Even odd puzzle
Tapio
Are there more odd than even numbers (including also infinite integers <
omega) ?
(Note: infinite integers are here rigorously defined as a sum of infinite
serie, analogous to reals (R), use dejagoogle for details)
Tapio
Steve Leibel
Well 1 = 1/2 + 1/4 + 1/8 + ... is the sum of an infinite series, does
that make 1 an infinite integer?
Virgil
If "here" means here, infinite integers have not been defined.
Having read the deja google "definition", I assert that they have not
been defined there either.
In the non-standard construction of the reals, as originated by Abraham
Robinson, there are infinite integers as well as infinite reals and
infinitesimal reals, but they are entirely diffierent.
Tapio
"Virgil" <vmh...@home.com> wrote in message
news:vmhjr2-E2DD32....@news2.rdc2.tx.home.com...> > A tiny midsummer puzzle ....
(cut)
>
> If "here" means here, infinite integers have not been defined.
repeating, a hint to use deja google was offered.
> Having read the deja google "definition", I assert that they have not
> been defined there either.
Your assertion is now an opinion, which is kindly noticed. It is an opinion
as long as you do not argue your assertation. Would you be a little bit more
analytical - please. Hopefully I can help you - too.
> In the non-standard construction of the reals, as originated by Abraham
> Robinson, there are infinite integers as well as infinite reals and
> infinitesimal reals, but they are entirely diffierent.
I assume my expression concerning the concept of infinite integers somehow
disturbs, because Abraham Robinson have used that concept earlier. I you
want we can use another expression that you prefer or - for example -
infinite notational expansion of finite integers. Let's mark it N(inf) or
something else that you cannot mix with other reserved symbols. OK?
Tapio
Johannes H Andersen
IMO you are getting away from your original question. If you are talking
of finite integers, then there is a bijection between even and odd
integers, that should do. Next, how do you define even and odd infinite
integers? I can even ask: are there infinite integers that are different?
or are they all the same by being infinite integers.
Johannes
Tapio
"Steve Leibel" <ste...@bluetuna.com> wrote in message
news:stevel-DC92FB....@nyctyp01.rdc-nyc.rr.com...> > A tiny midsummer puzzle ....
(cut)
> Well 1 = 1/2 + 1/4 + 1/8 + ... is the sum of an infinite series, does
> that make 1 an infinite integer?
1) The integer part of "infinite integer" was not defined that way.
2) I assume the name "infinite integer" somehow disturbs you? (see my reply
to Virgil).
3) You may express 1 ( and any other finite N) also in the format of
infinite notational expansion of finite integers.
Tapio
Tapio
"Johannes H Andersen" <jo...@madasafish.com> wrote in message
news:3B3DB04D...@madasafish.com...
>
>
> Tapio wrote:
> >
> > "Virgil" <vmh...@home.com> wrote in message
> > news:vmhjr2-E2DD32....@news2.rdc2.tx.home.com...
> > > In article <9hipoo$qnm$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi>
wrote:
> > >
> IMO you are getting away from your original question.
Hopefully not. I tried to answer Virgil愀 post.
> If you are talking
> of finite integers, then there is a bijection between even and odd
> integers, that should do.
Yes, without doubt.
>Next, how do you define even and odd infinite
> integers?
First, I am personally not very satisfied concerning the "naming" of
infinite integers. I have used it only because it is used earlier in this
ng. I would prefer (no better name available ?) N(inf).
Secondly, definition of even and odd N(inf) are the same as in the case of
finite integers.
>I can even ask: are there infinite integers that are different?
> or are they all the same by being infinite integers.
A good question! All the finite intergers are subset of N(inf). I wish you
may see the difference by the following descriptional examples:
Finite integer is simply a short notation format of N(inf). We call them (N)
finite integers, because we purposely do not write empty and non-significant
(containing zero) placeholders (a_n). Thus finite integers have a start (in
the placeholder a_1 and an end in some placeholder a_n). The simple reason
is that those zeros (on left side of terminating finite integer) do not
affect on counting operations. For example number one (1) written in the
form(at) of N(inf) is ...01, where ... means repeated infinitely. Thus,
there are zeros in the placeholders 2 ->oo.
As you hopefully notice, there may exist numbers so that all the
placeholders (a_n) (n 1->oo) are occupied by some meningful (or significant)
digit in the set {0,1,2,3,4,5,6,7,8,9}. The concept meaningfull or
significant expresses that those digits are significant in the counting
operations. An example about N(inf) with all significant placeholders is
...999 or ...111. It is clear that any placeholder (a_n) can contain (have)
some significant digit, not necessarely the same one. I assume you may agree
that those examples are not finite. They are infinite expressions.
Thus it should be evident N maps to Q and N(inf) maps to R.
Thus my original question should have some sense - hopefully.
Tapio
(Break of 24 hours, I return)
> Johannes
Johannes H Andersen
OK, I see now. An infinite integer is the same as an infinite sequence of
digits. In that case an even number is a sequence that from a certain
point onwards has only even digits. Similarly for odd numbers. Then there
must be infinite integers which are neither even nor odd.
Johannes
Johannes H Andersen
In that case there is also a bijection, by "rotating" all the digits:
0 <-> 1
1 <-> 2
2 <-> 3
...
9 <-> 0
Johannes
Virgil
> >I can even ask: are there infinite integers that are different?
> > or are they all the same by being infinite integers.
>
> A good question! All the finite intergers are subset of N(inf). I wish you
> may see the difference by the following descriptional examples:
> Finite integer is simply a short notation format of N(inf). We call them (N)
> finite integers, because we purposely do not write empty and non-significant
> (containing zero) placeholders (a_n). Thus finite integers have a start (in
> the placeholder a_1 and an end in some placeholder a_n). The simple reason
> is that those zeros (on left side of terminating finite integer) do not
> affect on counting operations. For example number one (1) written in the
> form(at) of N(inf) is ...01, where ... means repeated infinitely. Thus,
> there are zeros in the placeholders 2 ->oo.
> As you hopefully notice, there may exist numbers so that all the
> placeholders (a_n) (n 1->oo) are occupied by some meningful (or significant)
> digit in the set {0,1,2,3,4,5,6,7,8,9}. The concept meaningfull or
> significant expresses that those digits are significant in the counting
> operations. An example about N(inf) with all significant placeholders is
> ...999 or ...111. It is clear that any placeholder (a_n) can contain (have)
> some significant digit, not necessarely the same one. I assume you may agree
> that those examples are not finite. They are infinite expressions.
> Thus it should be evident N maps to Q and N(inf) maps to R.
> Thus my original question should have some sense - hopefully.
I can see a mapping of N(inf) to R that is neither a surjection nor an
injection. I suspect explicit construction of a bijection would be
difficult.
How does one extend the order relation of the finite integers to N(inf)?
What was your original question? I have lost track of where this all
started.
Tapio
(my text snipped)
> > > > Johannes
> >
> > OK, I see now. An infinite integer is the same as an infinite sequence
of
> > digits.
In other words infinite integer is an infinite sum (n 0->oo) a_n 10^n, where
a_n is some integer in the set {0,1,2,3,4,5,6,7,8,9}..., thus it is called
infinite because it is not finite.
>In that case an even number is a sequence that from a certain
> > point onwards has only even digits. Similarly for odd numbers. Then
there
> > must be infinite integers which are neither even nor odd.
> >
> > Johannes
I cannot follow your arguments above -sorry.
I assume N(inf) or finite integer is even if the last digit in the position
a_0 or a_0*10^0 = a_0 is even. Thus in the set {2,4,6,8}.Iff a_0 =0 then we
must have something else "on the left side" of the string, i.e. some digit
in the position a_1 etc. so that N(inf) or finite integer is even. (Assuming
the plain zero is neither even or odd). Iff zero is excluded, then the
integer is even if in the position a_0 (the last digit) we have some integer
in the set {0,2,4,6,8}
Odd numbers are those that are not even (excluding zero of course - as
above).
> In that case there is also a bijection, by "rotating" all the digits:
>
> 0 <-> 1
> 1 <-> 2
> 2 <-> 3
> ...
> 9 <-> 0
Sorry Johannes, I cannot follow how did you draw that conslusion in the case
of N(inf)? Maybe I have missed something.
Tapio
> Johannes
Tapio
>
> I can see a mapping of N(inf) to R that is neither a surjection nor an
> injection. I suspect explicit construction of a bijection would be
> difficult.
Ah... I assume there is now something I was not able to explain? The
problem must be at this end of modem. :-)
Let me try this way: a number in R is a sum of infinite serie: sum (n 1 ->
oo) (a_n)/10^n, where all a_n, which is a placeholder, have some integer in
the set {0,1,2,3,4,5,6,7,8,9}. Therefore we may express all the numbers in R
[0,1), if we have all possible combinations of a_n.
The infinite integer N(inf) is analogically: sum (n 0 -> oo) (a_n)10^n,
where all a_n, which is a placeholder, have some integer in the set
{0,1,2,3,4,5,6,7,8,9}.
The integer is finite, and therefore a special case of N(inf), if all
placeholders beginning from some arbitrary placehoder n>1 have filled by
zeros, i.e. on the left side placeholders.
Therefore we may express all the infinite integers in N(inf) [0,omega), if
we have all possible combinations of a_n.
Now, for every N(inf) [0,omega) there is some real in R [0,1) and vice
versa. This is one-to-one mapping. (Note finite integer 1 maps to omega -
above). Therefore it is also a clear consequence that there is no one-to-one
mapping from finite N [0,omega) to R [0,1).
The bijection is more evident if you express integers in N(inf) as "decimal
numbers" when the point of reference is omega zero instead of standard
zero. An example that hopefully clarifies this is the following:
Assume we have an infinite integer ...999 (or ...9). This is the infinite
sum (n 0->oo) (a_n)10^n, where all (or every) a_n equals to 9. It is
important to notice that the sum must result in ...9. If the sum does not
result in that infinite integer, then the sum must wrong. Usually the
infinite sum is defined as a limit, but we do not search the limit of that
sum! We search the sum instead of it's limit. The sum is correct, if it (the
sum) really results in the sum instead of limit.
Because all integers have a successor, we must be always able to add one.
But now all the possible placeholders are already occupied with the greatest
possible member of the set {0,1,2,3,4,5,6,7,8,9}. The next number must be
greater than any infinite integer N(inf). Therefore ...9 +1 or written in
the infinite form(at) ...9 + ...01 = omega. Thus, directly according to the
definition of omega. Therefore also ...9 is the greatest possible infinite
integer < omega. Note that ...9 is odd. If we add 1 (or ...01) omega must be
even. But omega is even only from the point of view that is the standard
point of reference, i.e. standard zero = the normal zero. Omega is the
first number one "behind" or "above" of N(inf), i.e. the first transinfinite
number. (Normally called transfinite). Because omega is the first
"transfinite number one", it is odd as we consider even and odd numbers
from the omega point of view.
Now let´s describe omega zero (0). Omega zero is the point of view that is
infinite far from standard zero point of view. The word "far" is used here
only as an illustrative meaning. If we write the infinite integer ...9 as a
"decimal number" when the point of reference is omega, then
...9 = 0,9... (or 0,999...) Note 0, is here omega zero reference point.
Note: we had to add ...01 or simply finite 1 to get omega.
Thus ...9 <omega. And ...9 divided by omega results in 0,999... Therefore
the infinite sum (n 0->oo) (a_n)10^n, where all (or every) a_n equals to 9,
results in ...9 but the limit of the the infinite sum (n 0->oo) (a_n)10^n,
where all (or every) a_n equals to 9, results in omega. Similarly 0.999...
<1 but the limit of 0.999... written as a infinite sum equals to 1.
