1.9999.... = 2
Jon LeVitre
Jye wrote:
: ok, i'm a sophomore in highschool. I'm pretty well versed in math,
: but my friend still does not believe 1.999999.....=2. Even though the
: teacher proved it to him several times!
Consider where the .999... came from.
If you got it by multiplying .333... by 3, then remember that
the .333... probably started out as 1/3.
Since .333... * 3 = 3 * 1/3 = 1.
jonl
____________________________________________________________________________
Jon LeVitre psu0...@odin.cc.pdx.edu
I support commerce on the Internet. If you send me spam, I'll gladly
proof read it for $250 per item. Sending any UCE to this account
constitutes acceptance of these terms.
Jye
ok, i'm a sophomore in highschool. I'm pretty well versed in math,
but my friend still does not believe 1.999999.....=2. Even though the
10n = 19.999...
n = 1.99999...
10n - n= 18
9n=18
n=2
is there another way to prove it? or is there a way to really explain
it well?
thanx
Jye
NATHANIEL SILVER
Jye wrote in article <34bd7c50....@news2.ibm.net>...
> but my friend still does not believe 1.999999.....=2. Even though the
> teacher proved it to him several times!
> it well?
Method 1. If your friend believes 2 > 1.999999... Ask your friend to give
a decimal expansion of a number strictly in-between them.
Method 2. Long division. We don't need proper quotients. 2 divided by 1.
_1.9...
1)2 .000
-1
______
1.0
-9 etc.
K. E. Pledger
In article <34bd7c50....@news2.ibm.net>, Jye wrote:
> ok, i'm a sophomore in highschool. I'm pretty well versed in math,
> but my friend still does not believe 1.999999.....=2. Even though the
> teacher proved it to him several times! he used the
>
> 10n = 19.999...
> n = 1.99999...
>
> 10n - n= 18
>
> 9n=18
> n=2
>
> is there another way to prove it? or is there a way to really explain
> it well?
>
Well, you could try this line of questioning, which makes good
mathematical sense.
You: Do you think 1.9999... is less than 2?
Friend: Yes.
You: O.K. _How_much_ less?
That should at least give something to think about.
Ken Pledger.
Unknown
Jye writes: > ok, i'm a sophomore in highschool. I'm pretty well versed in math,
> but my friend still does not believe 1.999999.....=2. Even though the
> teacher proved it to him several times! he used the
>
> 10n = 19.999...
> n = 1.99999...
>
> 10n - n= 18
>
> 9n=18
> n=2
>
> is there another way to prove it? or is there a way to really explain
> it well?
>
>
> Jye
We can see the problem more clearly if we consider the sum of a geometric progression:
1 + 9 x 1->ooSUMn(1/10^n) = 1.999(rec) = 1 + 9(1/10 + 1/100 + 1/1000 + ...)
The sum to infinity of (1/10^n) = 1->ooSUMn(1/10 - 1/10^n).10/(10-1)= 0.111(rec)
The problem arises in the term 1/10^n as n->oo
Mathematicians assume that this term is zero at infinity, although
for all values of n short of absolute infinity, it is an infinitessimal
amount which we can call iota, which the smallest conceivable quantity.
If we add this infinitessimal back on to the recurring decimal
we get the required result eg
1.999(rec) + iota = 2
Note that the value of iota is relative the order of infinity
at which we teminate the operation. Also, in the binary system
(say) the value of iota will be 2^w rather than 10^w in the decimal
system, where w is some infinite cardinal.
Jeff Evans
Jye wrote in message <34bd7c50....@news2.ibm.net>...
>ok, i'm a sophomore in highschool. I'm pretty well versed in math,
>but my friend still does not believe 1.999999.....=2. Even though the
>teacher proved it to him several times! he used the
Jye,
It's easier to think about this by showing that .99999.... = 1 (it will be
the same for all integer multiples)
First, the value for 1/9 = 0.1111....
And, 2/9 = 0.22222....
And, 3/9 = 0.333333....
...
8/9 = 0.8888888....
9/9 = 0.9999999.....
(If you notice the pattern, it would be logical to assume that 9/9 is
0.99999...)
However, you also have to consider this: Divide any number by itself - the
result is always one! For example:
6/6 = 1, 137/137 = 1, 1234/1234 = 1, etc.
This is due to the fact that when you multiply any number by one, the result
is always itself:
n*1 = n
(n*1)/n = n/n
1 = n/n
So, from the pattern of (n/9)'s, and from, 1=n/n, 0.99999... must equal 1.
Jeff.
Pat Fleckenstein
In article <34bd7c50....@news2.ibm.net> Jye writes:
>ok, i'm a sophomore in highschool. I'm pretty well versed in math,
>but my friend still does not believe 1.999999.....=2. Even though the
>teacher proved it to him several times! he used the
Consider this:
Try adding some positive number 'd' to 1.99999...
If d > 0, then 1.9999 + d > 2. If d is
0.000....0010000... where this '1' occurs at some
spot, then all of the 9's after this spot are left
unchanged and all of the 9's before this spot become
zero because of repeated carrying. The only hope is
to get that 0.000...0010000... to have the '1' past
the end of all of the nines. Have fun.
So, if d > 0, then 1.9999... + d > 2.
Similarly, then 2 - d < 1.9999...
From this, it is obvious that 1.9999... and 2 do
not differ by any amount since if they differed by
d, then 1.9999...+d would equal 2, but the above
showed that any positive 'd' would make 1.9999...+d
exceed 2.
alter,
pat
--
Tell a man there are 300 billion stars in the universe and he'll
believe you. Tell him a bench has wet paint on it and he'll have to
touch to be sure.
Jin-Soo KIM
Among above tries, the best is yours , I guess.
You hit the point!!!
Jye
On 15 Jan 1998 04:31:44 GMT, Verdigris <> wrote:
>Jye writes: > ok, i'm a sophomore in highschool. I'm pretty well versed in math,
>> but my friend still does not believe 1.999999.....=2. Even though the
>> teacher proved it to him several times! he used the
>>
>> 10n = 19.999...
>> n = 1.99999...
>>
>> 10n - n= 18
>>
>> 9n=18
>> n=2
>>
>> is there another way to prove it? or is there a way to really explain
>> it well?
>>
>> thanx
>>
>> Jye
>
>
>We can see the problem more clearly if we consider the sum of a geometric progression:
>
>1 + 9 x 1->ooSUMn(1/10^n) = 1.999(rec) = 1 + 9(1/10 + 1/100 + 1/1000 + ...)
>
>The sum to infinity of (1/10^n) = 1->ooSUMn(1/10 - 1/10^n).10/(10-1)= 0.111(rec)
>
>The problem arises in the term 1/10^n as n->oo
>
>Mathematicians assume that this term is zero at infinity, although
>for all values of n short of absolute infinity, it is an infinitessimal
>amount which we can call iota, which the smallest conceivable quantity.
>
>If we add this infinitessimal back on to the recurring decimal
>we get the required result eg
>
>1.999(rec) + iota = 2
>
>Note that the value of iota is relative the order of infinity
>at which we teminate the operation. Also, in the binary system
>(say) the value of iota will be 2^w rather than 10^w in the decimal
>system, where w is some infinite cardinal.
>
>
>
>
>
>
>
>
>
>
So, are you saying there is actually a difference.. but a difference
that is iota? Which would mean there really is no difference.. or ..
what are you trying to say?
I'm only in 10th grade.. :) , i can somewhat grasp what you're saying
, but not all.
please explain a little better please
thanx
Jye
Unknown
> So, are you saying there is actually a difference.. but a difference
> that is iota? Which would mean there really is no difference.. or ..
> what are you trying to say?
>
> I'm only in 10th grade.. :) , i can somewhat grasp what you're saying
> , but not all.
>
> please explain a little better please
>
> thanx
>
> Jye
We can add up the series 1/2 + 1/4 + 1/8 + 1/16 ... to a certain number of terms eg:
No of terms
1 1/2 = 1/2
2 1/2 + 1/4 = 3/4
3 1/2 + 1/4 + 1/8 = 7/8
4 1/2 + 1/4 + 1/8 + 1/16 = 15/16
Notice the pattern of the result is (2^n-1)/2^n
So, if we add up the geometric progression to 10 terms we get
(2^10-1)/ 2^n = 1023/1024 = 1 - 1/1024 = 0.999023438...
The same thing applies when we add up the progression:
9/10 + 9/100 + 9/1000 + 9/10000 ... = 0.9999 ...
When n becomes very large, the sum gets very close to 1 but there
is always a very small difference left over. We can call this
diference d.
Now, when n becomes infinite, that is to say greater than any finite
number, the difference d becomes 1 - 1/(an infinite number).
Mathematicians usually assume that the remainder is 1/oo = 0,
although they do not allow the term 1/oo to be stated as such
because there is no such number as oo (ie oo means absolute infinity).
Remember that we are not allowed to divide by zero in algebraic
expressions. Dividing by absolute infinity is also not allowed
because to say 1/oo = 0 leads us to say that 1/0 = oo.
One way out of this problem is to recognise that a small number
is always left over when we add up a geometric progression even if we
use an infinite sum. This residual number is not a specific number
but a class of numbers with infinite denominators. Such numbers are
called infinitessimals.
Those who say 1 = 0.999 (recurring) are allowing the validity of
dividing by absolute infinity. If we think that this is not good
mathematics, then we must recognise that 1 > 0.999 (recurring).
None of this matters for ordinary mathematics at high school level
and is only of interest to those interested in the foundations of
mathematics.
Given this, you can safely accept that 1= 0.999 (rec). However, you
should continue challenge anything like this that you find contradictory.
NATHANIEL SILVER
Jye wrote in article <34beba38....@news2.ibm.net>...
> On 15 Jan 1998 04:31:44 GMT, Verdigris <> wrote:
>
in >>math, but my friend still does not believe 1.999999.....= 2.
>>We can see the problem more clearly if we consider the sum of a geometric
>>progression: >>1 + 9 x 1->oo
>>SUMn(1/10^n) = 1.999(rec) = 1 + 9(1/10 + 1/100 + 1/1000 + ...)
>>The problem arises in the term 1/10^n as n->oo
A partial study list to help resolve problems:
Sequences, limits of sequences, least upperbound property of real numbers,
convergence of an infinite series defined as a limit of its sequence of
partial
sums, geometric series, decimal expansions as convergent geometric series.
NATHANIEL SILVER
A redone partial study list to help resolve problems:
Sequences, limits of sequences, least upperbound property of real numbers;
convergence of an infinite series; the sum of an infinite series,
defined as a limit of its sequence of partial sums; geometric series,
GSimp95605
When talking about the set of extended reals (where you throw in the
infinitesimals, numbers bigger than 0 but smaller than any positive real) do
you still define numbers as equivalence classes of Cauchy sequences? Is it the
same equivalence relation ({a}~{b} if the sequence {a-b} converges to 0)? If
not, then wouldn't it be possible to say that 1.9999 and 2.0 represent two
different sequences whose differences converge to some small infinitesimal?
-Gilberto
GSimp95605
Subject: Re: 1.9999.... = 2
From: Verdigris <>
Those who say 1 = 0.999 (recurring) are allowing the validity of dividing by
absolute infinity.
G: Not necessarily. That was only because your particular "proof" divided by
infinity. Another way to look at it is to say that 1=.99999.... because the
sequences {1,1,1,1,...} and {.9, .99, .999, .9999, .99999, .999999, .9999999,
...} are equivalent (according to a particular equivalence relation). Their
difference {1-.9, 1-.99, 1-.999, ...} is a Cauchy sequence which converges to
0.
G: Now, if you wanted to, you might be able to come up with a different
equivalence relation on Cauchy sequences.
Maybe two sequences are equivalent if, for some positive N, all the terms are
equal after the Nth term. But then you would probably get something different
from the set of reals.
Peace
Gilberto
NATHANIEL SILVER
GSimp95605 <gsimp...@aol.com> wrote in article
<19980116123...@ladder02.news.aol.com>...
To make infinitesimals rigorous, one can formally construct the reals as
before. Then formally construct the hyperreals as equivalence classes of
Cauchy sequences of reals. Abraham Robinson found a way to get canonical
representatives of each equivalence class by non-constructive methods.
Dave Seaman
In article <69k3fg$b...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>The problem arises in the term 1/10^n as n->oo
>
>for all values of n short of absolute infinity, it is an infinitessimal
>amount which we can call iota, which the smallest conceivable quantity.
No. For any n short of infinity, the value of 1/10^n is finite, not
infinitesimal. Also, there is a difference between assuming "that this
term is zero at infinity" and noticing that the limit is zero as n
approaches infinity.
By the way, even if you do insist on dragging in infinitesimals (which
are not needed here), there is no such thing as "the smallest
conceivable quantity". If iota is a positive infinitesimal, meaning
that 0 < iota < p/q for all positive integers p and q, then iota*iota
is a still smaller infinitesimal, and so on.
--
Dave Seaman dse...@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
Dave Seaman
In article <69mo93$1...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>When n becomes very large, the sum gets very close to 1 but there
>is always a very small difference left over. We can call this
>diference d.
>
>Now, when n becomes infinite, that is to say greater than any finite
>number, the difference d becomes 1 - 1/(an infinite number).
>
>Mathematicians usually assume that the remainder is 1/oo = 0,
>although they do not allow the term 1/oo to be stated as such
>because there is no such number as oo (ie oo means absolute infinity).
Quite the contrary. If x is an infinitely large element in an ordered
field F (meaning x > y for every rational y in F), then 1/x is
definitely not 0, but an infinitesimal.
>Remember that we are not allowed to divide by zero in algebraic
>expressions. Dividing by absolute infinity is also not allowed
>because to say 1/oo = 0 leads us to say that 1/0 = oo.
There is no such thing as "absolute infinity" in an ordered field. If
x is an infinitely large element in F, then x+1 and x^2 are still
larger. Division by zero is still undefined in any field, even a
non-Archimedean one.
>One way out of this problem is to recognise that a small number
>is always left over when we add up a geometric progression even if we
>use an infinite sum. This residual number is not a specific number
>but a class of numbers with infinite denominators. Such numbers are
>called infinitessimals.
If you allow infinitesimals in your field, then they are just as
specific as any other numbers in that field. Even an Archimedean field
may by composed of equivalence classes. The standard real numbers are
frequently defined as equivalence classes of Cauchy sequences of
rationals; there are no infinitesimals involved in this definition.
>Those who say 1 = 0.999 (recurring) are allowing the validity of
>dividing by absolute infinity. If we think that this is not good
>mathematics, then we must recognise that 1 > 0.999 (recurring).
Nonsense. The reason that 1 = 0.999 is that the Cauchy sequences
1, 1, 1, 1, ...
and
9/10, 99/100, 999/1000, ...
belong to the same equivalence class. No division by infinity and no
infinitesimals are involved. All that matters is that the relation
on Cauchy sequences of rationals given by
{p_k} ~ {q_k} iff lim(k->oo) (p_k - q_k) = 0
is reflexive, symmetric, and transitive, and therefore ~ is an
equivalence relation on the Cauchy sequences of rationals. The
resulting equivalence classes are the real numbers.
Jye
On 16 Jan 1998 04:38:59 GMT, Verdigris <> wrote:
>
>> So, are you saying there is actually a difference.. but a difference
>> that is iota? Which would mean there really is no difference.. or ..
>> what are you trying to say?
>>
>> I'm only in 10th grade.. :) , i can somewhat grasp what you're saying
>> , but not all.
>>
>> please explain a little better please
>>
>> thanx
>>
>> Jye
>
>We can add up the series 1/2 + 1/4 + 1/8 + 1/16 ... to a certain number of terms eg:
>
>No of terms
>
>1 1/2 = 1/2
>
>2 1/2 + 1/4 = 3/4
>
>3 1/2 + 1/4 + 1/8 = 7/8
>
>4 1/2 + 1/4 + 1/8 + 1/16 = 15/16
>
>Notice the pattern of the result is (2^n-1)/2^n
>So, if we add up the geometric progression to 10 terms we get
>(2^10-1)/ 2^n = 1023/1024 = 1 - 1/1024 = 0.999023438...
>
>The same thing applies when we add up the progression:
>
>9/10 + 9/100 + 9/1000 + 9/10000 ... = 0.9999 ...
>
>When n becomes very large, the sum gets very close to 1 but there
>is always a very small difference left over. We can call this
>diference d.
>
>Now, when n becomes infinite, that is to say greater than any finite
>number, the difference d becomes 1 - 1/(an infinite number).
>
>Mathematicians usually assume that the remainder is 1/oo = 0,
>although they do not allow the term 1/oo to be stated as such
>because there is no such number as oo (ie oo means absolute infinity).
>
>Remember that we are not allowed to divide by zero in algebraic
>expressions. Dividing by absolute infinity is also not allowed
>because to say 1/oo = 0 leads us to say that 1/0 = oo.
>
>One way out of this problem is to recognise that a small number
>is always left over when we add up a geometric progression even if we
>use an infinite sum. This residual number is not a specific number
>but a class of numbers with infinite denominators. Such numbers are
>called infinitessimals.
>
>Those who say 1 = 0.999 (recurring) are allowing the validity of
>dividing by absolute infinity. If we think that this is not good
>mathematics, then we must recognise that 1 > 0.999 (recurring).
>
>None of this matters for ordinary mathematics at high school level
>and is only of interest to those interested in the foundations of
>mathematics.
>
>Given this, you can safely accept that 1= 0.999 (rec). However, you
>should continue challenge anything like this that you find contradictory.
>
>
>
>
>
>
>
ah, but isn't it too stated that these infintesimels cannot exist. In
accordance to the Concept of Limits. Isn't it base knowledge in many
mathamatics that a smallest nor a biggest number exists? As proved
here:
.9(rec) + d = 1
.9(rec) > .9(rec) + d/2 > 1
you can always divide by two.
