Q: Is INFINITY+1=1+INFINITY?
Chea Chee Keong
same. Can anyone clarify on this?
:)
Virgil Hancher
Chee Keong) wrote:
It depends on whether you have an ordinal infinity or a cardinal infinity.
With cardinals, order doesn't count, and they are the same.
With some ordinals, such as the ordinal type of the natural numbers, it
makes a difference, since 1 + infinity has no last element, but infinity +
1 does.
Horst Kraemer
> Just wondering, whether 1+inf = inf+1? I was told that it is not the
> same. Can anyone clarify on this?
The question is meaningless unless you specify what "1+inf" or "inf+1"
mean (sic!) and especially what "inf" means.
In an algebraic context there should be no difference if "inf+1" is
defined at all.
If "inf" is something else, it is up to you to clarify.
Regards
Horst
Kurt Foster
. Just wondering, whether 1+inf = inf+1? I was told that it is not the
. same. Can anyone clarify on this?
.
Depends on whether it's cardinal (how many) or ordinal (ordering)
arithmetic.
In cardinal arithmetic, inf+1 = 1+inf = inf, all the "inf"s being the
*same* infinity.
But in ordinal arithmetic, addition is like concatenating strings, and
is not commutative. For example, taking the simplest infinite ordinal, w
(little omega, the ordinal type of the positive integers ordered by the
usual relation < ), and 1 the ordinal 1, then 1 + w = w, but w + 1 is NOT
w. A set of type w + 1 has a "greatest" element, but a set of type w
does not.
torqu...@my-dejanews.com
ckc...@iss.nus.sg (Chea Chee Keong) wrote:
> Just wondering, whether 1+inf = inf+1? I was told that it is not the
> same. Can anyone clarify on this?
In mathematics there is no one object called 'inf', 'infinity' or whatever.
In different branches of mathematics there is a wide variety of different
objects that you could think of as being kinds of infinity. For most of them
1+inf=inf+1 - but you must remember that 'inf' isn't a number in any normal
sense and that the operator '+' doesn't mean ordinary 'plus'. If someone just
says inf+1!=1+inf without defining 'inf' they are merely trying to blind you
with science rather than convey useful information.
-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
Christian Bau
> Just wondering, whether 1+inf = inf+1? I was told that it is not the
> same. Can anyone clarify on this?
First, you have to find out what you mean when you write "inf + 1", "1 +
inf" and "inf + 1 = 1 + inf". No, go one more step backward: Can you write
down precisely what you mean by INFINITY? For example, what does the
following mean?
lim 1/x = INFINITY
x->0
math...@hotmail.com
torqu...@my-dejanews.com wrote:
> In article <71m2at$iig$1...@godzilla.krdl.org.sg>,
> ckc...@iss.nus.sg (Chea Chee Keong) wrote:
>
> > Just wondering, whether 1+inf = inf+1? I was told that it is not the
> > same. Can anyone clarify on this?
>
> In different branches of mathematics there is a wide variety of different
> objects that you could think of as being kinds of infinity. For most of them
> 1+inf=inf+1 - but you must remember that 'inf' isn't a number in any normal
> sense and that the operator '+' doesn't mean ordinary 'plus'. If someone just
> says inf+1!=1+inf without defining 'inf' they are merely trying to blind you
> with science rather than convey useful information.
>
b_m_...@my-dejanews.com
> Just wondering, whether 1+inf = inf+1? I was told that it is not the
> same. Can anyone clarify on this?
First, mathematicians recognize several different notions of infinity.
The two on which arithmetic is most often performed are the infinite
cardinal numbers and the infinite ordinal numbers. Cardinal numbers
measure the 'size' of a set. For instance, aleph_0 is the 'size' (the
usual term is 'cardinality') of the set Z+ of positive integers. Two
sets have the same cardinality if there is a 1-1 correspondence between
them. For instance, the function f(n) = n + 1 provides a 1-1 correspondence
between the set N of non-negative integers and the set of positive integers,
so these sets are the same 'size'. If m and n are cardinal numbers, and
A and B are disjoint sets of cardinality m and n respectively, we define
m + n to be the cardinality of the union of A and B. And since the union
of A and B is the same as the union of B and A, m + n = n + m. In the
example N is the union of Z+ (of cardinality aleph_0) and {0} (of cardinality
1), so aleph_0 + 1 = 1 + aleph_0 = aleph_0.
Ordinal numbers, on the other hand, take into account the ordering of a
set. The natural numbers in their usual ordering give the ordinal number
omega (which I'll write 'w'): {0, 1, 2, 3, 4, ...}. But the positive
integers in their natural order have exactly the same order structure;
you can match them up 'side-by-side':
N: 0 1 2 3 4 ... n ...
Z+: 1 2 3 4 5 ... n+1 ...
If m and n are ordinal numbers, their ordinal sum is defined as the ordinal
number corresponding to the ordering m followed by the ordering n. (I'm
omitting a *lot* of details.) For instance, the 1-element trivially ordered
set {0} is the ordinal 1, and as we've seen, Z+ in its usual ordering
represents w, so 1 + w is represented by 0 followed by 1, 2, 3, ...; but
that's just N, which, like Z+, represents w. In this sense, then, 1 + w = w.
But w + 1 is represented by the ordering 1, 2, 3, 4, ..., 0, which looks very
different from N or Z+. For instance, it has a last element, and they don't.
Also, this last element has elements in front of it but no *immediate*
predecessor, whereas in N or Z+ every element with anything in front of it
does have an immediate predecessor. Thus, the ordinal number w + 1 is not
equal to the ordinal number 1 + w (which is equal to w).
In general if k is any infinite cardinal number, and '+' represents cardinal
addition, then k + 1 = 1 + k = k. But if k is an infinite ordinal number,
and '+' represents ordinal addition, then 1 + k = k < k + 1.
Brian M. Scott
wcal...@ptdprolog.net
ckc...@iss.nus.sg (Chea Chee Keong) wrote:
>Just wondering, whether 1+inf = inf+1? I was told that it is not the
>same. Can anyone clarify on this?
>
It depends what you mean by INFINITY. Here are two examples:
The ordinal number omega describes the infinite order type of the
natural numbers: 0,1,2,3,... The ordinal 1+omega=omega since
if we add a new element to the begining of the list (say -1)we
have -1,0,1,2,3,... which can be put in an order preserving
1-1 correspondence with the natural numbers. (Use x->x+1.) On
the other hand, omega+1 means adding a new element to
the end of the list (let's call it o) to get 0,1,2,3, ..., o.
(You should think of o as coming after ALL of the natural
numbers.) So omega+1 has a last element (o) which makes it
different from omega. Therefore 1+omega does not equal
omega+1.
The cardinal number aleph0 represents the infinite size of the
natural numbers as a set (no order involved). Cardinal
addition is commutative so 1+aleph0=aleph0+1. (In fact,
both are equal to aleph0.)
In general, 1+inf=inf+1 is false if you are referring to
an infinite ordinal number, but it is true if you are referring to an infinite
cardinal number.
-Bill Calhoun
-Math and CS Dept.
-Bloomsburg University
Zdislav V. Kovarik
Chea Chee Keong <ckc...@iss.nus.sg> wrote:
>Just wondering, whether 1+inf = inf+1? I was told that it is not the
>same. Can anyone clarify on this?
>
>:)
The one who told you so might have heard something about ordinal numbers
and their arithmetic. The rendition was sloppy, though.
It is true that if L is an infinite ordinal number then
1 + L = L but L + 1 > L
The sloppiness comes from:
not mentioning that the objects are ordinal numbers,
not mentioning that the addition is ordinal addition,
making the impression that "there is just one infinity" (the fact is that
there are so many infinite ordinal numbers that they do not fit into a
set)
just for starters.
A visualization of the non-commutativity can be made as follows:
The first infinite ordinal number (called omega) describes how positive
integer numbers are ordered:
1 < 2 < 3 < ....
and the ordering "type" is independent of the labeling of the entries; the
same ordinal number describes the ordered set of the following rational
numbers:
1/2 < 2/3 < 3/4 < 4/5 < ...
The ordinal number 1+omega describes sets ordered like
0 < 1/2 < 2/3 < 3/4 < ...
and it is the very same type of ordering as that described by omega
(luckily, a formula for re-labeling here can be : x becomes 2-1/x (check
it)).
In contrast, the ordering described by omega+1 is visualized by
1/2 < 2/3 < 3/4 < .... < 1
and is different from the order type 1+omega. Why:
Ordered sets of type omega (same as 1+omega) have no last entry, while
those of type omega+1 have a last entry.
There are more definitions and basic properties to absorb and check, so
that misunderstandings and incomplete, cute-sounding statements can be
avoided.
Cheers, ZVK(Slavek).
torqu...@my-dejanews.com
math...@hotmail.com wrote:
> blind you
> > with science rather than convey useful information.
> >
> SCIENCE ?????
Maybe you are unaware of the *idiomatic* expression "to blind with science".
Ian Goddard
>In article <71m2at$iig$1...@godzilla.krdl.org.sg>, ckc...@iss.nus.sg (Chea
>Chee Keong) wrote:
>
>> Just wondering, whether 1+inf = inf+1? I was told that it is not the
>> same. Can anyone clarify on this?
>>
>> :)
>
>It depends on whether you have an ordinal infinity or a cardinal infinity.
>
>With cardinals, order doesn't count, and they are the same.
>
>With some ordinals, such as the ordinal type of the natural numbers, it
>makes a difference, since 1 + infinity has no last element, but infinity +
>1 does.
IAN: It seems to me, and correct me if I'm wrong, that
we could also say that doing addition with the infinity
known as the continuum, or "c," is commutative, just
as is the case when doing addition with infinite
cardinal numbers, and thus: (1 + c) = (c + 1).
This seems to me to be the case since we can define c
as a span covering a finite length, such as one inch,
and across that span are an infinite number of zero-
sized points, and since we could always add 1 to a
one-inch span from any side, we have c and add 1.
Of course adding to c does not increase the infinite
density of c, such that c + c + c = c. That only other
case I know of where that can be said is 0 + 0 + 0 = 0.
**************************************************************
Visit Ian Williams Goddard --------> http://Ian.Goddard.net
______________________________________________________________
"A new scientific truth does not triumph by convincing its
opponents and making them see the light, but rather because
its opponents eventually die, and a new generation grows
up that is familiar with the idea from the beginning."
Max Plank - Nobel physicist
KSchmidt10
>>> same. Can anyone clarify on this?
when two cardinal numbers are added and at least one of them is infinite, the
sum is just the larger cardinal number.
So regardless of what size infinity you mean by "inf", commutativity holds.
Kris Schmidt
oho...@idirect.com
I...@Goddard.net (Ian Goddard) wrote:
> On Tue, 03 Nov 1998, vm...@frii.com (Virgil Hancher) wrote:
>
> >In article <71m2at$iig$1...@godzilla.krdl.org.sg>, ckc...@iss.nus.sg (Chea
> >Chee Keong) wrote:
> >
> >> same. Can anyone clarify on this?
> >>
The idea that infinity is a number leads to contradiction over
and over; we should stop the nonsense, there is no such possibility.
What (we) I mean by infinity is a process... with out end.
It makes no sense to talk about oo or 2^(oo) or any of those
cocepts. That one infinity is greater than another (Cantor)
displays the sillyness, even if you can demonstrate that N(0)
must be less than 2^N(0), from nonsense becomes nonsense. I am
reminded of Russell's sillyness in *Principia Mathematica* where
he claimed that 1/0 is a number (Volume 2).
I understand that opinions are free (agreeable or not).
--
Owen Holden
Herman Rubin
KSchmidt10 <kschm...@aol.com> wrote:
>>>> Just wondering, whether 1+inf = inf+1? I was told that it is not the
>>>> same. Can anyone clarify on this?
>when two cardinal numbers are added and at least one of them is infinite, the
>sum is just the larger cardinal number.
>So regardless of what size infinity you mean by "inf", commutativity holds.
This is assuming that they are comparable; that this is true for all
infinite cardinals is equivalent to the Axiom of Choice. If one assumes
that the cardinals are comparable, this is equivalent to n = 2n for all
infinite cardinals, and this is weaker than the Axiom of Choice, but
still not true in ZF or NBG.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Dave Seaman
>>In article <71m2at$iig$1...@godzilla.krdl.org.sg>, ckc...@iss.nus.sg (Chea
>>Chee Keong) wrote:
>>> Just wondering, whether 1+inf = inf+1? I was told that it is not the
>>> same. Can anyone clarify on this?
>>It depends on whether you have an ordinal infinity or a cardinal infinity.
>>With cardinals, order doesn't count, and they are the same.
>>With some ordinals, such as the ordinal type of the natural numbers, it
>>makes a difference, since 1 + infinity has no last element, but infinity +
>>1 does.
> IAN: It seems to me, and correct me if I'm wrong, that
> we could also say that doing addition with the infinity
> known as the continuum, or "c," is commutative, just
> as is the case when doing addition with infinite
> cardinal numbers, and thus: (1 + c) = (c + 1).
No, the same comments apply. Cardinal addition is always commutative:
1 + c = c + 1. Ordinal addition, however, is not commutative unless
there is an order-isomorphism between the results. The ordinal sum c+1
has a last element, but the ordinal sum 1+c does not. Therefore, they
are not the same as ordinals, but they have the same cardinality.
> This seems to me to be the case since we can define c
> as a span covering a finite length, such as one inch,
> and across that span are an infinite number of zero-
> sized points, and since we could always add 1 to a
> one-inch span from any side, we have c and add 1.
The points in an interval are not well-ordered according to the usual
linear ordering. For example, the half-open interval (0,1] fails to
have a least element. Therefore, you can't talk about c as an ordinal
until you have a well-ordering for it. This can be done in principle
(assuming the Axiom of Choice), but there is no explicit description of
such an ordering.
> Of course adding to c does not increase the infinite
> density of c, such that c + c + c = c. That only other
> case I know of where that can be said is 0 + 0 + 0 = 0.
When c is well-ordered, it ceases to be dense (except at the limit
ordinals, and then only on one side). Every ordinal has a successor,
which makes ordinals very different from the real numbers in the way
they are ordered. Density has nothing to do with deciding whether 1+c
= c+1.
--
Dave Seaman dse...@purdue.edu
Pennsylvania Supreme Court Denies Fair Trial for Mumia Abu-Jamal
<http://mojo.calyx.net/~refuse/altindex.html>
Ian Goddard
>> IAN: It seems to me, and correct me if I'm wrong, that
>> we could also say that doing addition with the infinity
>> known as the continuum, or "c," is commutative, just
>> as is the case when doing addition with infinite
>> cardinal numbers, and thus: (1 + c) = (c + 1).
>
>No, the same comments apply. Cardinal addition is always commutative:
>1 + c = c + 1.
IAN: You answer "no" and then repeat the statement
(1 + c) = (c + 1) as if it is not false, yet when
you said "no" it seems to imply it is false. ???
>Ordinal addition, however, is not commutative unless
>there is an order-isomorphism between the results. The ordinal sum c+1
>has a last element, but the ordinal sum 1+c does not.
IAN: The set of all real numbers from 0 to 1
is equal to c, and it has a last number, 1.
>> This seems to me to be the case since we can define c
>> as a span covering a finite length, such as one inch,
>> and across that span are an infinite number of zero-
>> sized points, and since we could always add 1 to a
>> one-inch span from any side, we have c and add 1.