If we devide ...9 by omega we will receive 0,999...
It follows infinite integers N(inf) [0,omega) and reals R [0,1) are well
ordered. The smallest member in N(inf) is finite 1 or (...01 written in the
infinite format) and the smallest member in R [0,1) is what we call epsilon,
which has digit 1 in a placeholder a_n as n ->oo.
It also follows: oo or infinity is not a number, which is consistent with
the standard theory. Numbers are digit strings in one dimension. oo
describes the "distance" of placeholders a_n in that string, if I somehow
want to describe the property of infinity in the dense placeholder line.
Actually - you hopefully observe - infinity is the concept relatated to the
point of reference. (Therefore I may say point of references are nested like
the peals of onion.)
By changing the point of reference you can always write any digit string R,
N(inf), Omega N(inf) etc. like reals or infinite integers. We call those
numbers by different name because we have a different point of reference,
i.e. "the decimal point". As an example omega zero point something
(0,xxx...) or standard zero dot something (0.xxx...). It follows- if we do
not specify the point of reference, then the presentations or expressions
are "formally" identical (compare - bijection in the case of 0.999... and
0,999... =...9).
If you now consider reals R [0,1) and N(inf) [0,omega) it should be evident
that there exist one-to-one mapping between N(inf) [0,omega) and R [0,1),
i.e. bijection.
> How does one extend the order relation of the finite integers to N(inf)?
This is a very good question! First we must always remember: a) there is
always a successor b) counting must be coherent without inconsistency c)
counting rolls over theory, i.e. the consistency over theory.
The order - iff you mean the successor - is of course standard and normal.
Please - remember finite integers have non-significant zeros on the left
hand side of digit string. The order (what is the successor, which is
greater or smaller) will be determined by the difference. ...2 >...1,
because the difference equals to ...1. ...02 >...01, because the difference
equals to 1 etc. The successor of an integer y is x , iff x-y=1, i.e.
y+1=x.
> What was your original question? I have lost track of where this all
> started.
My original question was written:
A start of copy
"A tiny midsummer puzzle (by warming up sauna):
Are there more odd than even numbers (including also infinite integers <
omega) ?
(Note: infinite integers are here rigorously defined as a sum of infinite
serie, analogous to reals (R), use dejagoogle for details)
Tapio"
The end of copy
I wish I been for your help!
A Hint: We have noticed above that the first odd integer is 1 or ...01
written in the infinite format. The greatest possible infinite integer
<omega equals to ...9 that is odd.
Tapio
Virgil
> "Virgil" <vmh...@home.com> wrote in message
> news:vmhjr2-492E6E....@news2.rdc2.tx.home.com...
> > In article <9hkf3p$ua$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi> wrote:
> (my txt snipped)
> >
> > I can see a mapping of N(inf) to R that is neither a surjection nor an
> > injection. I suspect explicit construction of a bijection would be
> > difficult.
>
> Ah... I assume there is now something I was not able to explain? The
> problem must be at this end of modem. :-)
>
> Let me try this way: a number in R is a sum of infinite serie: sum (n 1 ->
> oo) (a_n)/10^n, where all a_n, which is a placeholder, have some integer in
> the set {0,1,2,3,4,5,6,7,8,9}. Therefore we may express all the numbers in R
> [0,1), if we have all possible combinations of a_n.
> The infinite integer N(inf) is analogically: sum (n 0 -> oo) (a_n)10^n,
> where all a_n, which is a placeholder, have some integer in the set
> {0,1,2,3,4,5,6,7,8,9}.
Call a sequence, f:N -> {0,1,2,3,4,5,6,7,8,9}: n -> f_n a terminating
sequence if all but finitely many of the values f_n are equal to zero,
otherwise, call it non-terminating.
I can see how every sequence {a_n} maps to a real number in the
interval [0,1] as the limiting value of the convergent series
sum(a_n*10^(-n-1), n=0..oo), and that this mapping is a surjection but
not an injection.
Note that for every terminating sequence (except the all zero sequence)
there is a sequence with all but finitely many of its values equal to 9
which converges to the same limit. And the all nines sequence converges
to 1.
A similar mapping from the sequences of digits to integers can be
carried out explicitly only for terminating sequences by the mapping
{a_n} -> sum(a_n*10^n, n = 0..oo), since for non-terminating sequences
such series always diverge and do not determine any standard integer.
I understand your N(inf) to be the set os all sequences
f:N->{0,1,2,3,4,5,6,7,8,9}, with the terminating series corresponding to
the (finite) integers.
>
> The integer is finite, and therefore a special case of N(inf), if all
> placeholders beginning from some arbitrary placehoder n>1 have filled by
> zeros, i.e. on the left side placeholders.
> Therefore we may express all the infinite integers in N(inf) [0,omega), if
> we have all possible combinations of a_n.
Ibelieve I now understand the symbol N(inf), but what do you mean by
n(inf)[0,omega) ?
>
> Now, for every N(inf) [0,omega) there is some real in R [0,1) and vice
> versa. This is one-to-one mapping.
I can see a mapping f:N(inf) -> R[0,1], although it is many-to-one, not
one-to-one.
If your [0,omega) means [0,1) \union {omega}, the I can even deal with
g:N(inf) -> [0,omega), but it is still not one-to-one.
> (Note finite integer 1 maps to omega -
> above). Therefore it is also a clear consequence that there is no one-to-one
> mapping from finite N [0,omega) to R [0,1).
>
> The bijection is more evident if you express integers in N(inf) as "decimal
> numbers" when the point of reference is omega zero instead of standard
> zero. An example that hopefully clarifies this is the following:
> Assume we have an infinite integer ...999 (or ...9). This is the infinite
> sum (n 0->oo) (a_n)10^n, where all (or every) a_n equals to 9. It is
> important to notice that the sum must result in ...9. If the sum does not
> result in that infinite integer, then the sum must wrong. Usually the
> infinite sum is defined as a limit, but we do not search the limit of that
> sum! We search the sum instead of it's limit. The sum is correct, if it (the
> sum) really results in the sum instead of limit.
> Because all integers have a successor, we must be always able to add one.
> But now all the possible placeholders are already occupied with the greatest
> possible member of the set {0,1,2,3,4,5,6,7,8,9}. The next number must be
> greater than any infinite integer N(inf). Therefore ...9 +1 or written in
> the infinite form(at) ...9 + ...01 = omega.
A treatment more in line with the algorithms of standard arithmetic will
have ...9 + 1 = 0, so that ...9 = -1, with similar results for any
sequence having all but finitely many terms equal to 9.
In fact, this set of sequences becomes a group under placewise addition
with carrying, with every sequence having a negative.
> Thus, directly according to the
> definition of omega. Therefore also ...9 is the greatest possible infinite
> integer < omega.
Testing for equality is straightforward in N(inf), but how do you test
for greater than or less than in N(inf)?
> Note that ...9 is odd. If we add 1 (or ...01) omega must be
> even. But omega is even only from the point of view that is the standard
> point of reference, i.e. standard zero = the normal zero. Omega is the
> first number one "behind" or "above" of N(inf), i.e. the first transinfinite
> number.
Since non-terminating sequences are something other than finite, say
"uber-finite", if "omega" is larger than these, it must be
"super-uber-finite" or "trans-uber-finite".
(Normally called transfinite). Because omega is the first
> "transfinite number one", it is odd as we consider even and odd numbers
> from the omega point of view.
> Now let´s describe omega zero (0). Omega zero is the point of view that is
> infinite far from standard zero point of view. The word "far" is used here
> only as an illustrative meaning. If we write the infinite integer ...9 as a
> "decimal number" when the point of reference is omega, then
> ...9 = 0,9... (or 0,999...) Note 0, is here omega zero reference point.
I don't understand your notation. Does ...123 become 0,321...?
If you are saying that the "0," precedes ...9, you can't write 0,9...,
since there isn't a "first" 9 to put the "0," in front of. You would
have to write 0,...9 indicating infinitely many 9's to the left of the
visible 9.
From this point on, there are so many internal ambiguities, that I can't
follow it at all.
While this exposition may be quite creative, it just isn't internally
consistent.
Johannes H Andersen
Tapio wrote:
>
> "Johannes H Andersen" <jo...@madasafish.com> wrote in message
> news:3B3DD1EE...@madasafish.com...
> (my text snipped)
> > > > > Johannes
> > >
> > > OK, I see now. An infinite integer is the same as an infinite sequence
> of
> > > digits.
>
> In other words infinite integer is an infinite sum (n 0->oo) a_n 10^n, where
> a_n is some integer in the set {0,1,2,3,4,5,6,7,8,9}..., thus it is called
> infinite because it is not finite.
>
> >In that case an even number is a sequence that from a certain
> > > point onwards has only even digits. Similarly for odd numbers. Then
> there
> > > must be infinite integers which are neither even nor odd.
> > >
> > > Johannes
>
> I cannot follow your arguments above -sorry.
> I assume N(inf) or finite integer is even if the last digit in the position
> a_0 or a_0*10^0 = a_0 is even. Thus in the set {2,4,6,8}.Iff a_0 =0 then we
> must have something else "on the left side" of the string, i.e. some digit
> in the position a_1 etc. so that N(inf) or finite integer is even. (Assuming
> the plain zero is neither even or odd). Iff zero is excluded, then the
> integer is even if in the position a_0 (the last digit) we have some integer
> in the set {0,2,4,6,8}
>
I am was trying to find out how you define even and odd infinite integers.
So far I have understood your definition of infinite integers as a never
ending string of digits. You also mention somewhere that N(inf) is
equipotent with R, this seems plausible to me. Since we no longer can
determine even or odd by looking at the last digit, I had to think of
something else. The idea was to say that a number is even if all the
digits right of a certain point are even, Similarly with odd. I don't
know if it matches your ideas, but at least it is logical, it also
includes finite integers as a special case. However, it does not
split the numbers into even and odd since there will be numbers that
are neither even nor odd, e.g. 121212121212.....
> Odd numbers are those that are not even (excluding zero of course - as
> above).
I think this is premature as it is based on what you know from
finite integers. Why should we expect this to carry over? You
have said that N(inf) and R are equipotent. And R doesn't
split into even/odd however you define the partition.
>
> > In that case there is also a bijection, by "rotating" all the digits:
> >
> > 0 <-> 1
> > 1 <-> 2
> > 2 <-> 3
> > ...
> > 9 <-> 0
>
> Sorry Johannes, I cannot follow how did you draw that conslusion in the case
> of N(inf)? Maybe I have missed something.
>
Consistent with my definitions, you have:
12345675469468468468.................... 468 cont. (even) maps to
23456786570579579579.................... 579 cont. odd
What do you think about this? Or is there another way to define
even odd?
Johannes
Raymond Kristiansen
A slightly (way) off-topic response to your question: In Norway both Odd
and Even are common male names. A quick search in the Norwegian national
register shows that there are 26545 people in Norway with Odd as their
_first_ first name and 7032 with Odd as their _only_ first name. For
comparison, 4366 persons has Even as their _first_ first name and 3538
has it as their _only_ first name.
So in response to your question, I would say there are more odd than
even numbers :)
Regards
Ray
Peter L. Montgomery
Raymond Kristiansen <ray...@spam.no> writes:
>A slightly (way) off-topic response to your question: In Norway both Odd
>and Even are common male names.
They sound much like `Adam' and `Eve'.
Alas, one of these is rumored to be female.
--
The 21st century is starting after 20 centuries complete,
but we say someone is age 21 after 21 years (plus fetus-hood) complete.
Peter-Lawren...@cwi.nl Home: San Rafael, California
Microsoft Research and CWI
Zundark
> Tapio wrote:
[some stuff]
> I am was trying to find out how you define even and odd infinite integers.
