standard world. That
argument MIGHT work on some, but eventually it's going to crash and burn. It
does seem like one could concoct a system of "subfinite" values that act like
transfinite "numbers", but I'm not sure what the point would be. You could
declare a largest finite, alpha, with a special rule that alpha+1=alpha, like
omega-1=omega, and that wouldn't be any less pointless. But, no criticism
intended. You're in the process of figuring all this out. That's good.
>
> So 1/ 10^Omega cannot be halfed, but yet there exists numbers that are
> larger than
> 1-x and yet smaller than 1, these like 1/ [10^(Omega+1)] etc.
Sounds fishy..... ;)
>
> Virgil said that the summation formula that I have presented earlier
> and i will repeat it below is a false summation formula, does any body
> agree with him:
>
> [1/n] + [1/(n^2)] + [1/(n^3)] +......+[1/(n^k)] = [1/(n-1)] -
> [1/{(n-1)(n^k) }]
I think that looks okay. I don't see why anyone would argue about that, at
least for finite n and k.
>
> n= 2,3,4,5,............
> k= 1,2,3,4,............
>
> Of coarse as k grow bigger, [1/{(n-1)(n^k) }] becomes smaller, and
> thus the left side of the equation will coverge to 1/( n-1) , but it
> will never reach it, because as I said before
>
> the term [1/{(n-1)(n^k) }] cannot equal zero at finite or infinite
> level because zero summed by itself finitelly or infinitely will always
> equal zero.
Well, that's true of aboslute 0, unless you multiply by absolute oo, in which
case I consider that equal to 1. But, that's just me.....
>
> Accordingly my proove that 0.9999......... < 1 is more logical.
I think it's true in a sense and false in a sense. Depends whether
infinitesimals count or not, and it depends whether you have a sepcific
infinite number of digits, or simply an endless absolute infinity. In the
second case, it's pretty impossible to make an expression for the difference.
>
> But Tony as you said if standard mathematics looks at 0.999........ as
> representing a finite number , then there is no real( finite) number
> other than 1 which it could better symbolize.
Well, it's an infinite string representing a finite quantity less than or equal
to 1. In standard math, if the difference between tow values is less than any
finite value, then the two values are considered equal, but the standard system
only applies to the finite scale. On the infinitesimal level, those two values
may have a measurable difference and be considered distinct. That's what I've
been trying to explain to Virgil. But of course I'm WRONG!!! :)
>
> Zuhair
>
>
--
Smiles,
Tony
zuhair
still there is a problem, when you say converges to zero, at what
number of terms in that series this will happen, is it at Omega of c_i
or Omega+1 or 2^Omega.
Your Cauchy priniciple didn't mention that. It doesn't differentiate
between
0.9999........
which contains Omega of nines in it after the decimal point.
and 0.9999....... 9 which contains Omega+1 of nines in it after the
decimal point.
In reality I tend to think that it is at what Cantor called once as
"The Absolute Infinity" number of terms in those series that c_i will
be zero
or in simple words c_ Absolute infinity = 0
Anything less than that number will have c_i > 0.
Zuhair
Tony Orlow
Ugh, that's probably true, technically. I don't use omega, aleph_0, or
countable infinity in my theory. That's all phlogiston and spontaneous
generation to me. So, you're almost definitely right about that. I'm no expert
in transfinite "arithmetic".
>
> I refuted Virgils statements because in standard theories Omega/n were
> n is finite
> is always Omega itself.
Yup. I guess if you want to refute their arguments, it might be helpful to do
it on their terms. Personally, I've decided the standard system needs
renovating from the ground up. Omega's a different thing for me, beyond
mathematics.
Have a nice day!
>
> Smiles
>
> Zuhair
> >
> > Anyway, to address your questions, the power set of the continuum is the set of
> > all subsets, that is, the set of all unique sets of real numbers. So, your
> > combinations of these reals should be as sets, each pair being separated by
> > commas, not combined into a single string by concatentation. In fact, forget
> > the digital representation. That's a different kind of infinity.
> >
> > As an alternative image, consider c to be the 1 dimensional continuum, the real
> > line. Then c^n would be a continuous space of n dimensions. This is still
> > considered to be the same cardinality as c, as aleph_0^n is aleph_0 for finite
> > n. You could consider c^aleph_0 to be a countably infinite-dimensional space,
> > but this is equal to (2^aleph_0)^aleph_0 = 2^(aleph_0^2)=2^aleph_0=c. So, this
> > hyperdimensional space is considered to contain no more points than the 1D
> > line. In order to get to the next higher cardinality, you would have to have at
> > least 2^c, which can be considered the size of that set of all possible sets of
> > reals. If you want a geometric image of this, consider the number of vertices
> > on a c-dimensional hypercube, in an uncountably infinite-dimensional space.
> > That's 2^c. I'm sure that clarified your image, didn't it, Zuhair? heh.
> >
> > To all the standard bearers in the group, please correct me if I said anything
> > incorrect above, besides the opinions expressed of course. :)
> >
> >
> > --
> > Smiles,
> >
> > Tony
>
>
--
Smiles,
Tony
Arturo Magidin
Look at the DEFINITION:
>> >> A sequence of rationals is function from the natural numbers to the
>> >> rationals.
>> >> We say a sequence (a_0, a_1,...) (usually denoted {a_i}) converges to
>> >> the rational number Q if and only for every N>0 there exists M>0 such
>> >> that if n>M, then |a_n-Q|<1/N.
>still there is a problem, when you say converges to zero,
No, there is no problem when I say that. I am going EXACTLY by the
definition.
You may, perhaps, object to the definition, but then you are objecting
to all of mathematics.
>at what
>number of terms in that series this will happen,
This question is nonsense. I did NOT say that the sequence is
"eventually equal to zero". That is something else. I said the
sequence CONVERGES to zero. That is something else, and was EXPLICITLY
and CLEARLY defined for you above.
> is it at Omega of c_i
>or Omega+1 or 2^Omega.
None of them. There is no omega here. Sequences are indexed by NATURAL
numbers. Natural numbers are, BY DEFINITION, finite. Omega is NOT a
natural number. "The sequence converges to zero" is NOT equivalent to
"the sequence is eventually equal to zero".
>Your Cauchy priniciple didn't mention that.
Yes, it did. It gave ALL the definitions. You just seem unable to
parse them.
A sequence of rationals was defined to be a function from the NATURALS
to the rationals. We write it as (a_0, a_1, a_2, ..., a_n, ...), where
"a_n" is the value of the function at the natural number n. The only
terms that are defined are the terms indexed by NATURAL numbers, which
are FINITE. There is no "omega", there is no "omega + 1", there is no
"2^omega". There are only finite numbers.
The definition of "cauchy sequence" and "convergence" relates ONLY to
sequences; thus, it relates ONLY to finite indices. The definition of
convergence and of cauchy sequence is given entirely in terms of
FINITE indices and conditions on them.
Period.
> It doesn't differentiate
>between
>0.9999........
>which contains Omega of nines in it after the decimal point.
>
>and 0.9999....... 9 which contains Omega+1 of nines in it after the
>decimal point.
That expression does NOT represent a sequence, so it is not under
consideration. It is merely a symbolic string of symbols which does
not represent a real number.
>In reality I tend to think that it is at what Cantor called once as
>"The Absolute Infinity" number of terms in those series that c_i will
>be zero
In reality I tend to think that you are not bothering to
think. Certainly, you seem either unwilling or incapable of reading
simple mathematical definitions.
zuhair
I know that converges into zero is something else other than equal
zero.
The definition states that if the difference converges into zero then
the two Cauchy sequences are equivalent, I disagree with that.
They should be equivalent only when the difference REACHES zero.
and it should specify at which number of terms that would happen.
It is obvious that for any finite number of terms n , c_n is 1/10^n >
0
But if the number of terms in the sequence is infinite then there might
be a possibility of c_i equalling zero, but that would be imaginable at
absolute infinity only.
I agree with you in that you are working according to the definition.
But the definition itself is not convincing.
So I am questioning the definition itself.
Now I want you to answer that question please.
Define: 0.9999......... as 9/10 + 9/10^2 + 9/10^3 + ............. +
9/10^Omega
( 0.9999....... is defined to contain Omega of nines after the decimal
point )
would that number be eqivalent to 1.0000........
My point is that you now that for example the number 0.9999.......nth9
were n is finite , is never eqivalent to 1 , it is always lower than
one by 1/10^n.
Now for the number above which has Omega of nines in it after its
decimal point, would that number
be equivalent to 1. or you think that the difference 1/10^Omega is
larger than zero.
That was my question, can you answer it.
And suppose there is another number 0.9999....... 2^Omega th 9 , ie a
number which contains
2^Omega of nines after the decimal point.
Is that number different from the first number with Omega repeations of
9. or is the same
Is it eqivalent to number 1.0000........ or is lower than it.
