0.99(bar) cannot be proven to equal one (REVISED)
John Pimentel
Note: Typos/grammatical errors corrected from prior post.
What disturbs me about this whole affair is that I cannot go in reverse
to get back to where I started from. As I looked elsewhere to other
numbers it became apparent why. The problem with the approaches taken
thus far is that we cannot adapt the rules of mathematics to suit our
theories. If you wish to give PROOF for something then by don't stop when
it meets your needs, _prove_ that it is correct, or _prove_ that it is
not.
Ok, now that we are all on the defense we'll commence.
If 0.99(bar) equals 1.0 then these others must also be true:
0.88(bar) equals 0.9
0.77(bar) equals 0.8
0.66(bar) equals 0.7
0.55(bar) equals 0.6
0.44(bar) equals 0.5
0.33(bar) equals 0.4
0.22(bar) equals 0.3
0.11(bar) equals 0.2
0.00(bar) equals 0.1
and these too...
1.11(bar) equals 1.2
2.00(bar) equals 2.1
there are other absurdities that can be stated along these lines
but I will settle for these...
The formula is for proving that a non-terminating, repeating number is,
in fact, a rational number.
Let X = 0.16731673(bar)
The number of digits the decimal point needs to move before the next set
of repeating digits is 4, therefore for this 10^4 (10000) will be used.
10000X = 1673.16731673(bar)
- X = - 0.16731673(bar)
----------------------------
9999X = 1673
divide both side by 9999
X = 1673/9999 (which does not reduce)
if 9999 is divided into 1673 the result is 0.16731673(bar)
This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
For the next series of equations the number of digits the decimal point
needs to move before the next repeating digit is one, therefore 10^1.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 0.88(bar)
10X = 8.88(bar)
- X = - 0.88(bar)
---------------
9X = 8
divide both sides by 9
X = 8/9
if 9 is divided into 8 the result is 0.88(bar).
This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 0.77(bar) Let X = 0.66(bar)
10X = 7.77(bar) 10X = 6.66(bar)
- X = - 0.77(bar) - X = - 0.66(bar)
--------------- ---------------
9X = 7 9X = 6
X = 7/9 = 0.77(bar) X = 6/9 = 2/3 0.66(bar)
This equation is proven. This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 0.55(bar) Let X = 0.44(bar)
10X = 5.55(bar) 10X = 4.44(bar)
- X = - 0.55(bar) - X = - 0.44(bar)
--------------- ---------------
9X = 5 9X = 4
X = 5/9 = 0.55(bar) X = 4/9 = 0.44(bar)
This equation is proven. This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 0.33(bar) Let X = 0.22(bar)
10X = 3.33(bar) 10X = 2.22(bar)
- X = - 0.33(bar) - X = - 0.22(bar)
--------------- ---------------
9X = 3 9X = 2
X = 3/9 = 1/3 = 0.33(bar) X = 2/9 = 0.22(bar)
This equation is proven. This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 0.11(bar) Let X = 0.00(bar)
10X = 1.11(bar) 10X = 0.00(bar)
- X = - 0.11(bar) - X = - 0.00(bar)
--------------- ---------------
9X = 1 9X = 0
X = 1/9 = 0.11(bar) X = 0/9 = 0.00(bar)
This equation is proven. This equation is proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Summation to this point:
Does 0.88(bar) equal 0.9? It does not.
Does 0.77(bar) equal 0.8? It does not.
Does 0.66(bar) equal 0.7? It does not.
Does 0.55(bar) equal 0.6? It does not.
Does 0.44(bar) equal 0.5? It does not.
Does 0.33(bar) equal 0.4? It does not.
Does 0.22(bar) equal 0.3? It does not.
Does 0.11(bar) equal 0.2? It does not.
Does 0.00(bar) equal 0.1? It does not.
Humm...
Let's look elsewhere.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 1.11(bar) Let X = 1.99(bar)
10X = 11.11(bar) 10X = 19.99(bar)
- X = - 1.11(bar) - X = - 1.99(bar)
--------------- ---------------
9X = 10 9X = 18
X = 10/9 = 10 1/9 = 1.11(bar) X = 18/9 = 2.00(bar)
This equation is proven. This equation is NOT proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 2.00(bar) Let X = 2.99(bar)
10X = 20.00(bar) 10X = 29.99(bar)
- X = - 2.00(bar) - X = - 2.99(bar)
--------------- ---------------
9X = 18 9X = 27
X = 18/9 = 2.00(bar) X = 27/9 = 3.00(bar)
This equation is proven. This equation is NOT proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
If the whole number portion is removed from the equation by subtraction
and then added back in after the computation then the following is the
understanding:
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 1.11(bar) Let X = 1.99(bar)
Let Y = X - 1 = 0.11(bar) Let Y = X - 1 = 0.99(bar)
10Y = 1.11(bar) 10Y = 9.99(bar)
- Y = - 0.11(bar) - Y = - 0.99(bar)
--------------- ---------------
9Y = 1 9Y = 9
Y = 1/9 = 0.11(bar) Y = 9/9 = 1.00(bar)
add the one to Y which was taken add the one to Y which was taken
out out
Y + 1 = 1.11(bar) = X Y + 1 = 2.00(bar) != X
(!= means not equal for those that
don't know)
This equation is proven. This equation is NOT proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Let X = 3.00(bar) Let X = 2.99(bar)
Let Y = X - 3 = 0.00(bar) Let Y = X - 2 = 0.99(bar)
10Y = 0.00(bar) 10Y = 9.99(bar)
- Y = - 0.00(bar) - Y = - 0.99(bar)
--------------- ---------------
9Y = 0 9Y = 9
Y = 0/9 = 0.00(bar) Y = 9/9 = 1.00(bar)
add the three to Y which was taken add the two to Y which was taken
out out
Y + 3 = 3.00(bar) = X Y + 2 = 3.00(bar) != X
This equation is proven. This equation is NOT proven.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
Summation there is a limit at 1 which cannot be reached and _prove_ the
equation, therefore to solve the equation dealing exclusively with the
decimal portion is to operate within the range of:
0 <= |X| < 1
---
John Pimentel
pime...@ultranet.com
----
John Pimentel
pime...@roots.ultranet.com
Josh Boyd
WRONG... .88(bar) is OVBIOUSLY less than .89 --- Therefore .88(bar)
cannot equal .9
If you want to do that...
0.88(bar) = 0.89
The reason why .99(bar) = 1 is because of the very same argument.
Josh
Haran Pilpel
John Pimentel wrote:
> What disturbs me about this whole affair is that I cannot go in reverse
> to get back to where I started from. As I looked elsewhere to other
> numbers it became apparent why. The problem with the approaches taken
> thus far is that we cannot adapt the rules of mathematics to suit our
> theories. If you wish to give PROOF for something then by don't stop when
> it meets your needs, _prove_ that it is correct, or _prove_ that it is
> not.
>
> Ok, now that we are all on the defense we'll commence.
>
> If 0.99(bar) equals 1.0 then these others must also be true:
> 0.88(bar) equals 0.9
> 0.00(bar) equals 0.1
WHY??? When was this ever stated? How did you reach this conclusion
from
.9bar = 1? I'm afraid I don't follow your logic. In fact, as you
yourself
reach later:
.8bar = 8/9, .7bar = 7/9, ... .0bar = 0/9.
> The formula is for proving that a non-terminating, repeating number is,
> in fact, a rational number.
>
<LOTS of numbers snipped>
All right, if you want a completely formal proof, here it is.
Define the series a(n) = 1 - 10^n.
What is requested is: limit (n -> oo) a(n) = ?
Let e > 0. If e > 0 then obviously for all n>0, | a(n) - 1 | < 1 < e.
Therefore assume that 0 < e < 1. Then let N = 1/e. Then:
n > N --> |a(n) - 1| <= | 1-[1/(1/e)]^n - 1| = |e^n| < e
So by the definition of the limit of a series, the limit is 1.
Q.E.D.
Naturally this is tremendous overkill (my apologies to the group) but
for doubters, it is a complete, self-sufficient proof that .9bar is
indeed 1.
> ----
> John Pimentel
> pime...@roots.ultranet.com
Haran
--
e^(Pi*i) + 1 = 0 (Euler)
Dave B.
John Pimentel <pime...@roots.ultranet.com> wrote
> If 0.99(bar) equals 1.0 then these others must also be true:
> 0.88(bar) equals 0.9
Why?
0.88888... < 0.89 < 0.9
Clearly your assertion is false. 0.8repeat is not equal to 0.9. Are you
trolling?
However, 0.8999999... is equal to 0.9 in the same way that 0.999999... is
equal to 1.
Dave B.
John Pimentel
> Date: Thu, 06 Feb 1997 00:20:22 -0800
> From: Haran Pilpel <har...@ibm.net>
> Reply-to: har...@ibm.net
> To: pime...@roots.ultranet.com
> Subject: Re: 0.99(bar) cannot be proven to equal one (REVISED)
Yet another person that cannot understand.
OK, you say 0.99(bar) is 1.0 and it is so because there is no point
between 9.0 and 0.99(bar).
Humm, now find me a point between 0.88(bar) and 0.9, and let's expand
this even further find me a point between 0.11(bar) and 0.2 [now this
one should have plenty of possibilities, and I am only asking for a
point, just one, kindly assign a number to it].
Either, they are the same or you have no concept of limit as it
applies to 1.0.
Here's another. From 1.0 _go_back_ to 0.99(bar).
0 <= |X| < 1
----
John Pimentel
pime...@roots.ultranet.com
Tore August Kro
John Pimentel wrote:
>
Theorem:
Let a be a given number.
And let x be a number with the property that given any number e>0 then
the absolutevalue of the difference a - b is less than e.
|a-b|<e.
Proof:
Now assume that a =/= b, that is that a is not equal to b. Then we could
let e = |a-b|/2. Since a =/= b e must be positive. The statement above
then says that
|a-b| < e = |a-b|/2
And this is clearly a contradiction. Thus our assumption that a is
different from b false. And the only possibility left is that a = b.
Example:
Here is how to prove that 0.9999999... = 0.99(bar) = 1.
Now we could let a = 1. It is clearly seen that 0.99(bar) possesses the
property that given any small number e>0 that |1-0.99999999999..|<e.
Just expand 0.99(bar) as far as needed and subtract. Thus by the theorem
above 0.99(bar) = 1.
Tore August Kro
tor...@ifi.uio.no
Mathematics student at the University of Oslo.
Iain Davidson
>Example:
>Here is how to prove that 0.9999999... = 0.99(bar) = 1.
>Now we could let a = 1. It is clearly seen that 0.99(bar) possesses the
>property that given any small number e>0 that |1-0.99999999999..|<e.
>Just expand 0.99(bar) as far as needed and subtract. Thus by the theorem
>above 0.99(bar) = 1.
Surely, the point is that e can never become zero so you cannot say
1 and 0.9999... are equal in the normal sense of the word. You can
make e as close to zero as you want but that is not the same thing
as saying e actually becomes zero. The string of nines never ends so
e can never be zero.
1/n tends to zero, as n increases but there is no real number that
actually makes it zero .
2/n also tends to zero
Say * is the number that makes 1/n = 0, i.e. 1/* = 0 = 2/* = 0
so 1 = 0* = 2 = 0*. Surely anything that gives 1 =2 cannot be
consistent.
Iain Davidson Tel : +44 1228 49944
4 Carliol Close Fax : +44 1228 810183
Carlisle Email : ia...@stt.win-uk.net
England
CA1 2QP
Dave B.
John Pimentel wrote:
> Humm, now find me a point between 0.88(bar) and 0.9,
0.89 is easy enough.
> and let's expand
> this even further find me a point between 0.11(bar) and 0.2 [now this
> one should have plenty of possibilities, and I am only asking for a
> point, just one, kindly assign a number to it].
0.12 or 0.13 or 0.14 or 0.15 or 0.16 or 0.17 ...
>
> Either, they are the same or you have no concept of limit as it
> applies to 1.0.
I'm starting to suspect we're being trolled.