> So far I have understood your definition of infinite integers as a never
> ending string of digits. You also mention somewhere that N(inf) is
> equipotent with R, this seems plausible to me. Since we no longer can
> determine even or odd by looking at the last digit, I had to think of
> something else.
His "infinite integers" are never-ending on the left, not on the
right, so they do have a last digit (but not a first). So it's
easy to define what is meant by odd and even.
But most of his other claims about these "infinite integers"
either make no sense or are just wrong.
--
Zundark
Tapio
"Zundark" <zun...@dont-spam-me.com> wrote in message
news:9hpgrk$ivj$1...@taliesin.netcom.net.uk...
> Johannes H Andersen wrote:
>
> > Tapio wrote:
> [some stuff]
>
> > I am was trying to find out how you define even and odd infinite
integers.
> > So far I have understood your definition of infinite integers as a never
> > ending string of digits. You also mention somewhere that N(inf) is
> > equipotent with R, this seems plausible to me. Since we no longer can
> > determine even or odd by looking at the last digit, I had to think of
> > something else.
>
> His "infinite integers" are never-ending on the left, not on the
> right, so they do have a last digit (but not a first). So it's
> easy to define what is meant by odd and even.
Yes, exactly.
Johannes starts indexing from the omega point of reference. Thus J´s and my
indexing are just reversed - if I understood J´s mail correctly. Of couse
J´s indexing is also OK, if he want to define it so. J´s indexing of
placeholder goes from left to right as in the case of R [0,1). I assume it
is easy to see also in J´s case one-to-one mapping of the infinite integers
to R [0,1).
> But most of his other claims about these "infinite integers"
> either make no sense or are just wrong.
Your criticism will be considered and replied, if you would specify: why it
does not make sense and what is just wrong. I cannot see it easy, but I
assume you have something to learn me (us). Would you - please - show me the
errors.
Thanks!
Tapio
> --
> Zundark
Tapio
(snipped)
I think Zundark愀 comment had sense. You and me use different point of
reference, I index placeholders starting from the standard point of
zero,i.e. from right to left.. And I assume (correct me please) your index
starts from left to right, i.e. from omega point of reference.
You are correct, but I consider in your case there are maybe problems to
specify even and odd integers.OK?
>
> > Odd numbers are those that are not even (excluding zero of course - as
> > above).
>
> I think this is premature as it is based on what you know from
> finite integers. Why should we expect this to carry over? You
> have said that N(inf) and R are equipotent. And R doesn't
> split into even/odd however you define the partition.
Let me try to explain this way.
First your case: You wrote 121212121212.....
Iff I consider your infinite expression from my point of view, I can write
that number as an decimal number iff I choose and specify that the point of
reference is in your case omega. Applying the concept "omega zero point" the
above mentioned infinite number is written:
0,1212... or as a rational form(at) 4omega/33.
I see no problems to map one-to-one from N(inf) to R [0,1). For example:
0,1212... maps to 0.1212...
Iff I choose the standard point of reference (standard zero), which means on
the same time that the indexing of placeholders starts from right to left
then your number above would be written: ...1212.
Note: At this point people may say: Wait a minute, there is no end! They
would say: you cannot write 1212... as ...1212, because you do not know the
last digit or there is no last digit! This criticism is partially correct.
Let me try to illustrate:
Actually when an infinite integer is written, we know only that it must be
non-finite, i.e. infinite. If the number is infinite, then it cannot be
finite. Right?.
Now, lets observe 1212... and ...1212 are both infinite. If we allow to
accept that we can change the point of reference from omega to standard zero
(and vice versa), then - I assume - we may say that the both notations above
express exactly the same number. Iff we claim that those infinite numbers
are not the same, then we should show why they are different or
alternatively why they are not the same numbers. Can you show that?
I assume most of people think this way, which is also logical: infinite
expressions have another end or start, but there is no last numbers. Those
expression are - excuse me - one way infinite like R - for example. I do
understand that very well. But let me ask: Why we cannot have infinite expre
ssion with two ends (start and end) so that their "distance" is infinite?
Thus the start is infinite "far" from the end. I assume that also in this
case: the string can be infinite and there can be start and end. That is
also fact in the case of finite integers, but finite integers are not
infinite strings. Actually I cannot see why this is impossible: the string
can be infinite and there can be start and end. Can you explain why?
Iff we consider a simpler case like 111... (according to Johannes
expression) or ...111 (according to my analogous expression), then we should
know that all the placeholders are filled by digit 1. The number does not
change if we change the point of reference, i.e. the point of view, or does
it?
If you allow I would take a more radical (=non-orthodox or heretic) point of
view and write against all assumptions like this: 1...1, where ... means
that there are infinitely many digit 1 in every placeholder, thus
infinitely.
I know someone may criticize: ... means infinitely and your "last 1" is in
omega position and therefore it is not an infinite integer. Then I would
say: OK! if the notational agreement disturbs, let愀 write it this way:
1(continuously and infinitely)1. My point is here that all the placeholders
filled by 1 can form an infinite string, though it has start and end, which
are only the point how and wherefrom we look at the string.
> > > In that case there is also a bijection, by "rotating" all the digits:
> > >
> > > 0 <-> 1
> > > 1 <-> 2
> > > 2 <-> 3
> > > ...
> > > 9 <-> 0
> >
> > Sorry Johannes, I cannot follow how did you draw that conslusion in the
case
> > of N(inf)? Maybe I have missed something.
> >
>
> Consistent with my definitions, you have:
>
> 12345675469468468468.................... 468 cont. (even) maps to
> 23456786570579579579.................... 579 cont. odd
> What do you think about this? Or is there another way to define
> even odd?
I still cannot follow your thoughts, but I assume ...468 is even and ...579
is odd. Right?
> Johannes
Zundark
> > But most of his other claims about these "infinite integers"
> > either make no sense or are just wrong.
>
> Your criticism will be considered and replied, if you would specify:
> why it does not make sense and what is just wrong. I cannot see it
> easy, but I assume you have something to learn me (us). Would you -
> please - show me the errors.
You have claimed that a certain function from N(inf) to [0,1)
is one-to-one, even though it is clearly many-to-one and maps
...9999 to 1, which is not in [0,1).
Some of your other claims require a definition of addition on
N(inf), and also a definition of a less-than relation on N(inf),
neither of which you have provided, as far as I can see. If you
want me to make any useful comments on these other claims, you
will need to provide these definitions.
--
Zundark
Tapio
> In article <9hnqfj$1e6$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi> wrote:
>
> > "Virgil" <vmh...@home.com> wrote in message
> > news:vmhjr2-492E6E....@news2.rdc2.tx.home.com...
> > > In article <9hkf3p$ua$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi> wrote:
Virgil:
> Call a sequence, f:N -> {0,1,2,3,4,5,6,7,8,9}: n -> f_n a terminating
> sequence if all but finitely many of the values f_n are equal to zero,
> otherwise, call it non-terminating.
>
> I can see how every sequence {a_n} maps to a real number in the
> interval [0,1] as the limiting value of the convergent series
> sum(a_n*10^(-n-1), n=0..oo), and that this mapping is a surjection but
> not an injection.
I explained every Real [0,1) is an infinite sum. Every distinct sum, i.e. at
least one content in some placeholder (a_n) is different than in some other
infinite sum exactly in the same placeholders, presents a unic real [0,1).
For example:
0.1222.... is unic and different than 0.21222... that is unic string -too.In
other word, those are infinite sums and present only one real that is unic,
i.e. there is no other similar. There are infinite various combinations.
Each possible combinations expresses just one and only one real. Vice versa
every real in [0,1) has a specific string and it´s sum. There is only one
string that desribes exactly some real. What´s wrong with that?
Finite integers cannot map one-to-one to reals, because there are zeros on
the left hand side of terminating placeholder. We need all possible
combinations of placeholder content (just like in the case of reals [0,1))
and that is possible only and only, if we consider the concept of infinite
integers instead of finite integers. Then and only then we can find always
an infinite integers that has exactly the identical string, i.e. in the
corresponding placeholder there is exactly same digit content in the
corresponding real [0,1). For example ....111 written in the omega zero
form(at) is 0,111.... has the corresponding real R [0,1) written 0.111...
There is no other possibilities, this is the only one we can find. Therefore
we have exactly one-to-one mapping. What´s wrong with that?
> Note that for every terminating sequence (except the all zero sequence)
> there is a sequence with all but finitely many of its values equal to 9
> which converges to the same limit. And the all nines sequence converges
> to 1.
> A similar mapping from the sequences of digits to integers can be
> carried out explicitly only for terminating sequences by the mapping
> {a_n} -> sum(a_n*10^n, n = 0..oo), since for non-terminating sequences
> such series always diverge and do not determine any standard integer.
Yes, I assume, I understand your point. I would rather concentrate in sums
of digit strings (having placeholders) if you don´t mind. Consider some
infinite integer N(inf), it is clear that the sum diverges if the point of
reference is the standard zero. An important point is to change the point of
reference from the standard zero to omega zero. As a classic example
consider the infinite integer ...9 (or ...999). By diving this by omega
results in 0,999... Now this is convergent instead of divergent.
The limit of infinite integer ...999 is clearly omega, i.e 1, which is not
the finite integer 1, but the first transinfinite integer called omega.
Omega is the number that is greater than any infinite integer defined as a
infinite sum (see previous posts). If you remember - hopefully - we have to
add standard finite 1 (or written in the infinite format ...01) to ...999
before we can get omega. The simly reason was that all the placeholders were
infinitely occupied by digit nine (9). We can at image, we have carry over
infinitely, because we must be able always to add one to any number - finite
or infinite. Please note: All the infinite sums of infinite integers diverge
(from the standard reference point "normal zero"), but by changing the point
of reference to omega zero, you are able to get it to converge. I assume
that would be a kind of benefit - hopefully.
This is a simply consequence that all infinite integers written in the omega
zero format have a unic string that is exactly the same string as certain
and only one real within [0,1) can have (and vice versa - of course). Thus
every real within [0,1) maps to only one infinite integer. There is no other
similar infinite string.
> I understand your N(inf) to be the set os all sequences
> f:N->{0,1,2,3,4,5,6,7,8,9}, with the terminating series corresponding to
> the (finite) integers.
No, because the string is infinite. Finite integer has a unic infinite
string written in the infinite integer format, containing zeros on the left
hand side in the non-significant placeholders. An infinite integer (just
like a real [0,1)) have significant placeholders infinitely.
> > The integer is finite, and therefore a special case of N(inf), if all
> > placeholders beginning from some arbitrary placehoder n>1 have filled by
> > zeros, i.e. on the left side placeholders.
> > Therefore we may express all the infinite integers in N(inf) [0,omega),
if
> > we have all possible combinations of a_n.
>
> Ibelieve I now understand the symbol N(inf), but what do you mean by
> n(inf)[0,omega) ?
>
I try to explain. Please - ask if I cannot explain it clear enough.
Infinite integers cover all the integers from 1 to omega. Omega is excluded
because the finite integer 1 maps to omega (= first transinfinite integer
"after infinite integers" (See example ...999 + ...01 above). ...9 is the
largest infinite integer N(inf) because all the placeholders are occupied by
9 - infinitely. If we add one - because we must be able to do so, then we do
not have free room except in the transinfinite position (compare Hilbert's
hotel). This last addition of integer 1 causes, let´s say an overflow (of
carry) - if you allow that I express things a little bit concrete form.
There is another way to express one to one mapping that is perhaps easier to
understand (let´s call it - because of the simplicity - mirror mapping).
Here we map by swapping - i.e. by turning or reversing the indexing of all
the placholders:
Infinite integer [0,omega) Real [0,1)
...999 0.999...