Best,
Zuhair
zuhair
???????????? this is an escape rather than an answer.
Arturo Magidin
[.snip.]
>I know that converges into zero is something else other than equal
>zero.
Yet you asked for the index at which a sequence that converged to zero
had the terms equal to zero.
>The definition states that if the difference converges into zero then
>the two Cauchy sequences are equivalent, I disagree with that.
You cannot disagree with a definition. You can say that the definition
does not capture an idea that you have, or does not represent what you
think it should represent, but you cannot "disagree with a definition."
>They should be equivalent only when the difference REACHES zero.
That would be something else, and not the definition of equivalence
which is used to define real numbers. Change the definition if you
want, but then you are not working with what every one else in the
world calls the real numbers.
In other words: you are objecting to all of mathematics. You can play
with your own rules if you want, but be aware that what you are doing
has NOTHING to do with real numbers as they are understood by everyone
else in the world.
>and it should specify at which number of terms that would happen.
Then your definition of "real numbers" does not agree with the
definition of anybody else. You can call them "zuhair numbers" if you
want, but then you need to PROVE everything about them, since nobody
else works with them.
Good luck, and keep me out of it.
>So I am questioning the definition itself.
What you are doing is saying that you don't like the
definition. That's YOUR problem, not the problem of the definition. It
means you are unwilling to work with the standard rules as they relate
to real numbers. You can do that, but then you are NOT working with
real numbers.
>That was my question, can you answer it.
The answer is: you are not working with real numbers. You are working
with an invention of yours that are not what everyone else in the
world calls the real numbers. Calling them "real numbers" is either
confused or dishonest on your part.
Everything you have said is nonsense if we use the words "ordinal",
"cardinal", "natural", "omega", "real number", "decimal expansion",
etc. in their STANDARD usages.
Since you seem only interested in playing with yourself, do go ahead.
Arturo Magidin
[.snip.]
>> > It doesn't differentiate
>> >between
>> >0.9999........
>> >which contains Omega of nines in it after the decimal point.
>>
>> >
>> >and 0.9999....... 9 which contains Omega+1 of nines in it after the
>> >decimal point.
>>
>> That expression does NOT represent a sequence, so it is not under
>> consideration. It is merely a symbolic string of symbols which does
>> not represent a real number.
>
> ???????????? this is an escape rather than an answer.
Just because you don't like the answer does not mean it is not an
answer. Your question was the logical equivalent of "what kind of duck
is a fox terrier?" It would seem you would consider the answer "a fox
terrier is not a duck" to be "an escape rather than an answer."
A sequence is, BY DEFINITION, a map whose domain is the natural
numbers. A decimal expansion represents a sequence. Therefore, a
decimal expansion can only have digits at positions that correspond to
natural numbers. When one writes a decimal expansion, in the world
everyone but you plays in, one is using this expansion as shorthand
for a specific equivalence class of Cauchy sequences of rationals
(under the standard definition of equivalence, not the one you think
"should" be the definition). Thus, decimal expansions, BY DEFINITION,
contain digits only in locations correspoding to natural
numbers. Omega is not a natural number. You can define a function from
omega + 1 = omega \/ {omega}, whose range are digits, but in that case
that function is quite simply not a sequence, just as a fox terrier is
not a duck. So the symbolic string of digits which represents the
function from omega\/{omega} which has value 9 at all elements is NOT
a sequence, and not a real number.
It is not an escape. It is a precise, logical, correct answer. Don't
like it? Too bad.
zuhair
Arturo Magidin wrote:
> In article <1146165828....@u72g2000cwu.googlegroups.com>,
> zuhair <zalj...@yahoo.com> wrote:
> >
> >Arturo Magidin wrote:
>
> [.snip.]
>
>
> >> > It doesn't differentiate
> >> >between
> >> >0.9999........
> >> >which contains Omega of nines in it after the decimal point.
> >>
> >> >
> >> >and 0.9999....... 9 which contains Omega+1 of nines in it after the
> >> >decimal point.
> >>
> >> That expression does NOT represent a sequence, so it is not under
> >> consideration. It is merely a symbolic string of symbols which does
> >> not represent a real number.
> >
> > ???????????? this is an escape rather than an answer.
>
> Just because you don't like the answer does not mean it is not an
> answer. Your question was the logical equivalent of "what kind of duck
> is a fox terrier?" It would seem you would consider the answer "a fox
> terrier is not a duck" to be "an escape rather than an answer."
>
> A sequence is, BY DEFINITION, a map whose domain is the natural
> numbers.
Fair enough, this mean that the biggest number of terms in a sequence
would be Omega.
Or in other words a sequence either contains n terms when n is finite,
or Omega of terms if the number of terms in it is infinite.
Therefore the decimal expansion of 0.99999......... has Omega of 9 in
it.
and I repeat it would be lesser than 1 by 1/10^Omega.
Perhaps this is non sense. I don't know.
However you succeeded in illustrating to me that the symbole
0.9999.......9 is not a sequence
simply because the last 9 though a finite number but it is at a
transfinite position that is Omega+1
and therefore it is not a sequence.
well according to what you are saying then 1/10^Omega =0 according to
standard mathematics.
which something very strange as I see.
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>
>> Just because you don't like the answer does not mean it is not an
>> answer. Your question was the logical equivalent of "what kind of duck
>> is a fox terrier?" It would seem you would consider the answer "a fox
>> terrier is not a duck" to be "an escape rather than an answer."
>>
>> A sequence is, BY DEFINITION, a map whose domain is the natural
>> numbers.
>
>Fair enough, this mean that the biggest number of terms in a sequence
>would be Omega.
You really need to stop using standard mathematical words with your
own private meaning.
"Omega" is not a "quantity"; you do not speak about there being "omega
things". Omega is an ordinal. It corresponds to a well-ordering type.
And, no. It means EXACTLY that a sequence is a function whose domain
is the natural numbers. There are ->exactly<- one term for each
element of omega. Not "the biggest number of terms". There is exactly
one term for each element of omega.
>Or in other words a sequence either contains n terms when n is finite,
>or Omega of terms if the number of terms in it is infinite.
Will you ever learn to read?
A sequence is a function whose DOMAIN is the natural numbers. Not
whose domain is a subset of the natural numbers. There is always one
term for each natural number.
If you want to talk about "finite sequences", whose domains are the
sets I called "n" (where the set 0 is the empty set, and the set "n+1"
is the set {0, 1, ..., n}), then call them finite sequences.
>Therefore the decimal expansion of 0.99999......... has Omega of 9 in
>it.
No. Omega is not a quantity. omega is an ordinal.
>and I repeat it would be lesser than 1 by 1/10^Omega.
And it does not matter how many times you repeat it, it is still
nonsense. The decimal expansion represents a sequence. The sequence is
equivalent to the constant sequence 1, and therefore, since "1" and
"0.999..." are two distinct representations of the same equivalence
class, they are the SAME real number, by definition.
>Perhaps this is non sense. I don't know.
Yes, you do know. Because you have been told this many times. You just
ignore it and repeat it over and over and over. No matter how often
you repeat it, it is still nonsense. The symbol "10^omega" does not
represent a real number. You can define addition and multiplication of
ordinals, as well as exponentiation; under the standard definitions,
10^omega is the ordinal omega. And there is no standard definition of
"division of ordinals", so no matter how many times you repeat the
nonsense of writing "1/10^omega", it is still a symbol that lack any
meaning. It is nonsense.
Just because you can write down a symbol it does not, ipso facto,
grant it sense.
>However you succeeded in illustrating to me that the symbole
>0.9999.......9 is not a sequence
>simply because the last 9 though a finite number but it is at a
>transfinite position that is Omega+1
Actually, no. It would be in position omega, not position omega+1. The
order type of the digits is Omega+1; that is, your symbol represents a
function with DOMAIN omega+1, which has value 9 at each natural number
and value 9 at omega (recall that, BY DEFINITION,
omega+1 = omega \/ {omega} = {0, 1, 2, ..., n, ...} \/ {omega}
>and therefore it is not a sequence.
>
>well according to what you are saying then 1/10^Omega =0
No. It still has no meaning whatsoever.
>according to
>standard mathematics.
No. According to standard mathematics, the symbol you insist on
repeating is still complete and utter nonsense.
>which something very strange as I see.
That is quite plain.
zuhair
sorry I made a mistake the last 9 is of coarse 9/10^Omega+1 and this
is not a finite number
and its position of of coarse not finite.
but the number 0.9999......... satisfies the definition of sequence and
thus contains Omega
of terms in it, all terms are finite at finite positions, but the
number of these terms is transfinite
and it is Omega.
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> Arturo Magidin wrote:
> The definition states that if the difference converges into zero then
> the two Cauchy sequences are equivalent, I disagree with that.