>
>The points in an interval are not well-ordered according to the usual
>linear ordering. For example, the half-open interval (0,1] fails to
>have a least element. Therefore, you can't talk about c as an ordinal
>until you have a well-ordering for it. This can be done in principle
>(assuming the Axiom of Choice), but there is no explicit description of
>such an ordering.
IAN: Are the real numbers from 0 to 1 not well-ordered?
If, as you seem to say, c is not ordinal and ordinal addition
is not commutative, then what I said (1 + c) = (c + 1) is
always true, which makes your reply of "no" unclear.
>> Of course adding to c does not increase the infinite
>> density of c, such that c + c + c = c. That only other
>> case I know of where that can be said is 0 + 0 + 0 = 0.
>
>When c is well-ordered, it ceases to be dense (except at the limit
>ordinals, and then only on one side). Every ordinal has a successor,
>which makes ordinals very different from the real numbers in the way
>they are ordered. Density has nothing to do with deciding whether 1+c
>= c+1.
IAN: I didn't cite density for that reason,
but simply as an ancillary observation.
Brian M. Scott
>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>> IAN: It seems to me, and correct me if I'm wrong, that
>>> we could also say that doing addition with the infinity
>>> known as the continuum, or "c," is commutative, just
>>> as is the case when doing addition with infinite
>>> cardinal numbers, and thus: (1 + c) = (c + 1).
>>No, the same comments apply. Cardinal addition is always commutative:
>>1 + c = c + 1.
> IAN: You answer "no" and then repeat the statement
> (1 + c) = (c + 1) as if it is not false, yet when
> you said "no" it seems to imply it is false. ???
You snipped the context that makes sense of Dave's statement. He'd
previously explained that while cardinal addition is always
commutative, ordinal addition isn't always, using the ordinal type, w,
of the natural numbers as an example: w + 1 is an order type with a
last element, but 1 + w is identical in type to w and has no last
element, so the two are not the same.
Exactly the same occurs with the number c if (as is customary) it is
taken to be the first ordinal whose cardinality is that of the
continuum. Dave's 'no' means that whether 1 + c = c + 1 depends on
whether the addition in question is cardinal addition, in which case
the statement is true, or ordinal addition, in which case it's not,
for the same reason as in the previous example.
>>Ordinal addition, however, is not commutative unless
>>there is an order-isomorphism between the results. The ordinal sum c+1
>>has a last element, but the ordinal sum 1+c does not.
> IAN: The set of all real numbers from 0 to 1
> is equal to c, and it has a last number, 1.
But that set, though it's the same size, doesn't look a bit like the
ordinal c. An ordinal number is (among other things) a well-ordered
set; when we ask whether it has a last element or not, we're referring
to that particular well-ordering. The usual ordering of [0, 1] is not
a well-ordering.
>>> This seems to me to be the case since we can define c
>>> as a span covering a finite length, such as one inch,
>>> and across that span are an infinite number of zero-
>>> sized points, and since we could always add 1 to a
>>> one-inch span from any side, we have c and add 1.
>>The points in an interval are not well-ordered according to the usual
>>linear ordering. For example, the half-open interval (0,1] fails to
>>have a least element. Therefore, you can't talk about c as an ordinal
>>until you have a well-ordering for it. This can be done in principle
>>(assuming the Axiom of Choice), but there is no explicit description of
>>such an ordering.
> IAN: Are the real numbers from 0 to 1 not well-ordered?
They are not well-ordered in the usual ordering. In a well-ordered
set, every non-empty subset has a least element - this is the
definition of well-ordering, in fact - but as Dave pointed out, (0, 1]
has no least element in the usual ordering.
> If, as you seem to say, c is not ordinal and ordinal addition
> is not commutative, then what I said (1 + c) = (c + 1) is
> always true, which makes your reply of "no" unclear.
The symbol 'c' in this context usually denotes an the first ordinal
number that has the same cardinality as [0, 1]. Thus, it is both an
ordinal and a cardinal number and can participate in both ordinal and
cardinal addition.
[snip]
Brian M. Scott
Jeremy Boden
>
[cut of everything relating to addition of infinities...]
>The idea that infinity is a number leads to contradiction over
>and over; we should stop the nonsense, there is no such possibility.
>What (we) I mean by infinity is a process... with out end.
>It makes no sense to talk about oo or 2^(oo) or any of those
>cocepts. That one infinity is greater than another (Cantor)
>displays the sillyness, even if you can demonstrate that N(0)
>must be less than 2^N(0), from nonsense becomes nonsense. I am
>reminded of Russell's sillyness in *Principia Mathematica* where
>he claimed that 1/0 is a number (Volume 2).
>
>I understand that opinions are free (agreeable or not).
>--
>Owen Holden
This topic really deserves a thread of its own.
At the moment I'm reading Rudy Rucker's excellent book,
"Infinity and the Mind"
Your argument seems to be based around a distinction made by Aristotle
between the 'potentially infinite' and the 'actually infinite'
(see page 3). It should be obvious that this distinction is rather
difficult to maintain.
When you talk of number, I'm not sure exactly what you mean. So far as
I'm aware no one has claimed that infinity is a number in the same kind
of way that 42 is (and got away with it).
It is quite possible to extend the addition operation to objects which
aren't numbers. Open any school algebra book for an example.
In fact it makes a great deal of sense to talk of the comparative "size"
of infinite sets. Because a 1-1 mapping of one infinite set onto another
infinite set is not per se guaranteed we can end up with the situation
that some infinities are "bigger" than other infinities.
You can pretend this doesn't happen and it's "silly" in a rather
Pythonesque kind of way, but despite having your "free" opinion you are
*wrong*. Having a free opinion doesn't mean you are right and in this
case it just reveals that you haven't thought very deeply about the
subject.
Except in special circumstances, where the type of infinity is
predefined, it is actually silly and confusing to use the symbol oo.
Question; Is the infinity of positive integers <, = or > the infinity of
real numbers between 0 and 1?
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Ian Goddard
>>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>
>>>> IAN: It seems to me, and correct me if I'm wrong, that
>>>> we could also say that doing addition with the infinity
>>>> known as the continuum, or "c," is commutative, just
>>>> as is the case when doing addition with infinite
>>>> cardinal numbers, and thus: (1 + c) = (c + 1).
>
>>>No, the same comments apply. Cardinal addition is always commutative:
>>>1 + c = c + 1.
>
>> IAN: You answer "no" and then repeat the statement
>> (1 + c) = (c + 1) as if it is not false, yet when
>> you said "no" it seems to imply it is false. ???
>
>You snipped the context that makes sense of Dave's statement.
IAN: Well, to me, "No" doesn't
imply "Yes, but not always,"
even in the context it was.
> He'd previously explained that while cardinal addition is always
>commutative,
IAN: To be exact, he hadn't explained anything
previous to that sentence that I commented on.
>ordinal addition isn't always, using the ordinal type, w,
>of the natural numbers as an example: w + 1 is an order type with a
>last element, but 1 + w is identical in type to w and has no last
>element, so the two are not the same.
>
>Exactly the same occurs with the number c if (as is customary) it is
>taken to be the first ordinal whose cardinality is that of the
>continuum. Dave's 'no' means that whether 1 + c = c + 1 depends on
>whether the addition in question is cardinal addition, in which case
>the statement is true, or ordinal addition, in which case it's not,
>for the same reason as in the previous example.
IAN: OK, so, to be precise, by "No"
it was meant: "Yes, but not always."
>>>Ordinal addition, however, is not commutative unless
>>>there is an order-isomorphism between the results. The ordinal sum c+1
>>>has a last element, but the ordinal sum 1+c does not.
>
>> IAN: The set of all real numbers from 0 to 1
>> is equal to c, and it has a last number, 1.
>
>But that set, though it's the same size, doesn't look a bit like the
>ordinal c. An ordinal number is (among other things) a well-ordered
>set; when we ask whether it has a last element or not, we're referring
>to that particular well-ordering. The usual ordering of [0, 1] is not
>a well-ordering.
IAN: An ordered set is defined as "a set with an
order relation between the elements," and I think
that the set {0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1} has
"an order relation between the elements," even if
we were to include numbers falling between those.
Ernst Zermelo proved that any ordered set can be
a well-ordered set, and I can't see why the set of
all real numbers from 0 to 1 can't be well-ordered,
or why, if that set is well-ordered, we cannot add
the number 1 to it, and therefore have (c + 1) = c.
>>>> This seems to me to be the case since we can define c
>>>> as a span covering a finite length, such as one inch,
>>>> and across that span are an infinite number of zero-
>>>> sized points, and since we could always add 1 to a
>>>> one-inch span from any side, we have c and add 1.
>
>>>The points in an interval are not well-ordered according to the usual
>>>linear ordering. For example, the half-open interval (0,1] fails to
>>>have a least element. Therefore, you can't talk about c as an ordinal
>>>until you have a well-ordering for it. This can be done in principle
>>>(assuming the Axiom of Choice), but there is no explicit description of
>>>such an ordering.
>
>> IAN: Are the real numbers from 0 to 1 not well-ordered?
>
>They are not well-ordered in the usual ordering. In a well-ordered
>set, every non-empty subset has a least element - this is the
>definition of well-ordering, in fact - but as Dave pointed out, (0, 1]
>has no least element in the usual ordering.
IAN: 0 is not less than or before 1 in the usual
ordering? I've always used this order 0 1 2 3...
I don't see why the definition of a well-ordered
set cannot apply to a set of real numbers. Of
course I can't see everything, but I try :)
Ian Goddard
>Question; Is the infinity of positive integers <, = or > the infinity of
>real numbers between 0 and 1?
IAN: The infinity of positive integers is less
than the infinity of all reals between 0 and 1.
**************************************************************
Visit Ian Williams Goddard --------> http://Ian.Goddard.net
______________________________________________________________
" [T]he Indian sunya [zero] was destined to become
the turning point in a development without which
the progress of modern science, or commerce is
inconceivable. ... In the history of culture,
the invention of zero will always stand out
as one of the greatest single achi-
evements of the human race."
Tobias Dantzig
"Number, The Language of Science"
Macmillan Press, 1967
Dave Seaman
Ian Goddard <I...@Goddard.net> wrote:
>On Sat, 14 Nov 1998, sc...@math.csuohio.edu (Brian M. Scott) wrote:
>>>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>>>> IAN: It seems to me, and correct me if I'm wrong, that
>>>>> we could also say that doing addition with the infinity
>>>>> known as the continuum, or "c," is commutative, just
>>>>> as is the case when doing addition with infinite
>>>>> cardinal numbers, and thus: (1 + c) = (c + 1).
>>>>No, the same comments apply. Cardinal addition is always commutative:
>>>>1 + c = c + 1.
>>> IAN: You answer "no" and then repeat the statement
>>> (1 + c) = (c + 1) as if it is not false, yet when
>>> you said "no" it seems to imply it is false. ???
>>You snipped the context that makes sense of Dave's statement.
> IAN: Well, to me, "No" doesn't
> imply "Yes, but not always,"
> even in the context it was.
"No," in this case, means, "no, it is not true that that c, the
cardinality of the continuum, behaves differently from aleph_0 with
regard to commutativity of addition." I clarified by agreeing with a
previous poster that cardinal arithmetic is always commutative, but
ordinal arithmetic is not necessarily so. That's what I meant by, "the
same comments apply."
>> He'd previously explained that while cardinal addition is always
>>commutative,
> IAN: To be exact, he hadn't explained anything
> previous to that sentence that I commented on.
That's right. It wasn't my explanation that was snipped, but the
explanation that was snipped was quite correct and I was agreeing with
it.
>>ordinal addition isn't always, using the ordinal type, w,
>>of the natural numbers as an example: w + 1 is an order type with a
>>last element, but 1 + w is identical in type to w and has no last
>>element, so the two are not the same.
>>Exactly the same occurs with the number c if (as is customary) it is
>>taken to be the first ordinal whose cardinality is that of the
>>continuum. Dave's 'no' means that whether 1 + c = c + 1 depends on
>>whether the addition in question is cardinal addition, in which case
>>the statement is true, or ordinal addition, in which case it's not,
>>for the same reason as in the previous example.
> IAN: OK, so, to be precise, by "No"
> it was meant: "Yes, but not always."
No, to be precise, by "no" was meant "no, substituting c for a smaller
infinity does not change the argument." The very same distinction
between ordinal and cardinal arithmetic, as previously explained and
then snipped, still applies.
>>>>Ordinal addition, however, is not commutative unless
>>>>there is an order-isomorphism between the results. The ordinal sum c+1
>>>>has a last element, but the ordinal sum 1+c does not.
>>> IAN: The set of all real numbers from 0 to 1
>>> is equal to c, and it has a last number, 1.
>>But that set, though it's the same size, doesn't look a bit like the
>>ordinal c. An ordinal number is (among other things) a well-ordered
>>set; when we ask whether it has a last element or not, we're referring
>>to that particular well-ordering. The usual ordering of [0, 1] is not
>>a well-ordering.
> IAN: An ordered set is defined as "a set with an
> order relation between the elements," and I think
> that the set {0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1} has
> "an order relation between the elements," even if
> we were to include numbers falling between those.
Not every ordering is a well-ordering. By definition, a well-ordered
set is one with the property that every nonempty subset has a least
element. The set [0,1] with the usual ordering is not well-ordered.
The half-open interval (0,1] is an example of a nonempty subset that
does not have a least element.
The point is that the ordinal corresponding to c is a well-ordered
set. It also has the property that c + 1 has a last element, but 1 + c
does not. Therefore, ordinal addition is not commutative in this
case.
> Ernst Zermelo proved that any ordered set can be
> a well-ordered set, and I can't see why the set of
> all real numbers from 0 to 1 can't be well-ordered,
> or why, if that set is well-ordered, we cannot add
> the number 1 to it, and therefore have (c + 1) = c.
Nobody said [0,1] couldn't be well-ordered. In fact, if you accept the
Axiom of Choice, then *every* set (not just every ordered set) can be
well-ordered. The point is that the usual ordering of [0,1] is not a
well-ordering. It's true that [0,1] has a last element in the usual
ordering, but that doesn't change the fact that the well-ordered set
1+c does not have a last element, while c+1 does have a last element.
Therefore, ordinal addition does not commute in this case.
>>> IAN: Are the real numbers from 0 to 1 not well-ordered?
>>
>>They are not well-ordered in the usual ordering. In a well-ordered
>>set, every non-empty subset has a least element - this is the
>>definition of well-ordering, in fact - but as Dave pointed out, (0, 1]
>>has no least element in the usual ordering.
> IAN: 0 is not less than or before 1 in the usual
> ordering? I've always used this order 0 1 2 3...
But 0 is not a member of the set (0,1], which is shorthand for the
half-open interval { x : 0 < x <= 1 }. No matter what element of (0,1]
you choose, I can find a smaller member of the set. In fact, if you
choose a number x in (0,1], then x/2 is also in (0,1] and is smaller
than x. This proves that (0,1] has no smallest element in the standard
ordering.
> I don't see why the definition of a well-ordered
> set cannot apply to a set of real numbers. Of
> course I can't see everything, but I try :)
The real numbers can be well-ordered. It's just that the usual
ordering of the reals is not a well-ordering.
Brian M. Scott
>On Sat, 14 Nov 1998, sc...@math.csuohio.edu (Brian M. Scott) wrote:
[snip]
> IAN: Well, to me, "No" doesn't
> imply "Yes, but not always,"
> even in the context it was.