...998 0.899...
...997 0.799...
....
...11 0.11...
.....01 0.10...
Also in this special "mirror mapping" there is only one unic string
(infinite integer) that maps to another unic string within reals in [0,1).
I prefer the point of reference mapping:
Omega zero format Infinite integer (standard zero)
Real [0,1) Real [0,1) in omega format
omega
1 omega
0,999.... ...999
0.999... ...999
0,999...8 ...998
0.999...8 ...998
0,999...7 ...997
0.999...7 ...997
.......
0,00...1 ...01
0.00...1 (x) ....01
(x)= epsilon
Real [0,1) in omega format we have shifted the point of reference
infinitely "far away" from standard zero, but in this case to the right. I
guess what you may argue, but I leave it to the next mail (because this mail
is already quite long).
An example about real [0,1) in omega format. Let´s ask what is the mean
value of reals 0.999...8 and 0.999...7. Simply:
(0.999...8 + 0.999..7)/2 = 0.999...7,5 where 0.999...7 belong to the
standard real part and 0,5 belongs to the omega part, because 0,5 is
transinfinite. So, if two reals have nothing between, it does not mean that
they are equal as the standard explanation would possible explain.
Analogically - It does not means that integers 1 and 2 are equal, because
there is no integer between them. Therefore we introduce the standard
decimal point. The mean value 1+2 equals to 1.5. Therefore I introduced a
new decimal point -too. In this case the omega zero decimal point. What´s
wrong with that?
We can proof as seen during the discussion with Dave Seaman that 0.999...8
and 0.999..7 are not equal. The proof goes by writing them equal and
eliminating equal members from the both sides of the infinite sums (series).
7 is not 8 as it is well known. They cannot be equal. The proof is not
limited to certain placeholder a_n. Replace one digit in any placeholder and
the string and the sum is different. We can see it easy for example by
counting the difference of two infinite strings, say
...998 - ...997= ...01 = 1. Note: it is impossible to say did I count the
difference between two reals or between two infinite integers before it is
specified what was the point of reference: Standard zero, omega or maybe
something else.
> > Now, for every N(inf) [0,omega) there is some real in R [0,1) and vice
> > versa. This is one-to-one mapping.
>
> I can see a mapping f:N(inf) -> R[0,1], although it is many-to-one, not
> one-to-one.
I dropped out - sorry. Why many-to-one? Any string is unic. From the point
of view of the two different reference point, there are only one specific
and unic string: for example in infinite integer area or in the real area
[0,1). How those two unic string cases can map many-to-one ????
> If your [0,omega) means [0,1) \union {omega}, the I can even deal with
> g:N(inf) -> [0,omega), but it is still not one-to-one.
[0, omega) means the set of infinite integers from 0,1,2,3,... up to omega,
but omega excluded. All the integers - even the finite integers can be
written in the infinite format.
> > (Note finite integer 1 maps to omega -
Sorry I cannot follow your argumentation above. Let´s see 9+1=10, 99+1=100,
999+1=1000, but if we have infinite long string- note infinitely in every
and in all placeholders (9), then ...99999999999+1 simply overflow to omega
and not to zero.
> In fact, this set of sequences becomes a group under placewise addition
> with carrying, with every sequence having a negative.
??????
> > Thus, directly according to the
> > definition of omega. Therefore also ...9 is the greatest possible
infinite
> > integer < omega.
> Testing for equality is straightforward in N(inf), but how do you test
> for greater than or less than in N(inf)?
>
> > Note that ...9 is odd. If we add 1 (or ...01) omega must be
> > even. But omega is even only from the point of view that is the standard
> > point of reference, i.e. standard zero = the normal zero. Omega is the
> > first number one "behind" or "above" of N(inf), i.e. the first
transinfinite
> > number.
>
> Since non-terminating sequences are something other than finite, say
> "uber-finite", if "omega" is larger than these, it must be
> "super-uber-finite" or "trans-uber-finite".
This suits very well for me. I understand uber is German and would be
translated "over" . Over-finite or super-over-finite etc...
> (Normally called transfinite). Because omega is the first
> > "transfinite number one", it is odd as we consider even and odd numbers
> > from the omega point of view.
> > Now let´s describe omega zero (0). Omega zero is the point of view that
is
> > infinite far from standard zero point of view. The word "far" is used
here
> > only as an illustrative meaning. If we write the infinite integer ...9
as a
> > "decimal number" when the point of reference is omega, then
> > ...9 = 0,9... (or 0,999...) Note 0, is here omega zero reference point.
>
> I don't understand your notation. Does ...123 become 0,321...?
No, it does not. I depends now do you mean that 1 is repeated infinitely or
123 are repeated infinitely. This is a question of agreement. I would
prefer - but other formats are well OK - this:
...123 = 0,111.....123. There are infinite many 1 repeated.
...123123 would be written 0,123123... in the omega zero notation.
> If you are saying that the "0," precedes ...9, you can't write 0,9...,
> since there isn't a "first" 9 to put the "0," in front of. You would
> have to write 0,...9 indicating infinitely many 9's to the left of the
> visible 9.
This suits well OK -too, but remember all placeholders were occupied by 9.
In handled this in my reply toJohannes - if I remember right. It is namely
so that the only claim is that there are infinitely 9´s. It does not disturb
if there are infinitely nines with start and end, if the string is infinite-
not finite. Two end does not mean automatically finite. We assume it is
finite without further consideration. Tell me - please why not?
> From this point on, there are so many internal ambiguities, that I can't
> follow it at all.
>
> While this exposition may be quite creative, it just isn't internally
> consistent.
Hopefully I was helpfull. Therefore we discuss.
Tapio
Virgil
I will presume that "unic" means the same as "unique".
The string 0.1999... and the string 0.2000... are different, but
correspond to the same real number. In fact, any "terminating" decimal
except 0=0.000..., will also have a representation with all but
finitely many of its digits equal to 9.
Also, the string 0.999... evaluates to 1.
Therefore, the proposed mapping from digit strings to reals has image
[0,1], not [0,1), and it is many-to-one and not one-to-one.
That is what is wrong.
> Finite integers cannot map one-to-one to reals, because there are zeros on
> the left hand side of terminating placeholder. We need all possible
> combinations of placeholder content (just like in the case of reals [0,1))
> and that is possible only and only, if we consider the concept of infinite
> integers instead of finite integers.
I can accept terminating strings ( all but finitely many places are
zero) corresponding to ordinary integers. However, if we identify
non-terminating strings with "infinite integers", the correspondence
between infinite integers and reals in [0,1) leaves out ...999, and the
correspondence with reals in [0,1] is not one to one.
> Then and only then we can find always
> an infinite integers that has exactly the identical string, i.e. in the
> corresponding placeholder there is exactly same digit content in the
> corresponding real [0,1). For example ....111 written in the omega zero
> form(at) is 0,111.... has the corresponding real R [0,1) written 0.111...
> There is no other possibilities, this is the only one we can find. Therefore
> we have exactly one-to-one mapping. What´s wrong with that?
Answered above. some reals have 2 "infinite integers" corresponding.
>
> > Note that for every terminating sequence (except the all zero sequence)
> > there is a sequence with all but finitely many of its values equal to 9
> > which converges to the same limit. And the all nines sequence converges
> > to 1.
>
> > A similar mapping from the sequences of digits to integers can be
> > carried out explicitly only for terminating sequences by the mapping
> > {a_n} -> sum(a_n*10^n, n = 0..oo), since for non-terminating sequences
> > such series always diverge and do not determine any standard integer.
>
> Yes, I assume, I understand your point. I would rather concentrate in sums
> of digit strings (having placeholders) if you don´t mind. Consider some
> infinite integer N(inf), it is clear that the sum diverges if the point of
> reference is the standard zero. An important point is to change the point of
> reference from the standard zero to omega zero. As a classic example
> consider the infinite integer ...9 (or ...999). By diving this by omega
> results in 0,999... Now this is convergent instead of divergent.
This process of "division of an 'infinite integer' by 'omega'", has not
been explained clearly. Actually your definition of "omega" below is
also not too clear. How does "transfinite" differ from "infinite"?
Why is the limit of ...999 anything but the sequence ...999 itself?
If the "limit" of ...999 is "omega", what is the limit of ...123123?
> The limit of infinite integer ...999 is clearly omega, i.e 1, which is not
> the finite integer 1, but the first transinfinite integer called omega.
> Omega is the number that is greater than any infinite integer defined as a
> infinite sum (see previous posts). If you remember - hopefully - we have to
> add standard finite 1 (or written in the infinite format ...01) to ...999
> before we can get omega. The simly reason was that all the placeholders were
> infinitely occupied by digit nine (9). We can at image, we have carry over
> infinitely, because we must be able always to add one to any number - finite
> or infinite. Please note: All the infinite sums of infinite integers diverge
> (from the standard reference point "normal zero"), but by changing the point
> of reference to omega zero, you are able to get it to converge. I assume
> that would be a kind of benefit - hopefully.
> This is a simply consequence that all infinite integers written in the omega
> zero format have a unic string that is exactly the same string as certain
> and only one real within [0,1) can have (and vice versa - of course). Thus
> every real within [0,1) maps to only one infinite integer. There is no other
> similar infinite string.
>
> > I understand your N(inf) to be the set os all sequences
> > f:N->{0,1,2,3,4,5,6,7,8,9}, with the terminating series corresponding to
> > the (finite) integers.
>
> No, because the string is infinite. Finite integer has a unic infinite
> string written in the infinite integer format, containing zeros on the left
> hand side in the non-significant placeholders.
You must have snipped the section where I defined "terminating" as being
all zero past some point.
The above should be
Sequence Real in [0,1]
...999 0.999... = 1
...998 0.899... =9/10
...009 0.900... = 9/10
...997 0.799... = 8/10
...008 0.800... = 8/10
... ...
...111 0.111... = 1/9
...001 0.100... = 1/10
> Also in this special "mirror mapping" there is only one unic string
> (infinite integer) that maps to another unic string within reals in [0,1).
Wrong. for every rational in (0,1) which can be expressid in form
n/10^m, for non-negative integers n and m, there are _two_ "infinite
integer" strings corresponding to it. One is sequence co-finitely
a_n = 0 the other is co-finitely a_n = 9. NOTE: co-finite means all but
finitely many, so {a_n} giving ...33321 is co-finitely a_n = 3.
> I prefer the point of reference mapping:
>
> Omega zero format Infinite integer (standard zero)
> Real [0,1) Real [0,1) in omega format
> omega
> 1 omega
> 0,999.... ...999
> 0.999... ...999
> 0,999...8 ...998
> 0.999...8 ...998
> 0,999...7 ...997
> 0.999...7 ...997
> .......
> 0,00...1 ...01
> 0.00...1 (x) ....01
>
0,999...8 does not correspond to any possible _infinite_ sequence {a_n}
of digits, since the notation 0,999...8 assumes, contrary to fact, that
there is both a first and a last term in this infinite sequence. If you
are going to have an infinite string of digits, it must be open ended at
one end or the other (or both).
> (x)= epsilon
?
It will be all zeros with an overflow or "carry one". Is that your
"omega"? If so, you will now have to work out your "overflow" arithmetic
for the "carry registers".
>
> > In fact, this set of sequences becomes a group under placewise addition
> > with carrying, with every sequence having a negative.
>
> ??????
>
> > > Thus, directly according to the
> > > definition of omega. Therefore also ...9 is the greatest possible
> infinite
> > > integer < omega.
> > Testing for equality is straightforward in N(inf), but how do you test
> > for greater than or less than in N(inf)?