One cannot "disagree" with a definition in mathematics. One either
accepts it or rejects it.
>
> They should be equivalent only when the difference REACHES zero.
Using that as your equivalence relation, what you would get is only
another copy of the rational numbers, not the usual real numbers at all.
>
> Now I want you to answer that question please.
>
> Define: 0.9999......... as 9/10 + 9/10^2 + 9/10^3 + ............. +
> 9/10^Omega
>
> ( 0.9999....... is defined to contain Omega of nines after the decimal
> point )
>
> would that number be eqivalent to 1.0000........
>
> My point is that you now that for example the number 0.9999.......nth9
> were n is finite , is never eqivalent to 1 , it is always lower than
> one by 1/10^n.
Such objects are not any part of the real numbers at all, as 10^omega is
itself not a real number
>
> Now for the number above which has Omega of nines in it after its
> decimal point, would that number
> be equivalent to 1. or you think that the difference 1/10^Omega is
> larger than zero.
>
> That was my question, can you answer it.
Since it requires the existence of a real number which cannot be a real
number, the answer is that no such thing exists as a real number.
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
Since the question itself implies a condition contrary to fact, there
cannot be an answer to that question. One can only point out the
impossibility of a true answer.
It is like asking a bachelor
"Have you stopped beating your wife?"
Arturo Magidin
Virgil <vmh...@comcast.net> wrote:
>In article <1146165177.0...@e56g2000cwe.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
>> Arturo Magidin wrote:
>
>> The definition states that if the difference converges into zero then
>> the two Cauchy sequences are equivalent, I disagree with that.
>
>One cannot "disagree" with a definition in mathematics. One either
>accepts it or rejects it.
>>
>> They should be equivalent only when the difference REACHES zero.
>
>Using that as your equivalence relation, what you would get is only
>another copy of the rational numbers, not the usual real numbers at all.
I don't think you get "the rational numbers". Cauchy sequences that
converged to a rational number but were not eventually constant would
not be equivalent to the constant sequence corresponding to that
rational; and any such sequence would be non-equivalent to any shift
on that sequence. So the sequences
(1, 1, 1, ...)
(1/2, 3/4, 7/8, ..., (2^n-1)/2^n, ...)
(0, 1/2, 3/4, 7/7, ..., 2^{n-1}-1/2^{n-1}, ... )
would lie in distinct equivalence classes. Your equivalence classes
are just the classes of all sequence that are eventually equal, so I
think you would get essentially just the set of all Cauchy sequences
back (essentially, mind you).
Virgil
mag...@math.berkeley.edu (Arturo Magidin) wrote:
I spoke too quickly. The zero equivalence class would be the set of
sequences with finite carrier (zero for all but finitely many terms).
An equivalence class in general would be the set of all sequences which
differed from a given sequence at only finitely many terms.
The result would not even be a field, since there are obvious zero
divisors. e.g., the product of a sequence zero for all odd terms times a
sequence zero at all odd positions would always be the zero sequence.
zuhair
Arturo Magidin wrote:
> In article <1146169646.1...@g10g2000cwb.googlegroups.com>,
> zuhair <zalj...@yahoo.com> wrote:
> >
>
> >> Just because you don't like the answer does not mean it is not an
> >> answer. Your question was the logical equivalent of "what kind of duck
> >> is a fox terrier?" It would seem you would consider the answer "a fox
> >> terrier is not a duck" to be "an escape rather than an answer."
> >>
> >> A sequence is, BY DEFINITION, a map whose domain is the natural
> >> numbers.
> >
> >Fair enough, this mean that the biggest number of terms in a sequence
> >would be Omega.
>
> You really need to stop using standard mathematical words with your
> own private meaning.
>
> "Omega" is not a "quantity"; you do not speak about there being "omega
> things". Omega is an ordinal. It corresponds to a well-ordering type.
There should be some expression which can describe the quantity of
terms in a sequance
Since a sequence is a function who's domain is the natural numbers.
Then it should have
Aleph-0 of terms in it, because the multiplicity of terms in the
natural numbers set is descriped
by Aleph-0.
So 0.9999......... if it is to represent a real number then it should
be a cauchy sequence
and accordingly it should have Aleph-0 of nines in it.
Is that Correct?
zuhair
I wonder why subtraction and division are not defined for Transfinite
Ordinal numbers?
For example : n + Omega = Omega
but Omega + n > Omega
Now we can define two operators of subtraction , on is left subtraction
denoted as " L -" and the other is right subtraction denoted as "-R" as
followes
Define:
Omega L- n = Omega
Omega -R n < Omega
n is a finite number
Also about division we can also define two division operators in a
similar way L / and /R as below
Omega L / n= x is x: n.x = Omega
Omega /R n = z is z : z.n = Omega
For example Omega L / 3 = ( 1L / 3 ) Omega < Omega, because
3.Omega>Omega
While Omega / R 3 = Omega. 3 = 3+3+3+3+.......... = Omega
Now 1 L / Omega < Omega = infentismal = Left one omegath
but 1 / R Omega = ???
Any how if these are valid both right and left division operators
cannot be zero.
because even if the right division operator is not defined yet it
cannot be zero because
Omega of zeros is a zero and not one.
It is the left divion operator that might be interesting
1 L / Omega = 0 means 0.Omega =1 means Omega L - Omega = 1
now Omega L - Omega would be and indeterminate number like 0/0 .
While Omega -R Omega = Zero.
However all the above might seem to be loonatic to professional
mathematicians.
Anyhow I am still conviniced that we can form numbers like 1/2^Omega or
1/2^Aleph-x
And they cannot be zeros, but of coarse we should specify the division
operator.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
Learn to edit your replies and remove that which you are no longer
addressing. You followed up on 16 lines of text, and left 90 lines
which are no longer relevant. You are wasting space and resources.
>Arturo Magidin wrote:
>> In article <1146169646.1...@g10g2000cwb.googlegroups.com>,
>> zuhair <zalj...@yahoo.com> wrote:
>> >
>>
>> >> Just because you don't like the answer does not mean it is not an
>> >> answer. Your question was the logical equivalent of "what kind of duck
>> >> is a fox terrier?" It would seem you would consider the answer "a fox
>> >> terrier is not a duck" to be "an escape rather than an answer."
>> >>
>> >> A sequence is, BY DEFINITION, a map whose domain is the natural
>> >> numbers.
>> >
>> >Fair enough, this mean that the biggest number of terms in a sequence
>> >would be Omega.
>>
>> You really need to stop using standard mathematical words with your
>> own private meaning.
>>
>> "Omega" is not a "quantity"; you do not speak about there being "omega
>> things". Omega is an ordinal. It corresponds to a well-ordering type.
>
>There should be some expression which can describe the quantity of
>terms in a sequance
There is. But it is not the ordinal numbers. Ordinal numbers are about
->order<-, hence the name. What is used to describe quantities are the
cardinal numbers.
Of course, the fact that you do not know what either of them are (or
what the real numbers are, for that matter) goes a long way towards
explaining why you continue to spout nonsense.
>Since a sequence is a function who's domain is the natural numbers.
>Then it should have
>Aleph-0 of terms in it,
Yes; because the cardinality of the natural numbers is aleph_0. We use
cardinals when speaking about cardinality, ordinals when we are
dealing with well orders.
>So 0.9999......... if it is to represent a real number
No "if" about it. This is short hand for a Cauchy sequence of
rationals, and therefore ->does<- represent a (unique) real
number. Namely, 1, because the Cauchy sequence converges to 1.
>then it should be a cauchy sequence
It ->represents<- a Cauchy sequence of rationals. Decimal notation is
shorthand for a sequence. The sequence has as its d-th term (i.e., the
image of the natural number d) the rational number obtained by
truncating the decimal expression at the d-th decimal position. In
this case, the terms of the sequence are
(0, 0.9, 0.99, 0.999, 0.9999, ....)
or
(0, (10-1)/10, (10^2-1)/10^2, (10^3-1)/10^3, ..., (10^n-1)/10^n, ...)
>and accordingly it should have Aleph-0 of nines in it.
It has "aleph_0 nines" in it. A countably infinite number of nines in
it.
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
You quote over a hundred lines just so you can ignore them later. In
addition to your ignorance about mathematics, you also exhibit an
appalling ignorance about usenet ettiquette. That one is easier to
learn; why don't you?
>I wonder why subtraction and division are not defined for Transfinite
>Ordinal numbers?
I told you why: subtraction is really addition of additive
inverses. There are no additive inverses for ordinal numbers.
In other words: the symbol "-x" is the (unique) object which, when
->added<- to x, is equal to 0. By definition, "y-x" is short hand for
y + (-x), i.e., adding -x to y.
Alternatively, you can say that x-y = z if and only if z is the UNIQUE
object such that x = z+y, provided this exists.