It meant 'this is not entirely correct'.
>> He'd previously explained that while cardinal addition is always
>>commutative,
> IAN: To be exact, he hadn't explained anything
> previous to that sentence that I commented on.
I disagree, but it seems futile to argue about it now.
[snip]
>>>>Ordinal addition, however, is not commutative unless
>>>>there is an order-isomorphism between the results. The ordinal sum c+1
>>>>has a last element, but the ordinal sum 1+c does not.
>>> IAN: The set of all real numbers from 0 to 1
>>> is equal to c, and it has a last number, 1.
>>But that set, though it's the same size, doesn't look a bit like the
>>ordinal c. An ordinal number is (among other things) a well-ordered
>>set; when we ask whether it has a last element or not, we're referring
>>to that particular well-ordering. The usual ordering of [0, 1] is not
>>a well-ordering.
> IAN: An ordered set is defined as "a set with an
> order relation between the elements," and I think
> that the set {0,.1,.2,.3,.4,.5,.6,.7,.8,.9,1} has
> "an order relation between the elements," even if
> we were to include numbers falling between those.
And this particular set, with the particular order that you have in
mind, is (like any finite, linearly-ordered set) well-ordered. So
what?
> Ernst Zermelo proved that any ordered set can be
> a well-ordered set,
But not necessarily with the given ordering. Do you understand what a
well-ordering is? A linear order < on a set S is a well-ordering
of S if and only if each non-empty subset of S has a least element
with respect to <. This is not the case for [0, 1] if < is
interpreted in the usual way, since the set (0, 1] does not have a
least element with respect to <. Thus, < is not a well-ordering of
[0, 1]. In fact, there is no explicit well-ordering of [0, 1], and
it's consistent with the usual axioms of set theory (excluding the
axiom of choice) that [0, 1] has no well-ordering. With the axiom of
choice, of course, one can prove the well-ordering theorem, which
guarantees that any set -- not just one that already has some ordering
defined on it -- can be equipped with a well-ordering.
> and I can't see why the set of
> all real numbers from 0 to 1 can't be well-ordered,
> or why, if that set is well-ordered, we cannot add
> the number 1 to it, and therefore have (c + 1) = c.
That, I think, is at least partly because you don't understand what is
meant by ordinal addition. Essentially it consists in appending the
second ordered set to the first. Thus, if you take a set well-ordered
in type w (the type of the positive integers) and add a single
object at the end of it, you get a set that looks like this:
0 1 2 3 4 5 6 ... w,
which has the order type w + 1. This is a different ordering from
w, which looks like this:
0 1 2 3 4 5 6 ... .
Notice that the first has a last element; the second doesn't. This is
a structural difference in the orderings that shows that they aren't
the same. In cardinal addition, on the other hand, we don't care
about the ordering. And since these two sets can be paired up one for
one, they're the same size, and hence w + 1 = w if the addition is
understood to be cardinal addition.
[snip]
>>> IAN: Are the real numbers from 0 to 1 not well-ordered?
>>They are not well-ordered in the usual ordering. In a well-ordered
>>set, every non-empty subset has a least element - this is the
>>definition of well-ordering, in fact - but as Dave pointed out, (0, 1]
>>has no least element in the usual ordering.
> IAN: 0 is not less than or before 1 in the usual
> ordering? I've always used this order 0 1 2 3...
> I don't see why the definition of a well-ordered
> set cannot apply to a set of real numbers. Of
> course I can't see everything, but I try :)
Perhaps you didn't understand the symbol (0, 1]? It means the set of
all real numbers bigger than 0 but less than or equal to 1. But if
that set's a problem, we can use a different one. Consider the
following set:
{1, 1/2, 1/3, 1/4, 1/5, 1/6, ...}.
Note that it does *not* include the number 0. This set has no
smallest element in the usual ordering of the real numbers. That fact
_by_definition_ tells you that the real numbers aren't well-ordered by
the usual < relation.
Brian M. Scott
Ian Goddard
>>>> IAN: It seems to me, and correct me if I'm wrong, that
>>>> we could also say that doing addition with the infinity
>>>> known as the continuum, or "c," is commutative, just
>>>> as is the case when doing addition with infinite
>>>> cardinal numbers, and thus: (1 + c) = (c + 1).
>
>>>No, the same comments apply. Cardinal addition is always commutative:
>>>1 + c = c + 1.
>
>> IAN: You answer "no" and then repeat the statement
>> (1 + c) = (c + 1) as if it is not false, yet when
>> you said "no" it seems to imply it is false. ???
>
>"No," in this case, means, "no, it is not true that that c, the
>cardinality of the continuum, behaves differently from aleph_0 with
>regard to commutativity of addition."
IAN: OK, but I did ask at that point if addition
on inf(c) behaved "just as is the case when doing
addition with infinite cardinal numbers." I then
went on to posit a difference between the two,
which is what it seems your "no" refers to.
>The point is that the ordinal corresponding to c is a well-ordered
>set. It also has the property that c + 1 has a last element, but 1 + c
>does not. Therefore, ordinal addition is not commutative in this
>case.
IAN: The set of all real numbers from 0
to 1 = inf(c) and it has a last element,
1. That's the basis of what I'm saying
and seems to be what we disagree on.
If we add one number to the set of all
reals from 0 to 1 (R), as (R + 1) we do
not make R less than inf(c), as would be
the case in addition on infinite cardinals.
>But 0 is not a member of the set (0,1], which is shorthand for the
>half-open interval { x : 0 < x <= 1 }. No matter what element of (0,1]
>you choose, I can find a smaller member of the set. In fact, if you
>choose a number x in (0,1], then x/2 is also in (0,1] and is smaller
>than x. This proves that (0,1] has no smallest element in the standard
>ordering.
IAN: I've defined a set: all reals from
0 to 1. Is 0 not a member of that set?
Why is that set not a closed set?
Brian M. Scott
wrote:
[snip]
>That's right. It wasn't my explanation that was snipped,
Oops, sorry; it seems to have been Virgil Hancher's.
Brian M. Scott
Brian M. Scott
>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
[snip]
>>The point is that the ordinal corresponding to c is a well-ordered
>>set. It also has the property that c + 1 has a last element, but 1 + c
>>does not. Therefore, ordinal addition is not commutative in this
>>case.
> IAN: The set of all real numbers from 0
> to 1 = inf(c) and it has a last element,
> 1. That's the basis of what I'm saying
> and seems to be what we disagree on.
What do you mean by 'inf(c)'? The interval [0,1] with its usual
ordering is *not* well-ordered and is *not* order-isomorphic to the
ordinal number c, so its order properties (e.g., having a last
element) aren't relevant.
> If we add one number to the set of all
> reals from 0 to 1 (R), as (R + 1) we do
> not make R less than inf(c), as would be
> the case in addition on infinite cardinals.
I'm sorry, but I can make no sense of this at all. Adding the
cardinal number 1 to the cardinal number c gives the cardinal
number c; adding the ordinal number 1 to the ordinal number c gives
either the ordinal number 1 + c, which equals the ordinal number c,
or the ordinal number c + 1, which does not equal the ordinal number
c. Since R is neither a cardinal nor an ordinal number, it isn't
clear what you mean by R + 1, but I suspect that you're talking about
the cardinality of a set obtained by taking the union of R with some
one-element set. If so, you're talking about cardinal arithmetic, and
the appropriate statement is that |R U {x}| = |R|, where {x} is the
one-element set in question, and |A| is the cardinality of the set
A.
>>But 0 is not a member of the set (0,1], which is shorthand for the
>>half-open interval { x : 0 < x <= 1 }. No matter what element of (0,1]
>>you choose, I can find a smaller member of the set. In fact, if you
>>choose a number x in (0,1], then x/2 is also in (0,1] and is smaller
>>than x. This proves that (0,1] has no smallest element in the standard
>>ordering.
> IAN: I've defined a set: all reals from
> 0 to 1. Is 0 not a member of that set?
0 is indeed a member of that set, but this is beside the point. In
order for the usual ordering < to be a well-ordering of [0,1], EVERY
non-empty subset of [0,1] must have a smallest element with respect to
<. The set (0,1] is a non-empty subset of [0,1], but it does NOT have
a smallest element: for each x in (0,1], x/2 is a smaller member of
(0,1], so x can't be the smallest. There are many other non-empty
subsets of [0,1] that have no least element, and each of them shows
that [0,1] is not well-ordered by <.
> Why is that set not a closed set?
[0,1] *is* a closed set; so what?
Brian M. Scott
Ian Goddard
>> IAN: Well, to me, "No" doesn't
>> imply "Yes, but not always,"
>> even in the context it was.
>
>It meant 'this is not entirely correct'.
IAN: OK, so what part isn't correct? Here it is again:
IAN: It seems to me, and correct me if I'm wrong, that
we could also say that doing addition with the infinity
known as the continuum, or "c," is commutative, just
as is the case when doing addition with infinite
cardinal numbers, and thus: (1 + c) = (c + 1).
[ which was my point, and which we agree is true ]
This seems to me to be the case since we can define c
as a span covering a finite length, such as one inch,
and across that span are an infinite number of zero-
sized points, and since we could always add 1 to a
one-inch span from any side, we have c and add 1.
[ which is arguably irrelevent, but I don't think wrong ]
Of course adding to c does not increase the infinite
density of c, such that c + c + c = c. That only other
case I know of where that can be said is 0 + 0 + 0 = 0.
It seems that the fuss being made does not really apply to
what I said, but to a tangential digression into ordinals
that seems to be treated as a refutation of what I said.
>Perhaps you didn't understand the symbol (0, 1]? It means the set of
>all real numbers bigger than 0 but less than or equal to 1. But if
>that set's a problem, we can use a different one. Consider the
>following set:
>
> {1, 1/2, 1/3, 1/4, 1/5, 1/6, ...}.
>
>Note that it does *not* include the number 0. This set has no
>smallest element in the usual ordering of the real numbers. That fact
>_by_definition_ tells you that the real numbers aren't well-ordered by
>the usual < relation.
IAN: OK, thanks, now I see how the reals
do not satisfy the well-order definition.
I think that was the center of what we
were running around about. This means
that inf(c) does not fit a definition
of an ordinal number. Which makes me
wonder with respect to your statement:
>The interval [0,1] with its usual
>ordering is *not* well-ordered and is *not* order-isomorphic
>to the ordinal number c, ...
IAN: Is there an ordinal number c ?
**************************************************************
Visit Ian Williams Goddard --------> http://Ian.Goddard.net
______________________________________________________________
"He who pursues learning will increase every day;
he who pursues Tao will decrease every day."
Lao Tzu (Tao Te Ching)
Ian Goddard
>>> It also has the property that c + 1 has a last
>>> element, but 1 + c does not.
>> IAN: The set of all real numbers from 0
>> to 1 = inf(c) and it has a last element,
>> 1. That's the basis of what I'm saying
>> and seems to be what we disagree on.
>
>What do you mean by 'inf(c)'?
IAN: The infinity known as the
continunm, or "c," or inf(c).
Dave Seaman
Ian Goddard <I...@Goddard.net> wrote:
>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>The point is that the ordinal corresponding to c is a well-ordered
>>set. It also has the property that c + 1 has a last element, but 1 + c
>>does not. Therefore, ordinal addition is not commutative in this
>>case.
>
> 1. That's the basis of what I'm saying
> and seems to be what we disagree on.
No, that's not what we disagree on. You keep snipping the important
parts. What we disagree on is that Brian Scott and I keep telling you
the interval [0,1] with its standard order is not a well-ordered set.
You don't seem to understand that there is a difference between an
"ordered set" and a "well-ordered set", even though the difference has
been explained in this thread more than once.
An ordered set (i.e., a totally ordered set, or a linearly ordered set)
has the property that if a and b are members of the set, then the
trichotomy law holds: exactly one of the relations
1. a < b
2. a = b
3. a > b
holds. A well-ordered set, on the other hand, has a stronger
property: every nonempty subset of the given set has a least element.
It turns out that every well-ordered set is also a totally ordered set,
but the converse is false. I have given you an example: the interval
[0,1] with its standard ordering is a totally ordered set that is not
well-ordered.
The reason this is relevant is that ordinal addition concerns
well-ordered sets. When we talk about c as an ordinal, it is implicit
that c is well-ordered, since all ordinals have that property. When I
say that 1+c does not have a last element, I mean that it does not have
a last element with respect to that well-ordering, which is different
from the linear ordering that [0,1] has as a subset of the reals.
> If we add one number to the set of all
> reals from 0 to 1 (R), as (R + 1) we do
> not make R less than inf(c), as would be
> the case in addition on infinite cardinals.
I don't understand that sentence. Adding one to an infinite cardinal
never makes any difference in the size; it doesn't make it larger or
smaller. If you are only talking about the cardinality of a set, then
you are talking about cardinal addition, not ordinal addition, and I
have tried to make it clear from the beginning that cardinal addition
is commutative, but ordinal addition is not.
>>But 0 is not a member of the set (0,1], which is shorthand for the
>>half-open interval { x : 0 < x <= 1 }. No matter what element of (0,1]
>>you choose, I can find a smaller member of the set. In fact, if you
>>choose a number x in (0,1], then x/2 is also in (0,1] and is smaller
>>than x. This proves that (0,1] has no smallest element in the standard
>>ordering.
>
> IAN: I've defined a set: all reals from
> 0 to 1. Is 0 not a member of that set?
> Why is that set not a closed set?
You are talking about the set [0,1], not (0,1]. Those are two
different sets. Your set [0,1] is indeed closed, and I have not
claimed otherwise. My point is, [0,1] with its standard ordering is
not a well-ordered set. Do you remember the definition of a
well-ordered set? It means, by definition, that *every* subset has a
least element.
Do you agree that if I can produce a subset of [0,1] that does not have
a least element, then I have demonstrated that [0,1] is not
well-ordered according to the definition?
Do you agree that (0,1] is a subset of [0,1]?
Do you agree that (0,1] does not have a least element?
Do you agree that I have proved my claim?
Dave Seaman
Brian M. Scott <sc...@math.csuohio.edu> wrote:
>On Sat, 14 Nov 1998 22:09:22 GMT, I...@Goddard.net (Ian Goddard) wrote:
>>> He'd previously explained that while cardinal addition is always
>>>commutative,
>> IAN: To be exact, he hadn't explained anything
>> previous to that sentence that I commented on.
>I disagree, but it seems futile to argue about it now.
A clarification is needed. Looking back at the history of the thread,
it appears to me that it was Virgil Hancher who first explained the
difference between ordinal and cardinal addition. I don't have his
original posting, but I found followups that quote him. When I made my
statement that "No, the same comments apply.", I was referring to his
comments.
That statement that "the same comments apply" was my first statement in
this thread. Therefore, Ian is technically correct that I didn't
explain anything before that statement. However, I did include the
quoted context at the time and it should have been clear what I meant
by "the same comments."
Dave Seaman
Ian Goddard <I...@Goddard.net> wrote:
>On Sun, 15 Nov 1998, sc...@math.csuohio.edu (Brian M. Scott) wrote:
>
>>> imply "Yes, but not always,"
>>> even in the context it was.
>>
>>It meant 'this is not entirely correct'.
>
>
>IAN: OK, so what part isn't correct? Here it is again:
>
> IAN: It seems to me, and correct me if I'm wrong, that
> we could also say that doing addition with the infinity
> known as the continuum, or "c," is commutative, just
> as is the case when doing addition with infinite
> cardinal numbers, and thus: (1 + c) = (c + 1).
