0.99(bar) cannot be proven to equal 1.00
John Pimentel
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 approach taken
here is that we cannot adapt the rules of mathematics to suit our theory.
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.
then 0.88(bar) equals 0.9
then 0.77(bar) equals 0.8
then 0.66(bar) equals 0.7
then 0.55(bar) equals 0.6
then 0.44(bar) equals 0.5
then 0.33(bar) equals 0.4
then 0.22(bar) equals 0.3
then 0.11(bar) equals 0.2
then 0.00(bar) equals 0.1
and these too...
then 1.11(bar) equals 1.2
then 2.00(bar) equals 2.1
there are other absurdities that can be stated with 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 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 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) equals 0.9? It does not.
Does 0.77(bar) equals 0.8? It does not.
Does 0.66(bar) equals 0.7? It does not.
Does 0.55(bar) equals 0.6? It does not.
Does 0.44(bar) equals 0.5? It does not.
Does 0.33(bar) equals 0.4? It does not.
Does 0.22(bar) equals 0.3? It does not.
Does 0.11(bar) equals 0.2? It does not.
Does 0.00(bar) equals 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 - 1 = 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
Earle D. Jones
In article <1997020514...@lucius.ultra.net>,
pime...@roots.ultranet.com 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 approach taken
>here is that we cannot adapt the rules of mathematics to suit our theory.
>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.
>then 0.88(bar) equals 0.9
>then 0.77(bar) equals 0.8
>then 0.66(bar) equals 0.7
>then 0.55(bar) equals 0.6
>then 0.44(bar) equals 0.5
>then 0.33(bar) equals 0.4
>then 0.22(bar) equals 0.3
>then 0.11(bar) equals 0.2
>then 0.00(bar) equals 0.1
>
==========
In your attempt to "disprove" something that is true, you have uncovered
another approach to "proving" what you set out to "disprove".
>If 0.99(bar) equals 1.0 then these others must also be true.
This statement is false.
Let's consider part of your series in reverse order:
>then 0.11(bar) equals 0.2 False. 0.11(bar) equals exactly 1/9
(If you have a problem with this, divide 9 into 1 using long
division and stop when you see where it's headed.)
>then 0.22(bar) equals 0.3 False. 0.22(bar) equals exactly 2/9
>then 0.33(bar) equals 0.4 False. 0.33(bar) equals exactly 3/9
>then 0.44(bar) equals 0.5 False. 0.44(bar) equals exactly 4/9
>then 0.55(bar) equals 0.6 False. 0.55(bar) equals exactly 5/9
>then 0.66(bar) equals 0.7 False. 0.66(bar) equals exactly 6/9
>then 0.77(bar) equals 0.8 False. 0.77(bar) equals exactly 7/9
>then 0.88(bar) equals 0.9 False. 0.88(bar) equals exactly 8/9
And amazingly enough! 0.99(bar) equals exactly 9/9, or 1.
QED
If you believe that 0.99(bar) is not equal to 1.0, then it must either be
larger or smaller.
As many have said here before, if you believe that 0.99(bar) is smaller
than 1.0, then please write a number that is between 0.99(bar) and 1.0.
If you believe that 0.99(bar) is larger than 1.0, you should read
alt.science.fiction instead of this NewsGroup.
earle
=====
__
__/\_\
/\_\/_/
\/_/\_\ earle
\/_/ jones
Dave Weisbeck
>to get back to where I started from. As I looked elsewhere to other
>numbers it became apparent why. The problem with the approach taken
>here is that we cannot adapt the rules of mathematics to suit our theory.
>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.
>then 0.88(bar) equals 0.9
>then 0.77(bar) equals 0.8
>then 0.66(bar) equals 0.7
>then 0.55(bar) equals 0.6
>then 0.44(bar) equals 0.5
>then 0.33(bar) equals 0.4
>then 0.22(bar) equals 0.3
>then 0.11(bar) equals 0.2
>then 0.00(bar) equals 0.1
>and these too...
>then 1.11(bar) equals 1.2
>then 2.00(bar) equals 2.1
>there are other absurdities that can be stated with these...
>
>
What? How did you make these assumptions?
0.9 bar - 0.1 bar = 0.8 bar
1 - 0.1 bar = 0.8 bar
How do you get that 0.8 bar equals 0.9?
Further 0.8 bar = 0.8 * Sigma[n=0..infinity] (1/10)^n = 0.8/0.9 which
isn't 0.9 in the universe in which I do math. =)
[munched the rest since I assume it is based on the above...]
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
Dave Weisbeck
>to get back to where I started from. As I looked elsewhere to other
>numbers it became apparent why. The problem with the approach taken
>here is that we cannot adapt the rules of mathematics to suit our theory.
>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.
>then 0.88(bar) equals 0.9
>then 0.77(bar) equals 0.8
>then 0.66(bar) equals 0.7
>then 0.55(bar) equals 0.6
>then 0.44(bar) equals 0.5
>then 0.33(bar) equals 0.4
>then 0.22(bar) equals 0.3
>then 0.11(bar) equals 0.2
>then 0.00(bar) equals 0.1
>and these too...
>then 1.11(bar) equals 1.2
>then 2.00(bar) equals 2.1
>there are other absurdities that can be stated with these...
>
>
What? How did you make these assumptions?
J. Pimentel
In article <ejones-ya0235800...@206.15.64.37> ejo...@hooked.net (Earle D. Jones) writes:
> And amazingly enough! 0.99(bar) equals exactly 9/9, or 1.
Alright, since you think (why I can't image) that 0.99(bar) equals 1.00 then
prove it. BTW: Prove means that 1) You show the solution and 2) you
then work back to the problem.
>If you believe that 0.99(bar) is not equal to 1.0, then it must either be
>larger or smaller.
It is smaller, that's easy, this I can prove. If it can't be comprehended,
then that is the real problem.
If 0.99(bar) equals 1.0 then by the same exact absurd logic 0.88(bar)
equals 0.90. I have proven that 0.88(bar) does not equal 0.90 and
therefore 0.99(bar) does not equal 1.0, by the same method.
Look at this way the difference between 1.00 and 0.90 is 0.10; and the
difference between 0.90 and 0.80 is 0.10. The difference between
0.99(bar) and 0.88(bar) is 0.11(bar). These sets of numbers are spaced
evenly apart, but your position is that 0.99(bar), unlike it's counterpart
0.88(bar) somehow is exempt from the laws of mathematics and has the
extra value required to jump to the higher point.