Dave B.
David
"Dave B." <york...@sympatico.ca> wrote:
This is in response to the original article (if it exists. This
discussion seems to go on forever.)
The series .9, .99, .999, .9999 etc, *approaches* the value 1. It
*never* gets there. The number 1 is the *limiting* value of this
series. It is the smallest number *not part of the series* that is
thus approached. As the series increases, it also approaches values
larger than 1. However, 1 is the *smallest* value that the series
approaches as the number of terms increases.
The idea behind limits is that the limiting value of a series is
*not* the value of the series, but is the smallest value approached by
an increasing series.
The problem is no longer a dilemma. It has been solved, and is called
"The Calculus."
Period.
Lee Jaap
|>Note: Typos/grammatical errors corrected from prior post.
|>
|>What disturbs me about this whole affair is that I cannot go in reverse
|>to get back to where I started from. As I looked elsewhere to other
|>numbers it became apparent why. The problem with the approaches taken
|>thus far is that we cannot adapt the rules of mathematics to suit our
|>theories. If you wish to give PROOF for something then by don't stop when
|>it meets your needs, _prove_ that it is correct, or _prove_ that it is
|>not.
|>
|>Ok, now that we are all on the defense we'll commence.
|>
|>If 0.99(bar) equals 1.0 then these others must also be true:
|> 0.88(bar) equals 0.9
|> 0.77(bar) equals 0.8
|> 0.66(bar) equals 0.7
|> 0.55(bar) equals 0.6
|> 0.44(bar) equals 0.5
|> 0.33(bar) equals 0.4
|> 0.22(bar) equals 0.3
|> 0.11(bar) equals 0.2
|> 0.00(bar) equals 0.1
|>and these too...
|> 1.11(bar) equals 1.2
|> 2.00(bar) equals 2.1
Why? Prove that the first statement implies the rest.
You're using a common logical fallacy: If a then a implies b.
Just because the first statement is true, it is not obvious that
any of the rest are.
Broken.
--
J Lee Jaap <Jaa...@ASMSun.LaRC.NASA.Gov> +1 757/865-7093
employed by, not necessarily speaking for,
AS&M Inc, Hampton VA 23666-1340
Lee Jaap
|>> Date: Thu, 06 Feb 1997 00:20:22 -0800
|>> From: Haran Pilpel <har...@ibm.net>
|>> Reply-to: har...@ibm.net
|>> To: pime...@roots.ultranet.com
|>> Subject: Re: 0.99(bar) cannot be proven to equal one (REVISED)
|>
|>Yet another person that cannot understand.
|>
|>OK, you say 0.99(bar) is 1.0 and it is so because there is no point
|>between 9.0 and 0.99(bar).
|>
|>Humm, now find me a point between 0.88(bar) and 0.9, and let's expand
|>this even further find me a point between 0.11(bar) and 0.2 [now this
|>one should have plenty of possibilities, and I am only asking for a
|>point, just one, kindly assign a number to it].
.15 is between .1bar and .2. So are .12, .13, .14, .16, .17, .18, and
.19, to name a few.
|>Either, they are the same or you have no concept of limit as it
|>applies to 1.0.
Non sequitur. (Latin for "It does not follow.")
|>Here's another. From 1.0 _go_back_ to 0.99(bar).
Why? They're two names for the same number.
Have we given you enough rope yet? Or are you going to ignore
my post?
Tore August Kro
I have corrected a few typing errors in my last proof:
Theorem:
Let a be a given number.
And let b be a number with the property that given any number e>0 then
the absolutevalue of the difference a - b is less than e.
|a-b|<e. Then a=b.
Proof:
Now assume that a =/= b, that is that a is not equal to b. Then we could
let e = |a-b|/2. Since a =/= b e must be positive. The statement above
then says that
|a-b| < e = |a-b|/2
And this is clearly a contradiction. Thus our assumption that a is
different from b false. And the only possibility left is that a = b.
Example:
Here is how to prove that 0.9999999... = 0.99(bar) = 1.
Now we could let a = 1. It is clearly seen that 0.99(bar) possesses the
property that given any small number e>0 that |1-0.99999999999..|<e.
Just expand 0.99(bar) as far as needed and subtract. Thus by the theorem
above 0.99(bar) = 1.
Tore August Kro
Iain Davidson wrote:
>
>
>
> >Example:
> >Here is how to prove that 0.9999999... = 0.99(bar) = 1.
> >Now we could let a = 1. It is clearly seen that 0.99(bar) possesses the
> >property that given any small number e>0 that |1-0.99999999999..|<e.
> >Just expand 0.99(bar) as far as needed and subtract. Thus by the theorem
> >above 0.99(bar) = 1.
>
> 1 and 0.9999... are equal in the normal sense of the word. You can
> make e as close to zero as you want but that is not the same thing
> as saying e actually becomes zero. The string of nines never ends so
> e can never be zero.
Do you agree that 0.99(bar) has the property that given any number e>0,
no matter how small e is, then |1-0.99(bar)| < e ? If we define real
numbers in the normal way then this is a fact.
Now ASSUME that you are absolutely correct that 1 =/= 0.99(bar). That is
that in the normal sense of the words not equal. I shall prove that this
leads to an contradiction. And then there is only one option left,
namely that 1 is in fact equal to 0.99(bar), in the normal sense of the
word equal. In the normal sense 1 not equal to 0.99(bar) means that
there is a number e equal to the absolute value of the difference
between 1 and 0.99(bar). You surely must agree that when two numbers are
different that the difference of these numbers is different from 0,
don't you? And now from the paragraph above we have that
|1-0.99(bar)|<e. This is because 0.99(bar) has the property that given
any number e>0, no matter how small e is, then |1-0.99(bar)| < e.
But e was exacly |1-0.99(bar)|, so because we assumed that 1=/=0.99(bar)
we get the absurd expression |1-0.99(bar)|<e=|1-0.99(bar)| or if we
remove the e in the middle |1-0.99(bar)|<|1-0.99(bar)|. And no number
can be less than itself. So we have a CONTRADICTION. And this
contradiction contradicts that 1=/=0.99(bar). And when "1 is not equal
to 0.99(bar)" is a false statement, then "1 is equal to 0.99(bar)" is in
fact a true statement.
Now before you reply this message I suggest that you read what I have
written thoroughly, see if you can understand how things can be prooven
by assuming the oppostie and deducing a contradiction. Consult a
professor in mathematics if necessary.
> 1/n tends to zero, as n increases but there is no real number that
> actually makes it zero .
Yes, this is correct.
> 2/n also tends to zero
Yes, this is correct too.
> Say * is the number that makes 1/n = 0, i.e. 1/* = 0 = 2/* = 0
>
> so 1 = 0* = 2 = 0*. Surely anything that gives 1 =2 cannot be
> consistent.
So you have prooven by contradiction that there is no real number * that
makes 1/n = 0.
> Iain Davidson
Tore August Kro
tor...@ifi.uio.no
Albert Y.C. Lai
In article <1997020613...@cinna.ultra.net>,
Don't jump to that conclusion. Even though your responses to my
arguments haven't reached me, I will wait indefinitely and won't make a
similar judgement on you. Who is to say that you understand more than
other people?
>OK, you say 0.99(bar) is 1.0 and it is so because there is no point
>between 9.0 and 0.99(bar).
>Humm, now find me a point between 0.88(bar) and 0.9,
No. You find us a point between 0.9(bar) and 1.0.
--
Albert Y.C. Lai tre...@vex.net http://www.vex.net/~trebla/
Albert Y.C. Lai
In article <5ddn8o$ju5$1...@nntp.igs.net>,
mar...@cnwl.igs.net (David) wrote:
>The series .9, .99, .999, .9999 etc, *approaches* the value 1. It
>*never* gets there. The number 1 is the *limiting* value of this
>series.
And furthermore the notation 0.9bar denotes the limiting value of that
series.
>The problem is no longer a dilemma. It has been solved, and is called
>"The Calculus."
You almost talked me into believing that the slope of y=x^2 at x=0
approaches 0 but never gets there, that the area of the unit circle
approaches pi but never gets there.
Iain Davidson
>Do you agree that 0.99(bar) has the property that given any number e>0,
>no matter how small e is, then |1-0.99(bar)| < e ? If we define real
>numbers in the normal way then this is a fact.
Are you sure that 0.99(bar) is a real number, in fact an integer, if
it is equal to 1 , 1.0000000 ?
If what you are saying is that if I think of a small number greater
than zero then you can think of a smaller number I would, of
course, agree.
But what you are in effect saying is that |1 - 0.99(bar)| is the
smallest real number greater than zero. But there is no smallest
real number greater than zero. Assume it is s, but s/2 is less
than s, so there is no such real number. This means that the
premises of your argument are not correct so following through the
rest of your argument is pointless.
I could also say that
|sqrt(2) - (1 + 1/2 - 1/8 + 1/16 + ..| < e, e> 0
So that sqrt(2) is in fact equal to an actual infinite number of
additions and subtractions of rational numbers. But sqrt(2) is
irrational and the sum of any rational numbers is a rational
number. But an irrational number cannot equal a rational number.
The sum has the limit sqrt(2) but is it equal to sqrt(2)? It does
not seem to make any sense to say that a sum of rational numbers,
infinite or not, can equal an irrational number.
What you can say is
0.9 + 0.1 = 1
0.99 + 0.01 = 1
0.99999 .. 999 (n 9s) + 1/10^n = 1
If, in your terminology 0.99(bar) = 1, then 1/10^n = 0, for some n
But there is no n that makes 1/10^n = 0
What does make sense is to say that
| 1 - 0.999 ...99 (n 9s)| = 1/10^n
and that for any 1 > e > 0 there is an equality
1/10^n+1 < e < 1/10^n
Iain Davidson Tel : +44 1228 49944
uge...@aol.com
someone asked
>>Humm, now find me a point between 0.88(bar) and 0.9,
>
.01bar
proof:
.88bar + .01bar = 8/9 + 1/90 = 81/90 =9/10 = .9
about .9bar
.9bar is an infinite geometric series. The sum of an infinite geometric
series is a/(1-r). In the case of .9bar, a = .9 and r = 1/10
we get
.9/(1-1/10) = 1
If you won't accept that .9bar = 1 then you are back to Achilles and the
tortoise and have a lot of math to catch up on.
.
Dave B.
David <mar...@cnwl.igs.net> wrote in article <5ddn8o$ju5$1...@nntp.igs.net>...
> "Dave B." <york...@sympatico.ca> wrote:
>
> This is in response to the original article (if it exists. This
> discussion seems to go on forever.)
>
> The series .9, .99, .999, .9999 etc, *approaches* the value 1. It
> *never* gets there. The number 1 is the *limiting* value of this
> series. It is the smallest number *not part of the series* that is
> thus approached. As the series increases, it also approaches values
> larger than 1. However, 1 is the *smallest* value that the series
> approaches as the number of terms increases.
Why are you telling me this young fella? Have you gotten lost in the
plethora of names involved in posting to this now absurd issue? I am not
one of the crowd who believe that .9 repeat is not precisely one. I said
nothing which could be remotely construed as conceivably at odds with
anything you're trying to purport. Discussion on series with respect to
this issue is not necessary. I'm certainly not trying to encourage it.
Please take your assertions elsewhere.
Dave B.
Todd Harris
In article <32FB39...@ifi.uio.no>, Tore August Kro <tor...@ifi.uio.no>
wrote:
€Do you agree that 0.99(bar) has the property that given any number e>0,
€no matter how small e is, then |1-0.99(bar)| < e ? If we define real
€numbers in the normal way then this is a fact.
Please excuse my ignorance, but this seems to my simplistic way of thinking
to be assuming from the outset what it is you are trying to prove. Which
doesn't, of course, work out to much of a proof.
€Now ASSUME that you are absolutely correct that 1 =/= 0.99(bar). That is
€that in the normal sense of the words not equal. I shall prove that this
€leads to an contradiction. And then there is only one option left,
€namely that 1 is in fact equal to 0.99(bar), in the normal sense of the
€word equal. In the normal sense 1 not equal to 0.99(bar) means that
€there is a number e equal to the absolute value of the difference
€between 1 and 0.99(bar).