Jye
NATHANIEL SILVER
Dave Seaman <a...@seaman.cc.purdue.edu> wrote in article
<69nq2o$m...@seaman.cc.purdue.edu>...
> In article <69k3fg$b...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
> >The problem arises in the term 1/10^n as n->oo
> By the way, even if you do insist on dragging in infinitesimals (which
> are not needed here), there is no such thing as "the smallest
> conceivable quantity". If iota is a positive infinitesimal, meaning
> that 0 < iota < p/q for all positive integers p and q, then iota*iota
> is a still smaller infinitesimal, and so on.
Friday P.M.: back from the "trenches" - teaching high school math in
Brooklyn
To continue my thought. In the hyperreal numbers, taking a standard part to
get the sum of a series is unnecessary, as critics of infinitesimals would
be quick to point out. The reason is that sequences are discrete entities
as opposed to continuous. So, there's not much for the hyperreals to do.
ari...@mps.org
[snip]
>
> Now, when n becomes infinite, that is to say greater than any finite
> number, the difference d becomes 1 - 1/(an infinite number).
No, No, No, n doesn't become infinite---or at least there's no reason to
say that it does. As n becomes larger, the difference between the finite
sum and 1 becomes closer to 0, and you can get it as close to 0 as you
want by making n big enough. This is the _definition_ of an infinite
series summing to a finite number.
>
> Mathematicians usually assume that the remainder is 1/oo = 0,
> although they do not allow the term 1/oo to be stated as such
> because there is no such number as oo (ie oo means absolute infinity).
It isn't just that we don't allow it to be stated as such, it's that the
details involved in stating it are exceedingly touchy if you want to
avoid contradictions. That's why it's better just not to get into it,
and use the above definition for the sum of the series.
>
> Remember that we are not allowed to divide by zero in algebraic
> expressions. Dividing by absolute infinity is also not allowed
> because to say 1/oo = 0 leads us to say that 1/0 = oo.
>
> One way out of this problem is to recognise that a small number
> is always left over when we add up a geometric progression even if we
> use an infinite sum. This residual number is not a specific number
> but a class of numbers with infinite denominators. Such numbers are
> called infinitessimals.
This isn't a way out of the problem, it's a whole different line of
thought. The way out of the problem is to not even bring up the
1/infinity thing at all-- -just use the limit definition of the sum of
the infinite series.
>
> Those who say 1 = 0.999 (recurring) are allowing the validity of
> dividing by absolute infinity. If we think that this is not good
> mathematics, then we must recognise that 1 > 0.999 (recurring).
This is just surreal. You don't even need to bring infinite series into
the picture at all to see that 1=0.999... This has nothing to do with
infinitesimals. Even in nonstandard analysis this is true, since the
proof does not rely on the archimedean property of the real numbers, and
these are evidently both standard numbers (i.e. without infinitesimal
parts). Dividing by absolute infinity is _never_ good mathematics, and
yet 1=0.999... The issues are not related.
[snip]
Adrian
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
Horst Kraemer
On 16 Jan 1998 04:38:59 GMT, Verdigris <> wrote:
>
>The same thing applies when we add up the progression:
>
>9/10 + 9/100 + 9/1000 + 9/10000 ... = 0.9999 ...
>
>When n becomes very large, the sum gets very close to 1 but there
>is always a very small difference left over. We can call this
>diference d.
>
>Now, when n becomes infinite, that is to say greater than any finite
>number, the difference d becomes 1 - 1/(an infinite number).
In standard mathematics n does _not_ become infinite.
The symbol 0.999... denotes the 'limit' of the sequence
( 0.9 , 0.99 , 0.999 , 0.9999 , ... )
A number a is called the limit of the sequence
( a1,a2,a3,.... )
if for every arbitrary small positive number eps there is a "tail" of
this sequence
( a_n, a_n+1, ... , a_n+2 , .... )
such that all your d's : d = |a-a_i| will be smaller than eps for
every element of the tail.
In other words: The value of the symbol 0.999... is not something
which 'moves' and approaches 1, but it is _by_ _definition_ the number
which the sequence
( 0.9, 0.99, 0.999 , ... )
- the "something" - is approaching, i.e. it _is_ 1.
Regards
Horst
NATHANIEL SILVER
Royce <newa...@Worldnet.att.net> wrote in article
<69rbga$1...@bgtnsc01.worldnet.att.net>...
> However, whether or not 1.999999 = 2 is irrelevant - what is relevant
> is what happens when you use that amount in a calculation. Then it
> becomes clear that 1.999999 will not equal 2.
> For example 1.99999 x 1.99999 = 3.999996
> Or 1.99999 x 8 = 15.99992
> 1.99999 means nothing unless it is used in a math calculation.
>
It is very relevant to know, a priori, that 2 = 1.999...
When approximate values are used 1.999...9 in a
completed calculation almost cries out that it
could be 2, especially, for example, in floating-point
arithmetic or when measuring with a ruler.
Darrell Ryan
Royce wrote:
>
> However, whether or not 1.999999 = 2 is irrelevant - what is relevant
> is what happens when you use that amount in a calculation. Then it
> becomes clear that 1.999999 will not equal 2.
>
> For example 1.99999 x 1.99999 = 3.999996
Why are you rounding off? Do you understand that what we are talking
about is 1.999...(the 9's never stop)?
> 1.99999 means nothing unless it is used in a math calculation.
>
> Another thought - there is no such thing as 1 inch. It is merely a
> measurement which needs to be used in the context of a calculation.
...and what does this have to do with the price of tea in China?
--
Darrell Ryan http://www.cafes.net/darrell/
Darrell Ryan
Royce wrote:
>
> On 17 Jan 1998 23:39:21 GMT, "NATHANIEL SILVER"
> <mat...@worldnet.att.net> wrote:
>
> >Royce <newa...@Worldnet.att.net> wrote in article
> ><69rbga$1...@bgtnsc01.worldnet.att.net>...
> >> is what happens when you use that amount in a calculation. Then it
> >> becomes clear that 1.999999 will not equal 2.
> >> For example 1.99999 x 1.99999 = 3.999996
>
>
> conjunction with another value to equal a third value. But
> you are using it singly, as a result ( following the = sign)
> and then merely rounding it up to the appropriate whole
> number..
> Royce
Why do you say that? does this also mean we can't use any other
number, say 2 or pi or whatever, unless we attach an equal sign to it?
Using series is NOT an approximation. On the contrary, you seem to
want to approximate 1.99... by considering only a few decimal places.
The series .9+.09+.009+... converges to 1. In other words, the sum of
ALL the terms is 1. This is by _definition_.
Darrell Ryan
Royce wrote:
> >>
> >> I should have said "1.99999 means nothing unless it is used in
> >>
> >> conjunction with another value to equal a third value. But
> >> you are using it singly, as a result ( following the = sign)
> >> and then merely rounding it up to the appropriate whole
> >> number..
> >> Royce
> >
> >Why do you say that? does this also mean we can't use any other
> >number, say 2 or pi or whatever, unless we attach an equal sign to it?
> >
> not to use with other numbers.
Ah, but what IS a number? hint--it doesn't physically exist no more
than your 1 inch does. But I can think of all kinds of good things to
use numbers for. What good is "1 inch" if not used with other
"inches?"
>
> >Using series is NOT an approximation. On the contrary, you seem to
> >want to approximate 1.99... by considering only a few decimal places.
> >The series .9+.09+.009+... converges to 1. In other words, the sum of
> >ALL the terms is 1. This is by _definition_.
>
> By your reasoning, if .9 + .09 + .009 = 1, then pi = 3
Again, you are rounding off. Do you understand what the "..." tagged
on the end means? (you conveniently left this out).
Let me ask this: What is pi squared? Assume let's do this:
pi = 3.14
pi^2 = (3.14)(3.14) = 9.8596
So, pi^2 is exactly 9.8596. Can you see what's wrong with this
reasoning? If you see the error, notice that you are making the same
error if you think .9+.09+.009+... < 1. Comprehend "..."
I fear you don't understand what an infinite series is. You
conveniently chop off wherever you want to argue your case. Don't
"chop" anything off. 1.999 is less than 2, but 1.999... is equal to
2. Likewise.9+.09+.009 is less than 1, but .9+.09+.009+...is equal to
1.
For example. consider 1/3=.333...=.3+.03+.003+...
The fraction 1/3 is exactly equal to .333... ("..." means keep going,
don't stop). If you doubt that, perform the long division 1 divided
by 3 and see how many decimal places you have. If we decide to "chop
off" any of the 3's, then what we have is LESS than 1/3. Right? But
the notation .33... ,which is equivalent to .33 with a bar over it,
means that the 3's NEVER stop and in that case it is exactly EQUAL to
1/3. 3 times 1/3 equals 1 and .33... is another name for 1/3.
multiply .33... by 3 and you get .999..., right? Remember, we already
know 3(1/3)=1 and 3(1/3)=3(.33...)=.999... Therefore, .999...=1. See
the sci.math FAQ or go to http://www.cafes.net/darrell/99baris1.htm
The decimal .333... is equivalent to the series:
.3 + .03 + .003 + ... (the "..." means keep repeating the pattern)
If you believe this is exactly 1/3, then you must necessarily believe
.9+.09+.009+... is equal to 1. Just multiply each term in the series
by 3, remembering that (1/3)3=1.
Have you been exposed to the definition of a convergent series as the
limit of the partial sums?
Tony Thomas
Dave Seaman wrote:
>
> In article <69k3fg$b...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
> >The problem arises in the term 1/10^n as n->oo
> >
> >for all values of n short of absolute infinity, it is an infinitessimal
> >amount which we can call iota, which the smallest conceivable quantity.
>
> No. For any n short of infinity, the value of 1/10^n is finite, not
> infinitesimal. Also, there is a difference between assuming "that this
> term is zero at infinity" and noticing that the limit is zero as n
> approaches infinity.
*** Yes, when n is finite 1 - .9999... is finite. It follows that
.9999...
for all finite values of n is a proper fraction such that .9999... < 1.
This means that the statement 1 = .9999... is false for all finite
values of n.
If we consider a non-finite value w (say), the term 1/10^w may be
regarded as an infinitessimal. In this case 1/10^w > 0. Also, .9999...
remains fractional
for infinite numbers w1 < w2 < w3 < ... etc.
>
> By the way, even if you do insist on dragging in infinitesimals (which
> are not needed here), there is no such thing as "the smallest
> conceivable quantity". If iota is a positive infinitesimal, meaning
> that 0 < iota < p/q for all positive integers p and q, then iota*iota
> is a still smaller infinitesimal, and so on.
*** Try thinking of .9999... as the greatest definable fraction
constructable in
the decimal system of notation, so that 1 > .9999... follows from this
definition.
Then iota = 1 - .9999... being the least constructable decimal in the
system.
Note that Iota^2 is not wff in such a schema.
Regards
Verdigris
Tony Thomas
> >Mathematicians usually assume that the remainder is 1/oo = 0,
> >although they do not allow the term 1/oo to be stated as such
> >because there is no such number as oo (ie oo means absolute infinity).
>
> field F (meaning x > y for every rational y in F), then 1/x is
> definitely not 0, but an infinitesimal.
*** Yes, this what I said.
> >Remember that we are not allowed to divide by zero in algebraic
> >expressions. Dividing by absolute infinity is also not allowed
> >because to say 1/oo = 0 leads us to say that 1/0 = oo.
>
> There is no such thing as "absolute infinity" in an ordered field. If
> x is an infinitely large element in F, then x+1 and x^2 are still
> larger. Division by zero is still undefined in any field, even a
> non-Archimedean one.
*** I also said above "there is no such number as oo.
Absolute infinity does not arise if we represent the fractions as
an ordered set of infinitessimals of the form r/a^w (MODULO a)
> >One way out of this problem is to recognise that a small number
> >is always left over when we add up a geometric progression even if we
> >use an infinite sum. This residual number is not a specific number
> >but a class of numbers with infinite denominators. Such numbers are
> >called infinitessimals.
*** Either Lt n-> oo GP involves infitessimals or it involves dividing
by oo.
You can take your choice.
Tony Thomas
> > Now, when n becomes infinite, that is to say greater than any finite
> > number, the difference d becomes 1 - 1/(an infinite number).
Adrian replied:
> No, No, No, n doesn't become infinite---or at least there's no reason to
> say that it does. As n becomes larger, the difference between the finite
> sum and 1 becomes closer to 0, and you can get it as close to 0 as you
> want by making n big enough. This is the _definition_ of an infinite
> series summing to a finite number.
*** If n remains finite the 1 > .9999...
There can be no limit of a GP without considering what is meant by
the term
"as n tends to infinity the limit tends to L.
If we say 1 = L the we have agreed that n has reached infinity,
whatever
that might mean.
> > Mathematicians usually assume that the remainder is 1/oo = 0,
> > although they do not allow the term 1/oo to be stated as such
> > because there is no such number as oo (ie oo means absolute infinity).
>
> It isn't just that we don't allow it to be stated as such, it's that the
> details involved in stating it are exceedingly touchy if you want to
> avoid contradictions. That's why it's better just not to get into it,
> and use the above definition for the sum of the series.
*** I agree. What I am sayinig is that, in using the symbol .999... in
the
statement 1 = .999... we have assumed that Lt n-> oo (1/10^n) =
0.
I would prefer to say Lt n-> w (1/10^w) = L(w), where L(w) > 0.
W being some infinite number.
> > One way out of this problem is to recognise that a small number
> > is always left over when we add up a geometric progression even if we
> > use an infinite sum. This residual number is not a specific number
> > but a class of numbers with infinite denominators. Such numbers are
> > called infinitessimals.
>
> This isn't a way out of the problem, it's a whole different line of
> thought. The way out of the problem is to not even bring up the
> 1/infinity thing at all-- -just use the limit definition of the sum of
> the infinite series.
*** Ignoring the problem will not make it go away.
The 1/infinity thing is well established, the problem arises because
the term "infinity" is treated as if it were a singular entity
in the phrase "as n tends
to infinity". If we take this to mean that n remains finite, then 1
> .9999..
If we allow that n does not remain finite, then we need to decide
whether
1/a^n > 0 when (a < 1)and n takes up infinite values.
Justin or Stephanie Struby
Subtract one from both sides and the topic reduces to a problem already
discussed
in this newsgroup ad nauseum ;)
Dave Seaman
In article <34C27A...@uq.net.au>, Tony Thomas <EATh...@uq.net.au> wrote:
>> >Mathematicians usually assume that the remainder is 1/oo = 0,
>> >although they do not allow the term 1/oo to be stated as such
>> >because there is no such number as oo (ie oo means absolute infinity).
>> field F (meaning x > y for every rational y in F), then 1/x is
>> definitely not 0, but an infinitesimal.
>*** Yes, this what I said.
You contradicted yourself in the process. First you falsely claimed
that "mathematicians usually assume the remaineder is 1/oo = 0", and
then you denied it. Which is it? Would you care to name a few
mathematicians who make such an assumption?
>> >Remember that we are not allowed to divide by zero in algebraic
>> >expressions. Dividing by absolute infinity is also not allowed
>> >because to say 1/oo = 0 leads us to say that 1/0 = oo.
>> There is no such thing as "absolute infinity" in an ordered field. If
>> x is an infinitely large element in F, then x+1 and x^2 are still
>> larger. Division by zero is still undefined in any field, even a
>> non-Archimedean one.
>*** I also said above "there is no such number as oo.
> Absolute infinity does not arise if we represent the fractions as
> an ordered set of infinitessimals of the form r/a^w (MODULO a)
You say there is no such number as oo. The correct statement is that
oo is not a member of the standard real numbers. This doesn't rule out
the possibility that it could be a member of some non-Archimedean
field. If you are so sure oo is not a number, then why do you keep
trying to do arithmetic with it?
Besides, your reasoning is incorrect. You assumed a false statement
(1/oo = 0) in order to derive a contradiction, and then you claimed
that this disproved the existence of oo as a number. In fact, what it
disproves is your assumption that 1/oo = 0, since it would actually be
an infinitesimal.
>
>> >One way out of this problem is to recognise that a small number
>> >is always left over when we add up a geometric progression even if we
>> >use an infinite sum. This residual number is not a specific number
>> >but a class of numbers with infinite denominators. Such numbers are
>> >called infinitessimals.
>*** Either Lt n-> oo GP involves infitessimals or it involves dividing
>by oo.
>You can take your choice.
I choose neither. The standard definition for the limit of a sequence
mentions neither infinitesimals nor dividing by oo.
Dave Seaman
In article <34C280...@uq.net.au>, Tony Thomas <EATh...@uq.net.au> wrote:
[Re: the imagined "problem" of 1/oo"]
>*** Ignoring the problem will not make it go away.
Pretending a definition says something it doesn't will not make a problem
appear where it doesn't exist.
> The 1/infinity thing is well established, the problem arises because
>the term "infinity" is treated as if it were a singular entity
>in the phrase "as n tends
> to infinity".
Nonsense. The definition says that lim(n->oo) a_n = L if
for every epsilon > 0 there exists N > 0 such that
|a_n - L| < epsilon for every n > N. (*)
Show me where in the definition (*) I have treated the term "infinity"
as if it were a "singular entity", in your words. In fact, the
definition (*) doesn't mention infinity at all.
>If we take this to mean that n remains finite, then 1 > .9999..
Wrong. Each value of n that arises is indeed finite, but it is still a
fact that the real number 1 is precisely equal to .999.... More
specifically, let C be the ring of Cauchy sequences of rationals, and
let I be the ideal consisting of sequences that converge to zero. Then
the quotient ring C/I is, by definition the field of real numbers.