>
>[ which was my point, and which we agree is true ]
You left out some of the context in which you made that statement.
Here it is, excerpted from my first posting in the thread:
+In article <364ca3fd....@news.erols.com>,
+Ian Goddard <I...@Goddard.net> wrote:
+>On Tue, 03 Nov 1998, vm...@frii.com (Virgil Hancher) wrote:
[...]
+>>It depends on whether you have an ordinal infinity or a cardinal infinity.
+
+>>With cardinals, order doesn't count, and they are the same.
+
+>>With some ordinals, such as the ordinal type of the natural numbers, it
+>>makes a difference, since 1 + infinity has no last element, but infinity +
+>>1 does.
+
+> IAN: It seems to me, and correct me if I'm wrong, that
+> we could also say that doing addition with the infinity
+> known as the continuum, or "c," is commutative, just
+> as is the case when doing addition with infinite
+> cardinal numbers, and thus: (1 + c) = (c + 1).
+
+No, the same comments apply. Cardinal addition is always commutative:
+1 + c = c + 1. Ordinal addition, however, is not commutative unless
+there is an order-isomorphism between the results. The ordinal sum c+1
+has a last element, but the ordinal sum 1+c does not. Therefore, they
+are not the same as ordinals, but they have the same cardinality.
It appeared to me that you were trying to claim that Virgil Hancher's
explanation of the difference between ordinal and cardinal arithmetic
did not apply to c. The point of my posting was that Virgil Hancher's
comments still hold; substituting one cardinal number for another makes
no difference. In fact, though he said it makes a difference with
"some ordinals, such as the ordinal type of the natural numbers," I
will go further and state that it makes a difference with all infinite
ordinals, including c.
You asked to be corrected if you were wrong, and at least two people
have done that.
>It seems that the fuss being made does not really apply to
>what I said, but to a tangential digression into ordinals
>that seems to be treated as a refutation of what I said.
It's not a tangential discussion. As the above context shows, the
discussion was already about ordinals before you injected your comments
about c.
> IAN: OK, thanks, now I see how the reals
> do not satisfy the well-order definition.
> I think that was the center of what we
> were running around about. This means
> that inf(c) does not fit a definition
> of an ordinal number. Which makes me
> wonder with respect to your statement:
>>The interval [0,1] with its usual
>>ordering is *not* well-ordered and is *not* order-isomorphic
>>to the ordinal number c, ...
> IAN: Is there an ordinal number c ?
A set can be ordered in more than one way. The interval [0,1] is not
well-ordered in its usual order, but this doesn't mean the set can't be
well-ordered at all.
Every cardinal is "secretly" an ordinal, a fact that has already been
mentioned in this thread. Not every ordinal is a cardinal, however. A
cardinal is an ordinal that is not equivalent (does not have a
bijection with) any smaller ordinal. Therefore, c is both an ordinal
and a cardinal.
Jeremy Boden
<I...@Goddard.net> writes
>On Sat, 14 Nov 1998, Jeremy Boden <jer...@jboden.demon.co.uk> wrote:
>
>>Question; Is the infinity of positive integers <, = or > the infinity of
>>real numbers between 0 and 1?
>
> IAN: The infinity of positive integers is less
> than the infinity of all reals between 0 and 1.
>
I know.
I was really addressing the question to Owen Holden.
Herman Rubin
In the discussion, it has come up that there are differences
between cardinal arithmetic, which is commutative, associative,
multiplication distributive over addition, etc., and the
arithmetic of well-orderings, where one still has the
associative property, but no longer has commutativity of
either addition and multiplication, and has only one-sided
distributivity. It also has the one-sided cancellation
property.
It is not so well known, but there is a simple definition of
arithmetic on order types, of which well-orderings are a special
case. This precedes the work of Zermelo and Russell. If A and
B are disjoint representatives of order types a and b, the order
type a+b is defined to be that of the union of A and B, with the
orders in each set, but everything in A precedes everything in
B. The product of the order types is the order type of the
product with the lexicographical order.
Cancellation may not hold, but otherwise the properties are the
same as for well-ordered sets.
Ian Goddard
>>> IAN: To be exact, he hadn't explained anything
>>> previous to that sentence that I commented on.
>
>>I disagree, but it seems futile to argue about it now.
>
>A clarification is needed. Looking back at the history of the thread,
>it appears to me that it was Virgil Hancher who first explained the
>difference between ordinal and cardinal addition. I don't have his
>original posting, but I found followups that quote him. When I made my
>statement that "No, the same comments apply.", I was referring to his
>comments.
>
>That statement that "the same comments apply" was my first statement in
>this thread. Therefore, Ian is technically correct that I didn't
>explain anything before that statement. However, I did include the
>quoted context at the time and it should have been clear what I meant
>by "the same comments."
IAN: It's just that I'd answer a correct
statement with "Yes" not "No," as you did.
If there are ancillary points to be noted
I'd say "Yes, and these are the limits..."
Arthur L. Rubin
>
Well, many people seem to have confused cardinals and
ordinals, and perhaps order types.
Let me try some definitions, so we can agree on what
we're talking about:
omega (or omega_0) is the first infinite ordinal.
Aleph_0 is the first infinite cardinal, which
is the cardinal of N, or of omega.
c = 2^Aleph_0 = Beth_1 is the cardinality of the reals.
o(c) is the first ordinal with cardinality c.
eta is the order type of the rationals
lambda is the order type of the reals.
All ordinals are (representatives of) order types,
but not all order types are ordinals.
> IAN: OK, so what part isn't correct? Here it is again:
>
> IAN: It seems to me, and correct me if I'm wrong, that
> we could also say that doing addition with the infinity
> known as the continuum, or "c," is commutative, just
> as is the case when doing addition with infinite
> cardinal numbers, and thus: (1 + c) = (c + 1).
Cardinals: (1 + c) = (c + 1)
Ordinals: (1 + o(c)) not= (o(c) + 1)
Order Types: (1 + lambda) not= (lambda + 1)
...
> Of course adding to c does not increase the infinite
> density of c, such that c + c + c = c. That only other
> case I know of where that can be said is 0 + 0 + 0 = 0.
For any infinite cardinal x which is the cardinality of
a well-ordered set, x + x = x, but...
If alpha is an ordinal, then alpha + alpha = alpha only if
alpha = 0
eta + eta = eta
lambda + lambda not= lambda
lambda + 1 + lambda = lambda
--
Arthur L. Rubin 216-...@mcimail.com
Ian Goddard
>>On 14 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>>The point is that the ordinal corresponding to c is a well-ordered
>>>set. It also has the property that c + 1 has a last element, but 1 + c
>>>does not. Therefore, ordinal addition is not commutative in this
>>>case.
>>
>> IAN: The set of all real numbers from 0
>> to 1 = inf(c) and it has a last element,
>> 1. That's the basis of what I'm saying
>> and seems to be what we disagree on.
>
>No, that's not what we disagree on. You keep snipping the important
>parts.
IAN: You'll notice that most threads do not
result in continuously larger posts, since
most people, including me, comment on key
sentences and snip the rest, assuming that
others can review the record for the details.
I'm sorry that the effort has been made here
to frame my snipping as a disingenuous effort
to, I guess, claim that addition on infinity
c acts "just as" it does with addition on
infinite cardinals, which we all agree.
>What we disagree on is that Brian Scott and I keep telling you
>the interval [0,1] with its standard order is not a well-ordered set.
>You don't seem to understand that there is a difference between an
>"ordered set" and a "well-ordered set", even though the difference has
>been explained in this thread more than once.
IAN: The fact that I noted that there
is a difference between an ordered and
a well-ordered set by my observing that
Zermelo proved that any ordered set can
also be a well-ordered set proves that
what you say about me cannot be true.
>Do you agree that if I can produce a subset of [0,1] that does not have
>a least element, then I have demonstrated that [0,1] is not
>well-ordered according to the definition?
>
>Do you agree that (0,1] is a subset of [0,1]?
>
>Do you agree that (0,1] does not have a least element?
>
>Do you agree that I have proved my claim?
IAN: Yes to all. But it hasn't refuted anything
I said, I never said that inf(c) was an ordinal
number. My observation sought to point out that
the inf(c) that is all reals from 0 to 1 is uni-
que among Cantor's infinite numbers since unlike
those other infinite sets, it does have a last
number, 1. Your point about that such isn't the
case under the well-ordered-set definition since
a subset can be constructed that violates the
well-ordered definition is noted and appreciated.
So my proposition that addition on inf(c) works
just as it does in the case of any transfinite
cardinal number, stands correct, with the stip-
ulation that inf(c) is not an ordinal number,
which was a point not stipulated in my post.
Ian Goddard
>>>Question; Is the infinity of positive integers <, = or > the infinity of
>>>real numbers between 0 and 1?
>>
>> IAN: The infinity of positive integers is less
>> than the infinity of all reals between 0 and 1.
>>
>--Cut--
>
>I know.
>I was really addressing the question to Owen Holden.
IAN: Ooops, sorry. I simply define any public
post as a thing to which a public response is
not inappropriate, and I see no reason why I
should alter that view. BTW, you raised good
points in your case for the infinity concept.
Jeremy Boden
<I...@Goddard.net> writes
>On Sun, 15 Nov 1998, Jeremy Boden <jer...@jboden.demon.co.uk> wrote:
>
>>>>Question; Is the infinity of positive integers <, = or > the infinity of
>>>>real numbers between 0 and 1?
>>>
>>> IAN: The infinity of positive integers is less
>>> than the infinity of all reals between 0 and 1.
>>>
>>--Cut--
>>
>>I know.
>>I was really addressing the question to Owen Holden.
>
>
> IAN: Ooops, sorry. I simply define any public
> post as a thing to which a public response is
> not inappropriate, and I see no reason why I
> should alter that view. BTW, you raised good
> points in your case for the infinity concept.
>
>**************************************************************
>Visit Ian Williams Goddard --------> http://Ian.Goddard.net
>______________________________________________________________
>
Its just that I didn't want people to have the impression that I'm daft.
Jeremy Boden
<I...@Goddard.net> writes
--major cuts--
>On 15 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>Do you agree that if I can produce a subset of [0,1] that does not have
>>a least element, then I have demonstrated that [0,1] is not
>>well-ordered according to the definition?
>>
>>Do you agree that (0,1] is a subset of [0,1]?
>>
>>Do you agree that (0,1] does not have a least element?
>>
>>Do you agree that I have proved my claim?
>
> IAN: Yes to all. But it hasn't refuted anything
> I said, I never said that inf(c) was an ordinal
> number. My observation sought to point out that
> the inf(c) that is all reals from 0 to 1 is uni-
> que among Cantor's infinite numbers since unlike
> those other infinite sets, it does have a last
> number, 1. Your point about that such isn't the
> case under the well-ordered-set definition since
> a subset can be constructed that violates the
> well-ordered definition is noted and appreciated.
>
> So my proposition that addition on inf(c) works
> just as it does in the case of any transfinite
> cardinal number, stands correct, with the stip-
> ulation that inf(c) is not an ordinal number,
> which was a point not stipulated in my post.
>
>Visit Ian Williams Goddard --------> http://Ian.Goddard.net
Sigh...
Well-ordering is the vital bit.
A well ordered set has the property that *absolutely any* subset has a
least element. Here "least" must be interpreted extremely loosely; it
means least according to any definition of order which you decide to
choose.
So if you want to well-order [0,1] you can (provided you accept the
axiom of choice). However, do not expect the order relation to look
anything like the standard arithmetic < relation.
In fact my guess is that you would not find it possible to give a fully
finite definition of the relation - but that's only a guess.
Your problem is to redefine addition so that you can add ordinals (i.e.
order is important) but is also commutative (order is unimportant).
I think you can't do this.
Dave Seaman
Ian Goddard <I...@Goddard.net> wrote:
>On 15 Nov 1998, a...@seaman.cc.purdue.edu (Dave Seaman) wrote:
>>parts.
>
> IAN: You'll notice that most threads do not
> result in continuously larger posts, since
> most people, including me, comment on key
> sentences and snip the rest, assuming that
> others can review the record for the details.
I do that myself. You, however, snip the important parts and comment on
trivialities.
> I'm sorry that the effort has been made here
> to frame my snipping as a disingenuous effort
> to, I guess, claim that addition on infinity
> c acts "just as" it does with addition on
> infinite cardinals, which we all agree.
Since c *is* an infinite cardinal, I don't see why anyone would deny
that arithmetic on c is just like arithmetic on infinite cardinals. I
am not aware that anyone has done so.
It's just that every cardinal is also an ordinal, and arithmetic on
ordinals is different from arithmetic on cardinals.
>>What we disagree on is that Brian Scott and I keep telling you
>>the interval [0,1] with its standard order is not a well-ordered set.
>>You don't seem to understand that there is a difference between an
>>"ordered set" and a "well-ordered set", even though the difference has
>>been explained in this thread more than once.
> IAN: The fact that I noted that there
> is a difference between an ordered and
> a well-ordered set by my observing that
> Zermelo proved that any ordered set can
> also be a well-ordered set proves that
> what you say about me cannot be true.
It is apparent that you still do not understand the difference even
now, since Brian and I have pointed out that *every* set, ordered or
not, can be well-ordered. That assertion is equivalent to the Axiom of
Choice. Whenever either of us has pointed out that [0,1] is not
well-ordered with the usual ordering, you have responded with
irrelevant comments about the natural ordering on that set, seemingly
oblivious to the fact that the natural ordering is not a well
ordering.
>>Do you agree that if I can produce a subset of [0,1] that does not have
>>a least element, then I have demonstrated that [0,1] is not
>>well-ordered according to the definition?
>>
>>Do you agree that (0,1] is a subset of [0,1]?
>>
>>Do you agree that (0,1] does not have a least element?
>>
>>Do you agree that I have proved my claim?
>
>
> IAN: Yes to all. But it hasn't refuted anything
> I said, I never said that inf(c) was an ordinal
> number.
The very purpose of my original posting in this thread was to point out
that you *failed* to discuss the difference between ordinal and
cardinal arithmetic that had been explained by Virgil Hancher in the
paragraph you had quoted and then commented on. You missed the point
of his explanation.
>My observation sought to point out that
> the inf(c) that is all reals from 0 to 1 is uni-
> que among Cantor's infinite numbers since unlike
> those other infinite sets, it does have a last
> number, 1.
You have not said that before. If you had said it, I would have
pointed out that your statement was incorrect. The set of rational
numbers in [0,1] has cardinality aleph_0 and there is a last element,
1. Therefore, c is not unique in the way that you suggest. In fact,
there is even a well-ordered set w+1 that has the cardinality aleph_0
and also has a last element.
Brian M. Scott
>On Sun, 15 Nov 1998, sc...@math.csuohio.edu (Brian M. Scott) wrote:
[snip]
> This seems to me to be the case since we can define c
> as a span covering a finite length, such as one inch,
> and across that span are an infinite number of zero-
> sized points, and since we could always add 1 to a
> one-inch span from any side, we have c and add 1.
If you think of c in this fashion, you'll almost certainly just
confuse yourself. In this context c is a cardinal number, not a
segment of the real line. In the usual formulations of set theory,
this makes c a particular ordinal number. Specifically, c is a
transitive set well-ordered by the membership relation, and it is the
smallest such set admitting a bijection with the real numbers.