The difference between 1.00 and 0.99 by your logic and reasoning come to
a convergence point, and dismiss the notion that latter does not simply
approach the former, but that they become the same.
Then why does not this same logic apply to 0.90 and 0.88(bar)?
>As many have said here before, if you believe that 0.99(bar) is smaller
>than 1.0, then please write a number that is between 0.99(bar) and 1.0.
Here's the counter challenge find a number between 0.90 and 0.88(bar)
and write it here.
When you find a number there, then I will find a number between 1.00 and
0.99(bar).
Why is 0.88(bar) NOT equal to 0.90 _and_ what is the point that separates
this rational number from 0.90?
Until then you are fudging the laws of mathematics to suit your own
needs without understanding those laws.
>earle
>=====
As I have stated before there is a limit at one and when working with
only the decimal portion of the non-terminating, repeating number (and
it works with terminating, non-repeating numbers - see below):
a. 0 <= |X| < 1
Which means that the absolute value of X is equal to or greater than
zero and is less than one. -X works fine to however the equation sits
as:
b. -1 > (-X) =< 0
Put another way the equation sits as [0,1) or (-1,0] drawn on the number
line:
a. <------|------|------[------)------|----->
-2 -1 0 1 2
b. <------|------(------]------|------|----->
-2 -1 0 1 2
Here is a example of the conversation when applied to terminating,
none repeating number.
-=-=-=-=-=-=-=-=-=-=-=-=-=
X = 0.50
10X = 5.00
- X = - 0.50
-----------------------
9X = 4.5
90X = 45
X = 45/90 = 1/2 = 0.50
[Note: this is what constitutues a proof.]
-=-=-=-=-=-=-=-=-=-=-=-=-=
---
John Pimentel
pime...@ultranet.com
Albert Y.C. Lai
In article <1997020514...@lucius.ultra.net>,
"John Pimentel" <pime...@roots.ultranet.com> wrote:
>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.
Same to you. As I begin to see later, on the entire newsgroup minus
the spammers, you are the only one who hasn't followed that holy grail.
>If 0.99(bar) equals 1.0 then these others must also be true.
>then 0.88(bar) equals 0.9
Even implications must be proved. I am not going to talk about the
latter sentence "0.8bar = 0.9". I ask you, how do you prove the entire
statement "if 0.9bar = 1 then 0.8bar = 0.9"?
>The formula is for proving that a non-terminating, repeating number is
>in fact a rational number.
Ok, great. 0.9bar is a non-terminating, repeating number. If you
agree that it is a rational number (by virtue of being non-terminating
and repeating), then write it as a fraction x/y where x and y are
integers.
What amused me is that you applied a procedure to deduce that "if x =
1.11bar then x = 1.11bar" and you are happy with it; then you use the
procedure to deduce "if x = 1.99bar then x = 2.0bar" and now you are
not happy. Why do you accept what you need, and refuse what you
dislike? Didn't you say, don't stop at what you need? Either you must
never use the procedure (meaning, find another way to prove the
rationality of 1.11bar), or you must accept ALL consequences of the
procedure (meaning, accept that 1.9bar = 2). You can't have only one
side of the coin; you accept both sides or you throw away both.
To others: nevermind about the statement "0.8bar = 0.9", we know that
it is false, but John is attempting a proof by contradiction. If you
understand proofs by contradictions, you know that it is a Good Thing
to obtain a falsehood at the end. However, the intermediate steps
inside the proof must still be correct, that's what I am after.
If you argue that 0.8bar <> 0.9 then you are just strengthening his
argument. :)
--
Albert Y.C. Lai tre...@vex.net http://www.vex.net/~trebla/
Earle D. Jones
In article <pimentel.3...@ultranet.com>, pime...@ultranet.com (J.
Pimentel) wrote:
>In article <ejones-ya0235800...@206.15.64.37>
ejo...@hooked.net (Earle D. Jones) writes:
>> And amazingly enough! 0.99(bar) equals exactly 9/9, or 1.
>
>Alright, since you think (why I can't image) that 0.99(bar) equals 1.00 then
>prove it. BTW: Prove means that 1) You show the solution and 2) you
>then work back to the problem.
>
>>If you believe that 0.99(bar) is not equal to 1.0, then it must either be
>>larger or smaller.
>
>It is smaller, that's easy, this I can prove. If it can't be comprehended,
>then that is the real problem.
>
======clipsomestuff here=======
>
>>As many have said here before, if you believe that 0.99(bar) is smaller
>>than 1.0, then please write a number that is between 0.99(bar) and 1.0.
>
>Here's the counter challenge find a number between 0.90 and 0.88(bar)
>and write it here.
==========
******** Here is my answer:
I will write a number here that is between 0.90 and 0.88(bar)
************* 0.89 ***************
This number, 0.89 is *larger* than 0.88(bar) and *smaller* than 0.9.
You should have no problem understanding that.
Would you like three more such numbers?
************* 0.895 0.889 0.899 ************
Each of these is *larger* than 0.88(bar) and *smaller* than 0.9.
I wonder if you don't have some misunderstanding of the meaning of the term
(bar).
0.88(bar) means the same as 0.88888..........
which means an infinite number of eights after a decimal point.
We use (bar) and a row of dots .......... as a shorthand notation
meaning that there is an infinite number of eights (in this case).
********* Now it's your turn: ***********
>When you find a number there, then I will find a number between 1.00 and
>0.99(bar).
************ (Your answer goes here) **************
>
>Why is 0.88(bar) NOT equal to 0.90 _and_ what is the point that separates
>this rational number from 0.90?
********* I hope that the above answers this question. It is difficult to
answer such questions as why is 0.88(bar) NOT equal to 0.90. 0.88(bar) is
equal to something else. It is *exactly* equal to 8/9, which is definitely
less that 0.9 which is equal to 9/10.
Why isn't 2 + 2 = 22? Because it's equal to something else, namely 4.
Raymond E. Griffith
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 approach taken
. here is that we cannot adapt the rules of mathematics to suit our
theory.
. 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.
. then 0.88(bar) equals 0.9
Why? Your assumption is incorrect. There is no need to assume this.
We need to be careful with unwarranted assumptions. You will have to
specify *WHY* "IF 0.99(bar) = 1.0 THEN 0.88(bar) = 0.9."
0.88(bar) = 8/9 but 0.89999 (9 bar) = 0.9.
Here's how: (you show it later, the same way).
Let x = 0.88(bar)
then 10x = 8.88(bar)
subtract x from 10x to get
9x = 8
then x = 8/9
Now use exactly the same algebraic technique...