It's a field because the ideal I is maximal in C. The sequences
1, 1, 1, 1, ...
and
9/10, 99/100, 999/1000, ...
are two elements of C whose difference belongs to I, and therefore (by
the definition of a quotient ring) they map to the same element of
C/I = R.
> If we allow that n does not remain finite, then we need to decide
>whether
> 1/a^n > 0 when (a < 1)and n takes up infinite values.
The question does not arise, because n is always finite. Look again at
the construction of R = C/I just above and notice that the only place
where n is involved at all is as the implicit index for a Cauchy
sequence. There are infinitely many possible values for n, but each
value is finite.
Michael Greaney
Jye wrote in article <34bd7c50....@news2.ibm.net>...
> ok, i'm a sophomore in highschool. I'm pretty well versed in math,
> but my friend still does not believe 1.999999.....=2. Even though the
> teacher proved it to him several times! he used the
>
> 10n = 19.999...
> n = 1.99999...
>
> 10n - n= 18
>
> 9n=18
> n=2
>
> is there another way to prove it? or is there a way to really explain
> it well?
>
> thanx
>
> Jye
>
Express 1.999... as 1 + 0.999... The problem then becomes one of showing
that 0.999... = 1.
To show this, point out that 1/9 as a decimal is 0.111... Every multiple of
1/9, then, is a multiple of 0.111... For example, 3/9, ie 1/3, is 3 x
0.111... = 0.333... It follows then that 9/9 is 9 x 0.111... = 0.999...;
but 9/9 = 1 therefore 0.999... = 1.
Michael Greaney.
Jostein Herredsvela
I've always thought it was impossible to prove this. The continuum axiom
of calculus should tell us something about this. An axiom can't be
proved, though. But the proofs you provided are quite convincing.
So should we rely on the axiom, or should we prove it??
Jostein H.
Alex Brown
Isn't this just a little simplerr than everyone's taking it for? I mean,
1.99999...9 <repeating infinitely for those who are using approximations>
can be expressed as 1 + .99999...9 <repeating>.
A repeating decimal can be represented as a fraction as such:
_
.4 <the bar also designates repeating> = 4/9
__
.45 = 45/99
a repeating decimal is also represented as the repeating portion divided
by 10^<number of digits in the repeating portion> - 1.
Therefore:
_
.9 = 9/9 = 1, and 1.999999....9 = 1 + 9/9 = 2.
Then again, my logic might be flawed :).
--
Alex Brown Internet: amb...@spd.dsccc.com
**** The opinions expressed are not those of DSC Communications, Inc ****
RM Mentock
Royce wrote:
>
> However, whether or not 1.999999 = 2 is irrelevant - what is relevant
> is what happens when you use that amount in a calculation. Then it
> becomes clear that 1.999999 will not equal 2.
>
> For example 1.99999 x 1.99999 = 3.999996
I haven't seen a significant digits response to this, on my
newsreader yet.
When you are not being exact (i.e., 2 = 1.9999999999999999999 *etc.*),
then you should pay attention to the significant digits. Each of your
two factors has only 6 significant digits, so your answer should also
have only 6. So, 1.99999 x 1.99999 = 3.99999
--
D.
men...@mindspring.com
http://mentock.home.mindspring.com/index.htm
Dave Seaman
In article <34C37F...@fi.uib.no>,
Jostein Herredsvela <jost...@fi.uib.no> wrote:
>> Express 1.999... as 1 + 0.999... The problem then becomes one of showing
>> that 0.999... = 1.
>I've always thought it was impossible to prove this. The continuum axiom
>of calculus should tell us something about this. An axiom can't be
>proved, though. But the proofs you provided are quite convincing.
>So should we rely on the axiom, or should we prove it??
It depends on what you take as your starting point. For example, you
could define a square as a quadrilateral having four equal sides and
four equal angles. Then it becomes a theorem that all the angles are
right angles. On the other hand, you could adopt the definition that
all the angles of a square must be right angles, and the sides must be
equal. Then it becomes a theorem that a square has exactly four angles
and four sides. Does this mean you have proved an axiom?
The real numbers may be defined as an ordered field satisfying the
least upper bound axiom. Then you have do do some work to show that
such a field actually exists. On the other hand, you can define the
real numbers as the completion of the rationals with respect to the
distance metric. Then it becomes a theorem that the reals are an
ordered field and the least upper bound property is satisfied.
Unknown
a...@seaman.cc.purdue.edu (Dave Seaman) writes: > In article <34C280...@uq.net.au>, Tony Thomas <EATh...@uq.net.au> wrote:
>
> [Re: the imagined "problem" of 1/oo"]
>
> >*** Ignoring the problem will not make it go away.
>
> Pretending a definition says something it doesn't will not make a problem
> appear where it doesn't exist.
Thank you for your many responses. What you have said is consistent
with the impregnable edifice of mathematics. My divergent opinion arises
from the way the idea of infinity has been incorporated into the theory
of limits. This may be the only way of constructing the fortress but
it does have some drawbacks. I have tried to explain these in the following
, no doubt imperfect, argument.
______________________________________________________________________
Consider the simplest case where B is a binary fraction:
B = .111... = .1 + .01 + .001 + ... = SUM(r=1,oo,10^r) = lim r->oo [1 - 1/10^r]
In decimal notation: B = SUM(r=1,oo,2^r) = lim r->oo [1-1/2^r]
so B = 1 - lim r->oo [1/2^r]
B remains undefined until some meaning is attached to the RHS by means of
the definition: lim r->oo [1/2^r] = 0
In framing this definition a particular opinion about the meaning of
infinity, represented by the symbol oo, has been adopted.
The view is that infinity is singular. This follows from the idea that
infinity is boundless because, if there is more than one infinity,
one infinite entity would limit the other, which contradicts the
definition of infinity as boundless.
The consequence of this definition of infinity is that it precludes
the existence of a multitude of infinite entities and allows only
that there is one absolute entity. This ‘absolute infinity’ is
distinct from any hypothetical entities which are defined as
non-finite but greater than unity.
As an alternative to the definition B = 1 - lim r->oo [1/2^r] and
the interpretation of oo as an absolute entity, consider the following:
____________________________________________________________________
Hypothesis: the binary number .111... equals unity.
Let B = .111...
Finite Case
n is a finite positive integer
Let S1 be a binary string of finite length n, denoted by [111...1]n
Let S2 be a binary string of finite length n, denoted by [000...01]n
Let G1 be a binary fraction, corresponding to string S1, denoted by [.111...1]n
Let L1 be a binary fraction, corresponding to string S2, denoted by [.000...01]n
Then for all finite values of n:
G1 = SUM(r=1,r=n,2^-r) and L1 = 2^-n
Now G1 + L1 = 1 so (L1 > 0) => (G1 < 1)
2^-n > 0
Therefore [.111....1]n < 1
Non-finite Case
N is a non-finite integer where N > n for all finite values of n (NB: N is supernatural)
Let S3 be a binary string of length N, denoted by [111...1]N
Let S4 be a binary string of length N, denoted by [000...01]N
Let G2 be a binary fraction, corresponding to string S3, denoted by [.111...1]N
Let L2 be a binary fraction, corresponding to string S4, denoted by [.000...01]N
Then for all non-finite values of N
G2= SUM(r=1,r=N,2^-r) and L2 = 2^-N
Now G2 + L2 = 1 so (L2 > 0) => (G2 < 1)
2^-N > 0
Therefore [.111....1]N < 1
B = .111... = .1 + .01 + .001 + ...
Suppose B = SUM(r=1,r=T,2^r)
If B = 1 then B > G1 and B > G2
But SUM(r=1,r=T,2^r) < 1
So there is no number T such that B = SUM(r=1,r=T,2^r)
B is undefined and has no determinate value.
Therefore Either: (1) B has no definable sum and therefore no determinate value
or (2) B < 1 for all finite and non-finite integers.
Dave Seaman
In article <6a0t8m$4...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>Thank you for your many responses. What you have said is consistent
>with the impregnable edifice of mathematics. My divergent opinion arises
>from the way the idea of infinity has been incorporated into the theory
>of limits. This may be the only way of constructing the fortress but
>it does have some drawbacks. I have tried to explain these in the following
>, no doubt imperfect, argument.
What's confusing the issue is that this thread frequently switches back
and forth between standard and nonstandard analysis. In standard
analysis, there is no such thing as an infinitely large or infinitely
small number. Both of those concepts belong to nonstandard analysis.
As I have explained before, the standard notion of limits is based on
epsilon-delta analysis and does not require any actual infinities.
Since you seem to have a problem with the way "the idea of infinity has
been incorporated into the theory of limits", I suggest you stick to
the version that does not use infinities at all, namely, standard
analysis.
>B remains undefined until some meaning is attached to the RHS by means of
>the definition: lim r->oo [1/2^r] = 0
>
>In framing this definition a particular opinion about the meaning of
>infinity, represented by the symbol oo, has been adopted.
>The view is that infinity is singular. This follows from the idea that
>infinity is boundless because, if there is more than one infinity,
>one infinite entity would limit the other, which contradicts the
>definition of infinity as boundless.
Your use of the passive voice dodges an important question: WHO has
adopted such a view of infinity?
Are you talking about standard or nonstandard analysis? Either way,
your claim is false. In standard analysis, there are no infinitely
large numbers at all, while in nonstandard analysis there are
infinitely many. In either case, there is no such thing as a "singular
infinity".
If F is an ordered field, then an element x of F is said to be
"infinitely large" in F if x > y for every RATIONAL y in F (note:
*not* for *every* y in F). This is the definition of what it means for
a number to be infinitely large in an ordered field, and the definition
depends on the fact that every ordered field contains an isomorphic
copy of the rational numbers.
Let x be an infinitely large element in an ordered field F. Since the
positive elements of F are closed under addition and multiplication, it
follows that x+1, 2*x, and x^2 are all elements of F that are larger
than x. Clearly, it is not possible for F to contain a unique largest
element.
>The consequence of this definition of infinity is that it precludes
>the existence of a multitude of infinite entities and allows only
>that there is one absolute entity. This ‘absolute infinity’ is
>distinct from any hypothetical entities which are defined as
>non-finite but greater than unity.
Again, I ask: whose definition of infinity is this? The rest of your
argument is based on a faulty premise.
Unknown
>The consequence of this definition of infinity is that it precludes
>the existence of a multitude of infinite entities and allows only
>that there is one absolute entity. This ‘absolute infinity’ is
>distinct from any hypothetical entities which are defined as
>non-finite but greater than unity.
Again, I ask: whose definition of infinity is this? The rest of your
argument is based on a faulty premise.
--
Dave Seaman dse...@purdue.edu
Which premise?
Rabern
Verdigris wrote in article <6a3amc$a...@newsserver.trl.OZ.AU>...
1.999999 repeating forever does equal 2
here
x=1.999999 repeating
10x=19.9999999 repeating
so subtracts the equations
9x=18
x=2
doesn't that prove the point?
Landon Rabern
ric...@fau.edu
Darrell Ryan <dar...@cafes.net> wrote:
> The fraction 1/3 is exactly equal to .333... ("..." means keep going,
> don't stop). If you doubt that, perform the long division 1 divided
> by 3 and see how many decimal places you have. If we decide to "chop
> off" any of the 3's, then what we have is LESS than 1/3. Right? But
> the notation .33... ,which is equivalent to .33 with a bar over it,
> means that the 3's NEVER stop and in that case it is exactly EQUAL to
> 1/3. 3 times 1/3 equals 1 and .33... is another name for 1/3.
I don't see that. I started performing the long division 1 divided by 3.
How many decimal places am I supposed to have? When I stop, I do indeed
get a number that is less than 1/3. Why does that mean that if the 3's
never stop, then .33... is exactly equal to 1/3? It seems to me that it
is less than 1/3. Of course it would have to be, because, as you point
out, if it were equal to 1/3, then .999... would be equal to 1, which
we all know is not the case.
--Fred
Darrell Ryan
ric...@fau.edu wrote:
>
> I don't see that. I started performing the long division 1 divided by 3.
> How many decimal places am I supposed to have? When I stop, I do indeed
> get a number that is less than 1/3. Why does that mean that if the 3's
> never stop, then .33... is exactly equal to 1/3? It seems to me that it
> is less than 1/3. Of course it would have to be, because, as you point
> out, if it were equal to 1/3, then .999... would be equal to 1, which
> we all know is not the case.
>
> --Fred
Fred, yes .99...=1. See the sci.math fAQ or my page at
http://www.cafes.net/darrell/99baris1.htm
.33... is also equal to 1/3. The "..." means the 3's continue
indefinitely. This is _very_ elementary.
Dave Seaman
In article <6a3amc$a...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>
> >The consequence of this definition of infinity is that it precludes
> >the existence of a multitude of infinite entities and allows only
> >that there is one absolute entity. This ‘absolute infinity’ is
> >distinct from any hypothetical entities which are defined as
> >non-finite but greater than unity.
> Again, I ask: whose definition of infinity is this? The rest of your
> argument is based on a faulty premise.
>Which premise?
The one quoted above. The one that says the definition of infinity
"precludes the existence of a multitude of infinite entities..."
Counterexamples:
1. Infinite (transfinite) cardinal numbers.
2. Infinite (transfinite) ordinal numbers.
3. Infinitely large elements in an ordered field.
4. The compactification of the real numbers, in which a
distinction is drawn between +oo and -oo.
5. Points at infinity in projective geometry, which are
distinguished by their direction.
There are many ways to define "infinity" in mathematics, but it is
clear that uniqueness is not a necessary property.
Unknown
"Rabern" <lra...@hotmail.com> writes: >
Verdigris wrote in article <6a3amc$a...@newsserver.trl.OZ.AU>...
>
> >the existence of a multitude of infinite entities and allows only
> >that there is one absolute entity. This ‘absolute infinity’ is
> >distinct from any hypothetical entities which are defined as
> >non-finite but greater than unity.
>
> Again, I ask: whose definition of infinity is this? The rest of your
> argument is based on a faulty premise.
>
> Dave Seaman dse...@purdue.edu
>
>
> Which premise?
>
1.999999 repeating forever does equal 2
here
x=1.999999 repeating
10x=19.9999999 repeating
so subtracts the equations
9x=18
x=2
doesn't that prove the point?
Landon Rabern
It is simpler to say 1/9 = .111...
so 1 = 9 x .111... = .999...
and 1 + 1 = 2 = 1 + .999... = 1.999...
This effectively defines 1.999... = 2
In saying that 1/9 = .111...
reliance is being placed on a kind of induction
in the context of an infinite string. We conclude
that the string '.111...' only contains 1s.
An objection arises if we try to interpret
signs such as .111... as an infinite sum.
The interpretation of the sign requires the
(reasonable) assumption that lim->oo 1/2^n = 0
If the meaning of the signs in this equation are
uncertain, then the meaning of .111... is uncertain.
The meaning of lim->oo 1/2^n is uncertain if the meaning
of the individual signs is uncertain. In particular, this
applies to the orthodox definition of limits involving
the sign oo.
Consider the following questions:
What is the greatest proper fraction (GPF) that can be represented
in the decimal system (or in any other mode such as the
binary system)?
If this fraction GPF = .999... then .999... < 1, otherwise it would not
be a proper fraction. If GPF <> .999. how can it be represented?
Must we conclude that there is no greatest proper fraction or only
that it can not be represented in this way.
Unknown
a...@seaman.cc.purdue.edu (Dave Seaman) writes: > In article <6a3amc$a...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>
> >The consequence of this definition of infinity is that it precludes
> >the existence of a multitude of infinite entities and allows only
> >that there is one absolute entity. This ‘absolute infinity’ is
> >distinct from any hypothetical entities which are defined as
> >non-finite but greater than unity.
> Again, I ask: whose definition of infinity is this? The rest of your
> argument is based on a faulty premise.
The one quoted above. The one that says the definition of infinity
"precludes the existence of a multitude of infinite entities..."
Counterexamples:
1. Infinite (transfinite) cardinal numbers.
2. Infinite (transfinite) ordinal numbers.
3. Infinitely large elements in an ordered field.
4. The compactification of the real numbers, in which a
distinction is drawn between +oo and -oo.
5. Points at infinity in projective geometry, which are
distinguished by their direction.
There are many ways to define "infinity" in mathematics, but it is
clear that uniqueness is not a necessary property.
Dave Seaman dse...@purdue.edu
The argument (under the line below), previously posted, assumes that
there are non-finite 'integers' greater than any finite integer. It does
not mention infinity at all. the conclusion was that .111... remains
undefined from this point of view. In other words, If the GP referred
to below is summed to a finite integer or to a non-finite 'integer'
the result will always be less than unity. The point of saying this
is to remove the conjuring trick where we can never see the RHS of
the sign '.111...'.
The indeterminacy of .111... is avoided by the orthodox use of limits.
But the use of the 'infinite' limit limn->oo(1/a^n) = 0 (together with the
epsilon-delta rationalisation) can not be explained without considering
the meaning of the term 'infinity'. In this context, it is fairly clear
that the infinity referred to means 'that which is boundless in space,
time or number.
lim n->oo(1/a^n) = 0 simply means that when n increases without limit
the term 1/a^n is bounded below by zero. This is different to the
idea of 'infinite' numbers which can form a series ie when
u1 < u2 < u3 < ...<uN where one might have to disallow the notion of
infinity underlying the orthodox theory of limits. This would mean that
the application of the theory of infinite limits to non-standard
numbers would involve contradictions. However, if you maintain that the
theory of limits has nothing to do with infinity at all, then there would
be no inconsistency.
______________________________________________________________________
Let B = .111...