The symbol x^{-1} is, by definition, the (unique) object which, when
multiplied with x, is equal to 1, when it exists; by definition, "y/x"
is shorthand for y*(x^{-1}).
Alternatively you can say that "y/x = z" if and only if z is the
UNIQUE object such that y = z*x, provided such a unique element exists.
Ordinal addition is non-commutative. So is ordinal
multiplication. That, by itself is no obstacle. But ordinal addition
does NOT have additive inverses. And the equation x = z+y does not
always have a unique solution z. So for example, writing "omega - n =
alpha" would mean that alpha is the unique object such that omega =
alpha+n. But there is no such element at all, unless n = omega, in
which case the alpha is "any natural number", and therefore is not
unique. No matter how you cut it, you don't have a well defined
subtraction for ordinals.
Even if you go to cardinals, where addition and multiplication is
commutative, you still run into problems. For infinite cardinals,
kappa + lambda = kappa*lambda = max{kappa, lambda}, so again there are
no unique solutions to the equations, so no well defined addition nor
multiplication.
Just because you can write down a symbol does not mean this makes sense.
>For example : n + Omega = Omega
>
>but Omega + n > Omega
>
>Now we can define two operators of subtraction ,
No, you cannot. Because your equations do not represent unique solutions.
>on is left subtraction
>denoted as " L -" and the other is right subtraction denoted as "-R" as
>followes
>
>Define:
>
>Omega L- n = Omega
You can make any definitions you want, but this objects do NOT satisfy
the rules that relate addition with "subtraction", and this
"subtraction" does not have the usual properties of subtraction. As
such, it is foolish to call it subtraction, because it is nothing like
subtraction.
With usual subtraction, if a -x = a - y, then x=y. Here, no such
thing. Just one example of why it is foolish to try to call it subtraction.
>Omega -R n < Omega
>
>n is a finite number
The ONLY ordinals strictly less than omega are the natural numbers. So
here you would be saying that "right subtracting" a finite number from
omega yields a finite number. Which makes this "subtraction"
something that has absolutely nothing to do with addition or
subtraction. Which is why we do not use it.
The rest of your nonsense suffers form the same problems. You
"define", but you never bother to think about the consequences of such
"definitions."
>Now 1 L / Omega < Omega = infentismal = Left one omegath
There are no "infinitesimals" among the reals or among the
ordinals. This is all nonsense in that context.
Even in the context of nonstandard analysis, where infinitesimals
exist, it is still false that 0.9999... is different from 1.
zuhair
Hmmm...........
With usual addition if a+x = b+y , a<>b , then x <> y
While a + Omega = b + Omega , a <> b
and it is not foolish to try to call it addition!
> >Omega -R n < Omega
> >
> >n is a finite number
>
> The ONLY ordinals strictly less than omega are the natural numbers. So
> here you would be saying that "right subtracting" a finite number from
> omega yields a finite number. Which makes this "subtraction"
> something that has absolutely nothing to do with addition or
> subtraction. Which is why we do not use it.
>
> The rest of your nonsense suffers form the same problems. You
> "define", but you never bother to think about the consequences of such
> "definitions."
>
>
> >Now 1 L / Omega < Omega = infentismal = Left one omegath
>
> There are no "infinitesimals" among the reals or among the
> ordinals. This is all nonsense in that context.
>
> Even in the context of nonstandard analysis, where infinitesimals
> exist, it is still false that 0.9999... is different from 1.
>
>
>
> --
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> mag...@math.berkeley.edu
Let me see:
If we visualise Omega as a raw or stars *****.......
and the finite natural number n as ****....n*
Now I will attempt to visualize ordinal summation and then to derive
ordinal subraction from it.
***...n* + *****....... = *****...n*****......... = ***......
While ****.... + ***...n* = ***....... ***..n* > ****.....
>From that one can understand the following definitions of subtraction
( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
a unique result
While ****...... + (-***...n*) = ****.......000...n0 < ****...... a
unique result
Now multiplication is of two types
n.Omega = Omega + Omega+Omega +.......+ nth Omega > Omega
and
Omega .n = n+n+n+n+............. = Omega
So it follows that we should have two types of division of ordinals.
1) Omega/ n = z , z: n.z=Omega
here z< Omega and it equal one nth of Omega
and
2) Omega\n = z , z: z.n = Omega
here it is obvious that z= Omega
In a similar way
n/Omega = z , z: Omega.z= n
here z should be one Omegath of n.
and
n\ Omega = z , z: z.Omega= n
here the only chance for z is when it is zero. because Omega-Omega= n
can be true.
I don't see these definitions all together non sense.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
You still seem incapable of adhering to usenet ettiquette and edit
your replies. More evidence of laziness, I gather.
>> You can make any definitions you want, but this objects do NOT satisfy
>> the rules that relate addition with "subtraction", and this
>> "subtraction" does not have the usual properties of subtraction. As
>> such, it is foolish to call it subtraction, because it is nothing like
>> subtraction.
>>
>> With usual subtraction, if a -x = a - y, then x=y. Here, no such
>> thing. Just one example of why it is foolish to try to call it subtraction.
>
>Hmmm...........
>
>With usual addition if a+x = b+y , a<>b , then x <> y
>
>While a + Omega = b + Omega , a <> b
>
>and it is not foolish to try to call it addition!
We do not call it "addition", we call it "ordinal addition".
[.snip a lot of stuff that you should have removed yourself,
for instance my sig; physical laziness to complement your
mental one.]
>Let me see:
>
>If we visualise Omega as a raw or stars *****.......
>
>and the finite natural number n as ****....n*
>
>Now I will attempt to visualize ordinal summation and then to derive
>ordinal subraction from it.
You are not deriving "ordinal subtraction"; you are, at best, finding
possible answers for specific expressions involving ->specific<-
ordinals (to wit, omega and finite ordinals). You are still mssing a
lot of ordinals.
>***...n* + *****....... = *****...n*****......... = ***......
>
>While ****.... + ***...n* = ***....... ***..n* > ****.....
This is just a complicated way of saying: for ordinal addition,
n+omega = omega, and omega+n = omega + n <> omega.
>>From that one can understand the following definitions of subtraction
>
> ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
> a unique result
Nonsense in this notation. There are no "0"s. All you have are
"*"s. The finite ordinals were finite sequences of *s, the ordinal
omega was a (countably infinite) sequence of stars, AND THAT IS ALL
YOU DEFINED. Where did this idiocy of 0s come from?
Sloppy thinking (if any thinking went into it).
And, the problem is not the "unique" answer here. Once again, you
demonstrate that you DID NOT bother to read what was written, and
prefer instead to babble on. You can define it any way you want and
have a unique answer. I could define it as
(-n) + omega = apple tree
and it would be a unique answer.
But n + appletree is not equal to omega. and n is not the unique
thing to which you can add apple tree to get omega.
>While ****...... + (-***...n*) = ****.......000...n0 < ****...... a
>unique result
Your expression "***... 0000" is nonsense in this notation. All
you have are *s, where did this nonsense "0" come from?
>Now multiplication is of two types
No. You have lots of types. The ordinals do not end at omega.
>n.Omega = Omega + Omega+Omega +.......+ nth Omega > Omega
>
>and
>
>Omega .n = n+n+n+n+............. = Omega
This is at odds with the standard definition of ordinal
multiplication. Standard ordinal multiplication in these cases would
be that
Omega * 0 = 0
Omega * (n+1) = Omega*n + Omega
and in general
alpha * 0 = 0
alpha * (beta + 1) = (alpha*beta) + alpha
and for limit ordinals gamma,
alpha * gamma = union (alpha * beta) with the union ranging over all beta<gamma.
And still you confuse "being able to define" an expression with the
expression making sense in the context you wish to use it. You are
able to define anything any way you want. But your definitions do NOT
interact well with ordinal addition, ordinal multiplication (whether
the standard or your opposite convention), ordinal inequalities, and
the like.
Your major error consists of thinking that just because you can write
down an expression "omega - n", and just because you can write down
"omega - n = k" for some k, that automatically implies all the usual
theorems about inequalities and the like. They do not; those theorems
come from properties of these operations which your definitions simply
do NOT have.
>I don't see these definitions all together non sense.
Nobody said you could not give definitions. You can define every
ordinal addition and every ordinal multiplication to be anything you
want. You can define them to be a pony; you can define them to be
Santa Claus; you can define them to be all equal to 3.
But the definitions simply do not satisfy the properties you want. The
definitions will not necessarily satisfy such properties as "alpha -
beta < alpha for all beta>0". They will not necessarily satisfy that
if alpha < beta then alpha - gamma < beta - gamma. They will not
necessarily satisfy that if alpha and beta are ordinals then "alpha -
beta" will be an ordinal.