You have not answered my question: how do you determine order among yout
"infinite integers?
You have an infinite sequence of digits which you list in left to right
order to represent real decimals and in right to left order to represent
"infinite integers".
If the decimal 0.ddd... does not have a last digit, for d representing a
digit, then the corresponding "infinite integer" , ...ddd, cannot have a
first digit, and cannot be represented 0,d,,,ddd.
And a sequence, listed in either direction, which has both a first and a
last term is finite. There may be infinite objects with firsta and
lasts, but they are not sequences.
>
> > From this point on, there are so many internal ambiguities, that I can't
> > follow it at all.
> >
> > While this exposition may be quite creative, it just isn't internally
> > consistent.
>
> Hopefully I was helpfull. Therefore we discuss.
>
> Tapio
>
>
There are still too many inconsistencies.
Tapio
> There are still too many inconsistencies.
Thus, in e-mail I try concentrate in the mentioned inconsistencies. These
problem, which Virgil wrote in post, cover two basic problem groups:
"many-to-one mapping" and "first digit last digit" problem. (Some sub-titles
are added below)
"Virgil" <vmh...@home.com> wrote in message
news:vmhjr2-0E13FB....@news2.rdc2.tx.home.com...
> In article <9hqh73$fse$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi> wrote:
>
> > "Virgil" <vmh...@home.com> wrote in message
> > news:vmhjr2-4FF9E7....@news2.rdc2.tx.home.com...
> > > In article <9hnqfj$1e6$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi>
wrote:
> > >
> > > > "Virgil" <vmh...@home.com> wrote in message
> > > > news:vmhjr2-492E6E....@news2.rdc2.tx.home.com...
> > > > > In article <9hkf3p$ua$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi>
wrote:
> > (snip)
Tapio:
> > I explained every Real [0,1) is an infinite sum. Every distinct sum,
i.e. at
> > least one content in some placeholder (a_n) is different than in some
other
> > infinite sum exactly in the same placeholders, presents a unic real
[0,1).
> > For example:
> > 0.1222.... is unic and different than 0.21222... that is unic
string -too.In
> > other word, those are infinite sums and present only one real that is
unic,
> > i.e. there is no other similar. There are infinite various combinations.
> > Each possible combinations expresses just one and only one real. Vice
versa
> > every real in [0,1) has a specific string and it´s sum. There is only
one
> > string that desribes exactly some real. What´s wrong with that?
Virgil:
> I will presume that "unic" means the same as "unique".
Yes, indeed. Sorry
1) MANY-TO-ONE VERSUS ONE TO ONE PROBLEM
> The string 0.1999... and the string 0.2000... are different, but
> correspond to the same real number. In fact, any "terminating" decimal
> except 0=0.000..., will also have a representation with all but
> finitely many of its digits equal to 9.
>
> Also, the string 0.999... evaluates to 1.
>
> Therefore, the proposed mapping from digit strings to reals has image
> [0,1], not [0,1), and it is many-to-one and not one-to-one.
>
> That is what is wrong.
That (above) is just one of the basic question, which can be solved (IMHO)
with the aid of the concept of the "so called" infinite integers. If I have
understood correctly the problem is - among others:
1) many-to-one mapping instead of one-to-one mapping. The basic reason -
according to Virgil's post is that some reals in [0,1] have two string
presentations. Shortly: This problem has a connection to the popular 0.999..
thread.
2) [0,1] instead of [0,1) because of the same reason: many-to-one mapping
instead of one-to-one mapping.
Some earlier discussion as a introduction:
Tapio:
> > Finite integers cannot map one-to-one to reals, because there are zeros
on
> > the left hand side of terminating placeholder. We need all possible
> > combinations of placeholder content (just like in the case of reals
[0,1))
> > and that is possible only and only, if we consider the concept of
infinite
> > integers instead of finite integers.
Virgil:
> I can accept terminating strings ( all but finitely many places are
> zero) corresponding to ordinary integers. However, if we identify
> non-terminating strings with "infinite integers", the correspondence
> between infinite integers and reals in [0,1) leaves out ...999, and the
> correspondence with reals in [0,1] is not one to one.
Tapio:
> > Then and only then we can find always
> > an infinite integers that has exactly the identical string, i.e. in the
> > corresponding placeholder there is exactly same digit content in the
> > corresponding real [0,1). For example ....111 written in the omega zero
> > form(at) is 0,111.... has the corresponding real R [0,1) written
0.111...
> > There is no other possibilities, this is the only one we can find.
Therefore
> > we have exactly one-to-one mapping. What´s wrong with that?
Virgil:
> Answered above. some reals have 2 "infinite integers" corresponding.
> >
> > > Note that for every terminating sequence (except the all zero
sequence)
> > > there is a sequence with all but finitely many of its values equal to
9
> > > which converges to the same limit. And the all nines sequence
converges
> > > to 1.
> >
> > > A similar mapping from the sequences of digits to integers can be
> > > carried out explicitly only for terminating sequences by the mapping
> > > {a_n} -> sum(a_n*10^n, n = 0..oo), since for non-terminating
sequences
> > > such series always diverge and do not determine any standard integer.
Tapio:
> > Yes, I assume, I understand your point. I would rather concentrate in
sums
> > of digit strings (having placeholders) if you don´t mind. Consider some
> > infinite integer N(inf), it is clear that the sum diverges if the point
of
> > reference is the standard zero. An important point is to change the
point of
> > reference from the standard zero to omega zero. As a classic example
> > consider the infinite integer ...9 (or ...999). By diving this by omega
> > results in 0,999... Now this is convergent instead of divergent.
Approaching the problem "many-to-one" versus "one-to-one" mapping
I repeat: You claim many-to-one maping instead of one-to-one mapping,
because there are alt least two presentation (many) for each real like
0.999...=1 but only one presentation in the set of infinite integers.
Hopefully OK!
I would like to solve that problem by using following illustrative examples:
a) First, I notice we do not debate about the existence of infinite
integers. I assume there exist some kind of consensus that: at least it is
possible to define infinite integers as a infinite sum, which results in
infinite integer, (if we omit the discussion about convergence and limit and
possible inconsistencies that should be solved) At least I assume you have
quite well understood idea concerning infinite integers. I base this
observation on your replies, where you - among others - wrote:
Virgil (7/2/2001):
"If you are saying that the "0," precedes ...9, you can't write 0,9...,
since there isn't a "first" 9 to put the "0," in front of. You would
have to write 0,...9 indicating infinitely many 9's to the left of the
visible 9."
b) You wrote also in the same message:
"A treatment more in line with the algorithms of standard arithmetic will
have ...9 + 1 = 0, so that ...9 = -1, with similar results for any
sequence having all but finitely many terms equal to 9.In fact, this set of
sequences becomes a group under placewise addition
with carrying, with every sequence having a negative."
This was something I was not able to follow, therefore I repeat:
Assume some finite integer. Let´s say 99, because of simplicity. I assume
you may agree that 99<999<999<9999 etc.
Thus, the more we have nines in the finite string the greater is the
integer. Thus, it should follow as a consequence:
...9 > than any possible finite integer. More generally - infinite integers
are greater integers than finite integers, because infinite integers have
infinite significant palceholders and finite integers have finite
significant placeholders. (Note: we can write finite integer as infinite
integers as we add to the left hand side of the finite integer more
placeholders so that the string has infinitely placeholders. Now we fill
those placeholders with zeros, because those zeros indicates that the
placeholders are non-significant in counting operations. Therefore finite
integer is just a short-hand notation as the non-significant zeros are
purposely omitted (=not written).
Thus if ...9+1, then result must be greater than any infinite integer - as
seen earlier in previous posts. The result is omega. if we define that omega
is greater number than any infinite integer. Anyway, it is fact that the
integer (infinite or finite part) is zero, but the total sum of ...9+1 is no
way zero, because we add 1 the result is greater by one and all the
placeholders were occupied already by 9 in the string of ....9. The omega
can be written in the omega zero format like 1,000.... (or as you prefer
1,....0)
On the other hand ...9 is no way -1, because it is greater than any finite
string (=finite integer) of 9´s.
Therefore the infinite integers exist between omega and finite integers.
If I should somehow describe how those different positive numbers are
situated on a line, then the following illustrative picture may help:
<------------------W-------------------N---------------------1--------------
-----------0
<"trans-uber-finite-><-infinite integers-><----finite intgers----><---reals
[0,1]---------->
<-infinite and finite integers = all
integers--->
Infinite integers=uber-finite. W=omega.
c) You wrote in the last message (7/3/2001) (below):
Virgil:
"> Why is the limit of ...999 anything but the sequence ...999 itself?
> If the "limit" of ...999 is "omega", what is the limit of ...123123?"
This is a very important point and question!
As ...999 is the infinite integer and you can write it as a sum of infinite
serie, which diverges as you correctly mentioned earlier.
Why it should results in something else than the string (or sequence)
itself. Indeed why?. You certainly remember that I wrote that I have to add
1 to receive omega and now I say the limit of the sum of the infinite serie
is omega. This is at first glance a clear inconsistency. The explanation is
as follows:
1) We have to consider what does mean the sum of infinite serie. According
to the discussion with Dik Winter, he told something like this (he can
correct if I missed something):
There is no the limit of the sum of the infinite serie. We have only the sum
of the infinite serie and thst is the limit. This is the definition of the
sum of infinite serie. (We discussed about reals in the thread of the
problem 0.999... and 1).
(I jump over the details concerning tangential limits and asymtotical
limits.)
I think we have to separate two concepts: the limit of the sum of the
infinite serie and the sum of the infinite serie, because:
a) It is easy to imagine that we must have always right to add one to any
integer (Euclides).
If we consider ...999 exactly as you asked above, then the sum of this
divergent serie does not results in ...999 as espected, but it´s limit
omega. In this case we have to conclude ...999 presents omega, but not the
infinite integer ...999.
If we now add ....999+1 the result equals to omega +1.
b) If we now change the point of reference from standard zero to omega zero,
the we can write ...999 in the decimal format (according to your suggestion)
0,....9, which is the convergent serie like 0.999.... Why? The both
expressions are infinite convergent series. The infinite serie has a limit
omega (i.e. the first super-uber-finite number 1) and the last "classic" has
well known limit 1 as you pointed in your e-mail -too.
c) If we now keep rigorously about your evident and righterous observation:
"> Why is the limit of ...999 anything but the sequence ...999 itself?" then
we have inconsistency problem with infinite integer like ....999 (=0,...9).
We have surprisingly two limits namely that what I also prefer (with you)-
namely - ...999 and another - omega. We have two limits, because we changed
the point of reference. I assume, if we allow that (above) then we have
really inconsistency problems: Considering infinite integers we have two
limits and concerning reals [0,1] we have two expressions for the same real
(for example 0.999..=1 etc.).
d) Further: Considering two infinite integers ...998 and ...997, it should
be evident that their difference equals to 1. Namely: ...997+1 results in
...998. The same problem written in the format as the reference point is
omega zero: 0,...97+ 0,...01=0,...98 or 0,...98-0,...01=0,...97. Note that
surprisingly these calculations remind us about reals: the only difference
is that instead of "0," we have "0.".
I discussed about those two reals (above, 0.999...98 and 0.999...97) with
Dik Winter and Dave Seaman. My question was: If 0.999...=1, then what is the
next smaller real <0.999...). The answer I received was in principle this:
It cannot be 0.999..98 neither 0.999...97. Actually, nobody was able to
answer to my question, though we all knew that some reals must be samller
than 0.999.... The interesting point was that Dik Winter wrote 0.999..98 and
0.999..97 are equal because they have the same limit, namely 1 like 0.999...
have. The final result of discussion was that 0....98 and 0....97 are
different after setting them equal and eliminating equal members from the
both sides of equation, because 7 is not 8. So we had contradiction in the
definition of infinite sum.