Let x = 0.8999 (9 bar)
then 10x = 8.999 (9 bar)
so 9x = 8.1
and x = 0.9
The fact is that the real number system allows for dual representation
of numbers whenever we get a 9(bar) infinite repetition sequence. It is
simple algebra, following all the rules.
. then 0.77(bar) equals 0.8
No, 0.77(bar) = 7/9 , etc.
(Stuff deleted)
.
. 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 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.
Exactly! You have everything you need, right here!
.
. For the next series of equations the number of digits the decimal
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.
Just as I said.
. Let X = 1.99(bar)
. Let Y = X - 1 = 0.99(bar)
.
. 10Y = 9.99(bar)
. - Y = - 0.99(bar)
. ---------------
. 9Y = 9
.
. Y = 9/9 = 1.00(bar)
.
. add the one to Y which was taken out
.
. Y + 1 = 2.00(bar) != X
.
. (!= means not equal for those that don't know)
.
. This equation is NOT proven.
Why not? You have just shown by "proofs" done exactly the same way how
other decimal equivalences to fractions are calculated.
Why are you afraid to face the validity of your own proof?
Remember, if a = b and b = c then a = c (the transitive law of
equality).
You have admirably proven what you are trying to disprove!
-=-=-=
. 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
Wrong. You showed it yourself.
The decimal system does not claim unique representation all of the time.
But we deal with the problems of non-unique representation all the time
in algebra. That is why we "simplify." Why have a problem with it here?
I hope you will rethink your position and be guided by the algebraic
proofs you have so competently performed. You are almost to the
understanding you need.
Regards,
Raymond E. Griffith
Lee Jaap
|>In article <ejones-ya0235800...@206.15.64.37> ejo...@hooked.net (Earle D. Jones) writes:
|>> And amazingly enough! 0.99(bar) equals exactly 9/9, or 1.
|>
|>Alright, since you think (why I can't image) that 0.99(bar) equals 1.00 then
|>prove it. BTW: Prove means that 1) You show the solution and 2) you
|>then work back to the problem.
Well, it fits the pattern, which you conveniently editted from
Earle's quote of your post:
.1bar = 1/9
.2bar = 2/9
.3bar = 3/9
.4bar = 4/9
.5bar = 5/9
.6bar = 6/9
.7bar = 7/9
.8bar = 8/9
.9bar = 9/9
Or do you not believe that 9/9 = 1? I'm confused by your reply.
|>>As many have said here before, if you believe that 0.99(bar) is smaller
|>>than 1.0, then please write a number that is between 0.99(bar) and 1.0.
|>
|>Here's the counter challenge find a number between 0.90 and 0.88(bar)
|>and write it here.
Duh. It looks to me as if .8bar < .89 < .895 < .9.
I tink I done found two of them thar critters.
|>When you find a number there, then I will find a number between 1.00 and
|>0.99(bar).
Go right ahead.
|>Why is 0.88(bar) NOT equal to 0.90 _and_ what is the point that separates
|>this rational number from 0.90?
I done give ya two of them above. .898 is another.
Your question can be rephrased "Why is 8/9 not equal to 9/10 _and_
what is the point that separates them?" If two real numbers are
not the same, their arithmetic mean is between them. In this case,
(8/9 + 9/10) / 2 = 161/180 is between them.
I think you're the one who "proved" that .8bar = 8/9. Do you agree
that .9 = 9/10?
|>Until then you are fudging the laws of mathematics to suit your own
|>needs without understanding those laws.
It sure seems to me that *you* are the one who's fudging something.
Maybe a definition of a brain that understands math (in this universe).
This is true (aside from the speling erers :-) .
So what?
--
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
|>If you argue that 0.8bar <> 0.9 then you are just strengthening his
|>argument. :)
Sorry to disagree with you, Albert, but only on this point. I like
giving him enough rope to make a noose and stick his head through it
(figuratively speaking, of course).
I found a number (well, a few) between .8bar and .9. Then I
challenged him to find a similar number between .9bar and 1.
This is fun. Until it gets boring.
Dave Weisbeck
>Alright, since you think (why I can't image) that 0.99(bar) equals 1.00 then
>prove it.
I have yet to see why this proof is incorrect John. Maybe you can enlighten
me.
0.9 bar = 0.9 * Sigma[n=0..infinity] (1/10)^n = 0.9 * 1/0.9 = 1
>If 0.99(bar) equals 1.0 then by the same exact absurd logic 0.88(bar)
>equals 0.90.
Show me this logic please.
> I have proven that 0.88(bar) does not equal 0.90 and
>therefore 0.99(bar) does not equal 1.0, by the same method.
No, you have made a blatanly false assumption. You certainly haven't
proven anything.
>
>Look at this way the difference between 1.00 and 0.90 is 0.10; and the
>difference between 0.90 and 0.80 is 0.10. The difference between
>0.99(bar) and 0.88(bar) is 0.11(bar).
And just how do you assume that 0.1 = 0.11(bar)? The last time I checked, in
the universe I live in, 1/10 =/= 1/9. Maybe you are in another universe, in
which case a lot would be explained. Although it is a strange universe for 9
to equal 10.
>These sets of numbers are spaced
>evenly apart, but your position is that 0.99(bar), unlike it's counterpart
>0.88(bar) somehow is exempt from the laws of mathematics and has the
>extra value required to jump to the higher point.
Umm... no John. It has to do with the way the number system is built. I would
suggest a good course in analysis and proof at your local university to have
it fully explained.
>The difference between 1.00 and 0.99 by your logic and reasoning come to
>a convergence point, and dismiss the notion that latter does not simply
>approach the former, but that they become the same.
>Then why does not this same logic apply to 0.90 and 0.88(bar)?
You really need this explained? Ok, I guess it is obvious from what you
have written above that you do. If we assume that .9bar = 1 then subtracting
0.1bar from both sides gives us 0.8bar = 0.8bar. I am still confused how
you got your assumption that 0.88(bar) = 0.9.
>
>Here's the counter challenge find a number between 0.90 and 0.88(bar)
>and write it here.
Ok, how about 0.89. Next challenge please. You see John, 0.8(bar) is
equal to 0.8 * Sigma [n=0..infinity] (1/10)^n = 8/9 and 0.9 is equal to
9/10. So a number in between would be 89/100. If you can't see this then
I'll put them under a common denominator for you.
8/9 = 800/900; 9/10 = 810/900; 89/100 = 801/900. Since 800 < 801 < 810 I have
obviously found a number in between. Actually there are an infinitely many
numbers in between.