Finite Case
Non-finite Case
_______________________________________________________________________
ric...@acc.fau.edu
Darrell Ryan <dar...@cafes.net> wrote:
> Fred, yes .99...=1. See the sci.math fAQ or my page at
> http://www.cafes.net/darrell/99baris1.htm
Nice site! Maybe I'll make one too. I've never cared for the FAQ on this
thread. It is little more than an Apostles Creed for believers in the
equality. No space at all is given to any skeptical position.
In the meantime, let me address your argument that is based on 10x - x,
where x = 0.999... . Rather than a proof of the equality, it may be
viewed as a reductio ad absurdum of the idea that you can subtract x
and get a unique answer. This may be even clearer from the fact that
you can't cancel x under addition: x + x = 1 + x but x < 1.
Dave Seaman
In article <6a6h4q$6...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>The argument (under the line below), previously posted, assumes that
>there are non-finite 'integers' greater than any finite integer. It does
>not mention infinity at all.
What is your definition of a non-finite integer? If x is infinitely
large in an ordered field F, then I take it there must be some y,
x <= y < x+1, such that y qualifies as a "non-finite integer". Very
well. Which one is it?
One thought that comes to mind is that the ordinals could be regarded
as non-finite integers in some sense, but your later arguments pretty
much preclude this approach.
>In other words, If the GP referred
>to below is summed to a finite integer or to a non-finite 'integer'
>the result will always be less than unity. The point of saying this
>is to remove the conjuring trick where we can never see the RHS of
>the sign '.111...'.
The finite sum is of 2^(-n), where n is a positive (finite) integer.
If you want to extend the sum to "non-finite integers", whatever they
may be, you have a few questions to answer.
1. What is 2^(-n) for n a non-finite integer? The field axioms
don't assign a value to this.
2. If y is a non-finite integer, then what are the other
non-finite integers that are less than y? Is y/2 a
non-finite integer? How do we decide whether y is odd or
even? More generally, how about y/n, where n is a finite
integer? All of those numbers are certainly non-finite,
but which ones are non-finite integers? Surely y-n, for
every finite integer n > 0, qualifies as a non-finite
integer. However, all of those are larger than y/2, which
is larger than y/3, and so on.
3. Conventional infinite sums (sums over the natural numbers)
depend on the concept of a sequence of partial sums. The
basic idea is that you don't need to talk about infinity
in order to define the partial sum S_n for each n, and
therefore all that remains is to compute the limit of S_n
as n -> oo. What is the corresponding concept when summing
over "non-finite integers"? You can't even talk sensibly
about partial sums. Look again at question 2 in order to
appreciate the difficulty.
>So there is no number T such that B = SUM(r=1,r=T,2^r)
>B is undefined and has no determinate value.
>Therefore Either: (1) B has no definable sum and therefore no determinate value
>or (2) B < 1 for all finite and non-finite integers.
I'd say conclusion (1) looks like the winner.
What does any of this have to do with whether .1111... (binary) = 1?
The binary expansion represents an omega-sequence, not a sequence over
"non-finite integers".
Darrell Ryan
ric...@acc.fau.edu wrote:
>
> In the meantime, let me address your argument that is based on 10x - x,
> where x = 0.999... . Rather than a proof of the equality, it may be
> viewed as a reductio ad absurdum of the idea that you can subtract x
> and get a unique answer. This may be even clearer from the fact that
> you can't cancel x under addition: x + x = 1 + x but x < 1.
I'm not sure what your point is. You say x+x=1+x but x<1 which is a
contradiction. If you say x<1, then why did you "replace" x with 1?
Make up your mind, is x<1 or is x=1?
Remember, this specific method is not meant as "proof" that .99..=1.
It does, however, show that the "natural" operations performed lead to
the result, i.e. it's natural to assume .99..=1. The FAQ _is_ good
for something, IMHO. Before you can do the 10x-x thing (remember you
start with x=.99...) you have to first know what the notation .99...
means. A repeating decimal is shorthand for an infinite series. An
infinite series is shorthand for the limit of the sequence of partial
sums. Proving this limit is 1 is all you need to do. You have
already proved what you wanted to prove.
Unknown
Erratum.
The answer should read:
(Q2) What is 2^(-n) for n an NFI
An example would be 2^(-U1), which is a very small number.
From (D8) U(-2) = U1^(-2) = 1/U2 ie U(-r) = U1^(-r) with special
case U0 = U1^(0) = 1.
Unknown
*> The finite sum is of 2^(-n), where n is a positive (finite) integer.
*> If you want to extend the sum to "non-finite integers", whatever they
*> may be, you have a few questions to answer.
*>
*> 1. What is 2^(-n) for n a non-finite integer? The field axioms
*> don't assign a value to this.
*>
*> 2. If y is a non-finite integer, then what are the other
*> non-finite integers that are less than y? Is y/2 a
*> non-finite integer? How do we decide whether y is odd or
*> even? More generally, how about y/n, where n is a finite
*> integer? All of those numbers are certainly non-finite,
*> but which ones are non-finite integers? Surely y-n, for
*> every finite integer n > 0, qualifies as a non-finite
*> integer. However, all of those are larger than y/2, which
*> is larger than y/3, and so on.
*
*
*> Dave Seaman dse...@purdue.edu
Your questions are:
(Q1) What is your definition of a non-finite integer (NFI)
(Q2) What is 2^(-n) for n an NFI
(Q3) If y is an NFI what are the other NFIs that are less than y
(Q4) How do you decide whether y is odd or even
(Q5) Surely y - n qualifies an NFI
some preliminary definitions are required.
_____________________________________________________________________
Construction of non-finite numbers
Notes:
(i) The following definitions/axioms are in Polish notation
(ii) ‘e’ is used for set membership as in x e S
(iii) A constructible entity must be an n-ary string of non-finite length
(D1) S0 is the set of finite integers
(D2) S0f is the set of finite rational proper fractions
(D3) Ui is a number of order i where i is any integer
(D4) Si is the set of numbers of order i where Ui e Si
(D5) NSbKbeSib<Ui so meaning: Ui is the least member of Si
(D6) CaeSia+UieSi meaning: if a is in Si then a + Ui is in Si
(D7) CKaeSibeSi+1a<b meaning: all members of Si are less than all members of Si+1
(D8) Ui = U1^i where i is any integer
_____________________________________________________________________
(Q1) What is your definition of a non-finite integer (NFI)
Ui is an NFI for i > 0
From (D1) and (D5) it follows that U0 is the least member of S0
so that U0 = 1.
From (D7) CKaeS0beS1a<b means that every member of the set of finite
integers is less than every member of S1, so all the members of S1 are NFIs.
(Q2) What is 2^(-n) for n an NFI
An example would be 2^-U1, which is a very small number.
From (D8) U-2 = U1^-2 = 1/U1.
(Q3) If y is an NFI what are the other NFIs that are less than y
The NFIs defined above form a series : U1 < U2 < ... < Ui
As the the least member of the series U1 has non-finite fractions
U1/2 > U1/3 > ... U1/n where n e S0. However, these only exist subject to
(iii) above.
(Q4) How do you decide whether y is odd or even
If the parity is measured by a finite digit, then only U1 is odd, all the other
numbers in the series above are even in this sense.
However, we can construct hybrid numbers which have a finite component.
for example, aU1 + 3 (a e S0) is a hybrid constructed on the principle of complex
numbers. This is an odd number.
(Q5) Surely y - n qualifies an NFI
Yes, this is a hybrid NFI.
______________________________________________________________________________
Unknown
Erratum 2:
answer should read 'then only U0 is odd'
Fred Richman
> > In the meantime, let me address your argument that is based on 10x - x,
> > where x = 0.999... . Rather than a proof of the equality, it may be
> > viewed as a reductio ad absurdum of the idea that you can subtract x
> > and get a unique answer. This may be even clearer from the fact that
> > you can't cancel x under addition: x + x = 1 + x but x < 1.
> I'm not sure what your point is. You say x+x=1+x but x<1 which is a
> contradiction. If you say x<1, then why did you "replace" x with 1?
> Make up your mind, is x<1 or is x=1?
Sorry. Of course, x < 1. The reason x + x = 1 + x is because both sides
of that equation are equal to 1.999... . Clearly 1 + x = 1.999... ,
while x + x = 1.999... because at every place beyond the decimal point
you are adding 9 and 9, together with a carry of 1 from the next place.
--Fred
Darrell Ryan
Fred Richman wrote:
> Sorry. Of course, x < 1. The reason x + x = 1 + x is because both sides
> of that equation
<...>
Fred, you just did it again. You are saying two things which
contradict each other. If x=1, then x+x=1+x. If x does not equal 1,
then x+x<>1+x. This is very elementary algebra, and there is really
no point in discussing why 1.99..=2, .99..=1, etc. until you
understand why you are contradicting yourself.
x+x=1+x ...now solve for x
2x = 1+x
2x-x=1
x=1
Forget about x=.099.. or any of that for now. You need to first
understand that x=1 is the solution to x+x=1+x. Any x less than 1 is
not a solution to the equation. Algebra 1. But you are saying
x+x=1+x and at the same time you are saying x<1, which is a
contradiction.
Dave Seaman
In article <6a935p$k...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>Your questions are:
>(Q1) What is your definition of a non-finite integer (NFI)
>(Q2) What is 2^(-n) for n an NFI
>(Q3) If y is an NFI what are the other NFIs that are less than y
>(Q4) How do you decide whether y is odd or even
>(Q5) Surely y - n qualifies an NFI
I notice you left out my question about forming partial sums over NFI
sequences. How do you know that such sums exist?
>Construction of non-finite numbers
>
>Notes:
>(i) The following definitions/axioms are in Polish notation
>(ii) ‘e’ is used for set membership as in x e S
>(iii) A constructible entity must be an n-ary string of non-finite length
>
>(D1) S0 is the set of finite integers
>(D2) S0f is the set of finite rational proper fractions
>(D3) Ui is a number of order i where i is any integer
>(D4) Si is the set of numbers of order i where Ui e Si
>(D5) NSbKbeSib<Ui so meaning: Ui is the least member of Si
>(D6) CaeSia+UieSi meaning: if a is in Si then a + Ui is in Si
>(D7) CKaeSibeSi+1a<b meaning: all members of Si are less than all members of Si+1
>(D8) Ui = U1^i where i is any integer
In an earlier post you claimed that binary sequences [.111...1]N over
NFI's are less than 1, that [.000...01]N is positive for each N, and
that the sum of these two binary expansions is 1.
You still haven't addressed the issue of how such sums are to be
computed. However, your claims are mutually exclusive. The binary
omega-sequence .111... = sum(i=1 to oo) 2^(-i) is already precisely
equal to 1. I don't care how infinitesimally small your 2^(-N) is;
there is no room for it to fit between .111... and 1, since the size of
the gap is precisely zero.
Let a_k = 1 - s^(-k) for k = 0, 1, 2, .... Let I_k be the half-open
interval [a_(k-1),a_k) for k = 1, 2, 3, .... Then m_k = measure of I_k
= 2^(-k) for k = 1, 2, 3, ..., and by the countable additivity of
measure we have sum(i=1 to oo) 2^(-k) = measure([0,1)) = 1. That is a
precise equality.
Fred Richman
Darrell Ryan wrote:
> Fred, you just did it again. You are saying two things which
> contradict each other. If x=1, then x+x=1+x. If x does not equal 1,
> then x+x<>1+x. This is very elementary algebra, ...
> Forget about x=.099.. or any of that for now. You need to first
> understand that x=1 is the solution to x+x=1+x. Any x less than 1 is
> not a solution to the equation. Algebra 1.
Why are you so sure that 1 is the only solution to x + x = 1 + x?
We can see that
1 + 1 = 1 + 1
as both sides are equal to 2, and also that
.999... + .999... = 1 + .999...
as both sides are equal to 1.999... . So, unless .999... = 1, there
are clearly two solutions. Appealing to what you were taught in high
school algebra isn't very convincing because those same people teach
that .999... = 1. You are certainly correct in observing that if
.999... < 1, then high school algebra is going to have to undergo
some revision before it will apply to real numbers because you can't
cancel an x from each side of the equation x + x = 1 + x.
--Fred
Verdigris
Darrell Ryan wrote:
>
> >
> > Why are you so sure that 1 is the only solution to x + x = 1 + x?
> > We can see that
> >
> > 1 + 1 = 1 + 1
>
> 1's, i.e. x=1. Get it? Try replacing the x with any other number not
> equal to 1. 1 is the only solution. The reason I point you back to
> algebra 1 is because, quite frankly, you need it.
>
> x + x = 1 + x
>
>
> x + x - x = 1 + x - x
>
> x = 1
>
> Get it? If not, please consider studying algenbra 1 all over again,
> if you have even studied it at all. Frankly, I would like to discuss
> some math that is at least a _little_ more advanced than this!
>
> At any rate, I'm glad to see that you position now seems to be .99..=1
> which is what I've been telling you all along.Try Aleph-null + Aleph-null = 1 + Aleph-null
kem...@de.ibm.com
In <34CCB5D5...@fau.edu>, Fred Richman <ric...@fau.edu> writes:
>Darrell Ryan wrote:
>
>> Fred, you just did it again. You are saying two things which
>> contradict each other. If x=1, then x+x=1+x. If x does not equal 1,
>> then x+x<>1+x. This is very elementary algebra, ...
>
>> Forget about x=.099.. or any of that for now. You need to first
>> understand that x=1 is the solution to x+x=1+x. Any x less than 1 is
>> not a solution to the equation. Algebra 1.
>
>We can see that
>
> 1 + 1 = 1 + 1
>
>
> .999... + .999... = 1 + .999...
>
>as both sides are equal to 1.999... . So, unless .999... = 1, there
>are clearly two solutions. Appealing to what you were taught in high
>school algebra isn't very convincing because those same people teach
>that .999... = 1. You are certainly correct in observing that if
>some revision before it will apply to real numbers because you can't
Aside of the use of '=' as a equivalence relation sign and other nice
algebraic properties inherited in the definition of a field, not talking
about unique properties of continious functions (especially for R).....
How's about the solutions 0.999.... + any limit of a 0-sequence
like lim_(n-->00) ( [sum_(1 to n) (9/ 10^n) ] + a_n )
where a_n = 1/p_n with p_n the n-th prime ?
Darrell Ryan
Verdigris wrote:
> Try Aleph-null + Aleph-null = 1 + Aleph-null
Try sticking to standard analysis, which you know is the context of
this discussion. We're talking about someone who doesn't understand
that x=1 is the solution to
x + x = 1 + x
If someone doesn't understand that, it is quite obvious that they will
be even more blown away by wondering what aleph null, etc. are.
Do you think it's wise to introduce these other systems without the
student knowing his/her algebra 1? Of course not, so get with the
program.
Darrell Ryan
kem...@de.ibm.com wrote:
>
> Aside of the use of '=' as a equivalence relation sign and other nice
> algebraic properties inherited in the definition of a field, not talking
> about unique properties of continious functions (especially for R).....
> How's about the solutions 0.999.... + any limit of a 0-sequence
> like lim_(n-->00) ( [sum_(1 to n) (9/ 10^n) ] + a_n )
> where a_n = 1/p_n with p_n the n-th prime ?
Aside from all that, remember this is someone who says x<1 and at the
same time is a solution to
x + x = 1 + x
Get it? THIS IS ALGEBRA 1. I understand that this is sci.math where
more advanced topics are discussed, but ocassionally we stumble across
someone having trouble at a much more elementary level. Get it? The
person needs to understand that x=1 is a solution to
x + x = 1 + x
In other words, pretend you are in algebra 1. All the advanced
knowledge of mathematics that one can posess means absolutely nothing
if the person is tasked to help someone at such an elementary level,
and fails to "stoop" down to that level. Geez.
Sue Richman
Darrell Ryan wrote:
> > Why are you so sure that 1 is the only solution to x + x = 1 + x?
> > We can see that
> > 1 + 1 = 1 + 1
> Fred, surely you realize that what you did was replace the x's with
> 1's, i.e. x=1. Get it? Try replacing the x with any other number not
> equal to 1. 1 is the only solution.
Yes, I realize that. And then I replaced each x with .999..., showing
that there is another solution.
> The reason I point you back to
> algebra 1 is because, quite frankly, you need it.
> x + x = 1 + x
> .....now subtract x from both sides
> x + x - x = 1 + x - x
> x = 1
> Get it? If not, please consider studying algenbra 1 all over again,
> if you have even studied it at all. Frankly, I would like to discuss
> some math that is at least a _little_ more advanced than this!
What kind of argument is it to parrot something you were taught in
high school? Surely this discussion is beyond that. Do you think
that skeptics of the equality .999... = 1 are unaware of your dinky
argument? I doubt that your high school teachers showed you that
you can always subtract real numbers and get a unique answer. Your
argument above shows, in fact, that you cannot. (Take x = .999... .)
--Fred
Darrell Ryan
Sue Richman wrote:
>
> Darrell Ryan wrote:
>
> > > Why are you so sure that 1 is the only solution to x + x = 1 + x?
> > > We can see that
>
> > > 1 + 1 = 1 + 1
>
> > Fred, surely you realize that what you did was replace the x's with
> > 1's, i.e. x=1. Get it? Try replacing the x with any other number not
> > equal to 1. 1 is the only solution.
>
> Yes, I realize that. And then I replaced each x with .999..., showing
> that there is another solution.
...and you could also say x=1/1, x=2/2, x=1024/1024, x=cos(0),
x=.99..., etc which are all different expressions that represent the
_same_ number, namely 1.
x+x=1+x
x=1 <----- if that isn't obvious enough for you, you are
unteachable.
Get it? x=1. x is NOT less than 1. You are trying to say x<1 AND is
a solution to the equation. It can't be both.