We have to ask: Why they have the same limit and why those two reals are
equal, if in the case of infinite integers we can clearly say that
...998>...997. Especially if we write those two integers in the decimal
format as the point of reference is omega-zero instead the standard zero
(0,999...98 or 0,...98 and 0,999...97 or 0,...97) ? People see that
difference almost immediately because they think integers (even infinite
integers) as integers, but they do not see any difference in the case of
reals [0,1], though the string are indentical except we have different point
of reference. In the case of integers omega zero and in the case of reals -
standard zero.
I think the reason is very human: We think integers are the greater the more
we have digit from right to left. In the case of decimal numbers the right
most digits have less meaning the further to the right we look. In the both
cases we look at the numbers from the standard point of zero. Therefore they
also say there is no last digit.
(This small story may hopefully illustrate our situation:
If I now take the same attitude with imaginary people (somewhere in the
galaxy) who look at numbers from the point of omega zero and who have ever
heard about standard zero point of view. Let´s imagine I claim there is such
an tiny integer like 0,...01. (This is number one written in the infinite
format according to Virgil´s recommendation). They (those imaginary people)
would say: "Hey, wait a minute. There is no such a number you wrote. First
of all there is no last digit in our real world and there is no smallest
real in our real world." If I now tell them that there is a planet in this
galaxy called Earth and there are lots of mathematicians, who claim that
they have a different standard point of reference named what they call zero
and those people claim that this smallest number I wrote for you within your
"real world" is called an integer with the name "one". If I add that those
people in the planet Earth can handle even smaller numbers than that famous
"one" and if I tell they call them reals, I would expect something like "F..
You". If those imaginary people would have sci.math you may image what kind
of comment would be expected?) ;-)
I assume the most restricting point in our thinking is that we are too much
standard zero bounded. The change of point of reference is for most of
people out of question, though they use decimal dot quite flexible way in
their calculations. The most useful benefit of the change of point of
reference is that we can observe divergent series to converge. As an example
I solve your question."If the "limit" of ...999 is "omega", what is the
limit of ...123123?"
Let´s analyse your string ...123123. Assuming this is an infinite integer
with standard zero point of reference, then there exist a repeated string
fragment of 123 - infinitely, i.e. non-terminating fragment. Your number
written in the decimal format as the point of refrence is omega zero:
0,123123... or 0,...123 (as you prefer). It cannot be 0,23123123... neither
0,3123123... because in these cases the repeated fragment is not the same.
If we change the point of reference, then the number must be exactly the
same, .i.e the change of point of view does not affect on the presented
string pattern. Your number is clearly 41(omega)/333. But this is the plain
infinite sum as it should be. It is defined (the classic definition) that
the plain infinite sum is the same as the limit of infinite sum. The actual
limit is epsilon greater. (Epsilon environment). The epsilon in the case of
infinite integers is plain "number one" written in the infinite format
0,...01. Thus the limit is this much greater. Therefore the limit of
0,999... is omega. (See explanation below). (We can have even more precise
limit if we consider the decimal part.)
Everything is of course based on the fact: "For every epsilon there
exists...." And this true for any epsilon infinitely and without end - of
course. For infinite integers this fact is easy to see if we epand the
"epsilon" concept (in this case integer 1) to smaller "epsilons" namely to
standard decimal numbers, because there is always epsilon - infinitely,
according to the definition of epsilon-delta theory.
d) Let´s imagine two different strings like (because of simplicity) ...98
and ...81. The set question would be: What is their difference? Let´s
count: ...98-...87= ...1
You may assume I counted the difference of two infinite integers. Right?
Now I tell you I never mentioned the point of reference. Did you make some
assumption? Anyway it does not matter, because the purpose is totally
different. Actually, I would have counted above omega with trans-uber-finite
integers (the point of reference omega zero) or with trans-uber-finite reals
that are smaller than smallest real (let´s say some epsilon) (the point of
reference omega zero below reals. I can explain that later).
My point is that counting rolls over theory. Iff I have said my point of
reference was standard zero and I actually wanted to calculate the
difference of 0.99..98 and 0.88...81, then the standard reply would be:
there are NO such numbers called reals.
There is NO last digit etc.
So, why don't you claim that the sum of infinite serie must always produce
the expressed string, i.e. number? Why do you accept that it is is fear and
right in the case of infinite integers:"Why is the limit of ...999 anything
but the sequence ...999 itself?, but you accept in the case reals that
0.999...=1. You should claim that the sum of that infinite serie is exactly
0.999... and NOT 1? Exactly the same way as ...999 written in the decimal
format as the point of reference is omega zero, i.e ...999=0,...9=0,999...
The sums should be equal consistent and it´s not depending on the point of
reference. Can you accept that? Why do you use in the case of integers the
plain sum but in the case of reals it´s limit, i.e. the limit of the sum of
infinite serie? This kind of thinking (based on the definition) is something
I would like to call inconsistence. (Nothing personal - of course, Virgil).
Actually, you expressed exactly the same question as I expressed to myself:
"Why is the limit of ...999 anything but the sequence ...999 itself? And if
this is true it should be true also in the case of 0,...9 or 0....9 or
0,999... or 0.999... , because the point of reference cannot affect to the
sum, i.e. we must be able to turn or reverse the indexing (a_n) and the sum
must be exactly indentical.
To avoid any inconsistency and "NO exist" messages, we must have standard
calculation over any fields and over any point of reference.
e) Every terminating finite sum has exact result. There is no limit defined
for finite integer neither finite real [0,1].
Every non-teminating infinite sum concerning reals is considered to have
convergent limit. Every non-terminating infinite integer is a divergent sum
(without limit, i.e. plain sum), but iff it's written as a convergent sum as
the standard point of reference is changed to the omega zero point of
reference, then the sum has a limit (see above). If we change the point of
reference of real number from standard zero to the omega zero ( infinite
"far away" to the right hand direction) and if we write the sum that should
results in exactly the string number. In this case the sum is again
divergent (for example 0.999... = .....9)
If we want to avoid inconsistency we have to claim that the sum results in
the written number inspite of the point of the reference.
Now, bacause there are numbers greater than omega and smaller than any real
(compare the example of the mean value of two consecutive reals), then we
must accept that integers or reals are no way special cases. The same rules
should be valid for those numbers greater than omega or to the extensions of
reals.
If we write two divergent or convergent series equal, then it is easy to
show by eliminating equal members from the both sides of equation that those
choosen series are not equal iff one of the members (say in the any
palceholder (a_n) are different).
The point of reference must be taken account. For example by comparing
1.0000.... =0.999..... results in after simplification (carry over to the
right direction -->) to 9=10, that is not true. The number of members does
not affect on the elimination of equal members - it is just the same if
there are finite or infinite memebers in the strings.
When we compare series we have to solve that inconsistency.
A natural requirement are that
1) the sum must result in the written number
2) the change of the point of reference results in equal results, i.e. the
same string describing the number.
3) Those, who prefer the concept of limit, should consider the exact
definition of the limit and epsilon - delta theory and inconsistent
consequences versus the plain sum.
The plain infinite sum gives a right result. The limit of the infinite sum
(the standard definition of infinite sum including the limit concept)
results in the "next" neighbour string. This old definition fails as the
limit and the original string are set equal. It does not result even the
expected string. The limit (the classic sum of infinite serie) is good
approximation, but not the exact value (n 0 --> oo).
This arises directly from the epsilon-delta theory. It´s worth to recognize
the decimal expansion cannot always be exactly the fraction presentation
(Example 1/3 is about 0.333... simply because super-uber-finite decimals are
omitted. The same way epsilon theory does not consider super-uber-finite
decimals. The reason is exactly the same in the case of two consecutive
infinite integers. 1 is the smallest integer difference (as epsilon in the
case of reals), because we usually stop counting to integers. If we want to
be more precise, we should use decimal expansions.
The concept of plain sum resolves the many-to-one problem. We have only
one-to-one mapping.
2) THE PROBLEM OF "FIRST DIGIT - LAST DIGIT"
If we consider any terminating (finite) string that expresses some finite
integer or some terminating real (Q E R) and if we count the sum of that
serie, then we have to know what is the lenght of the string. In the case of
infinite string we know that the "lenght" of the string in endless,
non-terminating or infinite. In the both cases it is just enough to know how
"far" (infinite or infinitely) we have to count. The problems in the case of
infinite string are mainly menthal. We should see easy that it's a question
about menthal problem considering the following examples, where I purposely
concentrate in infinite integers:
a) ....8
b) 0,...8
c) 0,8...
a,b and c describes the same infinite integer. All of them are thus infinite
and every placeholder (a_n) contain a digit 8.
a) is divergent serie, bacause we have the standard zero as point of
reference. b) and c) are written in the decimal format, because we have in
these cases the omega zero point of reference. In all the cases a,b and c we
have to count the sum of infinite serie - not the limit, as you correctly
noticed and this sum must be the string that describes the number - exactly.
As you would see the result is independent if we write the first or the last
digit. In every case the sum equals to 8*omega/9.
So, why worry about first and last digit, because we have in every and in
all placeholders the digit 8 and the string are infinite in all cases? If
you think there exist inconsistency - please inform me - if you receive
different results in the case a,b or c.
Thus the problem of last-first digit arises from menthal barriers - not from
mathematics.
Consider the same strings above (a,b and c) but now as reals. Then we have
"0."- of course - instead of "0," The situation is exactly the same. The
infinite sums are the same inspite of "first or last digit problem". In the
case of a) the point of reference is omega zero (infinite far away to the
right from standard zero). Where is now the inconsistency?
In those cases we have a longer repeating string fragment, then we have to
count as seen earlier above, when I solved your example about ...123123. All
we have to consider is again here: infinite serie (string) and the string
fragment is repeated infinitely. The rest is pure counting, when the point
of reference is selected.
3) DIVISION QUESTION (OMEGA)
Some earlier conversation
Virgil:
> This process of "division of an 'infinite integer' by 'omega'", has not
> been explained clearly. Actually your definition of "omega" below is
> also not too clear. How does "transfinite" differ from "infinite"?
> Why is the limit of ...999 anything but the sequence ...999 itself?
> If the "limit" of ...999 is "omega", what is the limit of ...123123?
The division, multiplication, subtraction, addition and any other operation
are exactly similar around omega zero as around standard zero including
infinite sums as explained above. Omega zero is just another point of
reference. The standard zero point of reference is just one of those
reference points. There are infinitely various reference point above omega
and inside reals [0,1). You have to consider new point of reference if you
handle strings that are "longer" than infinite. For example consider number:
...123123.123123... This number has an infinite integer part (to the left
from decimal dot), then standard zero dot and "then normal" decimal part to
the right from decimal dot. The decimal part is infinite -too. If I write
this as omega zero decimal, then that number equals to:
0,...123123.123123.... . Because omega is number 1 (super-uber-finite), that
string number does not change because we divide by 1 if the string is
written already in the decimal format as the point of reference is omega
zero.
Situation is analogous to, if I divide real 0.999... by finite 1. It´s still
0.999...
Omega below or above were illustrative wording. Those expressions were not
good - sorry! Correct expressions would be greater than omega (above) and
smaller than omega (below).
"Transfinite" concept is usually (traditionally ?) bounded finite integers.
Ref. your text books. Look at my picture earlier: infinite integers, which
is also bad name, are actually extented finite integers with infinite string
of placeholders between omega and standard zero. Those numbers that are
greater than omega would be called traninfinite or something else. I do not
want to give new names. It is enough, if some useful old names are well
suitable. Your "uber-finite" and "super-uber-finite" are OK, if we
inderstand each others.