>When you find a number there, then I will find a number between 1.00 and
>0.99(bar).
>
Ok, go ahead. Oh, and I would like a common denominator too please. =)
>Why is 0.88(bar) NOT equal to 0.90 _and_ what is the point that separates
>this rational number from 0.90?
>
I think I already answered this one.
>Until then you are fudging the laws of mathematics to suit your own
>needs without understanding those laws.
>
Nope, I understand them perfectly well. No fudging either.
>
Believe it or not John it is widely known and accepted that 0.9 bar is 1. If
you would like I will search for a textbook that shows this proven so you can
read it for yourself? Although any analysis textbook will give you all the
tools to do this yourself.
Just in case you really need a very formal proof I will give you one.
Definition: A sequence, s(n), is said to _converge_ to the real number s
provided that:
for each e > 0 there exists a real number N such that n > N implies
that |s(n) - s| < e.
Ok, now that we have that under our belt, let us move on to our proof.
0.9(bar) = lim ( 1 - (1/10)^n ) = lim s(n) [these limits are n-> infinity]
i.e. s(1) = 0.9, s(2) = 0.99, ...
Now let us prove that this is indeed 1, or s(n) -> 1.
Given e > 0, let N = 1/e. Then for any n > N we have | 1 - 1/10^n - 1| =
= 1/10^n < 1/n < 1/N = e. Thus lim(1 - 1/10^n) = 1.
Note: the only detail I have left out is that 1/10^n < 1/n. This is obviously
true though, but I will gladly prove it by induction if you feel it is
necessary?
There, overkill, but bullet proof.
Have a nice day,
Dave Weisbeck
z2...@ugrad.cs.ubc.ca
Bart Willems
You can't always apply the "normal" algebraic rules when infinity is
involved.
E.g. in the "proof"
> Let x = 0.99(bar)
> then 10x = 9.99(bar)
> subtract x from 10x to get
> 9x = 9
> then x = 1
algebra is "abused":
10x - x = 9.99(bar) - .99(bar) is DEFINITELY NOT EQUAL to 9
!!!!!!!
The terms of inifinite sums may not be swapped, so
9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
- ...-9/(10^inf)
may NOT be rewritten as
9 + 9/10 - 9/10 + 9/100 -9/100 +...
as is explained in each advanced book on algebraic series and sums.
This might seem strange, but if we take the following expression:
Sum i + Sum(-i)
(i=0->inf) (i=0->inf)
= +inf + -inf AND THIS IS DEFINITELY NOT ZERO, which is what you
obtain by interchangeing terms.
Conclusion: one should be EXTREMELY CAREFUL when applying algebra to
situations in which "infinity" is involved.
.99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
It approaches 1 for an infinite number of terms, but there is a
difference between "being 1" and "approaching 1 asymptotically".
People who are not convinced of this should take a course on
differentials.
Bart
>
--
===================================================================
Bart Willems === Easics ===
ASIC Design Engineer === VHDL-based ASIC design services ===
===================================
mailto:ba...@easics.be http://www.easics.com/
(PLEASE DON'T REPLY TO sm...@easics.be !)
Tel: +32-16-298 408
Fax: +32-16-298 319 Kapeldreef 60, B-3001 Leuven, BELGIUM
be...@pacbell.net
Bart Willems wrote:
>
> You can't always apply the "normal" algebraic rules when infinity is
> involved.
>
> situations in which "infinity" is involved.
>
> .99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
> It approaches 1 for an infinite number of terms, but there is a
> difference between "being 1" and "approaching 1 asymptotically".
> People who are not convinced of this should take a course on
> differentials.
Nonsense! For anyone who *DEFINES* .9(bar) to be equal to the limit as n
approaches infinity, of the sum as x goes from 1 to n of 9/(10)^x,
.9(bar) = 1, even though *the sum* only approaches 1 asymptotically. Of
course, for people in this thread who are, implicitly or explicitly
(mostly implicitly, it appears to me, and also mostly incompletely),
defining .9(bar) differently, who knows what it is and isn't equal to?!
Of course, my comments assume that we are not using conflicting
definitions of "limit" or "approaches infinity" or "sum" or "goes from 1
to n", etc. But with the possible exception of the definition of
"limit", those terms aren't where the difficulties in this thread lie.
Alan Beban be...@pacbell.net
Lee Jaap
|>The terms of inifinite sums may not be swapped, so
|>
|> 9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
|>- ...-9/(10^inf)
|>
|>may NOT be rewritten as
|>
|>9 + 9/10 - 9/10 + 9/100 -9/100 +...
|>
|>as is explained in each advanced book on algebraic series and sums.
Read carefully, and I think you'll find that convergent series
can have their terms interchanged.
|>This might seem strange, but if we take the following expression:
|>
|>
|>Sum i + Sum(-i)
|>(i=0->inf) (i=0->inf)
|>
|>= +inf + -inf AND THIS IS DEFINITELY NOT ZERO, which is what you
|> obtain by interchangeing terms.
These are not convergent series. Therefore rearranging terms
cannot be directly justified (if at all).
|>Conclusion: one should be EXTREMELY CAREFUL when applying algebra to
|> situations in which "infinity" is involved.
Indeed. One must have a mathematically precise definition for
"infinity" before math can be done with it.
|>.99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
|>It approaches 1 for an infinite number of terms, but there is a
|>difference between "being 1" and "approaching 1 asymptotically".
|>People who are not convinced of this should take a course on
|>differentials.
If .9bar != 1, basic math says that there's a real number between
them. Tell me what it is. If you can't find one, they're (two
names for) the same number.
Please reread the definition of limit. If the limit exists, it
is a number, not a process.
Again, if you define .9bar as most mathematicians define it, it
is obviously the same value as 1. If you define it some other
way, we're not using the same language. Which is fine. Just
explain how your definition differs from the standard one.
Raymond E. Griffith
Bart Willems wrote:
>>>>>>>>>>>>>>>>>
You can't always apply the "normal" algebraic rules when infinity is
involved.
Let x = 0.99(bar)
then 10x = 9.99(bar)
subtract x from 10x to get
9x = 9
then x = 1
algebra is "abused":
10x - x = 9.99(bar) - .99(bar) is DEFINITELY NOT EQUAL to 9
!!!!!!!
The terms of inifinite sums may not be swapped, so
9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
- ...-9/(10^inf)
may NOT be rewritten as
9 + 9/10 - 9/10 + 9/100 -9/100 +...
as is explained in each advanced book on algebraic series and sums.