Here you set e equal to 1 - 0.99(bar), and I have no problem with that.
€You surely must agree that when two numbers are
€different that the difference of these numbers is different from 0,
€don't you?
I personally would,yes...
€And now from the paragraph above we have that
€|1-0.99(bar)|<e.
Here is where the argument starts to seem awfully circular to me...
€ This is because 0.99(bar) has the property that given
€any number e>0, no matter how small e is, then |1-0.99(bar)| < e.
This was the assumption from the beginning, but this is what you really
should be proving...
Seeming (to me) to leave the rest of your argument up in the air.
€But e was exacly |1-0.99(bar)|, so because we assumed that 1=/=0.99(bar)
€we get the absurd expression |1-0.99(bar)|<e=|1-0.99(bar)| or if we
€remove the e in the middle |1-0.99(bar)|<|1-0.99(bar)|. And no number
€can be less than itself. So we have a CONTRADICTION. And this
€contradiction contradicts that 1=/=0.99(bar). And when "1 is not equal
€to 0.99(bar)" is a false statement, then "1 is equal to 0.99(bar)" is in
€fact a true statement.
€
€Now before you reply this message I suggest that you read what I have
€written thoroughly, see if you can understand how things can be prooven
€by assuming the oppostie and deducing a contradiction. Consult a
€professor in mathematics if necessary.
I've read & re-read...and on Monday morning I intend to take this to my
Linalg prof. Mainly because I've been following this thread with some
fascination for a while now and I'd really like to get to the bottom of
this for my own peace of mind. At first glance it was "obvious" to me that
1 =/= 0.99(bar) but I've since noticed that the math-"heavies" all seem to
come down on the side of "does equal". Some of the proofs in previous posts
seem to be a bit more complete, but until I get better at reading ascii
equations they seem a bit inaccessible.
I do have to say, though, that claiming 1 = 0.99(bar) seems equivalent to
claiming you can count to infinity.
Enough mental simplicity on my part...anyone care to address this?
Todd Harris
--
Signature-File deleted by:
NetCensors Incorporated
"Your ignorance is our business"
Vincent Johns
(posted & emailed)
uge...@aol.com wrote:
>
> someone asked
> >>Humm, now find me a point between 0.88(bar) and 0.9,
> >
>
> .01bar
This looks like a number *much less than* 0.88(bar). You
may be thinking of the *difference* between those numbers.
> proof:
> .88bar + .01bar = 8/9 + 1/90 = 81/90 =9/10 = .9
> about .9bar
> .9bar is an infinite geometric series. The sum of an infinite geometric
> series is a/(1-r). In the case of .9bar, a = .9 and r = 1/10
> we get
> .9/(1-1/10) = 1
> If you won't accept that .9bar = 1 then you are back to Achilles and the
> tortoise and have a lot of math to catch up on.
IMHO, this is not a very compelling argument. A major purpose
of mathematics is to enable people to make sense of their surroundings
using their own wits, not just "accepting" something (like ".9bar = 1")
that someone else says is true. How does one know what the sum of an
infinite geometric series is? Probably a math. teacher said that, but
it's also necessary to have an idea of limits of infinite sequences
and such. Once one has some understanding of how they work, it's
fairly easy to start coming up with original ideas that the teacher
or textbook never specifically presented. It's not "I am really, really
sure that 0.9999... = 1" that gives you power, it's "I know how to show
that 0.9999... = 1 and much more besides that!"
Of course, some of this is not automatic. One needs to be comfortable
with using variables in addition to the constants of 1, 2, 3 that
were introduced in kindergarten before arguments about sequences make
much sense. Given knowledge of the tools needed for the proofs,
though, what one can do is greatly expanded.
BTW, I'm not opposed to arguments based on authority, even in
mathematics.
We build on what other people have done, and I can understand some of
Newton's, Gauss's, or Euler's work without having to rediscover it.
What I'm saying here is that any of us can go beyond quoting someone
else to use the tools to construct our own new reality (such as a
convincing argument that .9bar = 1, rather than just asking a person
to accept it on faith). Mathematics, science, and related disciplines
give us that power.
--
-- Vincent Johns
Please feel free to quote anything I say here.
Tore August Kro
Iain Davidson wrote:
>
>
>
> >Do you agree that 0.99(bar) has the property that given any number e>0,
> >no matter how small e is, then |1-0.99(bar)| < e ? If we define real
> >numbers in the normal way then this is a fact.
>
> it is equal to 1 , 1.0000000 ?
By 0.99(bar) I understand a zero then a decimal point then an infinite
number of 9's. This can be written as the limit of the sequence 0.9,
0.99, 0.999, 0.9999, 0.99999, ... ( Or alternatively as:
0.99(bar) = the sum from n=1 to infinity of 9/10^n. )
And by the completeness property of real numbers: Every bounded
monotonely increasing sequence of real numbers converges to a real
number. Thus 0.99(bar) is real.
> If what you are saying is that if I think of a small number greater
> than zero then you can think of a smaller number I would, of
> course, agree.
>
> But what you are in effect saying is that |1 - 0.99(bar)| is the
> smallest real number greater than zero.
By assuming that 0.99(bar) is different from 1 I can conclude that
|1-0.99(bar)| is the smallest real number greater than zero.
> But there is no smallest
> real number greater than zero. (...)
Exacly and therefore 1 = 0.99(bar)
> I could also say that
>
> |sqrt(2) - (1 + 1/2 - 1/8 + 1/16 + ..| < e, e> 0
And sqrt(2) - 1 = 1/(2+1/(2+1/(2+1/(2+1/(2+1/(2+1/(2....)))))))
[This looks really nice written out by hand]
> So that sqrt(2) is in fact equal to an actual infinite number of
> additions and subtractions of rational numbers.
Yes, you can write sprt(2) as an acutal infinite number of additions and
subtractions of rational numbers.
> But sqrt(2) is
> irrational and the sum of any rational numbers is a rational
> number. But an irrational number cannot equal a rational number.
This is also correct as long as you work with a finite number of
additions and subtractions.
BUT IF A SUM OF AN INFINITE NUMBER OF RATIONAL NUMBERS NEED NOT BE
RATIONAL.
> The sum has the limit sqrt(2) but is it equal to sqrt(2)? It does
> not seem to make any sense to say that a sum of rational numbers,
> infinite or not, can equal an irrational number.
As long as it is a sum of an infinite number of rational nubers it
works.
> What you can say is
>
> 0.9 + 0.1 = 1
>
> 0.99 + 0.01 = 1
>
> 0.99999 .. 999 (n 9s) + 1/10^n = 1
>
> If, in your terminology 0.99(bar) = 1, then 1/10^n = 0, for some n
>
> But there is no n that makes 1/10^n = 0
>
> What does make sense is to say that
>
> | 1 - 0.999 ...99 (n 9s)| = 1/10^n
>
> and that for any 1 > e > 0 there is an equality
>
> 1/10^n+1 < e < 1/10^n
Infinity does many weird things to numbers. Thus everything you have
learnt about the finite properties of real numbers isn't true for the
infinite case.
Read an real analysis book.
Tore August Kro
tor...@ifi.uio.no
Raymond E. Griffith
*Do you agree that 0.99(bar) has the property that given any number e>0,
*no matter how small e is, then |1-0.99(bar)| < e ? If we define real
*numbers in the normal way then this is a fact.
Todd Harris replied:
# Please excuse my ignorance, but this seems to my simplistic way of
# thinking to be assuming from the outset what it is you are trying to
# prove. Which doesn't, of course, work out to much of a proof.
-----------------------------
Todd, you are right. He is assuming what he is trying to prove. He is
trying to use the classical form of proof by contradiction. But, Tore
August Kro DOES have the essence of what we want.
Essentially, the question is, How far apart is
0.9999999999999999999999999999999999(9bar) and
1.0000000000000000000000000000000000(0bar)?
-----------------------------
Tore August Kro wrote:
*Now ASSUME that you are absolutely correct that 1 =/= 0.99(bar). That
*is that in the normal sense of the words not equal. I shall prove that
*this leads to an contradiction. And then there is only one option left,
*namely that 1 is in fact equal to 0.99(bar), in the normal sense of the
*word equal. In the normal sense 1 not equal to 0.99(bar) means that
*there is a number e equal to the absolute value of the difference
*between 1 and 0.99(bar).
Todd Harris replied:
# Here you set e equal to 1 - 0.99(bar), and I have no problem with
that.
Tore August Kro wrote:
*You surely must agree that when two numbers are
*different that the difference of these numbers is different from 0,
*don't you?
Todd Harris replied:
# I personally would,yes...
-----------------------------
Not just “personally.” There is solid math behind his statement.
Essentially, The trichotomy principle says that if a and b are any two
real numbers, then exactly one of these three things can happen:
a < b, or a = b, or a > b. If a < b or a > b then the *absolute value*
of the difference between them is greater than 0. So,
1) Assume that 1-0.99(9bar) <> 0.
2) Then. |1-0.99(9bar)| = some e>0
-----------------------------
Tore August Kro wrote:
*And now from the paragraph above we have that
*|1-0.99(bar)|<e.
Todd Harris replied:
# Here is where the argument starts to seem awfully circular to me...
-----------------------------
Yes, this is where his difficulties begin, but we will get out of them.
We have
From the above discussion, select (fix) some e (very very small) for the
difference between 1 and 0.99(9bar).
3) So |1-0.99(9bar)| = e fixed.
Now the form of the difference between 1.00(0bar) and 0.99(9bar) =
1/10^n for some n (integer)(if they are different). Then we have
4) 1/10^n = e and we can calculate an exact n.
Calculating an exact n for any fixed e>0 means that 0.99(9bar) has to
terminate after n nines.
5) n = -log(e)
6) Therefore 0.99(9bar) <= 1 - 1/10^n = Sum(9*10^i), i = 1
to n
But this can’t be! Since 0.99(9bar) repeats 9's infinitely (never
stops), then no matter what n we choose, n will always be too small (or,
no matter what e we choose, e will always be too big).
So, 7) Contradiction! |1 - 0.99(9bar)| < e for any positive e
you can
choose, no matter how small.
8) If the absolute value of the difference is nonpositive,
then it
has to be 0.
By the trichotomy principle (a < b or a = b or a > b), we are forced to
a contradiction. We assumed they were different and that we could find
some positive difference between them. We can't. We could, if there was
a difference. If there is no difference between them, they are equal.
9) So, since |1 - 0.99(9bar)| = 0, 1 = 0.99(9bar)
(stuff deleted)
-----------------------------
Todd Harris remarked:
*I've read & re-read...and on Monday morning I intend to take this to my
*Linalg prof. Mainly because I've been following this thread with some
*fascination for a while now and I'd really like to get to the bottom of
*this for my own peace of mind. At first glance it was "obvious" to me
that
*1 =/= 0.99(bar) but I've since noticed that the math-"heavies" all seem
to
*come down on the side of "does equal". Some of the proofs in previous
posts
*seem to be a bit more complete, but until I get better at reading ascii
*equations they seem a bit inaccessible.
*
*I do have to say, though, that claiming 1 = 0.99(bar) seems equivalent
to
*claiming you can count to infinity.
-----------------------------
Hmmm, you may have stumbled onto something.
Remember, infinity is not a “number”, but a term to imply that something
is unending. 0.99(9bar) does not “terminate” in infinity, nor is
0.99(9bar) in the process of adding 9’s to the end of it. The bar
indicates that it *has* an infinite number of nines.
There are actually two types of infinities! Countable infinities and
uncountable infinities. A countable infinity does not mean that you can
count to infinity, but that you can count infinitely, accounting for
every item on the way. A good example is the set of counting nnumbers.
Start counting, and keep going. As long as you never stop, you will be
accounting for every integer on your way, but still have left to account
for all those ahead (an infinite number). No matter how far you have
gone, there will be only a finite number of integers counted, and an
infinite number left to count (pretty heavy stuff, but well proven). An
example of an uncountable infinity is the set of real numbers.