Just giving a definition is not sufficient. Just writing down that you
will declare a string of symbols to be equal to another string of
symbols does not impart upon them the properties that the symbols have
in other contexts. And you insist not only in wanting to define the
symbols for "subtraction", you also want them to satisfy the
properties of subtraction such as alpha - beta < alpha. It is simply
assuming that such is true that is nonsense, and it is simply assuming
that there is an obvious definition which is also nonsense.
Arturo Magidin
I am tired of you not bothering to clean up your replies; of you
repeating the same thing; and of you ignoring what I write to continue
insisting on the same assertions even when I have explained what the
problems are. And this does not even touch on how tired I am of your
continued misuse of standard mathematical terminology.
I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it
even simpler: read the following book:
_Naive Set Theory_
by Paul R. Halmos.
Undergraduate Texts in Mathematics
Springer-Verlag
It is a small book: a mere 104 pages (including two in the index),
plus two of introduction, and one for a table of contents. It will
cover the basics of set theory, and the basics of natural numbers,
ordinals, and cardinals.
Then, and only then, return to these musings of yours. Until you
provide some evidence that you have bothered to attempt to acquire the
minimum of knowledge necessary to continue these discussions, and the
minimum of usenet ettiquette to do so without wasting time and space,
I will beg off.
zuhair
>From subtraction fool.
* - * = 0
got it, is it that difficult.
Dear Arturo it is obvious that you are idiot.
Zuhair
zuhair
You are very stuborn, you only wasted my time. I was looking forward to
discuss these matters with someone who is more enlightend than you,
some body like Tony.
Go and dwell in the darkness of your standard math. Stay their like a
rock in a mountine.
Zuhair
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>> >***...n* + *****....... = *****...n*****......... = ***......
>> >
>> >While ****.... + ***...n* = ***....... ***..n* > ****.....
>>
>> This is just a complicated way of saying: for ordinal addition,
>> n+omega = omega, and omega+n = omega + n <> omega.
>>
>> >>From that one can understand the following definitions of subtraction
>> >
>> > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
>> > a unique result
>>
>> Nonsense in this notation. There are no "0"s. All you have are
>> "*"s. The finite ordinals were finite sequences of *s, the ordinal
>> omega was a (countably infinite) sequence of stars, AND THAT IS ALL
>> YOU DEFINED. Where did this idiocy of 0s come from?
Can't even write without messing up the attributions, can you?
>
>From subtraction fool.
>
>* - * = 0
>
>got it, is it that difficult.
Nonsense. This is simply NOT contemplated in the notation YOU
introduced. The notation YOU introduced contained NOTHING but
"*"s. Then, suddenly, you introduce a new symbol which has NO meaning
in the notation YOU created. That makes it nonense.
>Dear Arturo it is obvious that you are idiot.
No, it is clear that ->YOU<- are a lazy idiot.
Either when you introduced the notation you failed to give it
COMPLETELY, or else you changed notation mid-stream. Either way, you
spoke nonsense.
Were ordinals expressed ONLY as "lists" of *s? That's what YOU DEFINED
THEM TO BE. Then putting "0"s makes it nonsense. Duh.
Arturo Magidin
zuhair <zalj...@yahoo.com> wrote:
>
>Arturo Magidin wrote:
>> In article <1146247083.7...@j33g2000cwa.googlegroups.com>,
>> zuhair <zalj...@yahoo.com> wrote:
>>
>> I am tired of you not bothering to clean up your replies; of you
>> repeating the same thing; and of you ignoring what I write to continue
>> insisting on the same assertions even when I have explained what the
>> problems are. And this does not even touch on how tired I am of your
>> continued misuse of standard mathematical terminology.
>>
>> I'll make it simple: GO READ A MATH BOOK ON THIS TOPIC. I'll make it
>> even simpler: read the following book:
>>
>> _Naive Set Theory_
>> by Paul R. Halmos.
>> Undergraduate Texts in Mathematics
>> Springer-Verlag
>>
>> It is a small book: a mere 104 pages (including two in the index),
>> plus two of introduction, and one for a table of contents. It will
>> cover the basics of set theory, and the basics of natural numbers,
>> ordinals, and cardinals.
>>
>> Then, and only then, return to these musings of yours. Until you
>> provide some evidence that you have bothered to attempt to acquire the
>> minimum of knowledge necessary to continue these discussions, and the
>> minimum of usenet ettiquette to do so without wasting time and space,
>> I will beg off.
>>
>
Right. Because I'm the one who repeated the same thing even after
agreeing it was nonsense as written, over and over again...
> you only wasted my time.
You wasted your own time, and you wasted mine by ignoring the replies
I wrote to you.
> I was looking forward to
>discuss these matters with someone who is more enlightend than you,
You mean, you were hoping someone would say you were right. Too bad
you were not.
>some body like Tony.
>
>Go and dwell in the darkness of your standard math. Stay their like a
>rock in a mountine.
Go buy a spellchecker.
zuhair
-***...n* + ****......... = 0000...n0***..........
The zeros denote the removed stars .
yet since 000...n0***..... can be bijected to ****........ then they
are equal
But ****........ + (-***...n* ) = ****....000...n0 = Omega-n and this
is a transfinite ordinal.
this number cannot be bijected to ****..... , in exactly the same why
****..... cannot be bijected
to ****..... ***...n* ( bijection her is respecting order )
The smallest transfinite ordinal that do not contain zeros in it after
the natural numbers is Omega
but ordinally speaking Omega is the smallest ordinal after the natural
numbers as Cantor thought
right finite subtracties of Omega are of smaller ordinal value than
omega but yet bigger ordinal value than any finite ordinal.
All of the above in absolutelly nonsense
Smiles
Just to wast you time Arturo
Zuhair
Virgil
If you are taking TO as your ideal, whatever you are interested in, it
is not mathematics and it is not knowledge.
Virgil
> Arturo Magidin wrote:
> > In article <1146247083.7...@j33g2000cwa.googlegroups.com>,
> > zuhair <zalj...@yahoo.com> wrote:
> > >>From that one can understand the following definitions of subtraction
> > >
> > > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
> > > a unique result
> >
> > Nonsense in this notation. There are no "0"s. All you have are
> > "*"s. The finite ordinals were finite sequences of *s, the ordinal
> > omega was a (countably infinite) sequence of stars, AND THAT IS ALL
> > YOU DEFINED. Where did this idiocy of 0s come from?
>
> >From subtraction fool.
>
> * - * = 0
>
> got it, is it that difficult.
One might make a case for
* - * = <the empty string>
OR
* - * = {}, (THE EMPTY SET} ",
but unless * represents some number, there is no case to be made for
* - * = 0
>
> Dear Arturo it is obvious that you are idiot.
>
> Zuhair
Arturo's only idiocy is continuing to try to teach one so invincibly
ignorant as Zuhair anything about mathematics.
>
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
Then you are essentially using a binary coding system based on two
symbols . "*" and either "0", or possible a space, rather than a unary
one based only on one symmbol.
>
>
> Just to wast you time Arturo
>
> Zuhair
Waste mostly snipped.
zuhair
Virgil wrote:
> In article <1146252352.8...@e56g2000cwe.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
> > Arturo Magidin wrote:
> > > In article <1146247083.7...@j33g2000cwa.googlegroups.com>,
> > > zuhair <zalj...@yahoo.com> wrote:
>
> > > >>From that one can understand the following definitions of subtraction
> > > >
> > > > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
> > > > a unique result
> > >
> > > Nonsense in this notation. There are no "0"s. All you have are
> > > "*"s. The finite ordinals were finite sequences of *s, the ordinal
> > > omega was a (countably infinite) sequence of stars, AND THAT IS ALL
> > > YOU DEFINED. Where did this idiocy of 0s come from?
> >
> > >From subtraction fool.
> >
> > * - * = 0
> >
> > got it, is it that difficult.
>
> One might make a case for
>
> * - * = <the empty string>
> OR
> * - * = {}, (THE EMPTY SET} ",
>
> but unless * represents some number, there is no case to be made for
How clever you are , it is obvious that * represents number one.
The pseudobinary numeral system is as follows
Zero = 0
One = *
Two = **
Three = ***
....
.....
....
.
.
.Omega= *****..........
Arturo say that subtraction cannot be defined for Omega, but this is
not true.
Left finite subtraction ( Removal of stars from the left ) from omega
will result in Omega itself, this is an obvious and unique result.
See that N= 1,2,3,4,...................
If I remove one from that series what I have is
2,3,4,5,........ which is bijectable to N. and therefore has the same
ordinal value as N.
The real problem lies in defining right subtraction from Omega. This is
difficult.
it would be more plausable to say that it also equals Omega
so ordinal subtraction of fintes from Omega would be commutative and
always resulting in Omega.
However ordinal subtraction from Omega+1 is non commutative.
so -1 + Omega+1 = Omega+1
While Omega+1 -1 = Omega
I think these can be easily visualized and intutively acceptable, but I
don't know if they can have a formal proof.