<<<<<<<<<<<<<<<<<<<<<<<<<<<< (break in his topic)
Actually, we can. See Principles of Mathematical Analysis, by Dr. Walter
Rudin, Professor of Mathematics, University of Wisconsin--Madison,
Chapter 3.
If sum(an) = A and sum(bn) = b then sum(an + bn) = A + B and
sum(c*an) = cA for any fixed c.
"Thus two convergent series may be added term by term, and the resulting
series converges to the sum of the two series" (pp. 72, 73).
In fact, your whole approach is rather strange. I believe you would
accept that
.999(9bar) - .999(9bar) = 0
and that
9 + .999(9bar) = 9.999(9bar).
Why then, do you object to 9 + .999(9bar) - .999(9bar) = 9?
------------------------------
Bart Willems continues:
>>>>>>>>>>>>>>>>>>>>>>
This might seem strange, but if we take the following expression:
Sum i + Sum(-i)
(i=0->inf) (i=0->inf)
= +inf + -inf AND THIS IS DEFINITELY NOT ZERO, which is what you
obtain by interchangeing terms.
<<<<<<<<<<<<<<<<<<<<<<<
You are wrong for this example. Adding (and subtracting) term by term is
allowed.
Perhaps you are confusing series summation with limits. There are a few
nasty critters out there, as
sum (-1)^i = 1 + (-1) + 1 + (-1) + (1) + ...
(i = 0 to inf)
This sum never converges. Rearrangements which do not converge give
problems. However, series which are absolutely convergent are no trouble
at all.
"If Sum(an) is a series of complex numbers which converges absolutely,
then every rearrangement of Sum(an) converges, and they all converge to
the same sum" (p. 78)
-------------------------
Bart Willems continues:
>>>>>>>>>>>>>>>>>>>>>>
Conclusion: one should be EXTREMELY CAREFUL when applying algebra to
situations in which "infinity" is involved.
<<<<<<<<<<<<<<<<<<<<<<
True. But it is even more important to know all of your theorems in the
right context. Your approach does not follow, since you were not careful
to apply the right context.
Bart Willems continues:
>>>>>>>>>>>>>>>>>>>>>>
.99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
It approaches 1 for an infinite number of terms, but there is a
difference between "being 1" and "approaching 1 asymptotically".
People who are not convinced of this should take a course on
differentials.
Bart
> >
> --
> ===================================================================
> Bart Willems === Easics ===
> ASIC Design Engineer === VHDL-based ASIC design services ===
> ===================================
> mailto:ba...@easics.be http://www.easics.com/
> (PLEASE DON'T REPLY TO sm...@easics.be !)
>
> Tel: +32-16-298 408
> Fax: +32-16-298 319 Kapeldreef 60, B-3001 Leuven, BELGIUM
<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
Again, you are applying (poorly) limiting ideas in the wrong context.
Furthermore, we do not say that the limit(sum(9*10^(-i)) approaches 1,
but that is equals 1.
.999(9bar) is a representation of that limiting value, and it does
indeed equal 1.
Get out your analysis texts again. You will be surprised!
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
Albert Y.C. Lai
In article <pimentel.3...@ultranet.com>,
pime...@ultranet.com (J. Pimentel) wrote:
>If 0.99(bar) equals 1.0 then by the same exact absurd logic 0.88(bar)
>equals 0.90.
Absurd logic. Finally you admitted it. We use sound logic, you use
absurd logic. You are a great creator of oxymoron terms.
Mike Housky
Bart Willems wrote:
>
> You can't always apply the "normal" algebraic rules when infinity is
> involved.
>
> E.g. in the "proof"
>
> > Let x = 0.99(bar)
> > then 10x = 9.99(bar)
> > subtract x from 10x to get
> > 9x = 9
> > then x = 1
>
> algebra is "abused":
>
> 10x - x = 9.99(bar) - .99(bar) is DEFINITELY NOT EQUAL to 9
> !!!!!!!
>
> The terms of inifinite sums may not be swapped, so
>
> 9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
> - ...-9/(10^inf)
>
> may NOT be rewritten as
>
> 9 + 9/10 - 9/10 + 9/100 -9/100 +...
>
> as is explained in each advanced book on algebraic series and sums.
I don't know where to stop quoting for sure, but this seems as good
a place as any. Yes, you may rearrange the terms of an infinite
series without disturbing the limit, PROVIDED that the series is
absolutely convergent. (This is a theorem due to Cauchy.) Absolute
convergence means that the sum of the absolute values of the terms
of the sequence converges to a finite limit.
> This might seem strange, but if we take the following expression:
>
>
> Sum i + Sum(-i)
> (i=0->inf) (i=0->inf)
>
> = +inf + -inf AND THIS IS DEFINITELY NOT ZERO, which is what you
> obtain by interchangeing terms.
Here's a classic example. The sum of the absolute values does not
converge. You can rearrange the terms of this series to get any integer
you desire.
This is not so with .9bar, since the terms are all positive. (0.9, 0.09,
0.009, etc.) If a limit exists, it is unique.
> Conclusion: one should be EXTREMELY CAREFUL when applying algebra to
> situations in which "infinity" is involved.
I agree. You need to corral them horses before you gets to branding 'em.
> .99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
> It approaches 1 for an infinite number of terms, but there is a
> difference between "being 1" and "approaching 1 asymptotically".
Do you believe in limits at all?
> People who are not convinced of this should take a course on
> differentials.
You might profit by a course in fundamental calculus, where the formal and
rigorous underpinning of the subject is developed. Some great mathematicians
(Riemann, Cauchy, Weierstrauss, and others) covered these questions well over
a century ago.
If .9bar has any meaning at all, it must be the limit of the series:
9/10 + 9/100 + 9/1000 + ...
Do you agree that this series has a limit? If it does, then since all terms are
positive it is automatically absolutely convergent and the itemwise manipulations
used to arrive at .9bar=1 are all valid. If you don't agree that this series has
a limit, go back to class and look up geometric series. That's what we have here
and the limit has been well-known for centuries.
Bon jour,
Mike.
Albert Y.C. Lai
In article <JAAPJL.97...@asmobj.larc.nasa.gov>,
jaa...@asmobj.larc.nasa.gov (Lee Jaap) wrote:
>Sorry to disagree with you, Albert, but only on this point. I like
>giving him enough rope to make a noose and stick his head through it
>(figuratively speaking, of course).
>
>I found a number (well, a few) between .8bar and .9. Then I
>challenged him to find a similar number between .9bar and 1.