(Without the axiom of choice there may not be such a set.) It comes
equipped with a built-in ordering that looks completely different from
the usual ordering of R (or any segment thereof).
[snip]
> This means
> that inf(c) does not fit a definition
> of an ordinal number. Which makes me
> wonder with respect to your statement:
Yes, it does: see above. The problem here is that you're using
'inf(c)' in a non-standard fashion.
Brian M. Scott
Ian Goddard
>> IAN: You'll notice that most threads do not
>> result in continuously larger posts, since
>> most people, including me, comment on key
>> sentences and snip the rest, assuming that
>> others can review the record for the details.
>
>I do that myself. You, however, snip the important parts and
>comment on trivialities.
IAN: Since I came to agree with your points, the
effort to frame standard snipping of them by me
as an effort by me to ignore them is meaningless,
except to the degree it can be used to degrade me.
>>>What we disagree on is that Brian Scott and I keep telling you
>>>the interval [0,1] with its standard order is not a well-ordered set.
>>>You don't seem to understand that there is a difference between an
>>>"ordered set" and a "well-ordered set", even though the difference has
>>>been explained in this thread more than once.
>
>> IAN: The fact that I noted that there
>> is a difference between an ordered and
>> a well-ordered set by my observing that
>> Zermelo proved that any ordered set can
>> also be a well-ordered set proves that
>> what you say about me cannot be true.
>
>It is apparent that you still do not understand the difference even
>now, since Brian and I have pointed out that *every* set, ordered or
>not, can be well-ordered. That assertion is equivalent to the Axiom of
>Choice. Whenever either of us has pointed out that [0,1] is not
>well-ordered with the usual ordering, you have responded with
>irrelevant comments about the natural ordering on that set, seemingly
>oblivious to the fact that the natural ordering is not a well
>ordering.
IAN: But you said I only commented on the trivial parts.
In fact I responded that I agree with your point, I even
agreed with you in that to which you reply above, so your
claim that whenever I responded to that I cited natural
ordering is manifestly false, unfair, and betrays an
effort to trash my person by any means possible.
Good points have been made in response to my post,
and I've agreed with them, thanked those who made
them and noted how they should be cited with regard
to my post about addition with inf(c), yet even after
that, the accusations and misrepresentations continue.
I don't have as much time on my hands to address them
as it seems others have to make them, so I'll just let
them vent their anger. I just wanted to introduce the
inf(c) into the discussion of addition on infinites in
hope of exploring how additions on inf(c) fit in this
area and to thereby stimulate constructive discourse.
**************************************************************
Visit Ian Williams Goddard --------> http://Ian.Goddard.net
Arthur L. Rubin
>
> On Sun, 15 Nov 1998 13:06:51 +0700, "Arthur L. Rubin"
> <216-...@mcimail.com> wrote:
>
> >Well, many people seem to have confused cardinals and
> >ordinals, and perhaps order types.
>
> since I started reading it.
Yes, you're probably correct. Ian actually had some good points,
except for that confusion.
Brian M. Scott
[snip]
> So my proposition that addition on inf(c) works
> just as it does in the case of any transfinite
> cardinal number, stands correct, with the stip-
> ulation that inf(c) is not an ordinal number,
> which was a point not stipulated in my post.
No, it doesn't. You've defined inf(c) to be the set [0,1]. This is
not (by any of the usual definitions) a cardinal number, and you've
not defined how it is to participate in the operation of addition.
And your final comment would be problematic even if you actually were
working with the cardinal number c: in ZFC every cardinal number *is*
an ordinal number.
Brian M. Scott
Brian M. Scott
<216-...@mcimail.com> wrote:
>Well, many people seem to have confused cardinals and
>ordinals, and perhaps order types.
Doubtless many people have, but in this thread only Ian has done so
since I started reading it.
Brian M. Scott
Ian Goddard
>The idea that infinity is a number leads to contradiction over
>and over; we should stop the nonsense, there is no such possibility.
IAN: Zero has a few oddities, such as n/0, n^0,
but it's still a number, and thus we can observe
that numbers can have unique attributes. Infinity
is certainly a mathematical entity
>What (we) I mean by infinity is a process... with out end.
IAN: An infinity can be contained in the way
that all the real numbers between 0 and 1 can
be contained within the interval from 0 to 1.
>It makes no sense to talk about oo or 2^(oo) or any of those
>cocepts. That one infinity is greater than another (Cantor)
>displays the sillyness, even if you can demonstrate that N(0)
>must be less than 2^N(0), from nonsense becomes nonsense. I am
>reminded of Russell's sillyness in *Principia Mathematica* where
>he claimed that 1/0 is a number (Volume 2).
>
>I understand that opinions are free (agreeable or not).
>--
>Owen Holden
>
>-----------== Posted via Deja News, The Discussion Network ==----------
>http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own
************************************************************
Visit Ian Williams Goddard ------> http://Ian.Goddard.net
------------------------------------------------------------
(+) Something can come from nothing, if, and only if, (-)
(-) that something is equal to nothing ((-)+(+) = 0). (+)
____________________________________________________________
"[I]n any closed universe the negative gravitational energy
cancels the energy of matter exactly. The total energy, or
equivalently the total mass, is precisely equal to zero."
- + - + Dr. Alan Guth (The Inflationary Universe) + - + -
Ian Goddard
Ooops, posting this was a blooper. It's been in "send later"
since whenever the thread was last active, half asleep I
accidently hit "yes" on a dialog box and...
Mike Deeth
Ian Goddard wrote:
> On Sat, 14 Nov 1998 12:07:46 GMT, oho...@idirect.com wrote:
>
> >The idea that infinity is a number leads to contradiction over
> >and over; we should stop the nonsense, there is no such possibility.
I agree. :-)
> IAN: Zero has a few oddities, such as n/0, n^0,
> but it's still a number, and thus we can observe
> that numbers can have unique attributes. Infinity
> is certainly a mathematical entity
Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT. I
propose that zero should be replaced by the underscore "_" so as to
eliminate confusion. All the natural numbers can repressent lengths on a
numberline but, zero only repressents a point. All Rational numbers k have
Rational inverses 1/k. Zero doesn't have a Rational inverse. _ = k/oo (for
all k) isn't math, its lunacy.
>What (we) I mean by infinity is a process... with out end.
> IAN: An infinity can be contained in the way
> that all the real numbers between 0 and 1 can
> be contained within the interval from 0 to 1.
Real numbers are processes.They don't exist as finished entities.
ie. What is the last digit of a Real number?
> >It makes no sense to talk about oo or 2^(oo) or any of those
> >cocepts. That one infinity is greater than another (Cantor)
> >displays the sillyness, even if you can demonstrate that N(0)
> >must be less than 2^N(0), from nonsense becomes nonsense. I am
> >reminded of Russell's sillyness in *Principia Mathematica* where
> >he claimed that 1/0 is a number (Volume 2).
> >
Nathan the Great
Age 11
Russell Steven Shawn O'Connor
Zero is defined as {}
{} is an ordinal because for all a in {} a is a subset of {} (vacuously
true because {} has no elements) and if a and b are distinct elements in {}
a is in b or b is in a (also vacuously true).
{} is a Natural Number because every nonempty subset B of {} has the
property that there exists m in B such that for all p in B m is not a
member of p (again vacuously true because {} has no elements).
>eliminate confusion. All the natural numbers can repressent lengths on a
>numberline
Why? Can you prove this or give me a reference for a proof?
>All Rational numbers k have Rational inverses 1/k.
Again can you point me to a proof of this. This statements seems to be
in sharp contrast to what I learned in my Rings and Fields class.
--
Russell O'Connor roco...@uwaterloo.ca
<URL:http://www.undergrad.math.uwaterloo.ca/%7Eroconnor/>
``And truth irreversibly destroys the meaning of its own message''
-- Anindita Dutta, ``The Paradox of Truth, the Truth of Entropy''
Virgil
roco...@undergrad.math.uwaterloo.ca (Russell Steven Shawn O'Connor)
wrote:
Whether or not zero is a natural number depends on your definition of
natural number.
It is perfectly possible to construct a satisfactory arithmetic either way.
When I was young, most mathematicians that I knew preferred the
construction in which one was taken as the first of the natural numbers. I
and many other young mathematics students of the day built successful
models of arithmetic up through at least the standard reals assuming that
1 was naturally the first natural number.
Some years later, the fashion in natural numbers changed, and most younger
mathematicians prefer to have zero as the first natural number. I believe
that this fashion is primarily based on a model of the natural numbers in
which the first natural number is identified with the empty set.
Arithmetics based on this model as neither more nor less satisfactory in
the long run that those based on the other model.
When God created the natural numbers, I think that (He/She) left enough
slack in them to start with zero or with one as each person pleases.
Otherwise we have a "big-endian" versus "little-endian" war.
AS a matter of general clarity, if you publish something which refers to
the first of the natural numbers as either zero or one, you should make
clear which school you belong to.
As you may have gathered by now, I prefer to start with one, but can get
along with those who prefer to start with zero.
Those who are totally inflexible in this matter, either in favor of
including or of excluding zero, I classify with Archimedes Plutonium and
Mike Deeth.
--
Virgil
vm...@frii.com
Jake Wildstrom
Well, zero is _sometimes_ a natural number. I've seen the Peano axioms done
both ways. But either way it's a number--if not a natural number, then an
integer at least. And eradicating it would not be nice. Closure of addition's
a good property and not really a gret one to lose.
And if you're planning to replace it with something else, it's pretty easy
to show that in any distributive and associative algebra, the additive identity
can't have a multiplicative inverse. That's just one of the rules. Deal with
it.
>All the natural numbers can repressent lengths on a
>numberline but, zero only repressents a point.
It represents the distance between a point and itself. This can sometimes be
useful.
>All Rational numbers k have
>Rational inverses 1/k. Zero doesn't have a Rational inverse. _ = k/oo (for
>all k) isn't math, its lunacy.
Which is why most people don't say it, except informally, as a shorthand for
lim_{x->oo} k/x=0. And all rational numbers _don't_ have inverses. That's how
fields are defined--they couldn't be defined any other way.
>>What (we) I mean by infinity is a process... with out end.
What you mean by infinity is your problem. Most of us can deal with it just
fine. THe problem is seeing things as "processes" and "unfinished". Again, the
definitions of infinity in mathematics are more often Platonic than
Aristotelian--it's finished, but in a nonobservable way.
>Real numbers are processes.They don't exist as finished entities.
Again, they're more Platonic than Aristotelian. We know pi and sqrt(2) don't
repeat and have an infinitude of digits after the decimal point. We know that
they're there, so we don't worry too much about them.
>ie. What is the last digit of a Real number?
That's like asking what the hypotenuse of a hexagon or the characteristic
equation of a vector is. Nobody said a real number had a last digit, and they
often don't. While you're at it, what's the last digit of 1/99? 0 or 1?
>> >It makes no sense to talk about oo or 2^(oo) or any of those
>> >cocepts. That one infinity is greater than another (Cantor)
>> >displays the sillyness, even if you can demonstrate that N(0)
>> >must be less than 2^N(0), from nonsense becomes nonsense.
Which is why Cantor never says that 2^(oo)>oo. What he says is that no set
can be mapped surjectively onto its power set. For finite sets, this would
translate to the trivial 2^n>n. For an infinite set, there's no way to express
this concept in terms of natural numbers alone, and saying 2^oo>oo captures
only the spirit, not the actual mathematical rigor, of Cantor's statement.
+--First Church of Briantology--Order of the Holy Quaternion--+
| A mathematician is a device for turning coffee into |
| theorems. -Paul Erdos |
+-------------------------------------------------------------+
| Jake Wildstrom |
+-------------------------------------------------------------+
Russell Steven Shawn O'Connor
Virgil <vm...@frii.com> wrote:
[A bunch of cool stuff on whether 0 is a natural number or not]
I agree with everything you say. One could take {} to represent 1 in my
construction, and I believe everything would be fine. The integers and
rationals etc. would have to be constructed in a slightly different
manner but I don't foresee any problems.
In high school I was of the belief that 1 was the first natural number.
0 was part of the "whole numbers". In first year I had a calculus course
and an algebra course. IIRC my calculus teacher believed that 1 was the
first natural number and my algebra teacher believed that 0 was the first
natural number. I kept my belief that 1 was the first natural number
until I read the axiomatic development of set theory. Then I felt it was
more "natural" to have 0 as the first natural number since {} was the
first natural number and {} has 0 elements.
In a similar manner I used to believe 0^0 was undefined. I mean if there
is disagreement over it's value, shouldn't it be undefined? Again by
reading the axiomatic development of set theory, I now believe that 0^0
should be 1, because there is indeed 1 function from the empty set to the
empty set, namely the empty function.
Of course all these are personal beliefs (like my belief in the Axiom of
Choice) and when one considers a theorem it has is validity (or not)
within the system the author is working in.
Richard Carr
:Date: 9 Feb 1999 21:32:59 GMT
:From: Jake Wildstrom <wil...@mit.edu>
:Newsgroups: sci.math
:Subject: Re: Addition on Infinities
:
:In article <36C03AB2...@ashland.baysat.net>,
:Mike Deeth <m...@ashland.baysat.net> wrote:
:>Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT. I
:
:
:Well, zero is _sometimes_ a natural number. I've seen the Peano axioms done
:both ways. But either way it's a number--if not a natural number, then an
:integer at least. And eradicating it would not be nice. Closure of addition's
:a good property and not really a gret one to lose.
:
:And if you're planning to replace it with something else, it's pretty easy
:to show that in any distributive and associative algebra, the additive identity
:can't have a multiplicative inverse. That's just one of the rules. Deal with
:it.
Except the trivial ring of course.
:
:>All the natural numbers can repressent lengths on a
:>numberline but, zero only repressents a point.
:
:It represents the distance between a point and itself. This can sometimes be
:useful.
:
:>All Rational numbers k have
:>Rational inverses 1/k. Zero doesn't have a Rational inverse. _ = k/oo (for
:>all k) isn't math, its lunacy.
:
:Which is why most people don't say it, except informally, as a shorthand for
:lim_{x->oo} k/x=0. And all rational numbers _don't_ have inverses. That's how
:fields are defined--they couldn't be defined any other way.
:
:>>What (we) I mean by infinity is a process... with out end.
:
:What you mean by infinity is your problem. Most of us can deal with it just
:fine. THe problem is seeing things as "processes" and "unfinished". Again, the
:definitions of infinity in mathematics are more often Platonic than
:Aristotelian--it's finished, but in a nonobservable way.
:
:>Real numbers are processes.They don't exist as finished entities.
:
:Again, they're more Platonic than Aristotelian. We know pi and sqrt(2) don't
:repeat and have an infinitude of digits after the decimal point. We know that
:they're there, so we don't worry too much about them.
:
:>ie. What is the last digit of a Real number?
:
:That's like asking what the hypotenuse of a hexagon or the characteristic
You could have a hypotenuse of a hexagon. The hypotenuse is just the
longest side- it is not restricted merely to right-angled triangles.
:equation of a vector is. Nobody said a real number had a last digit, and they
:often don't. While you're at it, what's the last digit of 1/99? 0 or 1?
:
:>> >It makes no sense to talk about oo or 2^(oo) or any of those
Yes, it does. It makes a lot of sense. The fact of the matter is that you
just aren't intelligent enough to understand it. Let's face it Nathan,
you're pretty dumb for an 11 year old, let alone for your real age.