------------------------------
Todd Harris remarked:
* Enough mental simplicity on my part...anyone care to address this?
*
* Todd Harris
*
* --
* Signature-File deleted by:
*
* NetCensors* Incorporated
* "Your ignorance is our business"
------------------------------
Not mental simplicity. It just isn’t “obvious” that the real numbers
should have more than one representation for any number. A good book in
real analysis details these things.
Hope this helps.
Raymond E. Griffith
________________________________________________
The educated man possesses six characteristics: He must be able to
1. communicate;
2. calculate;
3. see the world as it really is;
4. appropriately manipulate his environment;
5. accurately predict the effects of that manipulation;
6. admit the limits of his knowledge.
Raymond E. Griffith
Josh Boyd
> 4) 1/10^n = e and we can calculate an exact n.
>
> Calculating an exact n for any fixed e>0 means that 0.99(9bar) has to
> terminate after n nines.
Could you not say, though, that because .99(9bar) must terminate after n
nines, n must equal infinity (for the actual number .99(9bar)) because
there are an infinite number of nines?
Then it becomes...
1/10^(infinity) = e
And any constant divided by infinity ( 10^(infinity) = infinity ) is
necessarily zero ---
e = 0 <--- Which says that
|1 - .99(9bar)| = 0
1 = .99(9bar)
(Can you not do that? I don't see anything wrong with it...)
Josh
P.S. - Sorry if that's what you said... I didn't really read much of
this post.
Iain Davidson
>> >Do you agree that 0.99(bar) has the property that given any number e>0,
>And by the completeness property of real numbers: Every bounded
>monotonely increasing sequence of real numbers converges to a real
>number. Thus 0.99(bar) is real.
Great, at least we now know we are dealing with a number and not an
empty definition.
Why not go the whole hog and say that |1 - 0.9(bar)| >= 0
Then |1 - 0.9(bar)| is the smallest real number greater than or
equal to zero. The smallest real number => 0 is 0 so 1 = 0.9(bar)
>> So that sqrt(2) is in fact equal to an actual infinite number of
>> additions and subtractions of rational numbers.
>
>Yes, you can write sprt(2) as an acutal infinite number of additions and
>subtractions of rational numbers.
>
>> But sqrt(2) is
>> irrational and the sum of any rational numbers is a rational
>> number. But an irrational number cannot equal a rational number.
>
>This is also correct as long as you work with a finite number of
>additions and subtractions.
>
>BUT A SUM OF AN INFINITE NUMBER OF RATIONAL NUMBERS NEED NOT BE
>RATIONAL.
This seems more like theology than maths.
I don't see how adding rational numbers together either one by one
or all together or in some other way can effect this miraculous
transformation.
Does the process of addition suddenly change if you "overdo" it in
some way ?
But on the other hand, there is no reason why the limit or least
upper bound of a series of rationals should not be irrational.
>> What you can say is
>>
>> 0.9 + 0.1 = 1
>>
>> 0.99 + 0.01 = 1
>>
>> 0.99999 .. 999 (n 9s) + 1/10^n = 1
>>
>> If, in your terminology 0.99(bar) = 1, then 1/10^n = 0, for some n
>>
>> But there is no n that makes 1/10^n = 0
>>
>> What does make sense is to say that
>>
>> | 1 - 0.999 ...99 (n 9s)| = 1/10^n
>>
>> and that for any 1 > e > 0 there is an equality
>>
>> 1/10^n+1 < e < 1/10^n
>
>Infinity does many weird things to numbers.
Yes, or better the way we think about "infinity" leads to weird
conclusions.
What IS the fallacy in the argument ?
>Read an real analysis book.
This is very sound advice, one should always read widely.
So I had a look at "Mathematical Methods for Physicists" by George
Arfken, Chapter 5 Infinite Series.
He says
" Right at the start we face the problem of attaching meaning to
the sum of an infinite number of terms. The usual approach is by
partial sums. If we have an infinite sequence of terms u1, u2, u3,
.. we define the ith partial sum as
si = sum [n = 1 to i] un
This is a finite summation and offers no difficulties. If the
partial sums si converge to a finite limit as i -> oo
i -> oo lim si = S
the infinite series is said to be convergent and to have value S.
Note carefully that we reasonably, plausibly, but still ARBITRARILY
DEFINE the infinite series as equal to S."
So he is essentially saying that the difficulties associated with
infinite sums of rational numbers suddenly becoming irrational can
be avoided. No mysterious additions have to be made.
Would you say that was a reasonable assessment ?
Avital Pilpel
The problem here seems to be twofold:
1. first, the person who wrote that .9bar <> 1 seemed not to understand the
concept of the limit, that is, of course, that althought the series .9, .99,
.999 never "gets" to 1, it's limit is 1. Oh, well, even the ancient greeks
missed the whole concept.
2. Then he seems to assert something of the following sort: "OK, the series
converges to 1, but .9bar is not a series - it is a number...". this shows a
misunderstanding of what real numbers actually ARE. The cauchy definition of
a real numbers is as the equivalence class of all cauchy series of rational
numbers, when {an} and {bn} are equivalent if {an, bn} is also a cuachy
series. The SYMBOL .99999.. or 1.0000.. are merely, so to speak, a symbol for
these series. Sure, this goes agaisnt the ingrained intuition that a real
number is just "there" like a point on a line, but as long as we accept the
symbols .99999... and 1.0000... as indeed representing numbers, then we ARE
committed to this view of a series.
BTW, there IS a building of the reals on the rationals which treats them both
as geometrical points on a line - the Dedekind's cuts method. But there, the
symbol for the real number 1 would not be 1.0000.... or .9999.... but {A, B}
where A and B are certain (specific) sets of points. But this is no way out;
dedekind redefines, of course, the limit operation for the reals, and of
course the lim of .9, .99, .999... IS 1 in his system too. So once more, even
if we translate all our notation from .999... and 1.000... language to series
language and from there to dedkind cuts language, still the dedekind cut for
the limit of .9, .99, .999... is the dedekind cut for 1.
PS: There IS a esoteric field in analysis - non-standard analysis - that
indeed treats the infinitesimals in certain senses like "genuine" numbers. But
they too will disagree that .9999.... <>1, OK???!
--
Avital Pilpel.
=====================================
The majority is never right.
-Lazarus Long
=====================================
Vincent Johns
(posted & emailed)
Iain Davidson <ia...@stt.win-uk.net> wrote:
>
> [...]
> Tore August Kro <tor...@ifi.uio.no> wrote:
> >This is also correct as long as you work with a finite number of
> >additions and subtractions.
> >
> >BUT A SUM OF AN INFINITE NUMBER OF RATIONAL NUMBERS NEED NOT BE
> >RATIONAL.
>
> [...]
>
> Does the process of addition suddenly change if you "overdo" it in
> some way ?
>
> But on the other hand, there is no reason why the limit or least
> upper bound of a series of rationals should not be irrational.
Not having the physical means to add an infinite number of numbers
one by one, I usually consider the phrase "a sum of an infinite number
of rational numbers", or a similar expression, to refer to the
limit of a series. To me, it's just another way of expressing the
same thing.
Avital Pilpel
Josh Boyd wrote:
>
> > 4) 1/10^n = e and we can calculate an exact n.
> >
> > Calculating an exact n for any fixed e>0 means that 0.99(9bar) has to
> > terminate after n nines.
>
> nines, n must equal infinity (for the actual number .99(9bar)) because
> there are an infinite number of nines?
>
> Then it becomes...
>
> 1/10^(infinity) = e
>
> And any constant divided by infinity ( 10^(infinity) = infinity ) is
> necessarily zero ---
>
> e = 0 <--- Which says that
>
> |1 - .99(9bar)| = 0
> 1 = .99(9bar)
>
> (Can you not do that? I don't see anything wrong with it...)
>
> Josh
Well, indeed 1 = .9bar, but you can't do that. "Infinity" is not a real
number... The mix up is base on mixing two different senses of the word
infinity. The first is infinity as a cardinal, that is a quantity (to put is
VERY roughly) - such as "the 'number' of rational numbers" (aleph-null, if
you must know... :-)). The other, totally different sesne, is that of a
LIMIT, or a convergance of an infinite series like .9, .99, .999... we SAY
that the limit of this series as n APPROACHES infinity is 1, but the exact
definition of this term does NOT involve an infinite cardinal. It is an
infinite series, but it does not have the "element no. infinity", any more
than the fact thatthere is an infinite number of natural numbers means that
there is some "very big" natural number which is "equal" to infinity...
>
> P.S. - Sorry if that's what you said... I didn't really read much of
> this post.
--
Iain Davidson
>Given an infinite series of numbers x_1, x_2, x_3,... the usual way to
>define the sum of the series is: lim(n->oo) sum(i=1 to n) x_i.
That is it precisely. The sum of a series is DEFINED to be the
limit. This neatly avoids all metaphysical speculation about
sums of rationals becoming irrational. But a definition is not a
proof.
Avital Pilpel
Iain Davidson wrote:
>
>
> >> >Do you agree that 0.99(bar) has the property that given any number e>0,
> >> >no matter how small e is, then |1-0.99(bar)| < e ? If we define real
> >> >numbers in the normal way then this is a fact.
>
> >And by the completeness property of real numbers: Every bounded
> >monotonely increasing sequence of real numbers converges to a real
> >number. Thus 0.99(bar) is real.
>
> Great, at least we now know we are dealing with a number and not an
> empty definition.
>
> Why not go the whole hog and say that |1 - 0.9(bar)| >= 0
>
> Then |1 - 0.9(bar)| is the smallest real number greater than or
> equal to zero. The smallest real number => 0 is 0 so 1 = 0.9(bar)
>
>
>
> >> So that sqrt(2) is in fact equal to an actual infinite number of
> >> additions and subtractions of rational numbers.
> >
> >Yes, you can write sprt(2) as an acutal infinite number of additions and
> >subtractions of rational numbers.
> >
> >> But sqrt(2) is
> >> irrational and the sum of any rational numbers is a rational
> >> number. But an irrational number cannot equal a rational number.
> >
> >additions and subtractions.
> >
> >BUT A SUM OF AN INFINITE NUMBER OF RATIONAL NUMBERS NEED NOT BE
> >RATIONAL.
>
>
> I don't see how adding rational numbers together either one by one
> or all together or in some other way can effect this miraculous
> transformation.
>
> some way ?
no, not more than if you add 2 and 2 it is a "miracle" that you get 4...
Strictly speaking, of course, this fact - that the limit of real series
of rational numbers can be irrational - needs proof, but this is trivial
(think of 3, 3.1, 3.14, 3.141, 3.1415... it can be made exact easily).
Addition is not defined in the "regular" way for an "addition" of an
infinite number of numbers (what is known as an INFINITE SERIES). It is
not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
imfinite number of fractions".
Rather, this equasion, strictly speaking, is mere SHORTHAND for an exact
definition - that does NOT involve "adding an infinite number" - of the
fact that the limit of the parital sums is 2. People get confused
becase they assume that the "+" and "=" here mean the same thing as in
"2+2=4". It does not; it is an extention, so to speak, of the meaning
of these symbols in a new way.
Josh Boyd
> What does this mean? "Infinity" is not a number. With specific
> definitions it can be a shorthand for some more extensive math,
> but *definitions* are important.
Ok... We'll divide by a number "l" and take the limit as
l -> infinity.
> Divide by infinity? Please define.
You can consider infinity as a number, even though it is not... A
REALLY LARGE number. Thus, any number divided by a REALLY LARGE number
approaces zero... If you let l go to infinity (as mentioned above), the
division goes to 0.
Josh
Stan Armstrong
In article <4...@stt.win-uk.net>, Iain Davidson <ia...@stt.win-uk.net>
writes
>
>>BUT A SUM OF AN INFINITE NUMBER OF RATIONAL NUMBERS NEED NOT BE
>>RATIONAL.
counter examples. Mind you it has nothing to do with the stupid
Pimentel argument about .9(bar), which does equal 1 and I wish he and
all the others woudl stop talking about it. He will never accept
rational argument.