I see. This is another valid way to refute a flawed implication. He
claimed "S implies T", and to find out that he was wrong, you can
demonstrate that "S and not T". "S and not T" is the opposite of "S
implies T".
(My way is lazier; just ask him to rigourously justify "S implies T".
If he can't then end of story.)
Last time I was trying to say that "not T" is not enough to refute his
"S implies T", and might even strengthen his position if he framed it
in a tricky way.
>This is fun. Until it gets boring.
Um, my concern is to rectify the falsehood and confusion spread by
crackpots. Of course it is kind of undesirable to do the whole
argument again every time a new crackpot appears. Perhaps I should
compile a second FAQ some time. (Joe's FAQ deals with introductions to
this newsgroup, my FAQ-wanna-be will deal with crackpots (or more
precisely, crackpot falsehoods).)
Albert Yang
Lee Jaap (jaa...@asmobj.larc.nasa.gov) wrote:
: In article <32FB43...@easics.be> Bart Willems <ba...@easics.be> writes:
:
: |>The terms of inifinite sums may not be swapped, so
: |>
: |> 9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
: |>- ...-9/(10^inf)
: |>
: |>may NOT be rewritten as
: |>
: |>9 + 9/10 - 9/10 + 9/100 -9/100 +...
: |>
: |>as is explained in each advanced book on algebraic series and sums.
: Read carefully, and I think you'll find that convergent series
: can have their terms interchanged.
Only if it's absolutely convergent; you can't interchange terms on
a conditionally convergent series (e.g. 1 - 1/2 + 1/3 - 1/4 ...)
--
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
Lee Jaap
|>Lee Jaap (jaa...@asmobj.larc.nasa.gov) wrote:
|>: In article <32FB43...@easics.be> Bart Willems <ba...@easics.be> writes:
|>:
|>: |>The terms of inifinite sums may not be swapped, so
|>: |>
|>: |> 9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
|>: |>- ...-9/(10^inf)
|>: |>
|>: |>may NOT be rewritten as
|>: |>
|>: |>9 + 9/10 - 9/10 + 9/100 -9/100 +...
|>: |>
|>: |>as is explained in each advanced book on algebraic series and sums.
|>:
|>: Read carefully, and I think you'll find that convergent series
|>: can have their terms interchanged.
|>
|>Only if it's absolutely convergent; you can't interchange terms on
|>a conditionally convergent series (e.g. 1 - 1/2 + 1/3 - 1/4 ...)
OK. I goofed a little. Thanks, Albert, for the catch. However,
I *think* mentioning the term "convergent" should give enough hint
for the interested reader to find adequate discussion of the topic.
Cheers.
Lee Jaap
In article <32FCA6...@webworldinc.com> Mike Housky <mi...@webworldinc.com> writes:
|>Bart Willems wrote:
|>
|>> .99(bar) certainly is NOT equal to 1, just like 1/x never becomes 0.
|>> It approaches 1 for an infinite number of terms, but there is a
|>> difference between "being 1" and "approaching 1 asymptotically".
|>
|>Do you believe in limits at all?
Perhaps "believe" is not the right word. How about "Do you
understand the definition of 'limit'?"
When an "infinite" number of 9's is there, it is precisely
equal to 1.
1) What is the definition of .9bar?
2) What is the definition of "limit"?
Darrell Ryan
Lee Jaap wrote:
>
> When an "infinite" number of 9's is there, it is precisely
> equal to 1.
>
> 1) What is the definition of .9bar?
>
> 2) What is the definition of "limit"?
>
I would add this:
3) What is the definition of "=" ?
____________________________________________________________
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.
spamhater
Mike Housky wrote:
>
> Bart Willems wrote:
> >
> > You can't always apply the "normal" algebraic rules when infinity is
> > involved.
> >
> > E.g. in the "proof"
> >
> > > Let x = 0.99(bar)
> > > then 10x = 9.99(bar)
> > > subtract x from 10x to get
> > > 9x = 9
> > > then x = 1
> >
> > algebra is "abused":
> >
> > 10x - x = 9.99(bar) - .99(bar) is DEFINITELY NOT EQUAL to 9
> > !!!!!!!
> >
> >
> > 9.99(bar) - 0.99(bar) = 9 +9/10 + 9/100 + ..+9/(10^inf) - 9/10 -9/100
> > - ...-9/(10^inf)
> >
> > may NOT be rewritten as
> >
> > 9 + 9/10 - 9/10 + 9/100 -9/100 +...
> >
> > as is explained in each advanced book on algebraic series and sums.
>
> a place as any. Yes, you may rearrange the terms of an infinite
> series without disturbing the limit, PROVIDED that the series is
> absolutely convergent. (This is a theorem due to Cauchy.) Absolute
> convergence means that the sum of the absolute values of the terms
> of the sequence converges to a finite limit.
>
> > This might seem strange, but if we take the following expression:
> >
> >
> > Sum i + Sum(-i)
> > (i=0->inf) (i=0->inf)
> >
> > = +inf + -inf AND THIS IS DEFINITELY NOT ZERO, which is what you
> > obtain by interchangeing terms.
>
> Here's a classic example. The sum of the absolute values does not
> converge. You can rearrange the terms of this series to get any integer
> you desire.
>
> This is not so with .9bar, since the terms are all positive. (0.9, 0.09,
> 0.009, etc.) If a limit exists, it is unique.
>
> > Conclusion: one should be EXTREMELY CAREFUL when applying algebra to
> > situations in which "infinity" is involved.
>
> I agree. You need to corral them horses before you gets to branding 'em.
>
> > It approaches 1 for an infinite number of terms, but there is a
> > difference between "being 1" and "approaching 1 asymptotically".
>
> Do you believe in limits at all?
>
> > differentials.
>
> You might profit by a course in fundamental calculus, where the formal and
> rigorous underpinning of the subject is developed. Some great mathematicians
> (Riemann, Cauchy, Weierstrauss, and others) covered these questions well over
> a century ago.
>
> If .9bar has any meaning at all, it must be the limit of the series:
>
> 9/10 + 9/100 + 9/1000 + ...
>
> Do you agree that this series has a limit? If it does, then since all terms are
> positive it is automatically absolutely convergent and the itemwise manipulations
> used to arrive at .9bar=1 are all valid. If you don't agree that this series has
> a limit, go back to class and look up geometric series. That's what we have here
> and the limit has been well-known for centuries.
>
> Bon jour,
> Mike.
Yes, Mike, there is a limit.