The real problem comes when we subtract Omega from itself.
so Omega - Omega = indeteminate result.
In a similar manner not all division from omega is not all together
un-defined
In previous posts i made some trials to define them. and I see them
possible.
But it is the stubbornship of some people like Arturo that prevents
them from contemplating them.
Zuhair
zuhair
Virgil wrote:
> In article <1146252352.8...@e56g2000cwe.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
> > Arturo Magidin wrote:
> > > In article <1146247083.7...@j33g2000cwa.googlegroups.com>,
> > > zuhair <zalj...@yahoo.com> wrote:
>
> > > >>From that one can understand the following definitions of subtraction
> > > >
> > > > ( - ***...n* ) + *****....... = 0000...n0 ****....... = ****..........
> > > > a unique result
> > >
> > > Nonsense in this notation. There are no "0"s. All you have are
> > > "*"s. The finite ordinals were finite sequences of *s, the ordinal
> > > omega was a (countably infinite) sequence of stars, AND THAT IS ALL
> > > YOU DEFINED. Where did this idiocy of 0s come from?
> >
> > >From subtraction fool.
> >
> > * - * = 0
> >
> > got it, is it that difficult.
>
> One might make a case for
>
> * - * = <the empty string>
> OR
> * - * = {}, (THE EMPTY SET} ",
i think you mean if one wants to keep the system unary , then *-* empty
set.
You are right.
But i used a pseudobinary system were 0 symbolizes this empty set.
Zuhair
zuhair
What is then. Mathematology.
Zuhair
zuhair
Arturo Magidin wrote:
> In article <1145991892.1...@i39g2000cwa.googlegroups.com>,> >> In article <1145951829.2...@e56g2000cwe.googlegroups.com>,
> >> zuhair <zalj...@yahoo.com> wrote:
> >>
> >> [.snip.]
> >>
> >> >any real number can be represented as a specific sequence of numerals.
> >> >using the decimal numeral system for example it can be said that any
> >> >real number r can be represented by a string of numerals d_1 d_2 d_3
> >> >......... , were d_i = 0 or 1 or 2 or 3 or.... 9 or decimal dot, with
> >> >at most one of the d_i being a decimal dot. For example number 33 ,
> >> >here we have d_1 = 3 and d_2 =3 and d_3 = .
> >> >d_ j = 0 were j>3 .
> >>
> >> You once again forgot to exclude the possibility of non-unique
> >> representation (despite the fact that I mentioned it explicitly). Is
> >> there any point in my replying to your posts? Do you read the
> >> responses?
> >
> >Well you have the right, in order to exclude non-unique representation
> >I will add a condition to what I have said above that the d_i which is
> >the decimal point shouldn't be d_1.
> >
> >Second you think that number 1 can be represented as 0.99999........ (
> >decimal) , well I think that this is only an approximate representation
> >that is not correct.
>
> Then you are, quite simply, wrong. 0.9999... represents the infinite
> series 9/10 + 9/10^2 + ... + 9/10^n + ... which converges to 1, so
> 0.9999... is not an "approximate representation that is not correct",
> but an EXACT representation that is EXACTLY equal to 1. Period.
>
> >The only representation of 1 is 1.0000......... and to me
> >0.9999........... < 1.
>
> Then you are wrong. If that expression represented a number strictly
> smaller than 1, call it x, then (1+x)/2 would be a number strictly
> smaller than 1, and strictly larger than x.
>
> What, pray tell, is the decimal representation of this number?
How come I didn't read that post?
Anyhow I should answer it.
I will change your terminology.
Let x = 1- 0.9999..............
Now 1+ 0.99999... = 2 -x
Now (2 - x)/2 = 2/2 - x/2
now x/2= x
Then (2-x)/2 = 1 - x = 0.99999999999.........
So the answer to your question is that the result will be the same
number.
Zuhair
>
> >> You must also either exclude the possibility that there is some k>0
> >> such that d_i=9 for all i>k, or else the possibility that there is
> >> some k>0 such that d_i=0 for all i>k. Otherwise, your claim about
> >> "specific sequence of numerals" is false. For example, the real number
> >> 1 can be represented either as 1.000000..... or as
> >> 0.999999999999999999....
> >
> >wrong see above.
>
> Indeed, wrong. You are just wrong, completely, absolutely, and
> totally.
>
>
> >> >Now the set of all possible permutations of d_1 d_2 d_3 .............
> >> >will be the set R.
> >>
> >> No, it is not. For example, if the number you started with was "1",
> >> written as 1.000000... then the set of all possible permutations of
> >> these symbols does NOT yield the set of all real numbers. Numbers
> >> whose decimal expansion have exactly k occurrences of the digit 7 can
> >> only be obtained as permutations of each other, and no permutation of
> >> it will yield a number whose decimal expansion has exactly (k+1)
> >> occurrences of the digit 7 (or an infinite number of them).
> >
> >All of that is gibberish, I didn't say that the set of all real numbers
> >comes from rearrangment of
> >the symboles used in representing a single real number.
>
>
> Nonsense. You wrote: "the set of all possible permutations of
> d_1d_2d_3... will be the set R". And previously, you defined
> "d_1d_2d_3..." to be a specific string of numerals which represented a
> specific real number. So you did, indeed, state that the set of all
> real numbers comes from rearranging the symbols used in representing a
> single real number. If that is not what you ->MEANT<- to say, well,
> that's your problem for not saying what you mean and saying nonsense
> instead. However, I strongly suspect that what you meant to say was
> also nonsense, just a different kind of nonsense.
>
> >> >and this has the cardinality of c.
> >> >
> >> >Now 2^c is the cardinality of the power set of R.
> >> >
> >> >The power set of R is the set of all combinationas that can come from
> >> >R.
> >>
> >> No. The power set of R is the set of all SUBSETS OF R. Not the set of
> >> all "combinations of R", a term which you continue to use without
> >> saying just what it means.
> >
> >Their is no difference between these two terms , they are synonum.
>
> No, they are not. "Combination" has a very specific meaning; "subset"
> has a very specific meaning. They are not synonimous. Each subset
> corresponds to a "combination without repetition of objects taken form
> a set", but not every combination corresponds to a unique subset.
>
> >> >Since any member of R is represented by d_1 d_2 d_3...............
> >> >
> >> >Then the power set of R will have numerals that consists from a
> >> >combination of this representation .
> >>
> >> No. No. No. No. No.
> >>
> >> Is five times enough, or do you need me to repeat it even more times?
> >
> >Repeat what .
>
> That you are, and continue to, speak nonsense.
>
> > what I meant by combinations is what you call all
> >subsets.
>
> Still nonsense. A subset of R consists of a collection of elements of
> R. This is not a rearrangement of a presentation of a number, nor does
> it corresopnd to such a rearrangement, nor does it correspond to a
> concatenation of two such representations. You are speaking nonsense,
> over and over and over and over again.
>
> Here is a subset of R:
>
> {pi^r | r a positive real number}.
>
> Pray tell, what are the "numerals that consist from a combination of
> this representation" (->what<- representation?) that represents this
> subset and no other?
>
>
> >> The power set of R is not the set of all rearrangements of a
> >> sequence. It is the set of all SUBSETS of R; the set of all sets, each
> >> of which has all of its elements as real numbers.
> >
> >I agree, but you read me wrong.
>
> No, you speak wrong. So far, you either speak falsehoods or nonsense.
>
> >> >In order to describe this number we have 33. followed by Omega of zeros
> >> >which is followed by 1. which is followed by Omega of zeros. Now what
> >> >that sequence of numerals represent.
> >>
> >> It represents nonsense.
> >
> >Easy answer.
>
> Sometimes the easy answer is correct. What you wrote above is
> nonsense, and your assertions about representations of real numbers
> are alternating between nonsense and falsehoods.
>
> >> >That is my question.
> >>
> >> And the answer is: it represents nonsense. You are utterly confused:
> >> the reals cannot be obtained as the set of all rearrangements of a
> >> single real,
> >
> >I never said that
>
> You did say that. Exactly and explicitly. You perhaps did not MEAN
> that, but you did a rather poor job of saying what you meant in that
> case. What you said is exactly what I wrote above.
>
> >and the power set of the reals is not the "set of all
> >> combinations of reals".
> >
> >Yes they are. But you don't understand the world combinations.
>
> I understand it perfectly well. You, on the other hand, don't even
> know what a real is, apparently.
José Carlos Santos
>>> The only representation of 1 is 1.0000......... and to me
>>> 0.9999........... < 1.
>> Then you are wrong. If that expression represented a number strictly
>> smaller than 1, call it x, then (1+x)/2 would be a number strictly
>> smaller than 1, and strictly larger than x.
>>
>> What, pray tell, is the decimal representation of this number?
>
> How come I didn't read that post?
>
> Anyhow I should answer it.