>
>This seems more like theology than maths.
>
>
>This is very sound advice, one should always read widely.
>So I had a look at "Mathematical Methods for Physicists" by George
>Arfken, Chapter 5 Infinite Series.
>
>He says
>
>" Right at the start we face the problem of attaching meaning to
>the sum of an infinite number of terms. The usual approach is by
>partial sums. If we have an infinite sequence of terms u1, u2, u3,
>.. we define the ith partial sum as
>
>si = sum [n = 1 to i] un
>
>This is a finite summation and offers no difficulties. If the
>partial sums si converge to a finite limit as i -> oo
>
>i -> oo lim si = S
>
>the infinite series is said to be convergent and to have value S.
>Note carefully that we reasonably, plausibly, but still ARBITRARILY
>DEFINE the infinite series as equal to S."
>
>So he is essentially saying that the difficulties associated with
>infinite sums of rational numbers suddenly becoming irrational can
>be avoided. No mysterious additions have to be made.
>
>
>Would you say that was a reasonable assessment ?
>
series for Pi being irrational and Aitken would make no such claim.
>
--
Stan Armstrong
Stan Armstrong
In article <32FFD0...@vnet.net>, "Raymond E. Griffith"
<rgri...@vnet.net> writes
<Bobbitt)
I will not contribute to the 0.9(bar) = 1 argument, although I know it
does because it is clear that there are many there who sadly are either
incapable of understanding or, even more sadly, do not wish to
understand.
>
>
>------------------------------
>Not mental simplicity. It just isn’t “obvious” that the real numbers
>should have more than one representation for any number. A good book in
>real analysis details these things.
However, it is clear to me that 2, 4^(1/2), 8^(1/3) etc. are all equally
valid representation of two. We do not use them normally, except in
problems, just as we do not normally use 0.9(bar) to represent 1; we use
the most convenient and best understood representation.
>
>
>Hope this helps.
>
>Raymond E. Griffith
>
>________________________________________________
>
>The educated man possesses six characteristics: He must be able to
>1. communicate;
>2. calculate;
>3. see the world as it really is;
>4. appropriately manipulate his environment;
>5. accurately predict the effects of that manipulation;
>6. admit the limits of his knowledge.
>
>Raymond E. Griffith
--
Stan Armstrong
Michael V. Ziniti
Is there anything wrong with this?
x = .9(bar)
10x = 9.9(bar)
10x - x = 9.9(bar) - .9(bar)
9x = 9
x = 1
So, .9(bar) = 1.
Mike
Lee Jaap
|>> 4) 1/10^n = e and we can calculate an exact n.
|>>
|>> Calculating an exact n for any fixed e>0 means that 0.99(9bar) has to
|>> terminate after n nines.
|>
|>Could you not say, though, that because .99(9bar) must terminate after n
|>nines, n must equal infinity (for the actual number .99(9bar)) because
|>there are an infinite number of nines?
|>
|> Then it becomes...
|>
|>1/10^(infinity) = e
What does this mean? "Infinity" is not a number. With specific
definitions it can be a shorthand for some more extensive math,
but *definitions* are important.
|> And any constant divided by infinity ( 10^(infinity) = infinity ) is
|>necessarily zero ---
Divide by infinity? Please define.
|>e = 0 <--- Which says that
Be careful when you're talking about infinity. It is not a number.
Iain Davidson
>Addition is not defined in the "regular" way for an "addition" of an
>infinite number of numbers (what is known as an INFINITE SERIES). It is
>not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
>imfinite number of fractions".
You can add the limits of two series together
1 + 1/2 + 1/4 + .... (+) 1 + 1/3 + 1/9 + ...
Is the plus in the middle a "infinite" type plus or a "finite" type
one ?
If a limit is a finite number then you should b using a finite plus.
But when you add the series together you should be using the
infinite plus.
>Rather, this equasion, strictly speaking, is mere SHORTHAND for an exact
>definition - that does NOT involve "adding an infinite number" - of the
>fact that the limit of the parital sums is 2. People get confused
>becase they assume that the "+" and "=" here mean the same thing as in
>"2+2=4". It does not; it is an extention, so to speak, of the meaning
>of these symbols in a new way.
I thought symbols were supposed to be uniquely defined in maths.
Then it is in fact true that 0.99(bar) has been defined to be 1
rather than proved to be 1.
Tore August Kro
Iain Davidson wrote:
>
>
>
> >Given an infinite series of numbers x_1, x_2, x_3,... the usual way to
> >define the sum of the series is: lim(n->oo) sum(i=1 to n) x_i.
>
> That is it precisely. The sum of a series is DEFINED to be the
> limit. This neatly avoids all metaphysical speculation about
> sums of rationals becoming irrational. But a definition is not a
> proof.
>
> Iain Davidson
You say that "a definition is not a proof", but how can you prove
anything about something that is un defined?
Before anything can be proven about a mathematical object you need to
define this object in order to have a property of that object to start
the proof with. If not working with exact definitions one could run into
difficulties because you don't know what you are proving anything about.
Here is an example: You do have an intuitive feeling about what a set
is. It is some sort of collection of other mathematical objects. A set
can contain numbers, functions, relations and even other sets.
Now define the "set" S as the set containing all sets that does not
contain themselves. Let us first ask the question does S contain itself?
If so by the definition of S, S should not be contained in S. So we have
an contradiction. Let us then ask if S is not an element of S? Then S
should by the definition of S be contained in S. And we have another
contradiction.
We see that the problem is that we have on exact definition of a set. If
we did define a set, (which is done in set-theory) we would see that S
isn't a set.
So we need definitions. Another question is whether our definition is
good or bad.
Tore August Kro
tor...@ifi.uio.no
Iain Davidson
>>" Right at the start we face the problem of attaching meaning to
>>the sum of an infinite number of terms. The usual approach is by
>>partial sums. If we have an infinite sequence of terms u1, u2, u3,
>>.. we define the ith partial sum as
>>
>>si = sum [n = 1 to i] un
>>
>>This is a finite summation and offers no difficulties. If the
>>partial sums si converge to a finite limit as i -> oo
>>
>>i -> oo lim si = S
>>
>>the infinite series is said to be convergent and to have value S.
>>Note carefully that we reasonably, plausibly, but still ARBITRARILY
>>DEFINE the infinite series as equal to S."
>>
>>So he is essentially saying that the difficulties associated with
>>infinite sums of rational numbers suddenly becoming irrational can
>>be avoided. No mysterious additions have to be made.
>>
>>
>>Would you say that was a reasonable assessment ?
>>
>No, you must be misinterpreting Aitken. You cannot avoid Gregory's
>series for Pi being irrational and Aitken would make no such claim.
No you are misinterpreting both of us. We all agree that the LIMIT
of Gregory's series is irrational. What is IRRATIONAL is to say
that there is a point where ADDING rational to rational becomes
irrational. This is why the concept of limit was invented - to
circumvent absurd notions of infinity. This is why algebra alone is
not adequate to deal with "infinite" series.
he...@dorsai.org
this has gone on long enough..
is this an algebra help group or a place to just talk about nothing
Avital Pilpel
Iain Davidson wrote:
>
> >Addition is not defined in the "regular" way for an "addition" of an
> >infinite number of numbers (what is known as an INFINITE SERIES). It is
> >not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
> >imfinite number of fractions".
>
> You can add the limits of two series together
>
> 1 + 1/2 + 1/4 + .... (+) 1 + 1/3 + 1/9 + ...
>
> Is the plus in the middle a "infinite" type plus or a "finite" type
> one ?
>
> If a limit is a finite number then you should b using a finite plus.
>
> But when you add the series together you should be using the
> infinite plus.
Oh, dear...
You misunderstand the symbols that are used.
Where, exactly, is the (+) sign that you used supposed to be? after the
"last" member of the infinite series 1+1/2+1/4+... ? It can't be,
becasue there IS no last place. So what, exactly, does this "addition"
you wrote down MEAN?
If you meant it to mean "the addition of the two infinite series", then
once more - You do NOT define addition of series in this way, because it
is non-sensical, as I just shown; the DEFINITION af an addition of two
infinite (converging) series IS the FINITE addition of their two limits
(presuming that the limits exists).
To use the terms I used last time, you add two series NOT by combining
them into one "double" infinite series and then "adding all those
numbers together" as you suggest; it is a meaningless operation,
althought it LOOKS - on paper - as if it is meaningful (superficially).
Rather, you first computing the limit of both series separatly - using
what I called the "infinite sum" - and THEN separartly add the two
LIMITS. The limits ARE real numbers and are added in the usual (+) way.
My advice to you:
1. READ A GOOD INTRODUCTION TO ANALYSIS BOOK!!!!
2. SUSPEND DISBELIEF - If there is something there - like the above -
that you do not intuitevly understand, PLEASE, take their word for it
for the moment! I promise you that you WILL see later on that it DOES
make sense. OK?
Iain Davidson
>> >Given an infinite series of numbers x_1, x_2, x_3,... the usual way to
>> >define the sum of the series is: lim(n->oo) sum(i=1 to n) x_i.
>>
>> That is it precisely. The sum of a series is DEFINED to be the
>> limit. This neatly avoids all metaphysical speculation about
>> sums of rationals becoming irrational. But a definition is not a
>> proof.
>You say that "a definition is not a proof", but how can you prove
>anything about something that is un defined?
I'm not saying that you can't define things in terms of other
things as long as you can show that the definition is consistent.
But, proving something is usually taken to mean using rules of
deduction to derive theorems from certain axioms.
In geometry, say, you can't prove that parallel lines meet at
infinity or don't meet at infinity in terms of other assumptions.
If you assume parallel lines meet at infinity you get one type of
geometry, if you don't, you get another.
>Before anything can be proven about a mathematical object you need to
>define this object in order to have a property of that object to start
>the proof with. If not working with exact definitions one could run into
>difficulties because you don't know what you are proving anything about.
Yes, but deciding on your axioms is not proving them. You accept
them, provided they are consistent, and use them as a basis for
further deductions.
>We see that the problem is that we have on exact definition of a set. If
>we did define a set, (which is done in set-theory) we would see that S
>isn't a set.
>
>So we need definitions. Another question is whether our definition is
>good or bad.
Yes absolutely, a good definition is a logically consistent one and
one that is logically consistent with any other definitions you
use.
Dave Weisbeck
In article <4...@stt.win-uk.net>, ia...@stt.win-uk.net (Iain Davidson) wrote:
>
>I thought symbols were supposed to be uniquely defined in maths.
>
here long enough, but I won't. I _think_ that to say they are uniquely defined
within the context they apply to is accurate however. (e.g. in number theory
we use an equal sign with an extra line for congruence, and outside of number
theory it usually means "exactly equal to" or "defined to be".)
>Then it is in fact true that 0.99(bar) has been defined to be 1
>rather than proved to be 1.
>
You could argue that any mathematical proof is really a definition and not a
proof since math is based on definitions and so any proof could be taken down
to those elementary definitions. I don't really know what your point is?
Cheers,
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
eagleone
Avital Pilpel wrote:
>
> Iain Davidson wrote:
> >
> >
> > >> >Do you agree that 0.99(bar) has the property that given any number e>0,
> > >> >no matter how small e is, then |1-0.99(bar)| < e ? If we define real
> > >> >numbers in the normal way then this is a fact.
> >
> > >And by the completeness property of real numbers: Every bounded
> > >monotonely increasing sequence of real numbers converges to a real
> > >number. Thus 0.99(bar) is real.
> >
> > Great, at least we now know we are dealing with a number and not an
> > empty definition.
> >
> > Why not go the whole hog and say that |1 - 0.9(bar)| >= 0
> >
> > Then |1 - 0.9(bar)| is the smallest real number greater than or
> > equal to zero. The smallest real number => 0 is 0 so 1 = 0.9(bar)
> >
> >
> >
> > >> So that sqrt(2) is in fact equal to an actual infinite number of
> > >> additions and subtractions of rational numbers.