.9(bar) has a limit, and that limit is 1. That does not prove that
.9(bar) = 1, in fact, it proves that .9(bar) can NEVER equal 1. .9(bar)
represents an infinite amount of nines following the decimal point, as
the chain of nines approaches infinity the value approaches 1, but that
is all it can do, approach, never equal.
x = 0.999... x = 0.999...
10*x = 10*0.999... 10*x = 10*0.999...
10x = 9.999... 10x = 9.999...
10x -x = 9.999... -x 10x -0.999... = 9.999... -0.999...
9x = 9.999... -x 10x -0.999... = 9.000...
9x +x = 9.999... 10x = 9.000...
+0.999...
10x = 9.999... 10x = 9.999...
x = 9.999.../10 x = 9.999... /10
x = 0.999... x = 0.999...
x is an apple, 0.999... is an orange; need I say more.
This is supposed to be an algebra help NG. So let us help people not
confuse them.
Use some common sense.
Tore August Kro
Darrell Ryan wrote:
>
> Lee Jaap wrote:
>
> >
> > When an "infinite" number of 9's is there, it is precisely
> > equal to 1.
> >
> > 1) What is the definition of .9bar?
> >
> > 2) What is the definition of "limit"?
> >
>
> I would add this:
>
> 3) What is the definition of "=" ?
>
> ____________________________________________________________
> Darrell Ryan
1) The definition of .9bar is the limit of the sequence 0.9, 0.99,
0.999, 0.9999, 0.99999, ...
0.9bar = lim(n -> oo) x_n , where x_n = sum(i=1 to n) 9/10^i.
2) The definition of limit:
a) The limit of a function:
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.
b) The limit of a series:
Given a series x_n. The limit of x_n as n approaches infinity is a
number b such that for every real number e>0 there exists a natural
number N such that |x_n - b|<e when n>N.
3) The definition of "=" is trickier. I'll try to find something about
it, but i should be including that "=" is an equivalence relation: That
is that x=x, if x=y then y=x and if x=y and y=z then x=z.
Tore August Kro
tor...@ifi.uio.no
Stan Armstrong
In article <33028C...@cybermax.net>, spamhater <spam...@aol.com>
writes
>
>
>Yes, Mike, there is a limit.
> .9(bar) has a limit, and that limit is 1. That does not prove that
>.9(bar) = 1, in fact, it proves that .9(bar) can NEVER equal 1. .9(bar)
>represents an infinite amount of nines following the decimal point, as
>the chain of nines approaches infinity the value approaches 1, but that
>is all it can do, approach, never equal.
Damn it, I'm being drawn into this ridiculous argument again.
0.9(bar) does not have a limit; there is no "as n => 00" in the
expression. 1-10^(-n) certainly does have a limit as n tends to
infinity, by inspection it is 0.9(bar) and by analysis it is plainly
1.0.
I know it is over 40 years since i formally studied analysis but
developments in maths have not changed the conclusion of Cauchy and co.
--
Stan Armstrong
Mike Housky
spamhater wrote:
> Yes, Mike, there is a limit.
> .9(bar) has a limit, and that limit is 1. That does not prove that
> .9(bar) = 1, in fact, it proves that .9(bar) can NEVER equal 1. .9(bar)
> represents an infinite amount of nines following the decimal point, as
> the chain of nines approaches infinity the value approaches 1, but that
> is all it can do, approach, never equal.
In nearly all (I use nearly because I'm sure that someone can pipe up with a
reference to fringe text that claims something else is either standard or
superior) mathematical formulations of the real number field DEFINE a decimal
fraction as a series. If the fraction is nonterminating, the series is infinite.
All of these text that bother to discuss the point will DEFINE the value of an
infinite series as the limit of the sequence of partial sums of that series.
So, by standard usage, the value of .9bar (the whole thing, not any truncations
or partial sums) IS the limit. You have already agreed that the limit exists
and is 1 exactly. We shouldn't have anything else to talk about, except that
neither is an apple or an orange. However, as you noted:
[tabular nonsequitur snipped.]
> x is an apple, 0.999... is an orange; need I say more.
>
> This is supposed to be an algebra help NG. So let us help people not
> confuse them.
> Use some common sense.
I try to. The common part of "common sense" is the use of standards. Even when
they are not perfect, and few standards are, they do provide a basis for
communication of ideas. I can walk into any public college bookstore (and most
any private college bookstore) and find a text that defines these concepts in
terms compatible with what I have posted. If you insist that the value of .9bar
and the limit of .9bar are different, I doubt you'll find much support in
those same stores.
I'm no expert, but I have seen that there are alternate formulations of
mathematical systems. Most that survive long enough to get fairly well
developed are worth looking at, but most aren't the systems I would want to
teach to a first or perhaps even second-year algebra student. I was one of
the lucky victims of the New Math experiment in the '60s. I don't think that
exposure to different mathematical approaches is necessarily bad, and I don't
think it hurt me, but I heard that it was observed that achievemnent test scores
went down overall in disticts that used SMSG and other New Math texts.
Part of this is probably because instructors weren't trained to teach the new
curriculum. My first "New Math" teacher probably couldn't have integrated e^x
even with a CRC handbook to lean on. Also, there's a problem with language
and unless the recipient understands the language of the message, the message
is lost. It doesn't matter whether the instructor or the student lost it.
It's gone. Too bad. Maybe the next student will fare better.
I hope your course works for you.
Mike.
Lee Jaap
|>In article <33028C...@cybermax.net>, spamhater <spam...@aol.com>
|>writes
|>>
|>>
|>>Yes, Mike, there is a limit.
|>> .9(bar) has a limit, and that limit is 1. That does not prove that
|>>.9(bar) = 1, in fact, it proves that .9(bar) can NEVER equal 1. .9(bar)
|>>represents an infinite amount of nines following the decimal point, as
|>>the chain of nines approaches infinity the value approaches 1, but that
|>>is all it can do, approach, never equal.
|>
|>Damn it, I'm being drawn into this ridiculous argument again.
|>
|>0.9(bar) does not have a limit; there is no "as n => 00" in the
|>expression. 1-10^(-n) certainly does have a limit as n tends to
|>infinity, by inspection it is 0.9(bar) and by analysis it is plainly
|>1.0.
I'd say .9(bar) *is* a limit, by definition. It doesn't *have*
a limit.
A limit, if it exists, is a number (in this case, real).
Do you accept these premises, spamhater?
1) .9(bar) is a real number.
2) 1 is a real number.
3) .9(bar) != 1.