:>> >cocepts. That one infinity is greater than another (Cantor)
:>> >displays the sillyness, even if you can demonstrate that N(0)
:>> >must be less than 2^N(0), from nonsense becomes nonsense.
:
:Which is why Cantor never says that 2^(oo)>oo. What he says is that no set
:can be mapped surjectively onto its power set. For finite sets, this would
:translate to the trivial 2^n>n. For an infinite set, there's no way to express
:this concept in terms of natural numbers alone, and saying 2^oo>oo captures
:only the spirit, not the actual mathematical rigor, of Cantor's statement.
In fact, it is rigorous. Of course, you generally don't use the symbol oo,
but rather something like aleph_beta or kappa (I'd better put this
(kappa) in because otherwise Prof. Rubin will complain that I only
mentioned well-orderable cardinals etc.- he probably still will since he
did last time).
:
:+--First Church of Briantology--Order of the Holy Quaternion--+
:
:
oho...@idirect.com
I...@Goddard.net (Ian Goddard) wrote:
> On Sat, 14 Nov 1998 12:07:46 GMT, oho...@idirect.com wrote:
>
> >The idea that infinity is a number leads to contradiction over
> >and over; we should stop the nonsense, there is no such possibility.
>
> but it's still a number, and thus we can observe
> that numbers can have unique attributes. Infinity
> is certainly a mathematical entity
In my _opinion_ expressions such as 1/0 are not entities
of any kind let alone numbers. Lets try to define 1/0;
a/b defined (the x such that a=b*x)
1/0 defined (the x such that 1=0*x) given that 0*x=0 for all x
1/0 defined (the x such that 1=0) but 1=0 is contradictory
1/0 defined (the x such that (contradiction))
(the x such that (contradiction)) does not exist, (is not unique),
by Russell's theory of descriptions.
There is no value of x for which 1=0*x is true, therefore
(the x such that 1=0*x) is not unique i.e. it does not exist.
(0/0 also does not exist since (0=0*x) is true for all x, which
shows that (the x such that (0=0*x)) is not unique.
We cannot have any true statement which has 1/0 as subject,
such as 1/0=1/0, 1/0 exists, but we can have true statements
that contain (1/0). e.g.
1/0 does not exist, 1/0 is not equal to 1/0, etc..
We cannot say what 1/0 is, we can only say what it is not.
1/0 is not a number. (There is no thing (entity) which 1/0 is.)
Likewise; Infinity is not a mathematical entity.
Owen Holden
> >What (we) I mean by infinity is a process... with out end.
>
> IAN: An infinity can be contained in the way
> that all the real numbers between 0 and 1 can
> be contained within the interval from 0 to 1.
>
> >It makes no sense to talk about oo or 2^(oo) or any of those
> >cocepts. That one infinity is greater than another (Cantor)
> >displays the sillyness, even if you can demonstrate that N(0)
> >must be less than 2^N(0), from nonsense becomes nonsense. I am
> >reminded of Russell's sillyness in *Principia Mathematica* where
> >he claimed that 1/0 is a number (Volume 2).
> >
Mike Deeth
oho...@idirect.com wrote:
> In my _opinion_ expressions such as 1/0 are not entities
> of any kind let alone numbers. Lets try to define 1/0;
>
> a/b defined (the x such that a=b*x)
>
> 1/0 defined (the x such that 1=0*x) given that 0*x=0 for all x
>
> 1/0 defined (the x such that 1=0) but 1=0 is contradictory
>
> 1/0 defined (the x such that (contradiction))
>
> (the x such that (contradiction)) does not exist, (is not unique),
> by Russell's theory of descriptions.
>
> There is no value of x for which 1=0*x is true, therefore
> (the x such that 1=0*x) is not unique i.e. it does not exist.
>
> (0/0 also does not exist since (0=0*x) is true for all x, which
> shows that (the x such that (0=0*x)) is not unique.
>
> We cannot have any true statement which has 1/0 as subject,
> such as 1/0=1/0, 1/0 exists, but we can have true statements
> that contain (1/0). e.g.
> 1/0 does not exist, 1/0 is not equal to 1/0, etc..
>
> We cannot say what 1/0 is, we can only say what it is not.
>
> 1/0 is not a number. (There is no thing (entity) which 1/0 is.)
>
> Likewise; Infinity is not a mathematical entity.
>
> Owen Holden
>
Excellent Owen! :-)
Here is a true story: ;-)
A Cantorian who was being committed to a lunatic asylum inquired of his doctor
what method was employed to discover when he would be sufficiently recovered
to be discharged.
"Well," replied the doctor, "you see, it's this way. We have a big trough
of water and we turn on the tap. We leave it running and ask you to bail out
the water with a pail until it is emptied."
"How does that prove anything?" asked the Cantorian.
"Well," said the doctor, "if you're not crazy you will turn off the tap."
"Well I declare. I never would have thought of that," said the Cantorian.
:-)
Nathaniel Deeth
Age 11
brian tivol
> Here is a true story: ;-)
Oohh, it's a parable!
> A Cantorian who was being committed to a lunatic asylum inquired of
> his doctor what method was employed to discover when he would be
> sufficiently recovered to be discharged.
Hmm. I bet that the lunatic asylum represents Usenet; it seems
fitting.
> "Well," replied the doctor, "you see, it's this way. We have a big
> trough of water and we turn on the tap. We leave it running and ask
> you to bail out the water with a pail until it is emptied."
Ah, yes, the trough represents sci.math, and the spigot is the
connection to it from your Daddy's computer. The endless flow of
water droplets, evocative of the Chinese Water Torture, represents the
endless flow of inane, ill-conceived, egotistical drivel that you keep
posting to this newsgroup. Just as the water droplets will overfill
the trough, running onto the floor and irreparably damaging the tile
floor, so will your insipid postings swamp out sci.math and forever
discolor meaningful conversations.
> "How does that prove anything?" asked the Cantorian.
>
> "Well," said the doctor, "if you're not crazy you will turn off the tap."
Here the interpretation can vary. Indeed, the best course of action
would be to have your Daddy shot down your spigot-- maybe he could
force you to play outside and possibly even make a real-life friend or
two. However, it's not within the capability of most sci.math readers
to talk to your father; therefore, the most likely moral is:
*plonk*
> "Well I declare. I never would have thought of that," said the
> Cantorian.
>
> :-)
> Nathaniel Deeth
> Age 11
--brian
Steve Harris
True, infinity is not a number in the sense of the natural numbers or
the real numbers. But that does not mean it is of no use in
calculational expressions. For instance, when dealing with limits of
real-valued functions, it can be very useful--practical!--to have a
calculus of infinity. I can say,
"If f(x) grows arbitarily large as x approaches value, then f(x) + 2
grows arbitarily large as x approaches that same value"
or I can say,
"oo + 2 = oo"
Which do you think is going to be easier to recall and use in practise?
Similarly, I will tell my calculus students that oo + oo = oo, oo - oo
can be anyuthing (undefined, if you will), (oo)k = oo if k > 0, (oo)k =
-oo if k < 0, and (oo)0 can be anything.
If you are dealing only with real numbers, that's about all you need
(with similar statements about division and exponents)--and it's all
short-hand for statements about limits. But if you're dealing with
larger sets you will have to learn the calculus of higher cardinals.
It's all perfectly well-defined, just not of moment for most people's
purposes.
Steve
Carl R. White
] C. Holden,
]
] True, infinity is not a number in the sense of the natural numbers or
] the real numbers. But that does not mean it is of no use in
] calculational expressions. For instance, when dealing with limits of
] real-valued functions, it can be very useful--practical!--to have a
] calculus of infinity. I can say,
]
] "If f(x) grows arbitarily large as x approaches value, then f(x) + 2
] grows arbitarily large as x approaches that same value"
]
] or I can say,
]
] "oo + 2 = oo"
]
] Which do you think is going to be easier to recall and use in practise?
] Similarly, I will tell my calculus students that oo + oo = oo, oo - oo
] can be anyuthing (undefined, if you will), (oo)k = oo if k > 0, (oo)k =
] -oo if k < 0, and (oo)0 can be anything.
At least there's someone out there who thinks of infinity in the same way
that I do...
However: "Given any equation involving infinity as a 'largest value', the
infinities on the left hand side, although they look and act like
infinity, are in fact less than infinity. The exceptional circumstance
arising when not-a-numbers are more logical solutions for the RHS"
i.e. in "oo + 2 = oo", the first infinity is in fact conceptually less
than the second.
Your definition of "oo - oo" proves this. If they were the same infinity,
then the result would be zero. Some people would infer that the answer was
not-a-number. Then again, I suppose you could derive a solution from what
has gone before and say "oo - oo = 2" :)
One could infer that there is no 'largest number', just an infinity of
infinitely big numbers milling about, just beyond the horizon.
Carl
PS My computer tells me that "oo/oo = -nan". Where'd the '-' come from?!
But then again, it also tells me that -1e-308 > 0.
--
Carl R White -- Final Year Computer Science at the University of Bradford
E-mail........: cyrek-@-bigfoot.com -- Remove hyphens. Ta :)
URL...........: http://www.bigfoot.com/~cyrek/
Uncrackable...: "19.6A.23.38.52.73.45 25.31.1C 3C.53.44.39.58"
Hal Daume III
subject, and if anyone wants to correct me, please do) is in a more set
theoric point of view. If you have cardinal numbers a and b and find
sets X and Y such that |X|=a and |Y|=b and X and Y are disjoint (you
can prove that such sets exist), and |X| != n, for a natural
number n, and |Y| != n for a natural number n (i.e., X and Y are
infinite sets), then a+b is defined as the union of X and Y.
If |X| != |Y| (there exists no bijection from X to Y) then certainly the
sum of a and b cannot simultaneously equal |X| and |Y| (so, in a sense,
oo + oo != oo). On the other hand, if X and Y are countable sets (have
the same cardinality as the natural numbers), namely a = aleph_o and b =
aleph_o, then a + b = aleph_o + aleph_o = aleph_o = a = b.
The question then arises, for all infinite sets X, does |X| + |X| = |X|
for all X? The answer, basically, is yes and no, depending on whether
or not you accept the axiom of choice.
For more information on this, you might want to check out an elementary
set theory book.
(I hope I didn't get anything wrong here : ) )
- Hal
-----------------------------------------------------------------------
"Men are born ignorant, not stupid; they are made stupid by education."
- Bertrand Russell
-----------------------------------------------------------------------
Steve Harris
You write,
> in "oo + 2 = oo", the first infinity is in fact conceptually less
> than the second.
Not so. There is absolutely no conceptual difference between the two
instances of the symbol oo. You can think of the symbol oo as being an
extenstion of the real number system carrying a (partial) extension of the
algebraic operations (i.e., not all combinations involving oo are defined: oo
- oo is simply not defined). Or, alternatively, you can think of it all as
simply short-hand for statements about how limits of functions behave under
algebraic combinations of functions. This works, but is less intuitive,
hence, of less pedagogical value (and, for teaching calculus, pedagogy is of
high import).
> Your definition of "oo - oo" proves this. If they were the same infinity,
> then the result would be zero.
No: They are the same "infinity", but, since oo is not part of the real
number system, one must not expect that the algebraic operations work with
entire satisfaction on this symbol. In particular, subtraction is simply not
defined between oo and itself. Similarly, although -oo is defined, addition
is not defined between oo and -oo.
> Some people would infer that the answer was
> not-a-number. Then again, I suppose you could derive a solution from what
> has gone before and say "oo - oo = 2" :)
The easiest way to view it is that we're only *partially* extending addition
and subtraction to cover the symbols oo and -oo, not wholly so. The rules of
arithmetic for the real numbers simply don't all apply to the "extended
reals".
Steve
Keith Ellul
> The question then arises, for all infinite sets X, does |X| + |X| = |X|
> for all X? The answer, basically, is yes and no, depending on whether
> or not you accept the axiom of choice.
Do you need the axiom of choice for this? I thought that this was true
independent of AC.
-Keith!
---------------------------------------------------------------
Keith Ellul 4th Year Pure Math / Computer Science
kbe...@golden.net University of Waterloo
---------------------------------------------------------------
Hal Daume III
|X| + |X| = |X|, but if I recall correctly, the answer differed based on
some axiom, and I thought it was the axiom of choice, but I may be
mistaken.
Maybe someone can clarify this?
-----------------------------------------------------------------------
Jeremy Boden
c.uk>, Carl R. White <crw...@comp.brad.ac.uk> writes
...
>
>At least there's someone out there who thinks of infinity in the same way
>that I do...
>
>However: "Given any equation involving infinity as a 'largest value', the
>infinities on the left hand side, although they look and act like
>infinity, are in fact less than infinity. The exceptional circumstance
>arising when not-a-numbers are more logical solutions for the RHS"
>
>than the second.
No, it isn't.
You are referring to the cardinality of an infinite set (such as the set
of real numbers). It is a property of any infinite set that it is
equivalent (there is a bijection) between the set and a proper subset of
itself.
Hence card(oo + 2) = card(oo)
So the first and second oo are "conceptually" identical.
>
>Your definition of "oo - oo" proves this. If they were the same infinity,
>then the result would be zero. Some people would infer that the answer was
>not-a-number. Then again, I suppose you could derive a solution from what
>has gone before and say "oo - oo = 2" :)
No it doesn't.
oo is *not* a number, so subtraction isn't even defined.
>
>One could infer that there is no 'largest number', just an infinity of
>infinitely big numbers milling about, just beyond the horizon.
It's true that there is no largest number; for if N was such a beast
then N+1 would be even larger.
However, whilst there isn't an infinity of infinitely big numbers, there
is an indefinitely large number of indefinitely large numbers.
>
>Carl
>
>PS My computer tells me that "oo/oo = -nan". Where'd the '-' come from?!
> But then again, it also tells me that -1e-308 > 0.
>
Who can fathom what a computer might say?
How did you get it to recognise oo/oo?
At least the answer was half right - oo/oo is certainly 'not a number'
(nan).
Mike Oliver
Keith Ellul wrote:
>
> On Thu, 11 Feb 1999, Hal Daume III wrote:
>
> > The question then arises, for all infinite sets X, does |X| + |X| = |X|
> > for all X? The answer, basically, is yes and no, depending on whether
> > or not you accept the axiom of choice.
>
> Do you need the axiom of choice for this? I thought that this was true
> independent of AC.
Well, it is *true*, because AC is true, but you need AC to *prove* it.
Actually, I'm not sure you need full AC for the additive version; some
fragment might suffice. For the multiplicative version (i.e. |X^2| = |X| for all
infinite X) you *do* need full AC, and the proof is very pretty.
Suppose |X cross X| = |X| for all infinite X. For definiteness, we'll show
that you can wellorder R, the set of all reals; you can substitute any
other set for R.
Let Theta be the Hartog number of R; that is, Theta is the least ordinal
such that there is no surjection from R to Theta. This must exist by
Burali-Forti.
Now we're going to let X be the disjoint union of R and Theta; that is,
we assume we're coding reals and ordinals in such a way that no real
is an ordinal, and then we just take X = (R union Theta).
Now there must be a bijection f : X cross X --> X (actually all we
need is that f is injective). We're only going to consider
f restricted to R cross Theta; that is, we're only concerned
with values f(<x,alpha>) such that x is a real and alpha is
an ordinal less than Theta. For every such x and alpha,
f(<x,alpha>) will be either a real or an ordinal less than
Theta (and not both).