ANSWERS TO SOME OTHER QUESTION
(snip)
Tapio:
> > Also in this special "mirror mapping" there is only one unic string
> > (infinite integer) that maps to another unic string within reals in
[0,1).
>
> Wrong. for every rational in (0,1) which can be expressid in form
> n/10^m, for non-negative integers n and m, there are _two_ "infinite
> integer" strings corresponding to it. One is sequence co-finitely
> a_n = 0 the other is co-finitely a_n = 9. NOTE: co-finite means all but
> finitely many, so {a_n} giving ...33321 is co-finitely a_n = 3.
I disagree, see may explanation above. Note especially: You do not use two
expressions for infinite integers, iff the infinite integer is expressed in
decimal format as the point of refererence is omega zero
Virgil:
> 0,999...8 does not correspond to any possible _infinite_ sequence {a_n}
> of digits, since the notation 0,999...8 assumes, contrary to fact, that
> there is both a first and a last term in this infinite sequence. If you
> are going to have an infinite string of digits, it must be open ended at
> one end or the other (or both).
You know already my answer above: the counting of infinite serie is exactly
same. We can have "it must be open ended at
one end or the other (or both)", but also infinite with two ends. Sic! The
result is the same.
(snip)
Virgil:
> It will be all zeros with an overflow or "carry one". Is that your
> "omega"? If so, you will now have to work out your "overflow" arithmetic
> for the "carry registers".
The mathematics is just the same around the positive omega zero. If you add
9+1=10 , then the "tens" are carried over to the next position. If the
string is infinite as in the case ...9 when all the position of placeholders
are occupied by 9´s, then the carry goes into thenext free placeholder, in
this case to super-uber-finite position, that is omega =1 (the first
supe-uber-finite number). Reals [0,1) are the "first" infinite string area
(carry register goes to the "second" infinite area in the case of overflow),
infinite integers are the second infinite area and carry register
"overflows" to the third infinite area, the "third" infinite area is omega
area. The simple reason is that we do not have "tens", "hundreds" etc. to
have carry one, we have only "infinities" to have carry one mathematics in
the cases of decimal numbers, integers, omega etc..
(snip)
Virgil:
> You have not answered my question: how do you determine order among yout
> "infinite integers?
Shortest answer would be: see the picture above. The order of decimal
numbers is also standard inspite of the point of reference. Of-course 0,...9
>> 0...9.
Virgil:
> You have an infinite sequence of digits which you list in left to right
> order to represent real decimals and in right to left order to represent
> "infinite integers".
>
> If the decimal 0.ddd... does not have a last digit, for d representing a
> digit, then the corresponding "infinite integer" , ...ddd, cannot have a
> first digit, and cannot be represented 0,d,,,ddd.
> And a sequence, listed in either direction, which has both a first and a
> last term is finite. There may be infinite objects with firsta and
> lasts, but they are not sequences.
This was explained under the title "the problem of last -first digits" You
may have problems with sequences (?),but numbers that are only ordered
linear infinite strings with placeholders is very simple concistent concept.
We need only the points of references that are separated by inifinities.
Simpler than Cauchy sequences or Dedekind cut.
Any more inconsistencies? I´m here to help you!
Tapio
Tapio
mail. Now I´m waiting for your specifications or additional questions.
Tapio
"Zundark" <zun...@dont-spam-me.com> wrote in message
news:9hq15s$e3v$1...@taliesin.netcom.net.uk...
Zundark
> See my post today (replies to Virgil).
Your reply to Virgil is a mind-numbing heap of garbage.
> Your questions are answered in that mail.
No, they are not.
--
Zundark
Virgil
[ALL SNIPPED]
Tapio writes too great length, with much unneeded quoting of prior
postings.
If there is any gold among the dross, it is too thinly distributed for
me to mine.
Tapio
Tapio
"Zundark" <zun...@dont-spam-me.com> wrote in message
news:9hvgs5$cva$1...@taliesin.netcom.net.uk...
Tapio
answered? I (or we) would be pleased to read your rigorously analytic
conclusions instead of your opinions. (I have opinions about your opinions,
but those opinions do not help very much - especially you!)
Show your analytical touch. Argument against argument - please.
Tapio
"Tapio" <hurm...@dlc.fi> wrote in message news:...
Zundark
> We have read your opinions.So, what are your questions that are not yet
> answered? I (or we) would be pleased to read your rigorously analytic
> conclusions instead of your opinions. (I have opinions about your
> opinions, but those opinions do not help very much - especially you!)
> Show your analytical touch. Argument against argument - please.
I'm not interested in arguing with you. You asked me to show
you what was wrong with your idea of "infinite integers", and I
pointed out that I needed to know your definition of addition
on N(inf), and also how you define an ordering on N(inf). You
still haven't supplied these definitions, so I have nothing to
apply my analytical touch to.
--
Zundark
Tapio
> Tapio wrote:
(snip)
> I'm not interested in arguing with you. You asked me to show
> you what was wrong with your idea of "infinite integers", and I
> pointed out that I needed to know your definition of addition
> on N(inf),
Algebraic laws are normal: commutative, associative, distributive laws, laws
of sign. Ref. details in any text book.
and also how you define an ordering on N(inf). You
> still haven't supplied these definitions, so I have nothing to
> apply my analytical touch to.
Laws of order are also normal for infinite integers:
a<b, b<c => a<c
a<b <=> a+c<b+c
a<b, c>0 => ac<bc
a<b, c<0 => ac>bc
a>b <=> -a>-b
Carry operations are normal.
Infinite long digit string are separated by decimal marks.
If all placeholders in the infinite strings will be occupied with the
largest possible symbol depending on the used base system (base 10 system ,
bibary system etc.) then carry behaves exactly the same way as in the case
of any decimal system.
Tapio
> Zundark
The Scarlet Manuka
> "Zundark" <zun...@dont-spam-me.com> wrote in message
> news:9i037s$eha$1...@taliesin.netcom.net.uk...
> > I'm not interested in arguing with you. You asked me to show
> > you what was wrong with your idea of "infinite integers", and I
> > pointed out that I needed to know your definition of addition
> > on N(inf),
>
> Algebraic laws are normal: commutative, associative, distributive laws,
laws
> of sign. Ref. details in any text book.
>
> > and also how you define an ordering on N(inf). You
> > still haven't supplied these definitions, so I have nothing to
> > apply my analytical touch to.
>
> Laws of order are also normal for infinite integers:
> Carry operations are normal.
The problem is that all you have said is that these operations have
the usual properties. You were asked to actually define the operations:
for any two arbitrary infinite strings, what is the sum and how do I
determine which is smaller? It may be reassuring that you claim that
the results have the desired properties, but we'd still like to know
what the results *are*.
--
The Scarlet Manuka
Zundark
> "Zundark" <zun...@dont-spam-me.com> wrote
> > I'm not interested in arguing with you. You asked me to show
> > you what was wrong with your idea of "infinite integers", and I
> > pointed out that I needed to know your definition of addition
> > on N(inf),
>
> Algebraic laws are normal: commutative, associative, distributive
> laws, laws of sign.
Good, but that wasn't what I asked.
> > and also how you define an ordering on N(inf).
> Laws of order are also normal for infinite integers:
> a<b, b<c => a<c
> a<b <=> a+c<b+c
> a<b, c>0 => ac<bc
> a<b, c<0 => ac>bc
> a>b <=> -a>-b
This is not a definition of the ordering, it's just a list of
(alleged) properties of the ordering. If I have two elements
of N(inf), how do I tell which is the larger in your ordering?
> Carry operations are normal.
> Infinite long digit string are separated by decimal marks.
> If all placeholders in the infinite strings will be occupied with the
> largest possible symbol depending on the used base system (base 10 system ,
> bibary system etc.) then carry behaves exactly the same way as in the case
> of any decimal system.
This is really very vague, and could mean almost anything.
Can't you just give a formula?
--
Zundark
Virgil
> "Zundark" <zun...@dont-spam-me.com> wrote in message
> news:9i037s$eha$1...@taliesin.netcom.net.uk...
> > Tapio wrote:
> (snip)
>
> > I'm not interested in arguing with you. You asked me to show
> > you what was wrong with your idea of "infinite integers", and I
> > pointed out that I needed to know your definition of addition
> > on N(inf),
>
> Algebraic laws are normal: commutative, associative, distributive laws, laws
> of sign. Ref. details in any text book.
But you have failed to explain how to do addition, multiplication, etc.
on "infinite integers". For example, how does one add the "infinite
integer" corresponding to 15/37 (expressed decimally) and the "infinite
integer" corresponding to 5/7 (expressed decimally)?. How does one
multiply them?
THEN you must show that these rules for addition and multiplication obey
the algebraic rules you cite above. This does not come automatically.
>
> and also how you define an ordering on N(inf). You
> > still haven't supplied these definitions, so I have nothing to
> > apply my analytical touch to.
>
> Laws of order are also normal for infinite integers:
> a<b, b<c => a<c
> a<b <=> a+c<b+c
> a<b, c>0 => ac<bc
> a<b, c<0 => ac>bc
> a>b <=> -a>-b
But by what rules one compare two "infinite integers"?
Which is larger, ...010101 or ...101010, and why?
And does this same rule work for *all* "infinite integers"?
NOTE: comparing finite integers requires that each one has only finitely
many non-zero digits, but your "infinite integers" may have infintiely
many non-zero digits, so the sam rule won't work.
>
> Carry operations are normal.
They can't be "normal" for something like ...999 + 1, as there is
always a carry "left over". For finite integers, all carries are
eventualy accounted for, but not for "infinite integers".
> Infinite long digit string are separated by decimal marks.
This is not clear. Please give examples.
> If all placeholders in the infinite strings will be occupied with the
> largest possible symbol depending on the used base system (base 10 system ,
> bibary system etc.) then carry behaves exactly the same way as in the case
> of any decimal system.
But what of the carry unaccounted for?
>
> Tapio
>
> > Zundark
>
>
Tapio
discussion will bw hard to follow. I wish you follow my example.
On the other hand, I have not done a "number" by starting new title "Virgil
does not...", though you have not yet answered to my criticism and not yet
to my all questions. I understand you have in your mind a certain priority
concerning debated topics. I wish you appeciate fair play.
Now back to the business. Below some calculations as expected by request:
In article <9i5b9s$8fi$1...@tron.sci.fi>, "Tapio" <hurm...@dlc.fi> wrote:
>Virgil: HOW DO YOU DETERMINE ORDER?
>For example, given ...010101 = (01)01 and ...101010 = (10)10, where
>parentheses indicate infinite repetition of what is contained, which one
>comes first in your "increasing order"?
>This point is critical. You must explain HOW to to tell which of two
>labels comes first. And your rule must work for all "labels".
>Until you have resolved this issue, there is no point in proceeding.
Analysis of your problem
"given ...010101 = (01)01 and ...101010 = (10)10,where
parentheses indicate infinite repetition of what is contained"
1) You specify repetitions or period lengths (01) and (10). Note you do not
specify for example (1010) and terminating 1 and you neither specify
(101010) repeated. Thus those alternatives are rejected according to the
formulation of the set question.
2) ... dots is considered repeated infinitely.
3) You do not specify where is the decimal mark, but it is assumed you mean
infinite integers when the point of reference is standard zero, thus
immediately on the right side of your number.
Starting from those premises, without further given information, I give you
my solution to the problem. I assume the question was correctly understood.
Infinite integers written in decimal notation as the point of reference is
omega zero "0,"
0,(01) and 0,(10). Therefore 0,(10)>0,(01).
Discussion:
If the repetitions were specified in other way, see some examples (1) above,
then other solutions would be possible.