> > >
> > >Yes, you can write sprt(2) as an acutal infinite number of additions and
> > >subtractions of rational numbers.
> > >
> > >> But sqrt(2) is
> > >> irrational and the sum of any rational numbers is a rational
> > >> number. But an irrational number cannot equal a rational number.
> > >
> > >This is also correct as long as you work with a finite number of
> > >additions and subtractions.
> > >
> > >RATIONAL.
> >
> >
> > or all together or in some other way can effect this miraculous
> > transformation.
> >
> > Does the process of addition suddenly change if you "overdo" it in
> > some way ?
>
> no, not more than if you add 2 and 2 it is a "miracle" that you get 4...
>
> Strictly speaking, of course, this fact - that the limit of real series
> of rational numbers can be irrational - needs proof, but this is trivial
> (think of 3, 3.1, 3.14, 3.141, 3.1415... it can be made exact easily).
>
> infinite number of numbers (what is known as an INFINITE SERIES). It is
> not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
> imfinite number of fractions".
>
> definition - that does NOT involve "adding an infinite number" - of the
> fact that the limit of the parital sums is 2. People get confused
> becase they assume that the "+" and "=" here mean the same thing as in
> "2+2=4". It does not; it is an extention, so to speak, of the meaning
> of these symbols in a new way.
>
>
> =====================================
> The majority is never right.
> -Lazarus Long
> =====================================
Your equation 1 +1/2 +1/4 +1/8 +... = 2 is not correct. You are always
going half of the remaining amount to get to 2. You will never reach
it. This is where the theory applies, not walking. If you do not know
what I mean then I need not waste my time any longer. Please clean up
your spelling.
Darrell Ryan
Iain Davidson wrote:
>
> >You could argue that any mathematical proof is really a definition and not a
> >proof since math is based on definitions and so any proof could be taken down
> >to those elementary definitions. I don't really know what your point is?
>
> I always thought you started off with certain axioms or postulates
> and then used rules of deduction to prove theorems.
>
> I think "define" and "prove" do mean different things.
>
> If you introduce the concept of a limit you are defining it not
> proving it.
Here's something to think about.....
What is the DEFINITION of proof ?????
To me, IMHO, this is not a "chicken or the egg" kind of thing.
____________________________________________________________
Darrell Ryan
e-mail dr...@edge.net
personal website http://edge.edge.net/~dryan
business website http://www.edge.net/stmc
"Don't go squirrel hunting with an elephant rifle,
unless the squirrel doesn't believe that the shotgun
has already done the job."
D. R.
Josh Boyd
> You freaking idiots 1 cannot approach infinity. Get a life, find out
> what infinity is, and then start posting.
Why thank you for not posting an intelligent response... (Or counter
argument) I do know what infinity is. I have a very good "sence" of
infinity, even though I could never imagine it. ;)
Gee... If you couldn't approach infinity, I guess you couldn't do
Calculus, eh?
Josh
Iain Davidson
>You could argue that any mathematical proof is really a definition and not a
>proof since math is based on definitions and so any proof could be taken down
>to those elementary definitions. I don't really know what your point is?
I always thought you started off with certain axioms or postulates
and then used rules of deduction to prove theorems.
I think "define" and "prove" do mean different things.
If you introduce the concept of a limit you are defining it not
proving it.
Iain Davidson Tel : +44 1228 49944
eagleone
Josh Boyd wrote:
>
> > What does this mean? "Infinity" is not a number. With specific
> > definitions it can be a shorthand for some more extensive math,
> > but *definitions* are important.
>
> l -> infinity.
>
> > Divide by infinity? Please define.
>
> You can consider infinity as a number, even though it is not... A
> REALLY LARGE number. Thus, any number divided by a REALLY LARGE number
> approaces zero... If you let l go to infinity (as mentioned above), the
> division goes to 0.
>
> Josh
Tore August Kro
Iain Davidson wrote:
>
>
> >> >Given an infinite series of numbers x_1, x_2, x_3,... the usual way to
> >> >define the sum of the series is: lim(n->oo) sum(i=1 to n) x_i.
> >>
> >> That is it precisely. The sum of a series is DEFINED to be the
> >> limit. This neatly avoids all metaphysical speculation about
> >> sums of rationals becoming irrational. But a definition is not a
> >> proof.
>
> >You say that "a definition is not a proof", but how can you prove
> >anything about something that is un defined?
>
> I'm not saying that you can't define things in terms of other
> things as long as you can show that the definition is consistent.
Definitions are always consistent.
This is because a definition says:
Something with property P is called n.
Where P is some mathematical property and n is a name previously not
used. And since the name isn't used before there is nothing that this
definition can be inconsistent with. We are just giving things a name so
that it is easier talking about them. We do not have to give the
property P all the time.
Example:
Given a real function f. The limit of f(x) as x approaches y is a
number b such that for every real number e>0 there exists a real number
d>0 so that when 0<|y-x|<d then |f(x)-b|<e.
And when talking about mathematics it wouldn't be practical saying "the
number b such that for every real number e>0 there exists a real number
d>0 so that when 0<|y-x|<d then |f(x)-b|<e". Especially if we use this
alot. Therefore we give this property a name : The limit of f(x) as x
approaches y. But we did not have to give it the name limit, we could
have called it Blobb instead: The Blobb of f(x) as x approaches y. But
then again it is better haveing names that are easy to remember.
> But, proving something is usually taken to mean using rules of
> deduction to derive theorems from certain axioms.
>
> In geometry, say, you can't prove that parallel lines meet at
> infinity or don't meet at infinity in terms of other assumptions.
>
> If you assume parallel lines meet at infinity you get one type of
> geometry, if you don't, you get another.
>
> >Before anything can be proven about a mathematical object you need to
> >define this object in order to have a property of that object to start
> >the proof with. If not working with exact definitions one could run into
> >difficulties because you don't know what you are proving anything about.
>
> Yes, but deciding on your axioms is not proving them. You accept
> them, provided they are consistent, and use them as a basis for
> further deductions.
Axioms is relating mathematical objects to each other. So you need to
verify that axioms are consistent. But definitions are giving names.
> >> >Given an infinite series of numbers x_1, x_2, x_3,... the usual way to
> >> >define the sum of the series is: lim(n->oo) sum(i=1 to n) x_i.
So when I say this I say what is to be understood by the name an
infinite sum. Previous to this definition of an infinite sum this name
was unused so I am not changeing anything or stateing something about an
object we know anything about previously.
> >We see that the problem is that we have on exact definition of a set. If
> >we did define a set, (which is done in set-theory) we would see that S
> >isn't a set.
> >
> >So we need definitions. Another question is whether our definition is
> >good or bad.
>
> Yes absolutely, a good definition is a logically consistent one and
> one that is logically consistent with any other definitions you
> use.
>
> Iain Davidson
Now I challenge you to give your definitions of these terms:
a real number
a limit
a finite sum
a series
an infinite sum
0.9(9bar)
Now I have to go to my real analysis course.
Tore August Kro
tor...@ifi.uio.no
Iain Davidson
>Definitions are always consistent.
Definitions can be empty
SPOOKY is the biggest natural number
This follows your definition of definition, but does it have any
meaning ? Does SPOOKY even exist ?
>Axioms is relating mathematical objects to each other. So you need to
>verify that axioms are consistent. But definitions are giving names.
But can you prove an axiom ? Surely by definition an axiom is a
postulate something assumed and so is not capable of being proved.
Tore August Kro
Try to mark some distance on the floor in the room where you are at the
present. Then try to walk that distance. Observe that at one point you
must have passed half the distance, 1/2+1/4=3/4 of the distance,
1/2+1/4+1/8=7/8 of the distance,and so on. If you managed to cross the
distance then 1 +1/2 +1/4 +1/8 +... = 2, if you didn't then it's not.
To be more mathematical about this problem: The definition of 1 +1/2
+1/4 +1/8 +... is the limit as n approaches infinity of 1 +1/2 +1/4
+1/8 +... +1/(2^n). And this is 2.
Tore August Kro
tor...@ifi.uio.no
Tore August Kro
Tore August Kro
tor...@ifi.uio.no
Tore August Kro
You are correct that what you define doesn't allways exist.
This isn't any more mysterious than equations that doesn't have
solutions. Like this one. x -real number such that x<0 and x>5.
Iain Davidson wrote:
>
> >Definitions are always consistent.
>
> Definitions can be empty
>
> SPOOKY is the biggest natural number
>
> This follows your definition of definition, but does it have any
> meaning ? Does SPOOKY even exist ?
Property of natural numbers: For every natural number N there exists a
natural number M that is larger than N.
By the definition of SPOOKY: SPOOKY is a natural number.
Then there is a natural number M greater than SPOOKY.
Thus SPOOKY isn't the biggest natural number.
Theorem: SPOOKY does not exist.
> >Axioms is relating mathematical objects to each other. So you need to
> >verify that axioms are consistent. But definitions are giving names.
>
> But can you prove an axiom ? Surely by definition an axiom is a
> postulate something assumed and so is not capable of being proved.
>
> Iain Davidson
But you still havn't answered my challenge to you:
> Now I challenge you to give your definitions of these terms:
> a real number
> a limit
> a finite sum
> a series
> an infinite sum
> 0.9(9bar)
Tore August Kro
tor...@ifi.uio.no
Stan Armstrong
In article <33029D...@cybermax.net>, eagleone
<eagl...@cybermax.net> writes
>
>>
>> Addition is not defined in the "regular" way for an "addition" of an
>> infinite number of numbers (what is known as an INFINITE SERIES). It is
>> not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
>> imfinite number of fractions".
>
>
>
>going half of the remaining amount to get to 2. You will never reach
>it. This is where the theory applies, not walking. If you do not know
>what I mean then I need not waste my time any longer. Please clean up
>your spelling.
If you had read what Avital said and understood it you would realise
what he meant when he referred to an infinite series.
Also don't be so (a) patronising and (b) rude. I am fairly sure that
his English spelling is better than yours in Dutch. Indeed looking back
at some of your previous posts there are errors in your English.
--
Stan Armstrong
Iain Davidson
In article <330308...@ifi.uio.no>, Tore August Kro (tor...@ifi.uio.no) writes:
>You are correct that what you define doesn't allways exist.
>This isn't any more mysterious than equations that doesn't have
>solutions. Like this one. x -real number such that x<0 and x>5.
Quite, I didn't say there was any mystery.
>Iain Davidson wrote:
>>
>> >Definitions are always consistent.
>>
>> Definitions can be empty
>>
>> SPOOKY is the biggest natural number
>>
>> This follows your definition of definition, but does it have any
>> meaning ? Does SPOOKY even exist ?
>
>Property of natural numbers: For every natural number N there exists a
>natural number M that is larger than N.
>
>By the definition of SPOOKY: SPOOKY is a natural number.
>
>Then there is a natural number M greater than SPOOKY.
>
>Thus SPOOKY isn't the biggest natural number.
>
>Theorem: SPOOKY does not exist.
If you have a definition saying that SPOOKY does exist and then a
theorem saying that it doesn't, that seems to be pretty
inconsistent as they both can't be true.
.
>
>But you still havn't answered my challenge to you:
>> Now I challenge you to give your definitions of these terms:
>> a real number
>> a limit
>> a finite sum
>> a series
>> an infinite sum
>> 0.9(9bar)
I'm not quite sure what the point of the exercise would be.
These appear to be mathematical terms and I am not a mathematician.
Shouldn't you be able to define your own terms ?
If you have a definition of these terms please share them with me.
Would you be able to define a "Chomskian Hierarchy"
if I asked you ?.
Tore August Kro
Iain Davidson wrote:
>
>
> In article <330308...@ifi.uio.no>, Tore August Kro (tor...@ifi.uio.no) writes:
> >You are correct that what you define doesn't allways exist.
> >This isn't any more mysterious than equations that doesn't have
> >solutions. Like this one. x -real number such that x<0 and x>5.
>
> Quite, I didn't say there was any mystery.
>
> >Iain Davidson wrote:
> >>
> >> >Definitions are always consistent.