Suppose that for every real x, there is some alpha < Theta such that
f(<x,alpha>) is an ordinal rather than a real. Then for any
x, let g(x) = f(<x,alpha>) for the least alpha such that this
value is an ordinal. g is now an injection from the reals
into the ordinals; this easily gives you a wellordering of the reals.
The only other possibility is that there is some real x such
that for *every* alpha < Theta, f(<x,alpha>) is a real. But
then let h(alpha) = f(<x,alpha>) for this x. h is now an injection
from Theta into R, which by turning it around gives you a surjection
from R to Theta, contradicting the definition of Theta. Therefore
this can't happen, so we're in the situation of the previous paragraph
and the reals can be wellordered.
You can apply this argument to any other infinite set (though
you'll sometimes have to take care of the technical matter of
the disjoint union, which was automatic for the reals). Therefore
all sets can be wellordered, so AC holds.
Brian Pearson
infinity of points on the line segment represented by that inch. If we
have two inches, we have an infinity of points. IF we cut the inch in
half, we have an infinty of inches, since a point, by definition, is
infinitely small. It seems meaningless to say we can "add" infinities.
Infinity=Infinity.
-
BRIAN PEARSON PYW...@prodigy.com
ilias kastanas 08-14-90
Keith Ellul <kbe...@golden.net> wrote:
@On Thu, 11 Feb 1999, Hal Daume III wrote:
@
@> The question then arises, for all infinite sets X, does |X| + |X| = |X|
@> for all X? The answer, basically, is yes and no, depending on whether
@> or not you accept the axiom of choice.
@
@Do you need the axiom of choice for this? I thought that this was true
@independent of AC.
You do. Without AC, a Dedekind cardinal d might exist (i.e.
there are such models of ZF). Let k = d + aleph_0; then k + k > k.
But you don't need full AC; the statement "for all k, k + k = k"
holds in some models of ZF + ~AC, and is thus weaker than AC.
Ilias
Robert Low
>Carl wrote
>>PS My computer tells me that "oo/oo = -nan". Where'd the '-' come from?!
>> But then again, it also tells me that -1e-308 > 0.
>>
>Who can fathom what a computer might say?
>How did you get it to recognise oo/oo?
>At least the answer was half right - oo/oo is certainly 'not a number'
>(nan).
This is (nearly) IEEE standard arithmetic; 1/0 returns Inf, -1/0
returns -Inf, Inf * (+ve/-ve quantity) = +/- Inf, Inf*0, Inf/Inf, Inf-Inf
are all NaN, and so on. It's consistent and seems to produce a
result, but you should really regard Inf and NaN as two levels
of how badly you messed up the computation.
Carl R. White
] Carl R. White writes:
] ...
] >
] >i.e. in "oo + 2 = oo", the first infinity is in fact conceptually less
] >than the second.
] No, it isn't.
] You are referring to the cardinality of an infinite set (such as the set
] of real numbers). It is a property of any infinite set that it is
] equivalent (there is a bijection) between the set and a proper subset of
] itself.
]
] Hence card(oo + 2) = card(oo)
]
] So the first and second oo are "conceptually" identical.
Ok. I stand re-educated. (I bet you wish it was this easy with that Nathan
Deeth character... ;) )
] >One could infer that there is no 'largest number', just an infinity of
] >infinitely big numbers milling about, just beyond the horizon.
] It's true that there is no largest number; for if N was such a beast
] then N+1 would be even larger.
]
] However, whilst there isn't an infinity of infinitely big numbers, there
] is an indefinitely large number of indefinitely large numbers.
Careful! You nearly implied I might have done something right for a
change. :) :) :)
] >PS My computer tells me that "oo/oo = -nan". Where'd the '-' come from?!
] > But then again, it also tells me that -1e-308 > 0.
] >
] How did you get it to recognise oo/oo?
I use the Euphoria programming language. It allows the use of Inf (if you
can find out how to generate it) as an intermediate result. Useful for
arctangents and the like. [arctan(Inf) generates Pi/2]. (Un)Fortunately it
gives access to illogical functions such as Inf/Inf too.
] At least the answer was half right - oo/oo is certainly 'not a number'
] (nan).
As posted in another message, it's just the IEEE standard sniggering in
the corner that's causing all this (I knew this before I posted, but I
thought others might be interested). You'd have thought, though, that
electrical engineers would want sign consistency, even if Inf and NaN
*are* weird numbers.
Incidentally, according to IEEE, 1/Inf = 0, but not necessarily in any
other arrangement...
Carl, who's glad he knows the difference between infer and imply.
Ian Goddard
>> IAN: Zero has a few oddities, such as n/0, n^0,
>> but it's still a number, and thus we can observe
>> that numbers can have unique attributes. . . .
>
>Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
>I propose that zero should be replaced by the underscore "_" so as to
>eliminate confusion. . . .
IAN: Maybe it's not a natural number, but zero is
a number. A number is a digit that denotes a value,
and zero is a digit that denotes a value, that of
nothing, and therefore zero is a number. Zero's a
member of the set of all values. For example, the
value of all monies I've given you is zero; the
value of all energy in a closed universe = 0.
If we want numbers to express all values,
we need the number that we call "zero."
**********************************************************
Visit Ian Williams Goddard -----> http://Ian.Goddard.net
__________________________________________________________
" [T]he Indian sunya [zero] was destined to become
the turning point in a development without which
the progress of modern science, or commerce is
inconceivable. ... In the history of culture,
the invention of zero will always stand out
as one of the greatest single achi-
evements of the human race."
Tobias Dantzig
(Number: The Language of Science)
Virgil
Goddard) wrote:
>>Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
>>I propose that zero should be replaced by the underscore "_" so as to
>>eliminate confusion. . . .
Lets replace all numbers by underscores and elminate everything.
Zero is a number. Whether it is a natural number or not depends on your
understanding of the natural numbers. A good case can be made for having 0
as the first natural number, and a good case can be made for having 1 as
the first natural number.
Those who insist that 0 must never be a natural number and those who
insist that 0 must always be a natural number I find equally wrongheaded.
I can understand writings based on either assumption as long as the author
makes clear which assumption he/she is using.
It would be nice if there were some generally agreed upon convention to
distinguish between the two, such as "finite cardinals" inclusive of 0 and
"counting numbers" exclusive of zero, so we could avoid the ambiguity of
the "natural" numbers.
--
Virgil
vm...@frii.com
Lynn Killingbeck
>
> (snip)
>
> Zero is a number. Whether it is a natural number or not depends on your
> understanding of the natural numbers. A good case can be made for having 0
> as the first natural number, and a good case can be made for having 1 as
> the first natural number.
>
> Those who insist that 0 must never be a natural number and those who
> insist that 0 must always be a natural number I find equally wrongheaded.
>
> I can understand writings based on either assumption as long as the author
> makes clear which assumption he/she is using.
>
> It would be nice if there were some generally agreed upon convention to
> distinguish between the two, such as "finite cardinals" inclusive of 0 and
> "counting numbers" exclusive of zero, so we could avoid the ambiguity of
> the "natural" numbers.
>
> --
> Virgil
> vm...@frii.com
Removing the ambiguity is hopeless. One can only hope the individual
authors have the intelligence to recognize the ambiguity, and state
explicitly the assumption for their material. In the past couple of
weeks I have picked up two books, trying to see if a different
exposition could get me better understanding. One (1998) excludes 0. The
other (1999) includes 0 "...as a matter of convenience." Some of the
books don't bother with the definition, and the reader is left to divine
the intent. Sigh.
Lynn Killingbeck
Mike Deeth
Ian Goddard wrote:
> On Tue, 09 Feb 1999, Mike Deeth <m...@ashland.baysat.net> wrote:
>
> >> IAN: Zero has a few oddities, such as n/0, n^0,
> >> but it's still a number, and thus we can observe
> >> that numbers can have unique attributes. . . .
> >
> >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
> >I propose that zero should be replaced by the underscore "_" so as to
> >eliminate confusion. . . .
>
> IAN: Maybe it's not a natural number, but zero is
> a number. A number is a digit that denotes a value,
> and zero is a digit that denotes a value, that of
> nothing, and therefore zero is a number. Zero's a
> member of the set of all values. For example, the
> value of all monies I've given you is zero; the
> value of all energy in a closed universe = 0.
>
> If we want numbers to express all values,
> we need the number that we call "zero."
>
Could someone please give a non-supernatural example where adding,
subtracting, multipliing or dividing by zero is usefull.
Q: Why does a Cantorian only change his baby's diapers every month?
A: Because it says right on it "good for up to 20 pounds."
Mike Deeth
Virgil wrote:
> >>Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
> >>I propose that zero should be replaced by the underscore "_" so as to
> >>eliminate confusion. . . .
>
> Lets replace all numbers by underscores and elminate everything.
>
> Zero is a number. Whether it is a natural number or not depends on your
> understanding of the natural numbers. A good case can be made for having 0
> as the first natural number, and a good case can be made for having 1 as
> the first natural number.
Please submit:
(i) your 'good case' for zero being the first natural number.
(ii) an example where adding, subtracting, multiplying or dividing by 0 is
usefull.
Q: Why don't Cantorians double recipes?
A: The oven doesn't go to 700 degrees.
Brian Pearson
>
> Just to note, I didn't say that, the person I quoted said it.
>
>
>On Sun, 14 Feb 1999 11:48:22 -0600, vm...@frii.com (Virgil) wrote:
>
>>In article <36c6746d...@news.erols.com>, I...@Goddard.net (Ian
>>
>>>>Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a
DIGIT.
>>>>I propose that zero should be replaced by the underscore "_" so as
to
>>>>eliminate confusion. . . .
It would be kind of hard to write 10 or 100 or 1000, wouldnt it, without
zero. Zero is the difference between finger counting and a real number
system, IMO.
-
BRIAN PEARSON PYW...@prodigy.com
Richard Carr
:Date: Mon, 15 Feb 1999 21:51:32 -0600
:From: Mike Deeth <m...@ashland.baysat.net>
:Newsgroups: sci.math
:Subject: Re: Zero-out Zero ?
:
:
:
:Ian Goddard wrote:
:
:> On Tue, 09 Feb 1999, Mike Deeth <m...@ashland.baysat.net> wrote:
:>
:> >> IAN: Zero has a few oddities, such as n/0, n^0,
:> >> but it's still a number, and thus we can observe
:> >> that numbers can have unique attributes. . . .
:> >
:> >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
:> >I propose that zero should be replaced by the underscore "_" so as to
:> >eliminate confusion. . . .
:>
:> IAN: Maybe it's not a natural number, but zero is
:> a number. A number is a digit that denotes a value,
:> and zero is a digit that denotes a value, that of
:> nothing, and therefore zero is a number. Zero's a
:> member of the set of all values. For example, the
:> value of all monies I've given you is zero; the
:> value of all energy in a closed universe = 0.
:>
:> If we want numbers to express all values,
:> we need the number that we call "zero."
:>
:
:Could someone please give a non-supernatural example where adding,
:subtracting, multipliing or dividing by zero is usefull.
:
Computation of partial fractions and proper fractions is such an example.
Adding zero can very often be useful here.
:Q: Why does a Cantorian only change his baby's diapers every month?
:A: Because it says right on it "good for up to 20 pounds."
:
Where can you get these? I can make a fortune when I get back to England.
:Nathan the Abhorrent
:Age 11
:
:
:
:
William L. Bahn
foolish enough to issue you one) you will discover that there is such a
thing as finance charges. Commonly, how this is calculated is by multiplying
the average balance that you had on a given day by the daily periodic rate.
Since zero is not a value, how much money would you like to pay them each
month that you carry zero balance?
Your altitude above the ground in an airplane is the difference between your
altitude above mean sea level and the elevation of the ground below you. How
would you like to calculate your altitude above an area that is at sea
level? Which is less confusing - to have a bunch of rules for special cases
such as when your altitude is the same as sea level or when the ground's
elevation is the same as sea level or to just accept that the altitude and
elevation in those cases is zero (feet, meters, whatever) and use the same
rules and the same equations and the same procedures as you do for any other
altitude and elevation?
Mike Deeth wrote in message <36C8EB44...@ashland.baysat.net>...
>
>
>Ian Goddard wrote:
>
>> On Tue, 09 Feb 1999, Mike Deeth <m...@ashland.baysat.net> wrote:
>>
>> >> IAN: Zero has a few oddities, such as n/0, n^0,
>> >> but it's still a number, and thus we can observe
>> >> that numbers can have unique attributes. . . .
>> >
>> >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
>> >I propose that zero should be replaced by the underscore "_" so as to
>> >eliminate confusion. . . .
>>
>> IAN: Maybe it's not a natural number, but zero is
>> a number. A number is a digit that denotes a value,
>> and zero is a digit that denotes a value, that of
>> nothing, and therefore zero is a number. Zero's a
>> member of the set of all values. For example, the
>> value of all monies I've given you is zero; the
>> value of all energy in a closed universe = 0.
>>
>> If we want numbers to express all values,
>> we need the number that we call "zero."
>>
>
>Could someone please give a non-supernatural example where adding,
>subtracting, multipliing or dividing by zero is usefull.
>
>Q: Why does a Cantorian only change his baby's diapers every month?
>A: Because it says right on it "good for up to 20 pounds."
>
Mike Deeth
Brian Pearson wrote:
> >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a
> DIGIT.
> >I propose that zero should be replaced by the underscore "_" so as to
> >eliminate confusion. . . .
>
> It would be kind of hard to write 10 or 100 or 1000, wouldnt it, without
> zero. Zero is the difference between finger counting and a real number
> system, IMO.
Yes, the digit '0' does make a good place holder. :-)
Mike Deeth
Richard Carr wrote:
> :
> :Could someone please give a non-supernatural example where adding,
> :subtracting, multipliing or dividing by zero is usefull.
> :
>
> Computation of partial fractions and proper fractions is such an example.
> Adding zero can very often be useful here.
>
Please give an example.
Nathan
Age 11
Josh Kortbein
: Ian Goddard wrote:
: > On Tue, 09 Feb 1999, Mike Deeth <m...@ashland.baysat.net> wrote:
: >
: > >> IAN: Zero has a few oddities, such as n/0, n^0,
: > >> but it's still a number, and thus we can observe
: > >> that numbers can have unique attributes. . . .
: > >
: > >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a DIGIT.
: > >I propose that zero should be replaced by the underscore "_" so as to
: > >eliminate confusion. . . .
: >
: > IAN: Maybe it's not a natural number, but zero is
: > a number. A number is a digit that denotes a value,
: > and zero is a digit that denotes a value, that of
: > nothing, and therefore zero is a number. Zero's a
: > member of the set of all values. For example, the
: > value of all monies I've given you is zero; the
: > value of all energy in a closed universe = 0.
: >
: > If we want numbers to express all values,
: > we need the number that we call "zero."
: >
: Could someone please give a non-supernatural example where adding,
: subtracting, multipliing or dividing by zero is usefull.
"Non-supernatural?" Perhaps dividing by zero is esoteric and
conceptual, to you. It can be used, in the extended reals, to
talk about what happens to a rational function with a discontinuity -
divide by zero, and the function "goes to infinity" (yes, I know,
no such thing, blah blah blah). This is a notion which means,
in the language of limits, "the closer we get to 0, the bigger
the function gets." Non-supernatural applications abound, then.
Or do you not care about where functions have blowups? Ever heard
of the Tacoma Narrows Bridge?
Josh
--
Is that a real poncho or a Sears poncho?