There is only _one_ solution in the complex numbers. If we say x-1=0
we _know_ there is only one root by consequence of the fundamental
theorem of algebra (as if the statement x=1 was not obvious enough!).
It's 1st degree, so there is exactly one complex root. By showing x=1
is a solution, AND by showing x=.99.. is a solution, what we have
really shown is that 1=.99.. We _know_ there is only one solution, so
the two MUST be equal.
>
> What kind of argument is it to parrot something you were taught in
> high school? Surely this discussion is beyond that. Do you think
> that skeptics of the equality .999... = 1 are unaware of your dinky
> argument?
...now everyone knows what I'm taking about when I said we need to
"stoop" to this person's level and stay away from anything above an
algebra 1 level!
Hey, Fred was the one who said x<1 AND x+x=1+x. All I'm doing is
reminding him that the solution to x+x=1+x is x=1. I have yet to see
him confirm this. Fred, What are you afraid of? Are you denying that
x=1 is the solution? Let's get to the _real_ issue here. Are you
saying that .99..<1??? Is that why you think x<1 in the equation
x+x=1+x??? Hey, you (Fred) was the one who first brought that little
equation up.
>I doubt that your high school teachers showed you that
> you can always subtract real numbers and get a unique answer. Your
> argument above shows, in fact, that you cannot. (Take x = .999... .)
Ha! Ha! so you DO think .99..<1. Do you understand that 1/1, 2/2,
pi/pi, cos(0) are all equally correct ways of expressing the number
1? Unique answer does NOT mean there is only one way to WRITE the
answer.
Please, please, PLEASE study your algebra 1 !!!
Fred, you are very conveniently REFUSING to address the simple
equation that YOU brought up! x=1 is the solution, and it is the only
solution. Remember, thare are different WAYS to write the number 1,
so if you are convinced that x=.99... is also a solution, then you
MUST NECESSARILY believe that 1=.99... if you want to keep from
contadicting yourself. It's first degree, and as such has exactly one
solution in the complex numbers.
You signed this "Fred", but your identity has changed in the headers.
What are you afraid of, learning something? I'm done with this, so
find someone else to play games with. If you don't see that x=1 is
the solution to x+x=1+x, you need help man!
>
> --Fred
Verdigris
We are, of course, using different models. The orthodox model you are
using
can produce 'perfect' answers because of its refinement over time by the
best
available mathematical minds. However, the model I am using is more
general
and, therefore, has more explanatory power in non-finite regions. That
is to
say that it allows the usual laws of arithmetic to apply to NFIs, with
some
restrictions. In MB every number has an infinite construction eg the
number 3
belongs to the set bounded by U1, so 3 = w[0._011].
Let us call your model MA and my model MB.
Consider the following:
B = .111... = 1 (as before) in model MA.
b = 0[1.-.1]w = 1 - 2^w where n e S0 and n < w < U1 in model MB.
now T(b) = w[1.-.1]0 = 2^w - 1 where T(n) is binary 'reflection' about
the binary
point. So T(B) + 1 = 2^w. But what is T(B)? in model MA.
The question is what could be meant by the reflection ...111.000... ?
The natural interpretation is that it is the greatest representation
of a positive integer in binary notation where there is no terminating
digit on
the left. The justification for using this form is that it is analogous
to the construction of the symbol .111...
now ...111.000... = ...111 = 2^0 + 2^1 + 2^2 + ... = lim r->oo
SUM(1,oo,2^r)
We have, T(B) = ...111 by definition.
If T(B) is the symbol for the cardinal of the set N of positive finite
integers
Then T(B) + 1 = Ao + 1 = Ao = T(B). This is in contrast to T(b) + 1 >
T(b).
In model MB, T(b) + 1 = U1 = 2^w < U2 < U3 < ...
What is unclear in model MA is whether T(B) is an infinite measure which
can
be transcended or not. Given that 2^Ao is c > Ao, the answer would
appear to be yes.
In MA, therefore, we appear to have one absolute infinite entity
transcending another.
The advantage of model B is that this kind of 'mystical' construction is
avoided.
Thus is because there can never be an absolute infinity defined in MB,
but only
a series of infinite ordinals. The point of model MB is that it provides
a basis
for analysing the Cantorian construction of infinite decimals. One
consequence
is that in MB there is no distintion between rational an irrational
numbers defined
on non-finite decimals. This is because every permutation in an MB
number is clearly
a rational even though it has a non-repeating pattern.
Summing such non-finite entities is similar to summing in the finite
region.
One significant difference is that the sum of an NFI such as U1 over a
finite
range will always be less than the U1^2. Summing over an NF range would
be as
follows.
SUM(1,U1,r) = U1(U1 + 1)/2 = U1^2/2 + U1/2 < U2 where r has maximum
value
U1 = 2^w
Dave Seaman
In article <34CEF3...@uq.net.au>, Verdigris <EATh...@uq.net.au> wrote:
>We are, of course, using different models. The orthodox model you are
>using
>can produce 'perfect' answers because of its refinement over time by the
>best
>available mathematical minds. However, the model I am using is more
>general
>and, therefore, has more explanatory power in non-finite regions. That
>is to
>say that it allows the usual laws of arithmetic to apply to NFIs, with
>some
>restrictions. In MB every number has an infinite construction eg the
>number 3
>belongs to the set bounded by U1, so 3 = w[0._011].
I don't follow your notation here, but it appears to me that what you
are describing may be viewed as an ordered field. I think your notion
of NFI's is not particularly well-defined, however; more on that
later.
I'm not sure why you consider it so significant that every number has
an infinite construction; the same is true of the standard real
numbers.
>Let us call your model MA and my model MB.
>
>Consider the following:
>
>B = .111... = 1 (as before) in model MA.
>
>b = 0[1.-.1]w = 1 - 2^w where n e S0 and n < w < U1 in model MB.
Not so fast. Using your notation, let a = [.000...1]U1. What is the
binary expansion for 2*a? Before you can start adding and subtracting
these super-infinite binary strings, you need to have some idea what
the rules are. There is no binary zero immediately to the left of that
1, and therefore no place to carry into. Remember, you are claiming
that U1 is the smallest NFI. I presume you really mean U1 is the
smallest *positive* NFI, but it's still a dubious claim. You also
state that U1 is an even number. Why, then, is U1/2 not a still
smaller NFI?
Also, you haven't really explained what you mean by 2^n or 2^(-n),
where n is an NFI, except to say that 2^(-n) is "a very small number".
Yes, but which one is it?
>We have, T(B) = ...111 by definition.
>If T(B) is the symbol for the cardinal of the set N of positive finite
>integers
>Then T(B) + 1 = Ao + 1 = Ao = T(B). This is in contrast to T(b) + 1 >
>T(b).
Yes, cardinal arithmetic differs from arithmetic in non-Archimedean fields.
>In model MB, T(b) + 1 = U1 = 2^w < U2 < U3 < ...
>What is unclear in model MA is whether T(B) is an infinite measure which
>can
>be transcended or not. Given that 2^Ao is c > Ao, the answer would
>appear to be yes.
>In MA, therefore, we appear to have one absolute infinite entity
>transcending another.
By MA you apparently mean the standard real numbers. Aleph-null is not a
member of the standard reals.
>The advantage of model B is that this kind of 'mystical' construction is
>avoided.
What 'mystical' construction? The absence of infinitely large elements?
ric...@fau.edu
Darrell Ryan <dar...@cafes.net> wrote:
> x+x=1+x
> x=1 <----- if that isn't obvious enough for you, you are
> unteachable.
Perhaps you are the one who is unteachable if you cannot conceive of the
possibility that there is another solution. As was pointed out, any
infinite cardinal is a solution. Why couldn't there be another real
number that is also a solution?
> Get it? x=1. x is NOT less than 1. You are trying to say x<1 AND is
> a solution to the equation. It can't be both.
It can't be? I claim that .999... is such a number. It is obviously less
than 1, and when you add it to itself you get 1.999... , which is the
same thing you get if you add it to 1.
> There is only _one_ solution in the complex numbers. If we say x-1=0
> we _know_ there is only one root by consequence of the fundamental
> theorem of algebra (as if the statement x=1 was not obvious enough!).
That's pretty good, Darrell; I hadn't seen that argument before.
Actually, it's not the fundamental theorem of algebra (which asserts the
existence of roots of polynomials) that you need. What you need is the
theorem that there are at most n roots for a polynomial of degree n,
which is much more elementary. I would have to answer that the proof of
this theorem assumes that subtraction is unambiguously defined for real
numbers. Of course once we assume that, then your original argument shows
that .999... = 1. But the fact that .999... is less than 1 argues that
subtraction is not unambiguously defined for real numbers.
> > What kind of argument is it to parrot something you were taught in
> > high school? Surely this discussion is beyond that. Do you think
> > that skeptics of the equality .999... = 1 are unaware of your dinky
> > argument?
>
> ...now everyone knows what I'm taking about when I said we need to
> "stoop" to this person's level and stay away from anything above an
> algebra 1 level!
What is this preoccupation with algebra 1? In fact, to discuss this topic
intelligently, we have to get away from algebra 1.
> Hey, Fred was the one who said x<1 AND x+x=1+x. All I'm doing is
> reminding him that the solution to x+x=1+x is x=1. I have yet to see
> him confirm this. Fred, What are you afraid of? Are you denying that
> x=1 is the solution?
I'm denying that it is the only solution. Certainly it is a solution.
> Let's get to the _real_ issue here. Are you
> saying that .99..<1??? Is that why you think x<1 in the equation
> x+x=1+x??? Hey, you (Fred) was the one who first brought that little
> equation up.
Haven't I said that .999... < 1 often enough? Okay, I'll say it again:
..999... < 1. That would seem to be the alternative to .999... = 1. And no,
I don't think that x is necessarily less than 1 if x + x = 1 + x, but I do
think that it can be less than 1; it can also be equal to 1. There are two
solutions.
> Ha! Ha! so you DO think .99..<1. Do you understand that 1/1, 2/2,
> pi/pi, cos(0) are all equally correct ways of expressing the number
> 1? Unique answer does NOT mean there is only one way to WRITE the
> answer.
Yes, I understand that you can express the same number in different
ways. But I deny that .999... and 1 are different ways of expressing
the same number.
> Please, please, PLEASE study your algebra 1 !!!
Actually, I know algebra 1 perfectly well. That doesn't mean I have to
agree with it.
> You signed this "Fred", but your identity has changed in the headers.
> What are you afraid of, learning something?
Well, that was a blunder. I was home so I used my wife's account, but
forgot to change the identity to mine.
ari...@mps.org
In article <885998209...@dejanews.com>,
ric...@fau.edu wrote:
>
[snip]
>
> What is this preoccupation with algebra 1? In fact, to discuss this topic
> intelligently, we have to get away from algebra 1.
Yes, of course, we need to deal with algebra 0.9999999....
>
[snap]
A.
ENoether
>It can't be? I claim that .999... is such a number. It is obviously less
>than 1,
I am not an analyst, and I am not intimately familiar with some of the paradoxs
of modern set theory, but I have the following question :
You say that .999... is 'obviously' less than one. That is not at all obvious
to me. What ordering are you using on the real numbers to say that? Because if
so, what is 1 - .999... ?
To say 1 > .999..., by the fact that the reals are dense, you would have to
produce a number, say epsilon, such that
1 - .999... > epsilon.
Clearly such an epsilon does not exist. You may be able to get around the need
for such an epsilon by introducing a different ordering, but what would it be?
One more question :
If .999... is an algebraic number over the field of rationals, what is it's
minimal polynomial? I think it would have to be
f(x) = x - 1
n'est-ce pas?
(By the way : I am not claiming this as any kind of rigourous argument, and I
am not passionate about this point, so I will not get in a heated debate about
it. I am merely posing a couple of questions to shed some more light on the
subject).
Unknown
The notation in the text above was ad-hoc for the purpose of illustration.
The following notation may be more consistent.
________________________________________________________________________
Construction of Non-Finite Numbers
Notation
4(0...0)1 means the finite binary string (0000)
4(0...1)1 means the finite binary string (0001)
4(1...1)1 means the finite binary string (1111)
n(0...0)1 means the finite binary string (0...01)
n 1
w(0._.01)1 means the infinite binary string (0._.01)
w 1
w(0._.01)1 means the infinite binary string (0._.01)
w 1
w(10._.0)1 means the infinite binary string (10._.0)
w 1
The sign "..." indicates a finite range
The sign "._." indicates an infinite range
w[0._.01]1 means the infinite number [0._.01]
w 1
corresponding to the binary string w(0._.01)1
This number is not strictly the number 1 but its equivalent expressed as an infinite number.
This is analagous to defining a real number in the form of a complex number where
1 = 1+ 0i.
P2n is a binary power set of degree n and N2n is the equivalent numerical set.
If a e N2n then a = SUM(0,n,br2^r) where br = 0 or 1.
NB: The 0 and 1 in this definitions are 'meta numbers' borrowed from ordinary arithmetic.
In passing from the binary string (0001), say, to the binary number [0001] the symbol "1"
has been transformed from a permutational logical symbol into an arithmetic symbol.
All that is intended is that for every string s there is a number a which is constructible
as defined above. So, for every s e P2n there is an a e N2n.
If the binary system is understood, then Pn and Nn can represent the respective classes.
The 'cardinals' U_r form the series ._. U_(-2) < U_(-1) < U_0 < U_1 < U_2 ._.
Where U_(-1) = 1/U_1 = 2^(-w) and U_1 = 2^w etc.
We can concatenate two discrete sets to represent non-finite numbers with lage infinite,
finite and small infinite (infinitesimal) components.
For example:
._.] [0._.01] [0._.01] [0._.01] [0._.01] [ ._.
._.] 2w w+1 w 1 0 -w+1 -w -2w+1
This is the number 2^w + 1 + 2^(-w) + 2^(-2w) = U_1 + U_0 + U_(-1) + U_(-2)
The class which contains this number is N2w U Nw U N0 U N(-w)
Some other points:
The infinite power sets are not bottom-up Peano constructions but pre-exist in the
Pythagorean sense.
The numbers so defined lie within sets which are bounded above and below
meaning that they have a greatest and leasr member.
The set S0 of the finite positive integers is bounded below but has no greatest member.
S0 is a proper subset of Nw.
The motive for creating bounded infinite sets is so that logical operation can be applied
(with due care) to their logical equivalents. for example:
J w(0._.011)1 w(10._.01)1 = w(10._.010)1
The string length and order must be the same for such operations to be meaningful.
Note, in particular, that the scope of a negation must be a bounded set:
Thus N w(10._.0)1 = w(01._.1)1
This allows a useful monadic operator "*" to be applied to non-finite numbers:
eg *w[0._.01]1 = w[10._.0]1 which allows a finite number to be transormed into a non-finite
one and vice-versa. Clearly **n = n.
Kjell Fredrik Rogstad Pettersen
In article <885998209...@dejanews.com>, ric...@fau.edu writes:
> It can't be? I claim that .999... is such a number. It is obviously less
> same thing you get if you add it to 1.
> subtraction is not unambiguously defined for real numbers.
Gosh!!! Please explain to me
1. How do you define the real numbers?
2. How do you define addition on reals? (Is this ambigous too?)
3. Has every real number (with your definition) an additive inverse
(i.e. for any x, there is a (-x) such that x+(-x)=0) ?
4. If we define x-y = x+(-y) for any x,y, why is this not an unambiguos
definition?
I am extremely curious about your answers!
Kjell Fredrik
Frank Wappler
Fred <ric...@fau.edu> wrote:
> Darrell Ryan <dar...@cafes.net> wrote:
> > Do you understand that 1/1, 2/2, pi/pi, cos(0) are all equally
> > correct ways of expressing the number 1?
> > Are you saying that .99..<1???
> Yes, I understand that you can express the same number in different
> ways. But I deny that .999... and 1 are different ways of expressing
> the same number. [...]
> I claim that .999... [...] is obviously less than 1
> > Please, please, PLEASE study your algebra 1 !!!
> Actually, I know algebra 1 perfectly well.
> That doesn't mean I have to agree with it.
I don't know too much about algebra 1, but many responses seem to favor
.999 ... == 1 (at least wrt. what those two obviously different notations
are meant to denote). I was wondering how this relates to the "diagonal
argument" that (what exactly?) could not be counted:
Consider the following beginning of a list of all numbers between
0. and 1. (in the binary system; each blank digit means "0"):
.1
.001
.01
.11
.101
.011
.111
.0001
.1001
.0101
...
(continue in the "obvious way": as if counting whole numbers, but inverted)
Form a diagonalized number of that list, by picking the n-th digit from
the n-th entry of that list and altering it. In the binary system that
leaves no choice but to interchange "1" and "0", the diagonalized
number is uniquely determined by the list:
.011111...
Is this diagonalized number already in that list?
If yes (0.1 == .011111...) then -
is that list still "not complete" in any sense?
Thanks, Frank W ~@) R
ric...@fau.edu
kje...@sisyfos.uio.no (Kjell Fredrik Rogstad Pettersen) wrote:
> Gosh!!! Please explain to me
> 1. How do you define the real numbers?
> 2. How do you define addition on reals? (Is this ambigous too?)
> 3. Has every real number (with your definition) an additive inverse
> (i.e. for any x, there is a (-x) such that x+(-x)=0) ?
> 4. If we define x-y = x+(-y) for any x,y, why is this not an unambiguos
> definition?