> >>
> >> Definitions can be empty
> >>
> >> SPOOKY is the biggest natural number
> >>
> >> This follows your definition of definition, but does it have any
> >> meaning ? Does SPOOKY even exist ?
> >
> >Property of natural numbers: For every natural number N there exists a
> >natural number M that is larger than N.
> >
> >By the definition of SPOOKY: SPOOKY is a natural number.
> >
> >Then there is a natural number M greater than SPOOKY.
> >
> >Thus SPOOKY isn't the biggest natural number.
> >
> >Theorem: SPOOKY does not exist.
>
> If you have a definition saying that SPOOKY does exist and then a
> theorem saying that it doesn't, that seems to be pretty
> inconsistent as they both can't be true.
>
Well, the definition just gives a name to something. But having a
definition of something does by no mean imply that it exists. I could
talk about my mothers sister even though my mother does not have a
sister. Or about Atlantis, or the 56. sate of the United States, or
ghosts, ...
>
> >
> >But you still havn't answered my challenge to you:
> >> Now I challenge you to give your definitions of these terms:
> >> a real number
> >> a limit
> >> a finite sum
> >> a series
> >> an infinite sum
> >> 0.9(9bar)
>
> I'm not quite sure what the point of the exercise would be.
>
> These appear to be mathematical terms and I am not a mathematician.
> Shouldn't you be able to define your own terms ?
You should be able to define them if you want to discuss their
properties. Or at least you should accept the definitions given if you
can't come up with you own.
> If you have a definition of these terms please share them with me.
>
> Would you be able to define a "Chomskian Hierarchy"
> if I asked you ?.
>
> Iain Davidson
No, I can't define a "Chomskian Hierarchy", but we are not discussing
it, so I don't rally care.
Tore August Kro
tor...@ifi.uio.no
Mike Housky
eagleone wrote:
[this was aimed at Avital Pilpel]
> Your equation 1 +1/2 +1/4 +1/8 +... = 2 is not correct. You are always
> going half of the remaining amount to get to 2. You will never reach
> it. This is where the theory applies, not walking.
What theory? Math according to Zeno? Infinite geometric series should
have been a standard part of your education. These sums are well-defined
when the multiplier is less than 1, as the limit of the sequence of partial
sums.
Since this seems to have been omitted from your education, maybe some here
can help. Here's my attempt.
Let's define a sequence of numbers S as follows:
S = {S[0], S[1], S[2], ... }
S[0] = 1
S[1] = 1 + 1/2
S[2] = 1 + 1/2 + 1/4
S[n] = sum from k=0 to n of (2^-n)
= 1 + 1/2 + 1/4 + 1/8 + ... + 2^-n
Note that each member of the sequence is well defined for any finite
integer n >= 0. As n gets large you may notice that the value of S[n]
gets close to 2. Now, considrer the expression:
2*S[n] - S[n]
For finite n, this is clearly equal to S[n]. Also, using the defining
formula for S[n], this equals:
S[n] = 2*S[n] - S[n] = 2(1 + 1/2 + ... + 2^-n) - (1 + 1/2 + ... + 2^-n)
S[n] = (2 + 1 + ... + 2^(n-1)) - (1 + 1/2 + ... + 2^-n)
S[n] = 2 - 2^-n
This pretty much describes your "going half of the remaining amount".
The difference between 2 and S[n] is halved for each successive value
of n. Do you agree? Note that S is the sequence of partial sums of
the original series. By definition, the series has a limit if and only
if the sequence has a limit.
So, what's the limit of a sequence? We have a concise formula for the nth
member of the sequence S. As n gets large, S[n] appears to get very close
to 2. It appears to approach 2 as a limiting value, but that's not really
good enough for an exact definition of "limiting value" or "limit". What
can we do clarify this?
Here's a definition used by many mathematicians: A sequence S has a limit
s if and only if for each positive number e, there exists a number N, such
that |S[n]-s|<e for all n>N. Take a moment to swallow this definition.
It covers all the usual situations and clears up some of the murky ones.
Most importantly, what it requires is that no matter how small e is,
the members of S eventually get that close to s and stay there.
Well, for this sequence, and "guessing" that 2 is the limiting value,
e > |s - S[n]| = |2 - (2 - 2^-n)| ...or:
e > 2^-n
What's required is that for any positive e, no matter how small, there
is an N such that 2^-n is less than e for all n>N. This is easily done
by picking N > log2(1/e). This completes the proof.
So, according to this definition of limit, as hammered out by the likes
of Bolzano, Cauchy, Riemann and Weierstrauss, the sequence S has a limit
of 2. Exactly 2, not "two minus a smidge", even though all the sequence
members are less than 2. These and other great mathematicians are the
ones who put the calculus on a rigorous footing well over a century ago.
This style of definition of limit seems to me to be purposely designed
to avoid the sort of fuzzy thinking that leads people to think that
the power-of-1/2 series doesn't equal 2 or that .9bar doesn't equal 1.
Notice that the letter e as used here should not be confused with the
base of natural logarithms. I'm using it because textbooks tend to use
epsilon, the Greek equivalent of e, for this value.
> If you do not know
> what I mean then I need not waste my time any longer. Please clean up
> your spelling.
As Oscar Wilde wrote: "A gentleman is never unintentionally rude." I'm
fighting the inclination to be intentional. Have a nice life.
Cheers,
Mike.
Avital Pilpel
Michael V. Ziniti wrote:
>
> Is there anything wrong with this?
>
> x = .9(bar)
> 10x = 9.9(bar)
>
> 10x - x = 9.9(bar) - .9(bar)
> 9x = 9
> x = 1
>
> So, .9(bar) = 1.
>
> Mike
No0thing wrong in the result, but strictly speaking, in order to show
that if x=.9bar then 10x=9.9bar, you need the theorem that lim {an*c}
when c is a constant is equal to c*(lim {an}).
--
Avital Pilpel.
=====================================
The majority is never right.
-Lazarus Long
=====================================
eagleone
Josh Boyd <jb...@mail.coos.or.us> wrote in article
<3302B9...@mail.coos.or.us>...
> > You freaking idiots 1 cannot approach infinity. Get a life, find out
> > what infinity is, and then start posting.
>
> Why thank you for not posting an intelligent response... (Or counter
> argument) I do know what infinity is. I have a very good "sence" of
> infinity, even though I could never imagine it. ;)
>
> Gee... If you couldn't approach infinity, I guess you couldn't do
> Calculus, eh?
>
> Josh
>
Sorry about my language, I forgot to take my medication. The number 1 is a
finite number. A finite number, by its very definition, cannot approach
infinity. I cannot approach infinity. A variable can approach infinity.
1(one) is not a variable.
Yes, my previous post was irresponsible of me. Please excuse me. Thank You.
Lee Jaap
In article <330242...@columbia.edu> Avital Pilpel <ap...@columbia.edu> writes:
|>Iain Davidson wrote:
|>>
|>> >Addition is not defined in the "regular" way for an "addition" of an
|>> >infinite number of numbers (what is known as an INFINITE SERIES). It is
|>> >not that 1+1/2+1/4+1/8+... = 2 becasue someone bothered to "add an
|>> >imfinite number of fractions".
|>>
|>> You can add the limits of two series together
|>>
|>> 1 + 1/2 + 1/4 + .... (+) 1 + 1/3 + 1/9 + ...
|>>
|>> Is the plus in the middle a "infinite" type plus or a "finite" type
|>> one ?
|>>
|>> If a limit is a finite number then you should b using a finite plus.
|>>
|>> But when you add the series together you should be using the
|>> infinite plus.
|>
|>Oh, dear...
|>
|>You misunderstand the symbols that are used.
|>
|>Where, exactly, is the (+) sign that you used supposed to be? after the
|>"last" member of the infinite series 1+1/2+1/4+... ? It can't be,
|>becasue there IS no last place. So what, exactly, does this "addition"
|>you wrote down MEAN?
|>
|>If you meant it to mean "the addition of the two infinite series", then
|>once more - You do NOT define addition of series in this way, because it
|>is non-sensical, as I just shown; the DEFINITION af an addition of two
|>infinite (converging) series IS the FINITE addition of their two limits
|>(presuming that the limits exists).
Or you can create a new series with the ith term being the sum of
the ith terms of the original two series, then find the sum of
the new series.
If all the limits exist,
Sum a_i + Sum b_i = Sum (a_i + b_i)
There are two infinite sums on the left and one on the right.
Lee Jaap
In article <01bc1a00$8d85a740$2890...@cybermax.net.cybermax.net> "eagleone" <eagl...@cybermax.net> writes:
|>Sorry about my language, I forgot to take my medication. The number 1 is a
|>finite number. A finite number, by its very definition, cannot approach
|>infinity. I cannot approach infinity. A variable can approach infinity.
|>1(one) is not a variable.
|>
|>Yes, my previous post was irresponsible of me. Please excuse me. Thank You.
If you read carefully, the original poster wrote "l" (ell), not "1"
(one).
Of course, it's a poor choice, because it can lead to this confusion.
In most fonts I've seen, you can look closely to distinguish between
these two characters, but you often must look very closely.
Lee Jaap
|>Michael V. Ziniti wrote:
|>>
|>> Is there anything wrong with this?
|>>
|>> x = .9(bar)
|>> 10x = 9.9(bar)
|>>
|>> 10x - x = 9.9(bar) - .9(bar)
|>> 9x = 9
|>> x = 1
|>>
|>> So, .9(bar) = 1.
|>>
|>> Mike
|>This is correct if you ask me.
|>
|>Tore August Kro
|>tor...@ifi.uio.no
Correct, in a sense. Without a rigorous definition for the (bar)
operator, it isn't a rigorous proof.
Albert Yang
eagleone (eagl...@cybermax.net) wrote:
: Josh Boyd wrote:
: >
: > > What does this mean? "Infinity" is not a number. With specific
: > > definitions it can be a shorthand for some more extensive math,
: > > but *definitions* are important.
: >
: > Ok... We'll divide by a number "l" and take the limit as
: >
: > > Divide by infinity? Please define.
: >
: > You can consider infinity as a number, even though it is not... A
: > REALLY LARGE number. Thus, any number divided by a REALLY LARGE number
: > approaces zero... If you let l go to infinity (as mentioned above), the
: > division goes to 0.
: >
: > Josh
: what infinity is, and then start posting.
That was the letter "l", not the number "1". Try getting some manners
sometime.
--
Albert Yang |"Reports of my assimilation have
Internet: apy...@ucdavis.edu | been greatly exaggerated."
http://dcn.davis.ca.us/~albert/ | - Jean-Luc Picard, ST:FC
Iain Davidson
>Well, the definition just gives a name to something. But having a
>definition of something does by no mean imply that it exists. I could
>talk about my mothers sister even though my mother does not have a
>sister. Or about Atlantis, or the 56. sate of the United States, or
>ghosts, ...
Yes, I agree.
But that's a long way from whether or not you are actually proving
or defining the limit of an "infinite sum"
Basically what you seem to have been saying is that the limit of the
partial sums of a series is the smallest number not exceeded by any
of the partial sums. In the case of 0.99..,= 9/10 + 9/100 +..,= 1 -
1/10^n n -> big number, which must be a shorthand way of defining a
limit, 1 is the smallest number not exceeded by any of the partial
sums.
I agree this is a good way of interpreting or defining the phrase
"infinite sum" to make sense of it but it is not a proof.
Although you would have to demonstate that the definition was not
empty.
In propositional logic you can define "if.. then" in terms of a
truth table and that works very well with sentences like
"If the temperature of a piece of iron is increased, then its
volume increases"
"If you do not accept my definitions, I shall sulk"
"If Nero was a Christian saint, then 5 + 7 = 16"
But the last sentence, although true by definition, is almost
meaningless in a non-technical sense.
Iain Davidson
>> >But you still havn't answered my challenge to you:
>> >> Now I challenge you to give your definitions of these terms:
>> >> a real number
>> >> a limit
>> >> a finite sum
>> >> a series
>> >> an infinite sum
>> >> 0.9(9bar)
I must admit, having looked at it again I am warming to your
definition of limit.