Brian Pearson
>
>
>
>Brian Pearson wrote:
>
>> >Zero is NOT A NATURAL NUMBER. Zero is a PLACE HOLDER. Zero is a
>> DIGIT.
>> >I propose that zero should be replaced by the underscore "_" so as
to
>> >eliminate confusion. . . .
>>
without
>> zero. Zero is the difference between finger counting and a real
number
>> system, IMO.
>
>Yes, the digit '0' does make a good place holder. :-)
>
>Nathan the Great
>Age 11
I didnt say it was "just" a good place holder. I just said it would be
hard to do without it. Obviously, zero denotes a certain value. If
someone asked you how much money you have and you dont have any, you
could say "zero" money. But you certainly wouldnt say, "Hey, babe, I have
a great placeholder"<g>
-
BRIAN PEARSON PYW...@prodigy.com
Ian Goddard
>> IAN: Maybe it's not a natural number, but zero is
>> a number. A number is a digit that denotes a value,
>> and zero is a digit that denotes a value, that of
>> nothing, and therefore zero is a number. Zero's a
>> member of the set of all values. For example, the
>> value of all monies I've given you is zero; the
>> value of all energy in a closed universe = 0.
>>
>> If we want numbers to express all values,
>> we need the number that we call "zero."
>>
>
>Could someone please give a non-supernatural example where adding,
>subtracting, multipliing or dividing by zero is usefull.
IAN: Why is the sum of all energy in a closed
universe supernatural? Why is the sum of money
I own, of which your take is 0, supernatural?
************************************************************
Visit Ian Williams Goddard ------> http://Ian.Goddard.net
------------------------------------------------------------
(+) Something can come from nothing, if, and only if, (-)
(-) that something is equal to nothing ((-)+(+) = 0). (+)
____________________________________________________________
"[I]n any closed universe the negative gravitational energy
cancels the energy of matter exactly. The total energy, or
equivalently the total mass, is precisely equal to zero."
- + - + Dr. Alan Guth (The Inflationary Universe) + - + -
Mike Deeth
Brian Pearson wrote:
Brian, :-)
Where do you keep your 'zero amount' of money?
Do you hold it in your hand or is it out flying around the moon?
If something, as tangible as money, exists it should be located somewhere.
Have you ever misplaced your 'zero amount' of money?
Do you ever have to go searching for it?
When I spend my money ($5.38) I will be broke.
When I'm broke I will have NO amount of money.
I will say, "I DON'T HAVE ANY money."
But I will not say, "I have ZERO amount money".
Brian Pearson
>Brian, :-)
I will say I have zero amount of money, if I want to<g>. And I have. BTW,
I sometimes have negatives amounts of money that I keep in the bank<g>.
The bank charges me a fee of $25, then I am even more in the negative.
But yes, if I have zero amount of money, I keep it in the bank, or make
it less than zero....
-
BRIAN PEARSON PYW...@prodigy.com
Mike Deeth
William L. Bahn wrote:
> When you are old enough to have a credit card (if any company would be
> foolish enough to issue you one) you will discover that there is such a
> thing as finance charges. Commonly, how this is calculated is by multiplying
> the average balance that you had on a given day by the daily periodic rate.
> Since zero is not a value, how much money would you like to pay them each
> month that you carry zero balance?
I would like to pay NO money when I carry NO balance.
> Your altitude above the ground in an airplane is the difference between your
> altitude above mean sea level and the elevation of the ground below you. How
> would you like to calculate your altitude above an area that is at sea
> level? Which is less confusing - to have a bunch of rules for special cases
> such as when your altitude is the same as sea level or when the ground's
> elevation is the same as sea level or to just accept that the altitude and
> elevation in those cases is zero (feet, meters, whatever) and use the same
> rules and the same equations and the same procedures as you do for any other
> altitude and elevation?
What is longer, zero feet or zero miles?
Mike Deeth
> conceptual, to you. It can be used, in the extended reals, to
> talk about what happens to a rational function with a discontinuity -
> divide by zero, and the function "goes to infinity" (yes, I know,
> no such thing, blah blah blah). This is a notion which means,
> in the language of limits, "the closer we get to 0, the bigger
> the function gets." Non-supernatural applications abound, then.
You aren't *actually* dividing by zero. You're extrapolating on the value of a
function 1/x whose denominator approaches zero. Your example only describes
finite non-zero points of this function. You can't plot f(x) when x is 0.
Ingrid Voigt
> Nathan the Great
> Age 11
Using Aristotle you should be able to answer that for yourself.
Ingrid
Mike Deeth
Ian Goddard wrote:
> >Could someone please give a non-supernatural example where adding,
> >subtracting, multipliing or dividing by zero is usefull.
>
> IAN: Why is the sum of all energy in a closed
> universe supernatural? Why is the sum of money
> I own, of which your take is 0, supernatural?
Could someone please give a non-supernatural example where ADDING,
SUBTRACTING, MULTIPLYING or DIVIDING by zero is usefull?
Mike Deeth
Ingrid Voigt wrote:
> > Nathan the Great
> > Age 11
>
> Using Aristotle you should be able to answer that for yourself.
>
I would rather have you answer.
Ingrid Voigt
> Ingrid Voigt wrote:
>
> > Mike Deeth wrote:
>
> > > What is longer, zero feet or zero miles?
>
> > Using Aristotle you should be able to answer that for yourself.
> >
>
> I would rather have you answer.
>
> Nathan
If you insist... *sigh*
Zero is zero. Zero feet are just as long as zero miles
or zero lightyears. Or zero kilograms.
Ingrid
Carl R. White
] Could someone please give a non-supernatural example where adding,
] subtracting, multipliing or dividing by zero is usefull.
Going off on a tangent here... (Incidentally what is sin(2*Pi)?)
6502 Assembly:
CLC ; Clear carry flag
LDA 251 ; Load a one byte value from memory
ADC 252 ; Add another one byte value from memory
STA 253 ; Store the result in memory
LDA 0 ; Clear accumulator
ADC 0 ; Add 0
STA 254 ; Store in 254
Result? After some runs, 254 contains a one. Now work out (1)Why? and
(2)What would happen if we couldn't use zero...?
Carl R. White
] SUBTRACTING, MULTIPLYING or DIVIDING by zero is usefull?
<me mode=sucked_in>
In the computer programming language that I use (Euphoria 2.0),
multiplying most data structures by zero returns the base structure. This
then enables me to compare it with the true base structure I have created
and see where the errors are. Any non-zero values can get in the way...
In another post I've explained how adding zero (and subtracting if you
follow the principle through) is useful.
Dividing is a bit of a moot point. There's not much use dividing by zero
if you've got no concept for so called infinities, but then again, the
definition of the inverted (sec, csc, etc.) trig functions require a
divide by zero periodically...
Carl
PS What do you see when you turn on a simple calculator? What's 4+5-9?
David Kastrup
> On Mon, 15 Feb 1999, Mike Deeth wrote:
>
> ] Could someone please give a non-supernatural example where adding,
> ] subtracting, multipliing or dividing by zero is usefull.
>
> Going off on a tangent here... (Incidentally what is sin(2*Pi)?)
>
> 6502 Assembly:
>
> CLC ; Clear carry flag
> LDA 251 ; Load a one byte value from memory
> ADC 252 ; Add another one byte value from memory
> STA 253 ; Store the result in memory
> LDA 0 ; Clear accumulator
> ADC 0 ; Add 0
> STA 254 ; Store in 254
>
> Result? After some runs, 254 contains a one. Now work out (1)Why? and
> (2)What would happen if we couldn't use zero...?
We would write
SBB 255
instead?
Assuming that I remember the name of the opcode correctly.
Would this not have to be something like
LDA #0
or
LDA $0
or whatever that character was, BTW?
--
David Kastrup Phone: +49-234-700-5570
Email: d...@neuroinformatik.ruhr-uni-bochum.de Fax: +49-234-709-4209
Institut für Neuroinformatik, Universitätsstr. 150, 44780 Bochum, Germany
Ian Goddard
>Ian Goddard wrote:
>
>> >Could someone please give a non-supernatural example where adding,
>> >subtracting, multipliing or dividing by zero is usefull.
>>
>> IAN: Why is the sum of all energy in a closed
>> universe supernatural? Why is the sum of money
>> I own, of which your take is 0, supernatural?
>
>Could someone please give a non-supernatural example where ADDING,
>SUBTRACTING, MULTIPLYING or DIVIDING by zero is usefull?
IAN: By asking for a "non-supernatural x"
in reply to my message, you seemed to imply
that what I had stated on a somewhat
different angle was supernatural.
Zero is the value of all real numbers, and
thus all real numbers are the same as zero.
All useful operations are on parts of zero.
I enjoy seeing you challange traditional
views on infinity and was wondering what
you think about the infinity c? I noticed
that some of the discourse seemed to imply
that a problem with infinity is that we
never get there (that may not have been
your point), but the set of all reals
between 0 and 1 is all in "there" now.
Ian Goddard
>Brian, :-)
>
>Where do you keep your 'zero amount' of money?
IAN: He could keep it in an empty safe.
>Do you hold it in your hand or is it out flying around the moon?
IAN: I'm holding zero amount
of money in my hand right now.
>If something, as tangible as money, exists it should be located somewhere.
IAN: And where x is not, zero x is.
>Have you ever misplaced your 'zero amount' of money?
IAN: So much of the universe isn't
money, it's pretty hard to lose.
>Do you ever have to go searching for it?
>When I spend my money ($5.38) I will be broke.
>When I'm broke I will have NO amount of money.
>I will say, "I DON'T HAVE ANY money."
>But I will not say, "I have ZERO amount money".
IAN: Yes, grammatically that would be awkward,
but one could say "I have zero dollars," since
one could also say "I have five dollars." But
"I have zero amount of money" is as awkward
as saying "I have five amount of money."
Zero is a value, not just a useful place holder.
Carl R. White
] Carl R. White wrote:
] > 6502 Assembly:
] >
] > CLC ; Clear carry flag
] > LDA 251 ; Load a one byte value from memory
] > ADC 252 ; Add another one byte value from memory
] > STA 253 ; Store the result in memory
] > LDA 0 ; Clear accumulator
] > ADC 0 ; Add 0
] > STA 254 ; Store in 254
] >
] > Result? After some runs, 254 contains a one. Now work out (1)Why? and
] > (2)What would happen if we couldn't use zero...?
]
] We would write
] SBB 255
] instead?
SBC #255 IIRC :)
] Would this not have to be something like
] LDA #0
] or
] LDA $0
] or whatever that character was, BTW?
True, I was coding on the fly (something I do rather too often these
days). Just a teensy mistake.
However, you made the same mistake with the syntax of the SBC command :)
Then again, any mistakes that "ND" doesn't spot prove his lack of
knowledge...
Carl
PS To Nathan: Where are you when you jump?
Jac Cole
Mike Deeth wrote in message <36C8EEA7...@ashland.baysat.net>...
>Please submit:
>(i) your 'good case' for zero being the first natural number.
>(ii) an example where adding, subtracting, multiplying or dividing by 0 is
>usefull.
>Nathan the Great
>Age 11
>
>
1) The examples that I can think of where adding, multiplying, or dividing
by 0 is a useful thing to do might be things that you have not seen yet.
Also, they all center around 0's "algebraic" properties, i.e. that 0 is the
additive identity of the real numbers and that if a*b=0 then a=0 or b=0.
The fact that 0 is the additive identity, i.e. that x+0=x, is useful since
if I want to rewrite an expression, but I don't want to change its value,
then one of the things I can do is add 0. If you have gone through the
method of completeing a square, that is writing x^2+ax+b as (x+c)^2+d, then
you have seen a case where adding by 0, or in this case 0=a/2 - a/2. The
property that if a*b=0 then a=0 or b=0 allows us to solve equations in a
somewhat effective manner.
2) I have seen many people start their system of natural numbers with 0. I
have also seen many people start their set of natural numbers with 1. In
fact, the only convention that I have seen among mathematicians is that it
is one of those two sets, and they use the one that they need most often, or
that they personally like the most. The set of natural numbers are the
counting numbers, and I might agree that 1 should be the first natural
number. However, if you are looking at these the natural numbers as a set
with no structure, or only the amount of structure that you have talked
about in your posts (i.e. every element has a succesor), then I won't
disagree with you pick as your starting value, as long as you are
consistent.
Now, please do me a favor. A large part of the nature of mathematics rests
in the precise definitions that are placed on objects and ideas. You seem
to be using two different terms, but I have not seen you, Nathan the Great
Age 11, define them in a precise way. So, please define infinite, unlimited
(I believe that was the term you used in a previous post for the size, or
nature, of the set of natural numbers), and what it means for two sets to
have the same size. Also, add any other definitions that you think might be
useful. While it might be clear what the difference between infinite and
unlimited is, it will get everyone that wants to respond to your posts
thinking in the same way that you are thinking. Also, it may be clear to
you what it means for two sets to be the same size, but if you tell the rest
of the world the way you think of two sets having the same size, then at
least we will be on the same page in responding to you. If you can not give
precise definitions to these terms, then you should not be using them in a
mathematical argument, because without a precise definition, they mean
nothing.
Jac Cole
Mike Deeth
Ian Goddard wrote:
> >Brian, :-)
> >Where do you keep your 'zero amount' of money?
> IAN: He could keep it in an empty safe.
... or in his empty head.
> >Do you hold it in your hand or is it out flying around the moon?
>
> IAN: I'm holding zero amount
> of money in my hand right now.
No! We are all holding your zero amount of money in our hands right now.
> >If something, as tangible as money, exists it should be located somewhere.
>
> IAN: And where x is not, zero x is.
Where existence is not, zero existence is.
In other words, zero is non-existent.
Mike Deeth
Ian Goddard wrote:
> >> >Could someone please give a non-supernatural example where adding,
> >> >subtracting, multipliing or dividing by zero is usefull.
> >>
> >> IAN: Why is the sum of all energy in a closed
> >> universe supernatural? Why is the sum of money
> >> I own, of which your take is 0, supernatural?
> >
> >Could someone please give a non-supernatural example where ADDING,
> >SUBTRACTING, MULTIPLYING or DIVIDING by zero is usefull?
>
> IAN: By asking for a "non-supernatural x"
> in reply to my message, you seemed to imply
> that what I had stated on a somewhat
> different angle was supernatural.
>
You didn't add, subtract, multiply or divide by zero. You didn't *perform
an operation* with zero.
Mike Deeth
Carl R. White wrote:
Why do you post ancient 6502 opcodes to sci.math. The 6502's language was
dead before I was born. If you want me to know what you're talking about you
must translate that gobledeguk into BASIC.
ie. x=x+1: if x>255 then x=0
Ian Goddard
>> >If something, as tangible as money, exists it should be located somewhere.
>>
>> IAN: And where x is not, zero x is.
>
>Where existence is not, zero existence is.
>In other words, zero is non-existent.
IAN: Well, yes, and since some things
don't exist everywhere, we use zero to
denote the value of their nonexistence.
Motion for an object at rest is, for
example, nonexistent, and thus = 0.
>> >Where do you keep your 'zero amount' of money?
>>
>> IAN: He could keep it in an empty safe.
>
>... or in his empty head.
IAN: But that's ad hominem.
>> >Do you hold it in your hand or is it out flying around the moon?
>>
>> IAN: I'm holding zero amount
>> of money in my hand right now.
>
>No! We are all holding your zero amount of money in our hands right now.
IAN: Hay, it's a mutal situation. :)