1. As infinite decimals: 0.999... , 3.14159... , 1.000... . 2. More or
less digit by digit, with carries. Of course you can't start at the far
right, but there's at most a carry of one at each place, so there is no
real problem in defining addition unambiguously. 3. No. I'm really
thinking only of nonnegative real numbers. The fact that you can't
subtract unambiguously there makes the introduction of negative real
numbers somewhat problematic. There are several options available, but I
haven't found any that are really satisfactory. The problem is that the
positive decimal numbers are approximated from below, while the negative
ones are approximated from above. This makes them a little difficult to
mix. 4. It would be okay if we knew how to add a positive real number to
a negative real number. But it would not make subtraction unambiguous in
the sense that the equation a + x = b always has at most one solution.
That's because .999... + 1 = 1.999... and .999... + .999... = 1.999... .
Gene Scogin
ric...@fau.edu writes:
> Darrell Ryan <dar...@cafes.net> wrote:
>
> > x+x=1+x
> > x=1 <----- if that isn't obvious enough for you, you are
> > unteachable.
>
> Perhaps you are the one who is unteachable if you cannot conceive of the
> possibility that there is another solution. As was pointed out, any
> infinite cardinal is a solution. Why couldn't there be another real
> number that is also a solution?
>
> > Get it? x=1. x is NOT less than 1. You are trying to say x<1 AND is
> > a solution to the equation. It can't be both.
>
> It can't be? I claim that .999... is such a number. It is obviously less
> than 1, and when you add it to itself you get 1.999... , which is the
> same thing you get if you add it to 1.
OK then lets try proof by contridiciton:
I will, for the moment, assume that you are correct in your assertion
that 0.99999... is less than 1 and that 0.99999... is a solution to
the equation x + x = 1 + x
1) x = 0.99999...
2) y = 1 - x since we are assuming that x < 1 then y > 0, but
probably quite small
We can rewrite this as
3) x = 1 - y
Now lets bring back our old freind
4) x + x = 1 + x
Replace x with the value from equation 3
5) (1 - y) + (1 - y) = 1 + (1 - y)
Simplify
6) 2 - 2y = 2 - y
Subtract 2 from each side of the equation
7) -2y = -y
Divide both sides by -y
8) 2 = 1
So, IF 0.99999... is less than 1 AND 0.99999... is a solution
to x + x = 1 + x THEN 2 = 1
The fact "2=1" is inconsistant with the mathmatical system used to
generate that fact. Therefore the original premise that
"0.99999... is less than 1 AND 0.99999... is a solution to x + x = 1 +
x" is false.
Gene
--
DSC Communications Corporation Internet: gsc...@spd.dsccc.com
1000 Coit Road Plano, Texas 75075
**** The opinions expressed are not those of DSC Communications, Inc ****
Fred Richman
Gene Scogin wrote:
> OK then lets try proof by contridiciton:
> I will, for the moment, assume that you are correct in your assertion
> that 0.99999... is less than 1 and that 0.99999... is a solution to
> the equation x + x = 1 + x
> 1) x = 0.99999...
>
> 2) y = 1 - x since we are assuming that x < 1 then y > 0, but
> probably quite small
You can't subtract x from 1. That is the source of the contradiction. If
you could subtract 0.999... , then you could just subtract it from both
sides of the equation
0.999... + 0.999... = 1 + 0.999...
and conclude that 0.999... = 1.
--Fred
Fred Richman
Frank Wappler wrote:
That kind of diagonalization procedure need not work in binary if you
make the binary assumption that 0.111... = 1. If you have a list of (not
necessarily terminating) decimal numbers, then you can construct a
number whose n-th digit is different from the n-th digit of the n-th
entry in the list, and also different from 0 and 9. In that way you can
be sure that the newly constructed number is not equal to some number on
the list. And if you add it to the list, you can then construct a number
different from any number on the new list. And so on. But in binary you
run into the sort of trouble you ran into with your illustration. What
you might do there is make every odd digit of the new number equal to
zero, and the 2n-th digit different from the 2n-th digit of the n-th
entry in the list. I think that works if you don't allow any numbers on
the original list that end in all 1's.
--Fred
Unknown
[kje...@sisyfos.uio.no (Kjell Fredrik Rogstad Pettersen) writes: > In article <885998209...@dejanews.com>, ric...@fau.edu writes:
[> > It can't be? I claim that .999... is such a number. It is obviously less
[> > than 1, and when you add it to itself you get 1.999... , which is the
[> [snip]
[> > subtraction is not unambiguously defined for real numbers.
[> Gosh!!! Please explain to me
[>
[> 1. How do you define the real numbers?
[> 2. How do you define addition on reals? (Is this ambigous too?)
[> 3. Has every real number (with your definition) an additive inverse
[> (i.e. for any x, there is a (-x) such that x+(-x)=0) ?
[> 4. If we define x-y = x+(-y) for any x,y, why is this not an unambiguos
[> definition?
[> I am extremely curious about your answers!
[>
[> Kjell Fredrik
The definition of the reals is not perfect eg is
(2^(1/2))^1/2 a real number ?
c = a-b; a <= b is a problem requiring the inclusion
of zero as a number and the invention of the 'shadow set'
of negative numbers. Why is the form 0 - a; a +ve taken
to mean -a so that -a is wff and accorded the status of
an integer? How is it that "-" is used as a diadic operator,
a monadic operator and as part of a number symbol?
Why is there only one shadow set and not three or more? eg:
a Na NNa a Na NNa NNNa a N'a N'N'a N'N'N'a
-------- ------------- -------------------
0 1 0 0 1 2 0 0 2 1 0
1 0 0 1 2 0 1 1 0 2 1
2 0 1 2 2 1 0 2
Jan Stevens
[Posted and mailed]
In article <886106657....@dejanews.com>,
ric...@fau.edu writes:
> kje...@sisyfos.uio.no (Kjell Fredrik Rogstad Pettersen) wrote:
>
>> Gosh!!! Please explain to me
>
>> 1. How do you define the real numbers?
>
> 1. As infinite decimals: 0.999... , 3.14159... , 1.000... .
This definition is wrong! The problem of defining the reals is:
define the real numbers as extension of the rationals such that
the familiar operations also extend.
We have e.g. 1/3 + 2/3 = 1
so 0,333... + 0,666... = 0,999... = 1
Therefore we have to define 0,999... and 1,000... as different
representations of the same rational, just as the fraction 2/2 also
represents the integer 1.
Bye,
Jan.
Kjell Fredrik Rogstad Pettersen
In article <886106657....@dejanews.com>, ric...@fau.edu writes:
> kje...@sisyfos.uio.no (Kjell Fredrik Rogstad Pettersen) wrote:
> > 1. How do you define the real numbers?
>
> less digit by digit, with carries. Of course you can't start at the far
> right, but there's at most a carry of one at each place, so there is no
> real problem in defining addition unambiguously.
I totally agree that with YOUR definition of reals, 0.999... is different
from 1, and that addition is well-defined.
But you must realize that this is totally different from how reals are
defined in mathematics. There are several ways to do it, and the most
usual is by Cachy sequences over rationals. A Cachy sequence is a sequence
{a_1,a_2,...}
of rationals such that for any rational q>0 there is a number N such that
| a_n - a_m | < q for all n,m>N. Then we say that two Cachy sequences {a_n}
and {b_n} are equivalent if for any rational q>0, there is an N such that
| a_n - b_n | < q for all n>N. The reals are the set of all Cauchy sequences
modulo this equivalence relation.
By a decimal number 3.14159... we will mean the real number given by the
Cachy sequence {3/1, 31/10, 314/100, 3141/1000,...}. From this, one can
easily show that the Cachy sequence of 0.999... and 1.000... are equivalent,
so they give the same real number.
Define reals how you like it, it's a free world! But your definition is
practically useless for the normal purpose of reals.
> The fact that you can't
> subtract unambiguously there makes the introduction of negative real
> numbers somewhat problematic.
You make the problem. It is not problematic in the usual definition.
Kjell Fredrik
ric...@fau.edu
Jan Stevens wrote:
>> 1. As infinite decimals: 0.999... , 3.14159... , 1.000... .
> This definition is wrong!
Maybe. But that's the whole issue in this thread, isn't it? So it is
of some importance to explain exactly why this definition is wrong.
> The problem of defining the reals is:
> define the real numbers as extension of the rationals such that
> the familiar operations also extend.
That's a reasonable objective, but not everyone need espouse it.
> We have e.g. 1/3 + 2/3 = 1
> so 0,333... + 0,666... = 0,999... = 1
It looks like you think that 1/3 = 0,333... . If that were the case,
then clearly 1 = 0,999... (multiply both sides by 3). No skeptic of
the equality would agree that 1/3 = 0,333... . There is no solution
to the equation 3x = 1 in the (decimal) real numbers.
GSimp95605
Suppose someone wanted to insist on being "weird" and define the real numbers
as a finite sequence of digits (from 0 to 9) to the left of a decimal point
attached to a possibly infinite sequence of digits (from 0 to 9) to the right
of the decimal point.
In this case 1.999... really would be different from 2 because they are
different as sequences. How far could one get with such a definition. Is it a
group? Ring? Field? Can addition be uniquely defined? Division?
curious
Gilberto
Dave Seaman
I don't know where this thread is going, anymore. The discussion
started when you claimed that the standard model was deficient in the
way it deals with infinite limits. However, your model has exactly the
same defect. Even though an ordered field may contain infinitely large
(and infinitely small) elements, it still makes sense to talk about
limits of the form
lim f(x) = +oo
x->a
or
lim f(x) = b
x->oo
These expressions make sense because the elements of an ordered field
form a directed set, and therefore it makes sense to talk about whether
a net defined by f(x) on that directed set converges to anything, in
exactly the same way that the standard model talks about sequences. A
sequence is just a special case of a net. Therefore, I can't see that
you have changed anything in the way the model deals with an "absolute
infinity" that is not a member of the field under consideration.
In article <6ap1h0$i...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>________________________________________________________________________
>Construction of Non-Finite Numbers
>
>Notation
>
>4(0...0)1 means the finite binary string (0000)
>4(0...1)1 means the finite binary string (0001)
>4(1...1)1 means the finite binary string (1111)
>n(0...0)1 means the finite binary string (0...01)
> n 1
Surely you mean (0...0), with n zeroes, not (0...01). If 4(0...0)1 is
a string of four zeroes, then surely n(0...0)1 is a string of n
zeroes.
>w(0._.01)1 means the infinite binary string (0._.01)
> w 1
I take it the "w" is intended to represent the ordinal number omega.
If you have an omega-sequence of zeroes, followed by one more zero and
then a one, then the order type of your binary sequence is omega+2, not
omega, and therefore is seems you ought to write this as
(w+2)(0._.01)1.
>The sign "..." indicates a finite range
>The sign "._." indicates an infinite range
What kind of infinite range? An omega-sequence? Some of your later
notation seems to contradict this.
>w[0._.01]1 means the infinite number [0._.01]
> w 1
>
>corresponding to the binary string w(0._.01)1
>
>This number is not strictly the number 1 but its equivalent expressed as an infinite number.
Again, I see an (omega+2)-sequence here, not an omega-sequence. It
can't be an omega-sequence, because then it wouldn't have a last
digit. Do you also identify 1 with the (omega+1)-sequence
(w+1)(0._.1)1, as well as the (omega+3)-sequence (w+3)(0._.001)1, and
so on?
Perhaps this is how you handle carries -- a question I asked last time
that you still haven't addressed. Perhaps you mean the numbers can be
expressed in more than one way, as in
1 = (w+1)(0._.1)1 = (w+2)(0._.01)1
and therefore to compute 1+1 you use the second representation above in
order to avoid the problem that the carry has no place to go.
>We can concatenate two discrete sets to represent non-finite numbers with lage infinite,
>finite and small infinite (infinitesimal) components.
>
>For example:
>
> ._.] [0._.01] [0._.01] [0._.01] [0._.01] [ ._.
> ._.] 2w w+1 w 1 0 -w+1 -w -2w+1
Obviously, your notation differs from mine. Your numbering of the
digit positions doesn't seem to make sense. Do the sequences
represented by ._. above have a first element, a last element, both,
or neither? Perhaps the answer varies depending on which part of the
string you are looking at?
>
>This is the number 2^w + 1 + 2^(-w) + 2^(-2w) = U_1 + U_0 + U_(-1) + U_(-2)
>
>The class which contains this number is N2w U Nw U N0 U N(-w)
It appears that you have a representation for 2^w, but what is 3^w?
What about e^w = 1 + w + w^2/2! + w^3/3! + ...? Of what class is this
number?
>The numbers so defined lie within sets which are bounded above and below
>meaning that they have a greatest and leasr member.
Does that mean that U1-1 is undefined? How about U1/2?
This has gone far afield from the original 1.999...=2 discussion. Is there
supposed to be a connection here somewhere?
Frank Wappler
Fred Richman wrote:
> Frank Wappler wrote:
> > .1
> > .001
> > .01
> > .11
> > .101
> > .011
> > ...
> > (continue in the "obvious way" [...])
> > Form a diagonalized number of that list, by picking the n-th digit from
> > the n-th entry of that list and altering it. In the binary system that
> > leaves no choice but to interchange "1" and "0", the diagonalized
> > number is uniquely determined by the list: .011111...
> > If this diagonalized number already in that list (0.1 == .011111...)
> > then - is that list still "not complete" in any sense?
> If you have a list of (not necessarily terminating) decimal numbers,
> then you can construct a number whose n-th digit is different from the
> n-th digit of the n-th entry in the list, and also different from 0 and 9.
> In that way you can be sure that the newly constructed number is not equal
> to some number on the list. And if you add it to the list, you can then
> construct a number different from any number on the new list. And so on.
> [...]
> But in binary you run into the sort of trouble you ran into with your
> illustration. [...]
> That kind of diagonalization procedure need not work in binary if you
> make the binary assumption that 0.111... = 1.
> What you might do there is make every odd digit of the new number equal
> to zero, and the 2n-th digit different from the 2n-th digit of the n-th
> entry in the list. I think that works [...]
From the list above results .010101...01..., surely different
from every entry in the list, and not prone to ambiguity.
(Is this _really_ _strictly_ certain, even for _very_ large entry numbers?
Also implying: is even the way I constructed 0.11111... above _strict_?)
Either way, the fix you suggest seems to work ... or does it?
(To be honest about it:
As my earlier posts to this newsgroup indicate, I don't consider this
"troublesome" but "a theorem": numbers _are_ countable.
I guess I owe you for finally pointing me to an interesting technique ...)
> [...] if you don't allow any numbers on the original list that end
> in all 1's.
Here's the catch, IMHO.
The point is that the ambiguity in notation (01111... = 10000... )
doesn't merely "effect" the constructed diagonalized number but of
course also the entries in the list.
The question is:
Is there _any_ diagonalization procedure (for a given supposedly complete
list) which constructs a number that is not already in the list
(or equal to "0" or "1") when allowing for _all possible_ single or
multiple interchanges of notation?
Can _all observers_ prove (by such a procedure) that reals are more
than countable?
(The notion of "observer" is surely unusual in a mathematical context;
as in physics, I'd like it to indicate that every observer may use
any notation that him/her/it pleases, regardless of whether or not that
denotes the "same". Here "sameness" should be _measured_ by "equality, =".
I hope we agree that any proof should hold independent of the notation
in which it is cast. A proof is an automatic procedure to be performed
on axioms in their notation, whatever it is specificly.)
My main point is that it is not the notation that identifies any
number but its value, its context wrt. other numbers.
Since that context is usually developed starting from "zero" and
countable many axioms, I find that numbers ought to be countable as well
(contrary to "folklore", of course, unfortunately).
I figure that between this and "1.999... = 2" something has to give.
Best regards, Frank W ~@) R
Verdigris
Dave Seaman wrote:
>
> I don't know where this thread is going, anymore. The discussion
> started when you claimed that the standard model was deficient in the
> way it deals with infinite limits. However, your model has exactly the
> same defect. Even though an ordered field may contain infinitely large
> (and infinitely small) elements, it still makes sense to talk about
> limits of the form
>
> lim f(x) = +oo
> x->a
>
> or
>
> lim f(x) = b
> x->oo
>
> These expressions make sense because the elements of an ordered field
> form a directed set, and therefore it makes sense to talk about whether
> a net defined by f(x) on that directed set converges to anything, in
> exactly the same way that the standard model talks about sequences. A
> sequence is just a special case of a net. Therefore, I can't see that
> you have changed anything in the way the model deals with an "absolute
> infinity" that is not a member of the field under consideration.
>
> In article <6ap1h0$i...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
> >Construction of Non-Finite Numbers
> >
> >Notation
> >
> >4(0...0)1 means the finite binary string (0000)
> >4(0...1)1 means the finite binary string (0001)
> >4(1...1)1 means the finite binary string (1111)
> >n(0...0)1 means the finite binary string (0...01)
> > n 1
>
> a string of four zeroes, then surely n(0...0)1 is a string of n
> zeroes.
> >w(0._.01)1 means the infinite binary string (0._.01)
> > w 1
>
> If you have an omega-sequence of zeroes, followed by one more zero and
> then a one, then the order type of your binary sequence is omega+2, not
> omega, and therefore is seems you ought to write this as
No, it is defined by 2^w = 1 sand is a fraction relative the least whole
infinite number U1. ie w < U1/n; n e S0.
> (w+2)(0._.01)1.
>
> >The sign "..." indicates a finite range
> >The sign "._." indicates an infinite range
>
> What kind of infinite range? An omega-sequence? Some of your later
> notation seems to contradict this.
Each set Naw; a e S0 is a bounded set.
> >w[0._.01]1 means the infinite number [0._.01]
> > w 1
> >
> >corresponding to the binary string w(0._.01)1
> >
> >This number is not strictly the number 1 but its equivalent expressed as an infinite number.
>
> Again, I see an (omega+2)-sequence here, not an omega-sequence. It
> can't be an omega-sequence, because then it wouldn't have a last
> digit. Do you also identify 1 with the (omega+1)-sequence
> (w+1)(0._.1)1, as well as the (omega+3)-sequence (w+3)(0._.001)1, and
> so on?