I asked you before what was wrong with this argument.
0.9 + 1/10 = 1
0.99 + 1/10^2 = 1
0.99..99 + 1/10^n = 1
0.9999.... + 1/10^m = 1
If 0.999.. = 1 then this implies that 1/10^m = 0 but there is no
rational number with this property.
If you do accept the limit concept
the limit of 0.999... is 1
the limit of 1/10^n is 0
So everything works out neatly.
But on the other hand, because there is no rational number 1/10^m =
0, this seems to imply that you need some other kind of number to
make sense of limits, in the same way as x + 5 = 2 does not make
sense in the naturals. This must be true because certain rational
sequences have limits that are not rational. If irrational numbers
did not exist, these rational number sequences would not have a
limit. So you can't define a limit using just rational numbers.
So the question is how do you define the reals ?
If you define them in terms of limits then isn't there a touch of
circularity in the definition. You are saying that every real number
is a limit of a sequence and every limit of a sequence is a real
number and using each to define the other.
If you can define reals in another way then, of course, everything
you say follows.
Tore August Kro
Iain Davidson wrote:
>
> >> >But you still havn't answered my challenge to you:
> >> >> Now I challenge you to give your definitions of these terms:
> >> >> a real number
> >> >> a limit
> >> >> a finite sum
> >> >> a series
> >> >> an infinite sum
> >> >> 0.9(9bar)
>
> I must admit, having looked at it again I am warming to your
> definition of limit.
> So the question is how do you define the reals ?
>
> If you define them in terms of limits then isn't there a touch of
> circularity in the definition. You are saying that every real number
> is a limit of a sequence and every limit of a sequence is a real
> number and using each to define the other.
>
> If you can define reals in another way then, of course, everything
> you say follows.
>
> Iain Davidson
Let us say that one has Q, the rational numbers. One way to introduce
the real numbers is by equivalence classes of Cauchy-sequences.
A Cauchy sequence of rational numbers is a sequence (x_n) such that for
every rational number e>0, there is an integer N such that |x_n - x_m|<e
whenever n>=N and m>=N.
Now let S be the set of all Cauchy sequences of rational numbers.
Introduce an equivalence relation ~ between Cauchy sequences by
(x_n) ~ (y_n) if for every rational number e>0, there is an integer N
such that |x_n - y_n|<e whenever n>=N.
The equivalence class of an element (x_n) of S under the equivalence
relation ~ is the set [(x_n)] = {(y_n) in S such that (x_n) ~ (y_n)}.
The equivalence class of (x_n) is all elements of S that are equivalent
with (x_n) under ~.
And we see that S spilts into disjoint sets, the equivalence classes
under ~. That is if [(x_n)] =/= [(y_n)] then there is no element in both
[(x_n)] and [(y_n)].
Let S/~ be the set of all equivalence classes of S under ~.
That is S/~ = { [(x_n)] where (x_n) is a Cauchy sequence of rational
numbers }.
The set S/~ is the set of real numbers. We identify the rational numbers
as a subset of S/~; Q is identified with the set of equivalence classes
under ~ of all constant sequences of rational numbers.
Q = { [(x_n)] where x_i = r for all i and r rational number }.
Intuitively a Cauchy sequence represets the real number that it converge
to. But we still haven't defined convergence so we really can't talk
about it yet.
But we still are missing something essential: Addition, multiplication,
equality, inequality and so on. We have to define addition, ... on S/~
so that the properties we would like the real numbers to have does hold.
We should define addition + such that
for a,b,c elements of S/~
1) commutativity : a + b = b + a
[ We can talk about equality because that an element
[(x_n)] in S/~ is equal to [(y_n)] means that
(x_n) ~ (y_n). ]
2) assosiativity : a + ( b + c ) = ( a + b ) + c
3) zero : there exists an element z in S/~ such that
a + z = a
4) additive inverse : for every element a in S/~ there is an
element -a of S/~ such that a + -a = z
We should define multiplication * such that
for a,b,c elements of S/~
1) commutativity : a * b = b * a
2) assosiativity : a * ( b * c ) = ( a * b ) * c
3) unity : there is an element u of S/~ such that
a * u = a
4) reciprocals : for every element a =/= z there is an element
a^-1 such that a * a^-1 = u.
5) distributivity : a * ( b + c ) = ( a * b ) + ( a * c )
And we should define a relation <= such that
for a,b,c elements of S/~
1) reflexivity : a <= a
2) antisymetry : if a <= b and b <= a then a = b
3) transitivity : if a <= b and b <= c then a <= c
4) linear ordering : for every pair a,b either a <= b or b <= a
5) compability of <= and + : if a <= b then
a + c <= b + c for every c
6) compability of <= and * : if 0 <= a and 0 <= b
then 0 <= a * b
To define +, * and <= is starightforeward, but much work.
Trust me: It can be done.
Now let us pertend that we sucsessfully have defined +, * and <=.
And we also would like this property to hold:
Every sequence of elements of S/~ that is increasing and bounded
above converges to a limit in S/~.
This property does make sence if I define the terms "increasing
sequence", "bounded sequence" and "converge".
A sequence (a_n) of elements from S/~ (that is a sequence where each a_n
is an element of S/~) is said to be increasing if a_n <= a_(n+1) for
every natural number n.
A sequence (a_n) of elements from S/~ is bounded if there is an element
K of S/~ such that a_n <= K for every natural number n.
And here comes the definition of limit and converge:
A sequence (a_n) of elements from S/~ is said to converge to an element
b of S/~ if for every element e greater than 0 ( this means that z <= e
and e =/= z ) there is a natural number N such that -e <= a_n - b <=
whenever n >= N. We say that b is the limit of (a_n)
Now there is no circularity to the definition of real numbers.
Tore August Kro
tor...@ifi.uio.no
Tore August Kro
Iain Davidson wrote:
> I asked you before what was wrong with this argument.
>
> 0.9 + 1/10 = 1
> 0.99 + 1/10^2 = 1
>
> 0.99..99 + 1/10^n = 1
>
> 0.9999.... + 1/10^m = 1
>
> If 0.999.. = 1 then this implies that 1/10^m = 0 but there is no
> rational number with this property.
>
> If you do accept the limit concept
>
> the limit of 0.999... is 1
> the limit of 1/10^n is 0
>
> So everything works out neatly.
>
> But on the other hand, because there is no rational number 1/10^m =
> 0, this seems to imply that you need some other kind of number to
> make sense of limits, in the same way as x + 5 = 2 does not make
> sense in the naturals. This must be true because certain rational
> sequences have limits that are not rational. If irrational numbers
> did not exist, these rational number sequences would not have a
> limit. So you can't define a limit using just rational numbers.
>
> Iain Davidson
At one point you go from a finite number of 9's to an infinite number of
9's, but you still add 1/10^m even though the number m should be equal
to the number of 9's, and this isn't possible when m is a natural number
and the number of 9's are infinite.
> 0.99..99 + 1/10^n = 1
>
> 0.9999.... + 1/10^m = 1
Tore August Kro
tor...@ifi.uio.no
Iain Davidson
>> If you can define reals in another way then, of course, everything
>> you say follows.
{Impressive proof omitted}
> Now there is no circularity to the definition of real numbers.
I agree. Everything you say does follow
(I'll have to check the details though)
Iain Davidson
Surely the point is that there is no "infinite number" of 9s in the
sense of a well-defined number.
The partial sums go on forever. If you are saying that 0.999.. is
somewhere at the end of the list of partial sums then you are also
saying that there is a greatest natural number. But that cannot be.
In general, you have to go outside the rationals to find the limit
of the sequence of partial sums. Rational + rational = rational no
matter how far down the partial sums you go.
>> 0.99..99 + 1/10^n = 1
>>
>> 0.9999.... + 1/10^m = 1
Iain Davidson Tel : +44 1228 49944
altavoz
Josh Boyd <jb...@mail.coos.or.us>
> What does this mean? "Infinity" is not a number. With specific
> definitions it can be a shorthand for some more extensive math,
> but *definitions* are important.
Ok... We'll divide by a number "l" and take the limit as
l -> infinity.
> Divide by infinity? Please define.
You can consider infinity as a number, even though it is not... A
REALLY LARGE number. Thus, any number divided by a REALLY LARGE number
approaces zero... If you let l go to infinity (as mentioned above), the
division goes to 0.
Josh
altavoz: I don't agree . Infinity is larger than the largest
number ( Howard Anton calculus). It can not be thought of as
a number .
______End of text from altavoz___________
Jim Petty
jaa...@asmobj.larc.nasa.gov (Lee Jaap) wrote:
In a sense, the question cannot be properly asked without a rigorous
definition of the bar. The bar does seem to have a standard meaning,
however, since everyone here is useing ot to mean the same thing. It
may not be a rigorous proof, but I can't see picking apart the proof
other than that.
Jim Petty
Lee Jaap
In article <5ef9qc$l...@dfw-ixnews6.ix.netcom.com> jimp...@newreach.net (Jim Petty) writes:
|>jaa...@asmobj.larc.nasa.gov (Lee Jaap) wrote:
|>>In article <330305...@ifi.uio.no> Tore August Kro <tor...@ifi.uio.no> writes:
|>>|>Michael V. Ziniti wrote:
|>>|>>
|>>|>> x = .9(bar)
|>>|>> 10x = 9.9(bar)
|>>|>>
|>>|>> 10x - x = 9.9(bar) - .9(bar)
|>>|>> 9x = 9
|>>|>> x = 1
|>>|>>
|>>|>> So, .9(bar) = 1.
|>>|>>
|>
|>>operator, it isn't a rigorous proof.
|>
|>definition of the bar. The bar does seem to have a standard meaning,
|>however, since everyone here is useing ot to mean the same thing. It
|>may not be a rigorous proof, but I can't see picking apart the proof
|>other than that.
Proving that 0.9bar = 1 is about as useful, and makes as much sense,
as proving that 0.5 = 1/2, or that i^2 = -1. This is what I've meant
when I've said that with a mathematically useful definition for the
bar operator, there's no need for a proof, at least for this kind of
proof.
OTOH, I have said that this can be a useful _intuitive_ suggestion
that .9bar ought to be equal to 1. That's different from a proof.
Vincent Johns
(posted & emailed)
Iain Davidson <ia...@stt.win-uk.net> wrote:
> [...]
> In propositional logic you can define "if.. then" in terms of a
> truth table and that works very well with sentences like
>
> "If the temperature of a piece of iron is increased, then its
> volume increases"
>
> "If you do not accept my definitions, I shall sulk"
>
> "If Nero was a Christian saint, then 5 + 7 = 16"
>
> But the last sentence, although true by definition, is almost
> meaningless in a non-technical sense.
Even in everyday, non-technical speech, there are occasions
for using "if ... then" constructions in which both parts
are unconditionally false, or at least intended to be
considered to be false.
I've seen sentences like the Nero example used in this way,
as for example in "If Mr. Smith were at the meeting, he
would have voted for the measure," where speaker and listener
should both be aware that both the "if" and the "then" parts are
false (so the speaker is saying the Mr. Smith was *not* at the
meeting).
(Putting the verb into the subjunctive mood, as in "If Mr. Smith
were ..." instead of "If Mr. Smith was ...", states that I
know the "if" part is contrary to fact. Saying "If Mr. Smith
was at the meeting..." indicates that I don't know whether
he was there, and am saying what I expect to have happened
assuming that he was there.)
The construction used in the Nero example might be used as a
semi-polite way of saying "You're full of beans" or "I can't
believe what you just said."
For example, one might say, "If that were true, then 5 + 7 = 16".
(Usually the "then" part has some connection to the idea being
ridiculed; in this case, it might relate to someone's faulty
accounting.) If I wanted to severely denigrate someone's
character, I might say, "If he's a respectable person, then
Nero was a Christian saint." (BTW, in real life I usually try
to avoid language as strong as this!)
--
-- Vincent Johns
Please feel free to quote anything I say here.