>
> I will change your terminology.
>
> Let x = 1- 0.9999..............
That's a rather strange way to answer to Arturo, since he told you to
call _x_ to 0.9999... and you choose to call _x_ to 1 - 0.9999....
> Now 1+ 0.99999... = 2 -x
>
> Now (2 - x)/2 = 2/2 - x/2
>
> now x/2= x
How did you get that? But if you think (as I do) that this is a true
statement, then the conclusion is, of course, that x = 0. Since you
defined _x_ as 1 - 0.9999..., it follows that 1 = 0.999....
Now, what about answering Arturo's question. If x = 0.9999..., what is
the decimal representation of (1 + x)/2?
Best regards,
Jose Carlos Santos
zuhair
I answered Arturo's question.
Let me repeat it again but I will use Arturo's terminology.
x= 0.9999.......
Let y = 1-x
it follows that x= 1-y
Now 1 + x = 1+ 1- y = 2 - y
Now y/2 = y
Then ( 1+x)/ 2 = (2-y)/2 = 1 -( y/2) = 1- y = x = 0.99999...........
You see the same number.
Zuhair
José Carlos Santos
>>>>> The only representation of 1 is 1.0000......... and to me
>>>>> 0.9999........... < 1.
>>>> Then you are wrong. If that expression represented a number strictly
>>>> smaller than 1, call it x, then (1+x)/2 would be a number strictly
>>>> smaller than 1, and strictly larger than x.
>>>>
>>>> What, pray tell, is the decimal representation of this number?
>>> How come I didn't read that post?
>>>
>>> Anyhow I should answer it.
>>>
>>> I will change your terminology.
>>>
>>> Let x = 1- 0.9999..............
>> That's a rather strange way to answer to Arturo, since he told you to
>> call _x_ to 0.9999... and you choose to call _x_ to 1 - 0.9999....
>>
>>> Now 1+ 0.99999... = 2 -x
>>>
>>> Now (2 - x)/2 = 2/2 - x/2
>>>
>>> now x/2= x
>> How did you get that? But if you think (as I do) that this is a true
>> statement, then the conclusion is, of course, that x = 0. Since you
>> defined _x_ as 1 - 0.9999..., it follows that 1 = 0.999....
>>
>> Now, what about answering Arturo's question. If x = 0.9999..., what is
>> the decimal representation of (1 + x)/2?
>
> I answered Arturo's question.
>
> Let me repeat it again but I will use Arturo's terminology.
>
> x= 0.9999.......
>
> Let y = 1-x
>
> it follows that x= 1-y
>
> Now 1 + x = 1+ 1- y = 2 - y
>
> Now y/2 = y
Two questions here:
1) How do you know that y/2 = y?
2) How come you don't deduce from this that y = 0 and therefore that
1 = 0.9999...?
> Then ( 1+x)/ 2 = (2-y)/2 = 1 -( y/2) = 1- y = x = 0.99999...........
>
> You see the same number.
Best regards,
Jose Carlos Santos
zuhair
because y is an infinite number since, 0.9999...... is infinite, and if
you divide this
infinite number by any finite number you will get the same number.
1- 0.999........ = 1/(10^Aleph-0)
[1/(10^Aleph-0)]/n = 1/(10^Aleph-0)
This comes from the equation
[x/n] + [x/(n^2) ] + [x/ (n^3)] +.....+ [x/(n^k) ]= {x/(n-1)} - {
x/[(n^k)(n-1)]}
for n=2,3,4,5,........
K=1,2,3,4,.........
if we regard k as the number of terms in the sequence above.
Then when that sequence is infinite then = Aleph-0
so for x=9, n=10 and k=Aleph-0 we will have 0.9999...........= 1-
[1/(10^Aleph-0)]
Now whatever is the division operator in 1/(10^Aleph-0) is, the result
cannot equal zero.
and 1/(10^Aleph-0) > 0
So 0.9999........ <1
Zuhair
José Carlos Santos
I don't know what an "infinite number" is. Do you have a reference?
By the way, is 0,111111111111... an infinite number too?
> 1- 0.999........ = 1/(10^Aleph-0)
>
> [1/(10^Aleph-0)]/n = 1/(10^Aleph-0)
>
> This comes from the equation
>
> [x/n] + [x/(n^2) ] + [x/ (n^3)] +.....+ [x/(n^k) ]= {x/(n-1)} - {
> x/[(n^k)(n-1)]}
>
> for n=2,3,4,5,........
> K=1,2,3,4,.........
>
> if we regard k as the number of terms in the sequence above.
>
> Then when that sequence is infinite then = Aleph-0
>
> so for x=9, n=10 and k=Aleph-0 we will have 0.9999...........= 1-
> [1/(10^Aleph-0)]
>
> Now whatever is the division operator in 1/(10^Aleph-0) is, the result
> cannot equal zero.
>
> and 1/(10^Aleph-0) > 0
>
> So 0.9999........ <1
I have no idea what most of this means.
zuhair
According to my way of thinking , Yes. But these are only my dam
nonsensical misconceptions as standard mathematicians say. The whole
idea is not so serious.
>
> > 1- 0.999........ = 1/(10^Aleph-0)
> >
> > [1/(10^Aleph-0)]/n = 1/(10^Aleph-0)
> >
> > This comes from the equation
> >
> > [x/n] + [x/(n^2) ] + [x/ (n^3)] +.....+ [x/(n^k) ]= {x/(n-1)} - {
> > x/[(n^k)(n-1)]}
> >
> > for n=2,3,4,5,........
> > K=1,2,3,4,.........
> >
> > if we regard k as the number of terms in the sequence above.
> >
> > Then when that sequence is infinite then = Aleph-0
> >
> > so for x=9, n=10 and k=Aleph-0 we will have 0.9999...........= 1-
> > [1/(10^Aleph-0)]
> >
> > Now whatever is the division operator in 1/(10^Aleph-0) is, the result
> > cannot equal zero.
> >
> > and 1/(10^Aleph-0) > 0
> >
> > So 0.9999........ <1
>
> I have no idea what most of this means.
>
> Best regards,
>
> Jose Carlos Santos
Dear Santos. All of these ideas are only mine, they are not accepted by
standard mathematicians like Arturo , they consider them nonsense, Even
I don't have a solid believe in them, I am just figuring them out. So
don't take them seriouselly.
Yours
Zuhair
zuhair
I would like to thank Arturo Magidin for his beautiful remarks( except
the abuse ) , I confess he taught me a lot. I also want to thank the
others for their replies.
Also I want to apologise if any abuse has came from me.
Zuhair
José Carlos Santos
You will probably not read this, but even so I would like to call your
attention to the fact that 0,11111111... = 1/9. On the other hand, you
wrote that half of an infinite number is that same number. It follows
that (1/9)/2 = 1/9. Don't you see anything strange here?
José Carlos Santos
> Also I want to apologise if any abuse has came from me.
I have nothing to complain.
zuhair
My way of thinking is non standard, to me 0.11111............ < 1/9
0.1111..... = 1/10 + 1/10^2 + 1/ 10^3 + ........... = (1/9) - ( 1/
[9 (10^Aleph-0)] )
Since ( 1/ [9 (10^Aleph-0)] ) >0
you can imagine ( 1/ [9 (10^Aleph-0)] ) as a continuually decreasing
quantity
and thus 0.1111......... as continually approaching 1/9 but never
reaching it.
so 1/9 is the limit of the sequence above and not the upper max.bound.
Then 0.1111....... <1/9
Divide that number by itself will only result in itself, since it is
infinite?!
Zuhair
zuhair
Dear Santos,
You asked me before is 0.111......... is an infinite number? I answered
you yes.
I think I was wrong since 0.1111.... is a combination of a real number
and an infinite number.
see that 0.1111.....= 1/9 - ( 1/ [9 (10^Aleph-0)] )
1/9 is a real number while ( 1/ [9 (10^Aleph-0)] ) is an infinite
number.
Now halfing 0.111...... would be like the following
1/2 * 0.111...... = 1/2 * 1/9 - 1/2* ( 1/ [9 (10^Aleph-0)] )
= 1/18 - ( 1/ [18 (10^Aleph-0)] )=1/18 - ( 1/
[18 (19^Aleph-0)] )
since n^Aleph-0 = m^Aleph-0 , for m , n are finite numbers.
= 1/19 + 1/19^2 + 1/19^3 + ..................
Using the 19th numeral system (number system with base =19). this
number is 0.1111...........
Zuhair
zuhair
????
0.0555........= 5/9 - 5/10 - 5/ 10^Aleph-0 = 1/18 - 1/10^Aleph-0
result confirmed
Zuhair
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> Divide that number by itself will only result in itself, since it is
> infinite?!
>
> Zuhair
Any number that it is possible to divide by, divided by itself had
better be equal to 1, or all of mathematics will collapse.