Then you can tell us a number that is between .9(bar) and 1,
such as their aritmetic mean. (For any two distinct real
numbers, there is a real number between them.) If there is
no such number, then the two numbers are equal.
Robert Healey
You people are taking this much too seriously!
Look:
0.3(bar) = 1/3 right?
If we can all just agree that 3 x 1/3 = 1 then it follows that:
3 x 0.3(bar) = 0.9(bar) = 1
This also works with ninths:
1/9 = 0.1(bar)
2/9 = 0.2(bar)
3/9 = 0.3(bar)
4/9 = 0.4(bar)
.
.
.
8/9 = 0.8(bar)
9/9 = 0.9(bar) = 1
er...@selway.umt.edu
Of all of the ways to look at this annoying problem, this has been the
most convincing of them all; at least to me. Thank you for ending this
headache.
er...@selway.umt.edu
(((((((((((((((((((((((((((Seattle SuperSonics))))))))))))))))))))))))))))
BrianScott
Robert Healey <rhe...@ucsd.edu> wrote in part:
>You people are taking this much too seriously!
>Look:
>0.3(bar) = 1/3 right?
As has now been pointed out ad nauseam, the real
difficulty in the original question has nothing to do
with showing that 0.9bar = 1; it's establishing that
0.9bar has a meaningful definition and that one can
calculate with it in much the same was as one
calculates with, say, 0.625. Your argument merely
pushes this difficulty back a step. *If* a meaningful
definition of the non-terminating decimal 0.3bar is
already at hand, then there is of course no difficulty
defining and working with 0.9bar in the way that you
suggest. But although 0.3bar doesn't inspire the same
distrust among the mathematically naive that 0.9bar
does, it is just as difficult (or easy, depending on your
point of view) to define usefully and rigorously. Before
one can talk rigorously about 0.3bar, one has to
define convergence of infinite sequences and prove
some elementary theorems about it. *This* is the
serious mathematics that's involved.
Brian M. Scott
Do Not Use: brian...@aol.com
Always Use: sc...@math.csuohio.edu
Albert Y.C. Lai
In article <330C73...@ucsd.edu>,
Mathematics is supposed to be taken seriously and, "worse", rigorously.
If and when we need a break, we know which newsgroup to go to, we even
know to leave the computer and go out, thank you very much.
BrianScott
jimp...@newreach.net (Jim Petty) wrote:
>If two real numbers are not equal, then at some point in the chain of
>numbers following the decimal point they must differ. And since they
>must differ, there are an infinite number of real numbers between any
>two real numbers, no matter close they may be. However, .99[bar] has
>9's going on to infinity and 1.00 has zeros. Not only can't you put
>an infinite number of real numbers between the two, you can't even put
>one!!! Therefor, .99[bar] and 1.00 must equal.
>Note: Yu probably gathered from my terminology that I'm not a
>mathmetician, so be gentile if shooting me down. Still, I think I'm
>on to something here.
I wouldn't shoot you down at all. What you've written is admittedly
very informal and imprecise, but it seems to me that it essentially
contains the idea that 1 is the limit of the numbers 0.9, 0.99, 0.999,
etc. because you can't squeeze anything between 1.0 and *all* of
those numbers.
James M. Horner
The nonsense about this .9999...=1 is absurd.
Of course it can be proved if you know what it means !!!!
This isn't a statement of equality of two numbere represented by
different numerals!!! And, it can't be proved rigoursly using
algebra!!
The statement is a statement about the convergence of a certain
sequence.
It says that if one forms an infinite sequence {a(n)} where for
each positive integer n, a(n)=sum(from k=1 to n) (9/10^k),
[that is, a(1)=9/10
a(2)=9/10 + 9/100
a(3)=9/10 + 9/100 + 9/1000, etc.]
then the sequence {a(n)} converges to the real number designated by
the numeral 1.
Jim Petty
Here's a thought I just had on The Subject Which Wouldn't Die.
If two real numbers are not equal, then at some point in the chain of
numbers following the decimal point they must differ. And since they
must differ, there are an infinite number of real numbers between any
two real numbers, no matter close they may be. However, .99[bar] has
9's going on to infinity and 1.00 has zeros. Not only can't you put
an infinite number of real numbers between the two, you can't even put
one!!! Therefor, .99[bar] and 1.00 must equal.
Note: Yu probably gathered from my terminology that I'm not a
mathmetician, so be gentile if shooting me down. Still, I think I'm
on to something here.
Jim Petty
Iain Davidson
In article <19970222202...@ladder02.news.aol.com>, BrianScott (brian...@aol.com) writes:
>jimp...@newreach.net (Jim Petty) wrote:
>
>>If two real numbers are not equal, then at some point in the chain of
>>numbers following the decimal point they must differ. And since they
>>must differ, there are an infinite number of real numbers between any
>>two real numbers, no matter close they may be. However, .99[bar] has
>>9's going on to infinity and 1.00 has zeros. Not only can't you put
>>an infinite number of real numbers between the two, you can't even put
>>one!!! Therefor, .99[bar] and 1.00 must equal.
>
>>Note: Yu probably gathered from my terminology that I'm not a
>>mathmetician, so be gentile if shooting me down. Still, I think I'm
>>on to something here.
>
>very informal and imprecise, but it seems to me that it essentially
>contains the idea that 1 is the limit of the numbers 0.9, 0.99, 0.999,
>etc. because you can't squeeze anything between 1.0 and *all* of
>those numbers.
Yes, what you seem to be saying is that 1 is the smallest number
greater than any member of the sequence 0.9, 0.99, 0.999, ..., the
general term of which could be written 1 - 1/10^n. This shows that
no matter how many terms you take, you will never reach 1, in other
words no term becomes 1, yet any number less than 1 can be reached
and exceeded after a finite (no matter how large) number of terms.
So 1 is the smallest number greater than any term in 0.9, 0.99,
0.999, ...
The misleading thing about the 0.99.. notation is that it seems to
suggest that it is somewhere in the sequence 0.9, 0.99, 0.999, ...
extended endlessly - the "final term" - but there is no end to the
sequence, so 0.99.. is shorthand for something external to the
sequence, the smallest number greater than any term in the sequence
or its limit 1.
So 0.99.. is defined to be the limit of the sequence 0.9, 0.99,
0.999, .. which is equal to 1 and so 0.99.. must equal 1
Iain Davidson Tel : +44 1228 49944
4 Carliol Close Fax : +44 1228 810183
Carlisle Email : ia...@stt.win-uk.net
England
CA1 2QP