>
> Perhaps this is how you handle carries -- a question I asked last time
> that you still haven't addressed. Perhaps you mean the numbers can be
> expressed in more than one way, as in
This is a hierarchy, arithmetical operations on the LHS of any set can
never
reach the RHS of the set.
> 1 = (w+1)(0._.1)1 = (w+2)(0._.01)1
>
> and therefore to compute 1+1 you use the second representation above in
> order to avoid the problem that the carry has no place to go.
>
> >We can concatenate two discrete sets to represent non-finite numbers with lage infinite,
> >finite and small infinite (infinitesimal) components.
> >
> >For example:
> >
> > ._.] [0._.01] [0._.01] [0._.01] [0._.01] [ ._.
> > ._.] 2w w+1 w 1 0 -w+1 -w -2w+1
>
> Obviously, your notation differs from mine. Your numbering of the
> digit positions doesn't seem to make sense. Do the sequences
> represented by ._. above have a first element, a last element, both,
> or neither? Perhaps the answer varies depending on which part of the
> string you are looking at?
I think there was a typographical error in the previous posting.
The definition is:
a = SUM(0,w,br x 2^(r-1))
>
> >
> >This is the number 2^w + 1 + 2^(-w) + 2^(-2w) = U_1 + U_0 + U_(-1) + U_(-2)
> >
> >The class which contains this number is N2w U Nw U N0 U N(-w)
>
> It appears that you have a representation for 2^w, but what is 3^w?
> What about e^w = 1 + w + w^2/2! + w^3/3! + ...? Of what class is this
> number?
3^w is undefined, 3^w is the least infinite whole number in N3w.
> >The numbers so defined lie within sets which are bounded above and below
> >meaning that they have a greatest and leasr member.
>
> Does that mean that U1-1 is undefined? How about U1/2?
I have already said that U1-1 is the greates member of Nw, where
U1-1 = w[1._.1] so U1/2 = w[01._.0]. U1 is the least member of N(w+w).
> This has gone far afield from the original 1.999...=2 discussion. Is there
> supposed to be a connection here somewhere?
>
> --
> Dave Seaman dse...@purdue.edu
My original point was that such symbols as .999... are undefined because
they
depend, in one way or another, on the concept of infinity because they
require
symbols such as lim f(x) = +oo to explain them.
n->oo
I have argued that the meaning of 'infinity' in this context is
different from the
more definite idea of an infinite ordinal set which is bounded above and
below
by a member of that set. The finitistic application of the operator '+'
to the 'finite'
members of this set will always map to finite members. The symbol
.999... could
therefore mean the unlimited repetition of the same digit without
producing a symbol
of infinite length. This is on the principle that a finite number of
operations on a finite number can never result in an infinite number.
Alternatively, the operator '+' applied to the upper bound of the set
will jump out of
the set. This can be generalised in the idea G + L = T , where T is
never in the base
set. Thanks to the intractable nature of absolute infinity, we can
always construct
a new disjoint set in which T is the least member. It is clear from this
process that the operator "+" carries with it the seeds of infinity. The
foundation of arithmetic on this operator leads inevitably, therefore,
to a consideration of the meaning of infinity.
he infinite process can be curtailed by the use of a finite or infinite
modulus, so that
mapping within the set is maintained, but this is not applicable to the
symbol .999...
The binary power set P2w, where w is a non-finite number, includes a
greatest member
w(1._.1) and a least member w(0._01). These symbols, represent infinite
entities which
can be used in a consistent logical calculus. The type of infinity used
here does not lead to indefinite symbols.
The idea of an infinite sequence in arithmentic is finitistic, in the
sense that the
result of the process will always remain within the set I have called
S0, the set
of finite positive integers. Alternatively, an infinite sequence can be
viewed as
having no upper bound whatever. If .999... is the first type of
construction,
it can never have a finite upper bound but it can have an infinite upper
bound
which is not a member of the set. In the second type of construction it
can never
have either a finite or infinite bound. This latter case produces
entities which
are not properly defined.
The questions you have not answered are:
(A) Does the symbol .999... involve the idea of infinity in any way ?
If the answer is Yes:
(B) How is the idea of infinity used to be defined ?
ie please provide a satisfactory definition of infinity in this context.
NB: The epsilon-delta definition involves the finitistic type of
infinity described above
and is subject to the ojections I have already made.
If the answer is No:
(C) How can the symbol .999... be defined without directly of indirectly
introducing
the idea of infinity.
In summary, If you use an absolutely unbounded infinity, the result
will always be
undefined but if you use a bounded infinity, you have to accept that
.999... < 1. If you
define .999... = 1 you surreptitiously introduce the concept of absolute
infinity and all the consequences which follow from using a concept
which is inherently indefinite in meaning.
It may be the case that the means used to avoid the incorporation of an
absolute infinity into numbers defined as infinite sequences has been
perfectly accomplished. However, I have not yet seen a convincing
demonstration of this.
Fred Richman
Kjell Fredrik Rogstad Pettersen wrote:
> I totally agree that with YOUR definition of reals, 0.999... is different
> from 1, and that addition is well-defined.
>
> But you must realize that this is totally different from how reals are
> defined in mathematics.
I don't know about "totally". I think Courant in his calculus book
introduced reals as infinite decimals---it is not an unheard of
definition. The difference lies only in the treatment of infinite
decimals that end in all 9's. That is precisely the subject of this
thread.
> There are several ways to do it, and the most
> usual is by Cachy sequences over rationals. A Cachy sequence is a sequence
>
> {a_1,a_2,...}
>
> of rationals such that for any rational q>0 there is a number N such that
> | a_n - a_m | < q for all n,m>N. Then we say that two Cachy sequences {a_n}
> and {b_n} are equivalent if for any rational q>0, there is an N such that
> | a_n - b_n | < q for all n>N. The reals are the set of all Cauchy sequences
> modulo this equivalence relation.
Well, sure. If you put that equivalence relation on the Cauchy sequences
you will conclude that .999... = 1. In fact, you are simply making the
equality .999... = 1 true by definition. You can certainly do that;
"it's a free world", as you said. Obviously the sequence .9, .99, .999,
... converges to 1. The question is whether that justifies saying that
.999... = 1. It is a bit like Zeno's paradox: it keeps getting closer,
but never quite gets there.
--Fred
Fred Richman
Frank Wappler wrote:
> Fred Richman wrote:
>
> > [...] if you don't allow any numbers on the original list that end
> > in all 1's.
>
> Here's the catch, IMHO.
> The point is that the ambiguity in notation (01111... = 10000... )
> doesn't merely "effect" the constructed diagonalized number but of
> course also the entries in the list.
Is that a problem? All you are trying to do is to construct a number
that is not equal to any number on the list. Does it hurt to replace a
number on the list like .01111... by the equal number .10000... ?
> I hope we agree that any proof should hold independent of the notation
> in which it is cast.
I'm not so sure of that. In any event, I don't think it has to be
independent of the notation it uses for the objects it deals with. For
example, I could prove that 1578465 is divisible by 9 by observing that
1 + 5 + 7 + 8 + 4 + 6 + 5 = 36 is divisible by 9. This proof depends
heavily on the use of base 10 notation.
--Fred
Kjell Fredrik Rogstad Pettersen
In article <6ar130$p...@newsserver.trl.OZ.AU>, Verdigris <> writes:
> The definition of the reals is not perfect eg is
> (2^(1/2))^1/2 a real number ?
Yes, of course. Why not?
> c = a-b; a <= b is a problem requiring the inclusion
> of zero as a number and the invention of the 'shadow set'
> of negative numbers.
Why is it a problem?
> How is it that "-" is used as a diadic operator,
> a monadic operator and as part of a number symbol?
It is common to use it for several purposes as long as it is clear
what is what.
> Why is there only one shadow set and not three or more? eg:
This is more philosophical than practical when talking about
mathematics. Does there exist any number at all? What is 1
actually?
We create hammers and saws not by following one unique "true" way
to do it, a truth that existed before man, but because we find it
practical to do it so. It works the way we do it. The basic idea
is the same about mathematics. There are several ways to define
what you could call numbers, but by not allowing negative numbers
and subtraction, it has a rather limited application. There is a
common way to regard reals (OK, several ways to define it, but
they all lead to the same) such that it is a nice extention of
rationals, carring all basic properties with it.
Kjell Fredrik
Kjell Fredrik Rogstad Pettersen
In article <6av2nv$f...@bgtnsc02.worldnet.att.net>, Fred Richman <ric...@fau.edu> writes:
> Kjell Fredrik Rogstad Pettersen wrote:
>
> > I totally agree that with YOUR definition of reals, 0.999... is different
> > from 1, and that addition is well-defined.
> >
> > But you must realize that this is totally different from how reals are
> > defined in mathematics.
>
> I don't know about "totally".
OK, a too strong word. Wy point was that the result you come up with has a
very important difference, namely that you do not accept 0.999...=1. It
seems to be minor, but have a lot of serious consequences.
Of course, there are several ways to define reals. One way is by
Cauchy-sequences, another is, as you say, to
> introduce reals as infinite decimals---it is not an unheard of
> definition. The difference lies only in the treatment of infinite
> decimals that end in all 9's. That is precisely the subject of this
> thread.
As I have said earlier, you can define it the way you like. My philosophy
is that you can't say that a definition is "wrong", but you can point out
its usefulness or uselessness. I calim that your definition of reals is
very difficult to use for any good purpose.
1. You can anly do addition, not subtraction.
2. It is impossible to describe it as an extension of the rationals. As
somebody pointed out in another posting: You will have problems with
1/3. With your definition, there is no real number x such that
x+x+x=1. (If I'm wrong, what's the number?)
3. It is not even an extension of integers, since you seem to have problems
with negative numbers. Only natural numbers could be included.
4. You will have a lot of problems with limits and you can not do calculus.
So tell me: Why do you define reals as you do? Can it be used for anything?
When I define reals, I want a number system that
- extends the rationals such that addition, subtraction, multiplication and
non-zero division is the same. I want a field with rationals as subfield.
- Has the supremum property: For any set S with an upper bound, there is a
real number x such that x>=a for all a in S and for all y with the same
property, we have x<=y. This is fundamental for doing calculus, limits etc.
> --Fred
Kjell Fredrik
Fred Richman
Kjell Fredrik Rogstad Pettersen wrote:
> As I have said earlier, you can define it the way you like.
That is very tolerant of you.
> My philosophy
> is that you can't say that a definition is "wrong", but you can point out
> its usefulness or uselessness. I calim that your definition of reals is
> very difficult to use for any good purpose.
That is certainly the way to attack it. But I doubt if there is much
that can't be done using only decimal fractions, which is all we need in
practice. Moreover, at least for the positive real numbers, the decimal
expansion approach actually gives a richer system. One can talk about
equality, and also about equality in the limit. As for the negative
numbers, I know how to incorporate them by approximating from below,
just as for the positive numbers. It just doesn't seem very satisfactory
to me because the natural way, suggested by the notation -3.14159... ,
is to approximate them from above.
The purpose of the definition is to accommodate the intuition that
.999... < 1, just as one of your purposes is to be able to subtract.
> 1. You can anly do addition, not subtraction.
Certainly true, at least in general. Some subtractions are possible.
> 2. It is impossible to describe it as an extension of the rationals. As
> somebody pointed out in another posting: You will have problems with
> 1/3. With your definition, there is no real number x such that
> x+x+x=1. (If I'm wrong, what's the number?)
No, you are right. In fact, I believe I pointed that out in response to
someone who wrote as if 1/3 = .333... . Is that a serious problem for
you? I find it interesting, a natural consequence of our choice of 10 as
a number base.
> 3. It is not even an extension of integers, since you seem to have problems
> with negative numbers. Only natural numbers could be included.
Negative integers aren't a problem. The decimal fractions are
cancellable, so you can introduce additive inverses for them without
difficulty.
> 4. You will have a lot of problems with limits and you can not do calculus.
I don't think there is any problem with limits. In fact, .999... is
equal to 1 in the limit.
> When I define reals, I want a number system that
> - Has the supremum property: For any set S with an upper bound, there is a
> real number x such that x>=a for all a in S and for all y with the same
> property, we have x<=y. This is fundamental for doing calculus, limits etc.
Fundamental? I don't think so. I much prefer calculus without the
supremum property, which is highly nonconstructive. Anything worthwhile
in the calculus can be done without invoking the supremum property.
--Fred
Frank Wappler
Fred Richman wrote:
> > > What you might do there is make every odd digit of the new number
> > > equal to zero, and the 2n-th digit different from the 2n-th digit
> > > of the n-th entry in the list.
> Frank Wappler wrote:
> > The point is that the ambiguity in notation (01111... = 10000... )
> > doesn't merely "effect" the constructed diagonalized number but of
> > course also the entries in the list.
> Is that a problem? All you are trying to do is to construct a number
> that is not equal to any number on the list. Does it hurt to replace a
> number on the list like .01111... by the equal number .10000... ?
Yes, it "hurts": then your suggested diagonalization gives 0.000...
I guess if you admit 0.000... and 1.000... = .111... as entries
in the list - why not as the first two - then the diagonalization
"proof" fails.
> > I hope we agree that any proof should hold independent of the
> > notation in which it is cast.
> I'm not so sure of that.
I find this of central importance. Any comments welcome.
> I could prove that 1578465 is divisible by 9 by observing that
> 1 + 5 + 7 + 8 + 4 + 6 + 5 = 36 is divisible by 9. This proof depends
> heavily on the use of base 10 notation.
The proof for the theorem
"(Ex | 1578465_10 = 9*x) <-> Ey | 1 + 5 + 7 + 8 + 4 + 6 + 5 = 9*y"
starting from Peano axioms (I guess) certainly does not.
That's my point:
You can play all kinds of neat/stupid tricks with notation
(Goedel springs to mind ...), but the justification, the meaning,
that what can be communicated to _any_body else is not in the
notation but in the structure of the axioms and how they may be
legally manipulated. The concept of "number" includes more than
some specific symbols.
Frank W ~@) R
Verdigris
Kjell Fredrik Rogstad Pettersen wrote:
>
Thanks.
On the first point, I was quoting from "Real Variable" J.M Hyslop,
Oliver & Boyd.
page 11. where the following symbols are stated as undefined in
analysis:
0^(-2), 5/0, 2^(2^1/2)), (-1)^(1/3), (-2)^(1/2)
Your principle of practicality is a good from the point of practical
application,
but is this a respectable principle as far as pure mathematics is
concerned?
See Hardy's "A Mathematicians apology, chapter 28.
NATHANIEL SILVER
(2^(1/2))^1/2 = 2^(1/4), fourth root of 2.
As for 2^(2^1/2), consider the following sequence, which converges to
2^1/2 = 1.41421356...
a(1) = 2^1, a(2) = 2^(14/10), a(3) = 2^(141/100), a(4) = 2^(1414/1000),...
Define 2^(2^1/2) as the limit (as n goes to infinity) of 2^(a(n)).
Verdigris <EATh...@uq.net.au> wrote in article
<34D4D0...@uq.net.au>...
> Kjell Fredrik Rogstad Pettersen wrote:
> >
> > In article <6ar130$p...@newsserver.trl.OZ.AU, Verdigris writes:
> >
> > > The definition of the reals is not perfect eg is
> > > (2^(1/2))^1/2 a real number ?
> >
> > Yes, of course. Why not?
> >
Fred Richman
> > Frank Wappler wrote:
>
> > > The point is that the ambiguity in notation (01111... = 10000... )
> > > doesn't merely "effect" the constructed diagonalized number but of
> > > course also the entries in the list.
>
> > Is that a problem? All you are trying to do is to construct a number
> > that is not equal to any number on the list. Does it hurt to replace a
> > number on the list like .01111... by the equal number .10000... ?
>
> Yes, it "hurts": then your suggested diagonalization gives 0.000...
> I guess if you admit 0.000... and 1.000... = .111... as entries
> in the list - why not as the first two - then the diagonalization
> "proof" fails.
Of course it hurts my suggested diagonalization to have .01111... on the
list. That's why I want to replace it by .10000... . The question is
whether it hurts the theorem which says that I can construct a number
that is not equal to any number on the list.
> > > I hope we agree that any proof should hold independent of the
> > > notation in which it is cast.
>
> > I'm not so sure of that.
>
> I find this of central importance. Any comments welcome.
For one thing, I'm not sure it makes any sense. What does it mean for a
proof to be independent of the notation in which it is cast? What is an
example of an alleged proof that is not independent of the notation in
which it is cast? What about the proof I give below?
> > I could prove that 1578465 is divisible by 9 by observing that
> > 1 + 5 + 7 + 8 + 4 + 6 + 5 = 36 is divisible by 9. This proof depends
> > heavily on the use of base 10 notation.
>
> The proof for the theorem
> "(Ex | 1578465_10 = 9*x) <-> Ey | 1 + 5 + 7 + 8 + 4 + 6 + 5 = 9*y"
>
> starting from Peano axioms (I guess) certainly does not.
What if it doesn't? That's another theorem, and another proof. Moreover,
your theorem itself is *about* decimal notation---that's the only thing
that makes it interesting. I've given a proof that 1578465 is divisible
by 9 that depends heavily on the notation. I think it's a reasonable
proof despite that. Maybe what you mean is that there should alway be
*some* proof of the theorem that is independent of the notation, not
that *every* proof should be.
For the case at hand, the diagonal argument showing that, given a list
of real numbers, you can construct another one that is not on the list,
you can get a proof that doesn't use decimal representation (or binary
representation), but the one that does is particularly attractive
because we are so used to decimal representation.