Zuhair claims to have "numbers" x such that x/x = x.
Not even TO is that far in the bag.
David W. Cantrell
> In article <1146420387.5...@y43g2000cwc.googlegroups.com>,
> "zuhair" <zalj...@yahoo.com> wrote:
>
> > Divide that number by itself will only result in itself, since it is
> > infinite?!
> >
> > Zuhair
>
> Any number that it is possible to divide by, divided by itself had
> better be equal to 1, or all of mathematics will collapse.
Virgil, I didn't think you were prone to hyperbole.
It depends, among other things, on what you mean by "number". In interval
arithmetic, an interval is sometimes considered to be an "interval number".
Let's consider an example in which an interval number is divided by itself:
[1,2] / [1,2] = [1/2,2]
Hmm. I don't think anything collapsed. (Perhaps you'd have liked the answer
to have "collapsed" to the degenerate interval number [1,1]. But at least
the correct answer does _contain_ 1.)
> Zuhair claims to have "numbers" x such that x/x = x.
In the reals, the only such number is x = 1, of course. But there are
systems (such as wheels) in which an element, other than 1, divided by
itself can yield that same element. Of course, whether you want to call
such an element a "number" is a different matter.
Regards,
David
Virgil
David W. Cantrell <DWCan...@sigmaxi.org> wrote:
> Virgil <vmh...@comcast.net> wrote:
> > In article <1146420387.5...@y43g2000cwc.googlegroups.com>,
> > "zuhair" <zalj...@yahoo.com> wrote:
> >
> > > Divide that number by itself will only result in itself, since it is
> > > infinite?!
> > >
> > > Zuhair
> >
> > Any number that it is possible to divide by, divided by itself had
> > better be equal to 1, or all of mathematics will collapse.
>
> Virgil, I didn't think you were prone to hyperbole.
>
> It depends, among other things, on what you mean by "number".
'When I use a word, it means just what I choose it to mean -- Neither
more nor less.'
'The question is, which is to be master -- that's all!'
> In interval
> arithmetic, an interval is sometimes considered to be an "interval number".
> Let's consider an example in which an interval number is divided by itself:
>
> [1,2] / [1,2] = [1/2,2]
>
> Hmm. I don't think anything collapsed. (Perhaps you'd have liked the answer
> to have "collapsed" to the degenerate interval number [1,1]. But at least
> the correct answer does _contain_ 1.)
>
> > Zuhair claims to have "numbers" x such that x/x = x.
>
> In the reals, the only such number is x = 1, of course. But there are
> systems (such as wheels) in which an element, other than 1, divided by
> itself can yield that same element. Of course, whether you want to call
> such an element a "number" is a different matter.
I don't, particularly when speaking to Zuhair.
>
> Regards,
> David
José Carlos Santos
>> You will probably not read this, but even so I would like to call your
>> attention to the fact that 0,11111111... = 1/9. On the other hand, you
>> wrote that half of an infinite number is that same number. It follows
>> that (1/9)/2 = 1/9. Don't you see anything strange here?
>
> My way of thinking is non standard, to me 0.11111............ < 1/9
Then please tell me what's the decimal representation of 1/9.
zuhair
1/9 , the only decimal numeral which could be used to represent that
number is 0.1111....... but this is in reality only as approximation.
Like one who cannot utter kh and so he repeates k,k,k,... and at each
repeatition he becomes nearer to sound of kh but never reach it.
so kh=k,k,k,k,.............
If you want to represent the number 1/9 you should use another numeral
system, like the one to the base 9, as such it will be 0.1.
In other words , there is no single number system ( except the
pseudobinary system that I have invented) that can represent all
numbers without some approximated.
Best regards,
Zuhair
zuhair
Ok , dear Virgil.
Aleph-0 * Aleph-0 = Aleph-0
so it is possible to say that Aleph-0/Aleph-0 = Aleph-0
Of coarse I now that Aleph-0/Aleph-0 = |x| : x<=Aleph-0
so division of Aleph-0 by itself don't reveal number one as you
thought. It results in a variable and not a number.
However Divsion of Aleph-0 by a finite number n yields Aleph-0 itself.
Same thing to be applied to heigher cardinals like 2^Aleph-0.
So for example [1/(2^Aleph-0)] /2 = 1/2*(2^Aleph-0) = 1/(2^Aleph-0)
So this small infinite number cannot be halfed
Zuhair
José Carlos Santos
>>>> You will probably not read this, but even so I would like to call your
>>>> attention to the fact that 0,11111111... = 1/9. On the other hand, you
>>>> wrote that half of an infinite number is that same number. It follows
>>>> that (1/9)/2 = 1/9. Don't you see anything strange here?
>>> My way of thinking is non standard, to me 0.11111............ < 1/9
>> Then please tell me what's the decimal representation of 1/9.
> The decimal numeral system ( base 10) cannot represent a number like
> 1/9 , the only decimal numeral which could be used to represent that
> number is 0.1111....... but this is in reality only as approximation.
This is false. If you think in a nonstandard way or you use a
nonstandard notation, that's one thing. But saying that, in the decimal
system, 0,1111... is not 1/9 is just plain false.
> Like one who cannot utter kh and so he repeates k,k,k,... and at each
> repeatition he becomes nearer to sound of kh but never reach it.
>
> so kh=k,k,k,k,.............
>
> If you want to represent the number 1/9 you should use another numeral
> system, like the one to the base 9, as such it will be 0.1.
>
> In other words , there is no single number system ( except the
> pseudobinary system that I have invented) that can represent all
> numbers without some approximated.
Am I missing something, or this is a vastly arrogant statement?
For my part, this thread is over.
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
What do you mean by a "small infinite number"?
Most people regard infinite as meaning large, not small.
And why do you say that a number that you have just halved cannot be
halved?
Is there any part of your psuodo-number-system that makes any sense?
Virgil
"zuhair" <zalj...@yahoo.com> wrote:
> José Carlos Santos wrote:
> > zuhair wrote:
> >
> > >> You will probably not read this, but even so I would like to call your
> > >> attention to the fact that 0,11111111... = 1/9. On the other hand, you
> > >> wrote that half of an infinite number is that same number. It follows
> > >> that (1/9)/2 = 1/9. Don't you see anything strange here?
> > >
> > > My way of thinking is non standard, to me 0.11111............ < 1/9
> >
> > Then please tell me what's the decimal representation of 1/9.
> The decimal numeral system ( base 10) cannot represent a number like
> 1/9
The decimal system that everyone except Zuhair uses can. The standard
decimal system can represent any rational number.
> the only decimal numeral which could be used to represent that
> number is 0.1111....... but this is in reality only as approximation.
In \reality\ it is 1/9, it is only in zuhairity that it is not.
>
> In other words , there is no single number system ( except the
> pseudobinary system that I have invented) that can represent all
> numbers without some approximated.
Misguided arrogance.
zuhair
José Carlos Santos wrote:
> zuhair wrote:
>
> >>>> You will probably not read this, but even so I would like to call your
> >>>> attention to the fact that 0,11111111... = 1/9. On the other hand, you
> >>>> wrote that half of an infinite number is that same number. It follows
> >>>> that (1/9)/2 = 1/9. Don't you see anything strange here?
> >>> My way of thinking is non standard, to me 0.11111............ < 1/9
> >> Then please tell me what's the decimal representation of 1/9.
> > The decimal numeral system ( base 10) cannot represent a number like
> > 1/9 , the only decimal numeral which could be used to represent that
> > number is 0.1111....... but this is in reality only as approximation.
>
> This is false. If you think in a nonstandard way or you use a
> nonstandard notation, that's one thing. But saying that, in the decimal
> system, 0,1111... is not 1/9 is just plain false.
You are not understanding me, everything depends on what you want from
the decimal system to do, If you want it to represent real numbers then
there is no real number other than 1/9 that 0.1111......... can
represent. But this is not the sole truth. The reality is 0.1111......
is approximation of 1/9 and not exactly equal 1/9 , 1/9 cannot be
represented in the decimal system in a full manner .
>
> > Like one who cannot utter kh and so he repeates k,k,k,... and at each
> > repeatition he becomes nearer to sound of kh but never reach it.
> >
> > so kh=k,k,k,k,.............
> >
> > If you want to represent the number 1/9 you should use another numeral
> > system, like the one to the base 9, as such it will be 0.1.
> >
> > In other words , there is no single number system ( except the
> > pseudobinary system that I have invented) that can represent all
> > numbers without some approximated.
>
> Am I missing something, or this is a vastly arrogant statement?
It is a vastly arrogant statement of mine. The usual non sensical
statement I always post to that forum.
I told you from the beginning that most of my thinking is regarded as
non sense by standard mathematicians you chose to wast your time with
me.
zuhair
I mean it cannot be halved to result in a number that is smaller than
itself.