(snip)
>If the problem is so damned trivial, deal with it NOW instead of writing
>a lot of junk that is meaningless until the problem is solved.
Your opinion about priority accepted above.
I assume you have enjoyed education how to debate. This is your change to
lift the sci.math. profile. :-). Even you can make mistakes, as me -too, but
I do not name your mistakes.
>Tapio, you are concentrating on irrelevancies and avoiding the
>essentials.
I just tried to explain more carefully the principles before going into
deaper details to avoid harmful misunderstandings.
>A lot is still obscure, such as how to find the product of ...9 = (9).
with itself.
The repeated unit is (9). Solution is smaller than omega, because ...9 is
smaller than omega. Therefore (...9.)^2= (9)8.
>For example, how does one add the "infinite
>integer" corresponding to 15/37 (expressed decimally) and the "infinite
>integer" corresponding to 5/7 (expressed decimally)?. How does one
>multiply them?
Addition:
(15/37) +(5/7) =1,11969(111969)
Multiplication:
15/37 = 0,(405)
5/7=0,(714285)
(15/37)*(5/7)= 75/259=0,(289575)
Any further questions?
Note: Starting from Monday I cannot reply during a week.
Tapio
Virgil
> >Virgil: HOW DO YOU DETERMINE ORDER?
>
> >For example, given ...010101 = (01)01 and ...101010 = (10)10, where
> >parentheses indicate infinite repetition of what is contained, which one
> >comes first in your "increasing order"?
>
> Analysis of your problem
It is not my problem, it is your problem, Tapio. You are asserting that
you have constructed something, but you have failed to give essential
details necessary in that construction.
When asked for general rules, you respond with mostly useless specific
examples.
When asked for specifics, you respond with generalities.
>
> "given ...010101 = (01)01 and ...101010 = (10)10,where
> parentheses indicate infinite repetition of what is contained"
>
> 1) You specify repetitions or period lengths (01) and (10). Note you do not
> specify for example (1010) and terminating 1 and you neither specify
> (101010) repeated. Thus those alternatives are rejected according to the
> formulation of the set question.
> 2) ... dots is considered repeated infinitely.
> 3) You do not specify where is the decimal mark, but it is assumed you mean
> infinite integers when the point of reference is standard zero, thus
> immediately on the right side of your number.
>
> Starting from those premises, without further given information, I give you
> my solution to the problem. I assume the question was correctly understood.
Apparently something was not understood by you, as your "answers" are
useless.
>
> Infinite integers written in decimal notation as the point of reference is
> omega zero "0,"
> 0,(01) and 0,(10). Therefore 0,(10)>0,(01).
>
> Discussion:
> If the repetitions were specified in other way, see some examples (1) above,
> then other solutions would be possible.
This solves only one case, and does so in a way which does not extend to
arbitrary "infinite integers".
Suppose that we have two "infinite integers" which do not become
repeating, for example, the infinite integers corresponding to sqrt(1/2)
and sqrt(1/3), which are irrationals. The "infinite integer"
corresponding to the real number sqrt(1/2) starts something like
...581187601707, and the "infinite integer" corresponding to sqrt(1/3)
starts something like ...981962053775, but neither of them repeats as
the infinite integers corresponding to proper rationals do.
How do you add such "infinite integers"?
How do you multiply such "infinite integers"?
How do you sort such "infinite integers" by "size"?
If these operations are based on the operations applied to the
corresponding decimal fractions, you have some problems.
Some additions become impossible, for example,
decimals 0.7777... and 0.888... add to give 1.666... .
But the "infinite integer" <--> decimal correspondence only works for
decimals between 0 and 1, so sometimes "infinite integrals" cannot be
added to each other by adding their decimal counterparts.
[snip]
> >For example, how does one add the "infinite
> >integer" corresponding to 15/37 (expressed decimally) and the "infinite
> >integer" corresponding to 5/7 (expressed decimally)?. How does one
> >multiply them?
>
> Addition:
> (15/37) +(5/7) =1,11969(111969)
This is ordinary addition of (15/37) +(5/7), then converted, improperly
as it happens, to an "infinite integer".
(15/37) +(5/7) = 290/259 = 1.119691(119691),
But the correspondence between "infinite integers" and reals was only
defined for reals between 0 and 1, so the decimal 1.119691(119691),
being greater than 1, has no corresponding "infinite integer".
>
> Multiplication:
> 15/37 = 0,(405)
> 5/7=0,(714285)
>
> (15/37)*(5/7)= 75/259=0,(289575)
You are again defining an operation on "infinite integers" by the
corresponding operation on proper decimal fractions (between 0 and 1),
Tapio
> repeating, for example, the infinite integers corresponding to sqrt(1/2)
> and sqrt(1/3), which are irrationals. The "infinite integer"
> corresponding to the real number sqrt(1/2) starts something like
> ...581187601707, and the "infinite integer" corresponding to sqrt(1/3)
> starts something like ...981962053775, but neither of them repeats as
> the infinite integers corresponding to proper rationals do.
>
> How do you add such "infinite integers"?
> How do you multiply such "infinite integers"?
> How do you sort such "infinite integers" by "size"?
There is know something I was not able to explain for you. Your examples
above are "mirror mapped" strings. (reversed or transponed order, maybe you
have better English word). That was not the original idea of infinite
string. I try to clarify:
Finite integer example:
12345= ...(0)12345 =(0)12345 (standard zero point reference)
Infinite "far" to left there exist omega point of reference. The same finite
integer written from omega point of reference is:
0,...(0)12345 =0,(0)12345.
Note: the string is no reversed.
"Infinite integers" are those numbers, which do not have (0) on the left
side of terminating finite string. Inifinite integers can have can any other
combination of {0,1,2,3,4,5,6,7,8,9} except (0). Infinite integers are
infinite strings, finite integers are finite strings. Finite integers can be
written as infinite integers by writing (0) on the left hand side of finite
string. Therefore, all the integers are actually infinite integers. (Let愀
say we are too lazy to remind in every context that (0) is omitted.) Finite
integers are just called simply interegs (N) because we want to tell that we
use now short notation by omitting (0) that are usually in practice
non-significant and because we use the standard zero point of reference.
Analocigally terminating rational (for example 0.12345(0) is simple written
0.12345, because the non-significant (0) are omittet purposely. Rationals
(Q) are the subset of reals (R). (Of-course rational can have infinite
string with repeated string fragment as seen in earlier examples). But if
you consider integers (finite integers) there is nothing corresponding to
infinitely repeated. (This would be one of the reasons, we simply have to
consider such a possibility - therefore infinite integers. (Another reason
would be that those reals without repeating period.)
I described earlier the use of several points of reference instead of the
standard zero decimal mark. The applications of these marks between "stacks"
(with infinite and all combinations of possible strings based on the
selected base system), means that all stacks are similar copies, therefore
one-to-one mapping. Copy maps to copy.
Because the stacks contain all the possibilities, of course we can find
"mirror string" within the same stack. but also within preceding and
successing stacks. Now you have erraneously (because of my poor explanation)
understood that the mirror image of reals describes the defined infinite
integers. This is not what I meant. There is mirror mapping (inversed
copying) but also direct mapping ( plain copying)
The mapping is one-to-one between the two successive point of references (x)
x{all infinitely possible strings}x{all infinitely possible strings}x
A B
from A to B and from B to A, because B is the copy of A.
Unfortunately ( ;-) it is typically so that A is called "integers" without
further knowlegde that A contains all infinite integers (including short
notations called finite integers). B is called decimal extension ( real part
after "0.something").
Actually xA is exactly the similar as xB (copy) or Ax is exactly similar
than Bx.
If we write xA, we see infinite integers as decimal numbers (example
0,(9) ). If we write, Ax we see the same pattern string, actually number, as
integer (example (9) ). The same is true for the copy B or for any copy.
Actually, because we have infinite similar copies separated by
> If these operations are based on the operations applied to the
> corresponding decimal fractions, you have some problems.
>
> Some additions become impossible, for example,
> decimals 0.7777... and 0.888... add to give 1.666... .
Therefore, we have carry as you just expressed between infinite "stacks".
You had to use the next stack on the left side. The operation is exactly the
same between "tens and hundreds" than between infinite stacks. You just have
two infinite stacks separated by mark (decimal mark). The next stack on the
left side of infinite integers is "omega stack" that is greater than any
infinite integer. Therefore I introduced omega zero "0," earlier. I have no
reason to consider than the "omega stack" behaves somehow different way.
Why, because all the stacks werecopies! Altough I didn't mention, let愀 say
I also copied the counting rules (if that somehow helps).
By using your example:
0,(7) and 0,(8) add to give 1,(6). Note: if I want to be more precise I can
consider what happens at the point of standard zero:
Nothing very special - the string just continues to the next stack that is
also infinite.
1,(6).(6).
> But the "infinite integer" <--> decimal correspondence only works for
> decimals between 0 and 1, so sometimes "infinite integrals" cannot be
> added to each other by adding their decimal counterparts.
In principle quite exactly, but not perfectly! Every stack (base 10 system)
have infinite alternatives labeled from (0)1 to (9).
Therefore, more exactly I would say "only works for decimals between (0)1
and (9)" (smallest and greatest possibilities within infinite alternatives).
Now, to deal with "smaller units" we divided (0)1 into infinite part and
copied the labels. Therefore decimal mark was invented, because we dont want
to mix or overlap with two copies. And, as the decimal marks was once
invented, we apply the same mark on the left side of the "infinite integer"
stack, because it is nothing else than copy of "omega stack". Namely (0)1,
that is the smallest "unit" in "omega stack" was divided into infinite parts
and labels were copied and so we created infinite integers (including short
notations called finite integers). The concept "integers" or "decimals" just
reflect the point of reference,i.e. do we look the infinite string from left
to right or vice versa, - nothing else.
Thus the applied calculations depend on the selected point of reference. For
example a kind of illusion is this example:
0,5*0,5=0,25 (0,25 <0,5), but 5*5=25 25>5. The fact is that both results in
"string fragment" 25. We just used the different point of refrence.
> [snip]
> > >For example, how does one add the "infinite
> > >integer" corresponding to 15/37 (expressed decimally) and the "infinite
> > >integer" corresponding to 5/7 (expressed decimally)?. How does one
> > >multiply them?
> >
> > Addition:
> > (15/37) +(5/7) =1,11969(111969)
>
> This is ordinary addition of (15/37) +(5/7), then converted, improperly
> as it happens, to an "infinite integer".
>
> (15/37) +(5/7) = 290/259 = 1.119691(119691),
>
> But the correspondence between "infinite integers" and reals was only
> defined for reals between 0 and 1, so the decimal 1.119691(119691),
> being greater than 1, has no corresponding "infinite integer".
Yes it has: 1,11969(111969), because "1," is in omega stack and
11969(111969) (the decimal part) is in the infinite integer stack. Observe -
the number from omega-infinite integer area maps one-to-one to infinite
integer-decimal area
1,11969(111969) maps to 1.11969(111969). Mapping is no way limited to two
consecutive stacks. Why? All stacks were copies.
> > Multiplication:
> > 15/37 = 0,(405)
> > 5/7=0,(714285)
> >
> > (15/37)*(5/7)= 75/259=0,(289575)
>
> You are again defining an operation on "infinite integers" by the
> corresponding operation on proper decimal fractions (between 0 and 1),
(see above) My point of reference was in this case omega zero. More
rigorously: Forget, omegas, integers,decimals. See, stacks are similar
copies separated by agreed special mark. You may equally say: infinitely
coupled (linked) decimals stacks with separator (The point of view from left
to right) or infinitely linked integer stacks (the point of view from right
to left).
Tapio