--Fred
Frank Wappler
Fred Richman wrote:
> That's why I want to replace .01111... by .10000... .
> construct a number that is not equal to any number on the list.
My point is that if you don't get to choose if and which to replace
but want to be applicable to any listing of the same _values_
(where "same" means "="), i.e. _allowing_ any replacement,
then no diagonalization might work at all, there might be no proof for
that "string of symbols" (which is therefore not - yet - a theorem.)
> What is an example of an alleged proof that is not independent of
> the notation in which it is cast?
Notation 1:
Theorems:
(1 + 4 = 5), (2 + 3 = 5), (A = C) & (B = C) -> (A = B)
Proof technique:
use the last theorem as template;
assign the symbols "1 + 4" to "A", "5" to "C", "2 + 3" to "C";
connect the theorems with symbols "&" and "->"
Resulting proven theorem:
(1 + 4 = 5) & (2 + 3 = 5) & ((A = C) & (B = C) -> (A = B)) -> 1 + 4 = 2 + 3
Notation 2:
Replace the symbol "(" with "quwyirbqo";
replace the symbol "1" with "&@^%$_#)@^*_";
etc.
> I've given a proof that 1578465 is divisible by 9 that depends heavily
> on the notation.
You (or some reliable translator) must be able to transform your proof
to the "standard notation", which uses only the symbols "Zero", "Successor",
"(", ")", "Exist", "All", "Not", "VariableSymbol", "Prime", "Plus", and
"Times" (as far as I recall them).
All the many theorems that you implied and used in the proof you gave can
be translated into that notation; that's how everybody can determine that
those _are_ theorems and you indeed gave a proof instead of just making
something up.
Someone who doesn't recognize the particular notation ("decimal") that
you chose will not be able to recognize what you presented as a proof
at all; you have to supplement a symbolic translation table and
the steps in which the transformations are to be applied.
> That's another theorem, and another proof.
It's the same, only in another notation; they are "(...??)-morph",
"equivalent under transformation".
> For the case at hand, the diagonal argument showing that, given a list
> of real numbers, you can construct another one that is not on the list,
> you can get a proof that doesn't use decimal representation (or binary
> representation), but the one that does is particularly attractive
> because we are so used to decimal representation.
What if I for one were _very_ comfortable with binary notation?
Should that allow me to use different mathematical theorems than
yourself?
Frank W ~@) R
Frank Wappler
Frank Wappler wrote:
> Notation 1:
> [...]
> (1 + 4 = 5) & (2 + 3 = 5) & ((A = C) & (B = C) -> (A = B))
> -> 1 + 4 = 2 + 3
> Notation 2:
> Replace the symbol "(" with "quwyirbqo";
> replace the symbol "1" with "&@^%$_#)@^*_";
(there was a typo: of course it should be `assign symbol "2 + 3" to B')
> Fred Richman wrote:
> > What is an example of an alleged proof that is not
^^^^^^^^
> > independent of the notation in which it is cast?
Sorry, I had misread your question. Obviously, that example
is one of a proof which _is_ independent of notation;
as anything that I would consider "a proof" at all.
The issue of being _not_ independent seems to come up with the
diagonalization/non-countability "proofs" we discussed;
if it wouldn't work in binary notation, why should what one
may derive in any other notation be accepted as "proof"?
Frank W ~@) R
Unknown
Fred Richman <ric...@fau.edu> writes: > > > Frank Wappler wrote:
[> For the case at hand, the diagonal argument showing that, given a list
[> of real numbers, you can construct another one that is not on the list,
[> you can get a proof that doesn't use decimal representation (or binary
[> representation), but the one that does is particularly attractive
[> because we are so used to decimal representation.
[> --Fred
The flaw in the diagonal argument is that it starts with
a false assumption, namely that there is a diagonal which
spans the complete list.
(1) The assumption that there is a diagonal implies that the
list of numbers comprises a square matrix of digits,otherwise
there could be no diagonal in the intended sense of the proof.
(2) The number of columns of digits in the list (whether finite or not)
is the same as the number of digits in a square matrix.
(3) The exhaustive permutation of s symbols in n columns is s^n > n.
(4) The matrix containing all such permutations is, therefore,
a rectangle of dimension s^n x n.
(5) There is no diagonal, in the intended sense, in such
a rectangular matrix.
(6) The fact that the permutations cannot be listed in a square
matrix is irrelevant to the question as to whether they can be listed
at all.
The conjuror invites us to look at the square so that we cannot see
that the rectangle of unequal sides does contain all the numbers.
Kjell Fredrik Rogstad Pettersen
> On the first point, I was quoting from "Real Variable" J.M Hyslop,
> Oliver & Boyd.
> page 11. where the following symbols are stated as undefined in
> analysis:
>
> 0^(-2), 5/0, 2^(2^1/2)), (-1)^(1/3), (-2)^(1/2)
2^(2^(1/2)) should be unproblemtic to define. You can go via exp and
ln, and those are trivially defined.
(-1)^(1/3) is also possible. But you might get problems when extending
to complex.
The others are impossible, without making conflicts with the most
basic rules for exponents.
> Your principle of practicality is a good from the point of practical
> application,
> but is this a respectable principle as far as pure mathematics is
> concerned?
When I mean practiality, I still think about pure mathematics, not only
applied. I think this is an important principle when defining anything.
Of course, I do not claim that other definitions are objectively wrong,
but I like to see some use for it. The definition of reals used today
seems to be the best one from this point of view.
But if others like to do it differently just for fun, don't let me stop
them.
Kjell Fredrik
Kjell Fredrik Rogstad Pettersen
In article <6b0avm$8...@bgtnsc03.worldnet.att.net>, Fred Richman <ric...@fau.edu> writes:
> Moreover, at least for the positive real numbers, the decimal
> expansion approach actually gives a richer system.
Why richer? Because it seems to contain more?
> The purpose of the definition is to accommodate the intuition that
> .999... < 1, just as one of your purposes is to be able to subtract.
My intuition is that .999... = 1.
On non-extention of rationals, you ask:
> Is that a serious problem for you?
I really like to bring with me the rational numbers, if not you can not do
elementary things in calculus (see further down)
> Negative integers aren't a problem. The decimal fractions are
> cancellable, so you can introduce additive inverses for them without
> difficulty.
With that extension, you can not have both addition and associativity of
addition. Just try to compute
( .999... + (-1) ) + 1 and .999... + ( (-1) + 1 )
You will carry with you a bunch of bracets you can't resolve. I will not
call that simple.
> I don't think there is any problem with limits. In fact, .999... is
> equal to 1 in the limit.
So if a limit converges, it does not have a unique real number as the
limit? OK, that is possible in topology in general, but you will indeed
have problems with being precise in calculus.
> Fundamental? I don't think so. I much prefer calculus without the
> supremum property, which is highly nonconstructive. Anything worthwhile
> in the calculus can be done without invoking the supremum property.
Can you differentiate? (How, definition)
Integrate? (Try integrate x^2, and remember division by 3 is forbidden)
Have the mean value theorem?
( f(x)=3*x, f(0)=0 and f(1)=3, find c between 0 and 1 s.t. f(c)=1 )
Kjell Fredrik
Jan Stevens
In article <886170104....@dejanews.com>,
ric...@fau.edu writes:
> Jan Stevens wrote:
[...]
>
>> The problem of defining the reals is:
>
>> define the real numbers as extension of the rationals such that
>> the familiar operations also extend.
>
> That's a reasonable objective, but not everyone need espouse it.
>
Yes, everyone has to. You are free to define your decimal `reals' but
you should not use the term real numbers for them. Eudoxos had already
a clear understanding of the reals. Dedekind even wrote a popular book
`Was sind und sollen die reelle Zahlen?'.
The real numbers can be represented on the number line.
It is possible to cut a unit segment into three parts, showing
that the equation 3x = 1 has a solution.
It is perfectly possible to define the reals without using the decimal
system. In fact it is rather arbitrary to use decimals
(or any other fixed base). One can also use continuous fractions.
>> We have e.g. 1/3 + 2/3 = 1
>> so 0,333... + 0,666... = 0,999... = 1
>
> It looks like you think that 1/3 = 0,333... .
This is just long division!
Jan.
ric...@fau.edu
kje...@sisyfos.uio.no (Kjell Fredrik Rogstad Pettersen) wrote:
> Fred Richman <ric...@fau.edu> writes:
> > Moreover, at least for the positive real numbers, the decimal
> > expansion approach actually gives a richer system.
> Why richer? Because it seems to contain more?
It *does* contain more. When you identify .999... with 1, you get a
quotient of the set of decimal numbers.
> > The purpose of the definition is to accommodate the intuition that
> > .999... < 1, just as one of your purposes is to be able to subtract.
> My intuition is that .999... = 1.
I guessed as much. But other people have other intuitions.
> > Negative integers aren't a problem. The decimal fractions are
> > cancellable, so you can introduce additive inverses for them without
> > difficulty.
>
> With that extension, you can not have both addition and associativity of
> addition. Just try to compute
> ( .999... + (-1) ) + 1 and .999... + ( (-1) + 1 )
>
> You will carry with you a bunch of bracets you can't resolve. I will not
> call that simple.
That's your prerogative. But don't tell me the associative law doesn't
hold. It does. The fact that you can't write ( .999... + (-1) ) in any
simpler form is no more disturbing than that you can't write the fraction
3/5 in any simpler form. When I multiply 3/5 by 5, I get 3. When I add 1
to ( .999... + (-1) ), I get .999... .
> > I don't think there is any problem with limits. In fact, .999... is
> > equal to 1 in the limit.
>
> So if a limit converges, it does not have a unique real number as the
> limit? OK, that is possible in topology in general, but you will indeed
> have problems with being precise in calculus.
I don't think so. We have equality, and we have equality in the limit.
We use whichever one is appropriate.
> > Fundamental? I don't think so. I much prefer calculus without the
> > supremum property, which is highly nonconstructive. Anything worthwhile
> > in the calculus can be done without invoking the supremum property.
>
> Can you differentiate? (How, definition)
Of course. The definition is the same as the one you are familiar with.
> Integrate? (Try integrate x^2, and remember division by 3 is forbidden)
Hold it. I thought we were talking about getting along without invoking
the supremum property. If you're still worried about decimal numbers, the
integral of x^2 is (.333... )x^3. Integrals and derivatives are limits,
so the appropriate context is equality in the limit.
> Have the mean value theorem?
> ( f(x)=3*x, f(0)=0 and f(1)=3, find c between 0 and 1 s.t. f(c)=1 )
I believe that is the intermediate value theorem, not the mean value
theorem. For the function you give, of course the intermediate value
theorem holds. For an arbitrary uniformly continuous function, I would
say it does not. The problem is the sensitivity of c to the data that
describes f. The problem is ill posed. This precludes being able to
construct c in general. What is true is that you can always find c such
that f(c) is as close to 1 as you please.
ric...@fau.edu
fw7...@csc.albany.edu (Frank Wappler) wrote:
> The issue of being _not_ independent seems to come up with the
> diagonalization/non-countability "proofs" we discussed;
> if it wouldn't work in binary notation, why should what one
> may derive in any other notation be accepted as "proof"?
That seems a little odd. If we wanted to prove theorems about the
properties of a certain parabola, we would normally choose a coordinate
system tailored to the problem. Would you then claim that nothing we
proved was valid because the exact same proof wouldn't work in some
other coordinate system? Choosing a notation for real numbers is much
the same as choosing a coordinate system.
Dave Seaman
In article <6b3fq9$4...@newsserver.trl.OZ.AU>, Verdigris <> wrote:
>The flaw in the diagonal argument is that it starts with
>a false assumption, namely that there is a diagonal which
>spans the complete list.
A list is an omega-sequence of decimal expansions. Each decimal
expansion is an omega-sequence of digits. Therefore, the diagonal
exists and is indexed by omega.
What the proof shows is that, given any omega-list of decimal (or
binary) expansions, there is a real number not represented by any
expansion in the list. You have to be careful, of course, about the
fact that the mapping of decimal (binary) expansions to real numbers is
not one-to-one, but that turns out to be a minor problem.
Dave Seaman
In article <34D3A5...@uq.net.au>, Verdigris <EATh...@uq.net.au> wrote:
>My original point was that such symbols as .999... are undefined because
>they
>depend, in one way or another, on the concept of infinity because they
>require
>symbols such as lim f(x) = +oo to explain them.
> n->oo
The symbol .999... depends on infinity, but not the way you claim. A
decimal expansion such as .999... is basically a sequence of digits.
By definition, a sequence (meaning an omega-sequence) is a mapping
whose domain is the set of natural numbers. If we don't have the
(infinite) set of natural numbers available, then the symbol .999...
does not even exist, and therefore the question in the subject line of
this thread cannot even be asked.
My point is that in order to answer the question of whether .999... =
1, it is sufficient to have access to the very same infinity that is
needed in order to ask the question in the first place. No other
infinities are required; no non-finite integers and no "absolute"
infinities.
[...]
>The binary power set P2w, where w is a non-finite number, includes a
>greatest member
>w(1._.1) and a least member w(0._01). These symbols, represent infinite
>entities which
>can be used in a consistent logical calculus. The type of infinity used
>here does not lead to indefinite symbols.
It doesn't? What is an upper bound for all such infinities?
Perhaps you will say that there is no such upper bound in your model.
That is my point, exactly. There is no upper bound for the rationals
within the standard real numbers, either.
>The questions you have not answered are:
>
>(A) Does the symbol .999... involve the idea of infinity in any way ?
>
>If the answer is Yes:
>
>(B) How is the idea of infinity used to be defined ?
>ie please provide a satisfactory definition of infinity in this context.
>
>NB: The epsilon-delta definition involves the finitistic type of
>infinity described above
>and is subject to the ojections I have already made.
>
>If the answer is No:
>
>(C) How can the symbol .999... be defined without directly of indirectly
>introducing
>the idea of infinity.
>
>In summary, If you use an absolutely unbounded infinity, the result
>will always be
>undefined but if you use a bounded infinity, you have to accept that
>.999... < 1. If you
>define .999... = 1 you surreptitiously introduce the concept of absolute
>infinity and all the consequences which follow from using a concept
>which is inherently indefinite in meaning.
>
>It may be the case that the means used to avoid the incorporation of an
>absolute infinity into numbers defined as infinite sequences has been
>perfectly accomplished. However, I have not yet seen a convincing
>demonstration of this.
Let's start with the basics. One of the axioms of ZF is the axiom of
infinity, which says that there is a set that contains 0 and is closed
under the successor operation. From the other axioms of ZF it can be
shown that there is a set omega, which is the smallest set having those
properties. This is the set of natural numbers, and is also called the
smallest transfinite ordinal number. Since larger infinities exist,
omega is obviously not an "absolute infinity".
As I explained previously, without omega you can't even formulate a
symbol .999... in the first place. The set omega is the only infinity
that I need in order to define the real numbers and show why .999... =
1. No more tools are needed to answer the question, than to ask it.
Given a digit sequence
n . d_1 d_2 d_3 d_4 ...,
we may associate it in a natural way with a sequence of rational
numbers { a_k }, where
a_0 = n,
a_1 = a_0 + d_1 / 10,
a_2 = a_1 + d_2 / 100,
a_3 = a_2 + d_3 / 1000,
and so on. The resulting sequence { a_k } is clearly a Cauchy sequence
of rationals. Let C be the set of all Cauchy sequences of rationals,
and let + and * be defined componentwise on C. Then (C,+,*) is a
commutative ring.
Next, let I be a particular subset of C, defined as follows: a
sequence { a_k } belongs to I iff for each epsilon > 0, there exists N
> 0 such that |a_n| < epsilon for every n > N.
It is easy to see that the set I is closed under addition in C, and
that it is absorbing with respect to multiplication (that is, if { a_k }
is in I and { b_k } is in C, then the product { a_k * b_k } is in I).
Therefore, I is an ideal in the ring C.
Moreover, the ideal I is maximal in C: if { a_k } is in the set
difference C-I, then there exists { b_k } in I such that the sum
{ a_k + b_k } has an inverse in C.
Next, we form the quotient ring C/I, which is known to be a field
because of the maximality of I. This is the field of real numbers.
To sum up: when we ask whether .999... = 1, we are talking about
objects that have more than one meaning. The left-hand side may be
viewed as a decimal expansion, a member of a set that I will call D.
The right-hand side may be viewed as an integer, but for our purposes
it is more useful to view it as a rational number, a member of the set
Q. We have the following diagram:
.999... --> D
|\
| \
| \
| \
| \
| \
V V
C---> C/I = R
^ ^
| /
| /
| /
| /
| /
|/
1 --> Q
Therefore, the question of whether .999... = 1 really means, if you
follow the arrows through the above diagram, starting from .999... in D
at the top of the diagram, or from 1 in Q at the bottom, do you end up
at the same element of the field C/I = R at the right-hand side of the
diagram?
Starting from the top, we have the decimal expansion .999... in D,
which maps to the Cauchy sequence
0, 9/10, 99/100, 999/1000, ... (1)
in C. Starting at the bottom, we have the rational number 1, which
maps to the constant (therefore Cauchy) sequence
1/1, 1/1, 1/1, 1/1, ... (2)
in C. These are different elements of C, The question is whether
sequences (1) and (2) map to the same element of C/I = R. By the
definition of a quotient ring, the images in C/I of the sequences (1)
and (2) are equivalence classes in C, in which two sequences are
members of the same class iff their difference is a member of the ideal
I. Therefore, we construct the difference of the sequences (1) and
(2), obtaining
-1/1, -1/10, -1/100, -1/1000, ...
which is easily seen to possess the criterion needed for membership in
I. Therefore, by the definition of a quotient ring, .999... and 1
represent the same element in C/I = R.
No absolute infinities in sight.