FINAL PROOF: 0.9(bar) = 1!!!!!!!!!!!!!!!!!!
Jonas Hartwig
I've seen that some of you have difficulties in understanding
the fact that 0.9(bar) infact is the same number as 1. Here
comes a proof:
we let x = 0.9(bar)
then 10x = 9.9(bar) since the 9s goes on for ever
10x-9 = 0.9(bar)
10x-9 = x (see above)
9x = 9
x = 9/9
x = 1
0.9(bar) = 1
.. and there you have it!
/Jonas Hartwig
yl...@dataphone.se
Daniel A. Markham
yl...@dataphone.se (Jonas Hartwig) writes:
>I've seen that some of you have difficulties in understanding
>the fact that 0.9(bar) infact is the same number as 1. Here
>comes a proof:
>we let x = 0.9(bar)
>then 10x = 9.9(bar) since the 9s goes on for ever
Nope.
10x < 9.9(bar)
0.9(bar) ^ infinity = 0.0(bar)
1 ^ infinity = 1
therefore 0.9(bar) is not equal to 1.
--
* Dan Markham * Peanut butter on a *
* bla...@wwa.com * sand tire gets frozen *
* Game Programmer * on the rusty star of *
* Konami Computer Entertainment * a gravy pop tart. *
Stan Armstrong
In article <5ebir4$t...@shoga.wwa.com>, "Daniel A. Markham"
<bla...@sashimi.wwa.com> writes
>
>Nope.
Yup!
>10x < 9.9(bar)
>
>0.9(bar) ^ infinity = 0.0(bar)
are seeking to prove. You have proved nothing by this circular
argument..
>
>1 ^ infinity = 1
>therefore 0.9(bar) is not equal to 1.
>
>
>
--
Stan Armstrong
Lee Jaap
In article <5ebir4$t...@shoga.wwa.com> bla...@sashimi.wwa.com (Daniel A. Markham) writes:
|>yl...@dataphone.se (Jonas Hartwig) writes:
|>
|>>I've seen that some of you have difficulties in understanding
|>>the fact that 0.9(bar) infact is the same number as 1. Here
|>>comes a proof:
|>
|>>we let x = 0.9(bar)
|>>then 10x = 9.9(bar) since the 9s goes on for ever
|>
|>Nope.
|>10x < 9.9(bar)
Prove it.
|>0.9(bar) ^ infinity = 0.0(bar)
Prove it.
|>1 ^ infinity = 1
One out of three correct so far.
|>therefore 0.9(bar) is not equal to 1.
You can't prove something if your assumptions are wrong.
--
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
Brad Ballinger
Here's a fresh approach to the problem that I haven't seen yet.
What if you were to define .(9bar) recursively, thereby avoiding the entire
issue of infinity, infinite sums, etc.?
This is what I mean: Let's define a new symbol, a, which will function just
like a digit but isn't an already defined thing. For instance,
.a means a/10,
.0a means a/100,
.9a means .9 + a/100
...and so on.
In particular, we'd like it to have the property that .a = .9a, because it
would then follow that .a = .9a = .99a = .999a = ...
So let's suppose such a thing exists. Then we'd have:
.a = .9a Multiply by 100;
10 * a = 90 + a
9 * a = 90
a = 10 (1)
It can be verified that if a = 10, then in fact .a = .9a = .99a = ...
and so such a thing is well-defined.
Let b = 9.(9bar). Observe that .b = .(9bar) = .9(9bar) = .9b, and so by
the argument above, b = 10.
Equivalently, .(9bar) = 1 .
This proof does not use infinity, so there shouldn't be any objections to it
on that ground. I suppose I could have spent more time giving a really really
solid exposition, in which all the notation was carefully laid out, but that
would reduce to the same core argument. Modulo that, I believe this is a
proof, independent of any underlying notions of limiting values or whatever.
Yours,
Brad Ballinger
Out-To-Luncher Extraordinaire
Robin E. Brown
Daniel A. Markham wrote:
>
> yl...@dataphone.se (Jonas Hartwig) writes:
>
> >I've seen that some of you have difficulties in understanding
> >the fact that 0.9(bar) infact is the same number as 1. Here
> >comes a proof:
>
> >we let x = 0.9(bar)
> >then 10x = 9.9(bar) since the 9s goes on for ever
>
> Nope.
> 10x < 9.9(bar)
>
>
>
> * Dan Markham * Peanut butter on a *
> * bla...@wwa.com * sand tire gets frozen *
> * Game Programmer * on the rusty star of *
> * Konami Computer Entertainment * a gravy pop tart. *
r^infinity = 0 if and only if abs(r) < 1. Thus stating .9(bar)^infinity
= 0 assumes .9(bar) does not equal 1. You can't prove a fact by assuming
its true!
Robin
Ben J. Jacobs
Jonas Hartwig wrote:
>
> I've seen that some of you have difficulties in understanding
> the fact that 0.9(bar) infact is the same number as 1. Here
> comes a proof:
>
> we let x = 0.9(bar)
> then 10x = 9.9(bar) since the 9s goes on for ever
> 10x-9 = x (see above)
> 9x = 9
> x = 9/9
> x = 1
>
> 0.9(bar) = 1
>
> .. and there you have it!
>
> /Jonas Hartwig
> yl...@dataphone.se
And oh my god .... we are back to square one :)
arpit...@gmail.com
Hope if you could help me by giving some of the details you know....
Barb Knox
> Exactly...... I am on the same mission!!!
>
> Hope if you could help me by giving some of the details you know....
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | ,008015L080180,022036,029037
| B B a a r b b | ,047045,L014114L4.
| BBB aa a r bbb |
-----------------------------
b.a.ch...@gmail.com
1=(9/9)=[1-0.1(bar)]= 0.9999999999999999...(0.9(bar))
2=(18/9)=2(1-0.1(bar))=1.8888888888888888...(1.8(bar))
3=(27/9)=3(1-0.1(bar))=2.7777777777777777...(2.7(bar))
4=(36/9)=4(1-0.1(bar))=3.6666666666666666...(3.6(bar))
5=(45/9)=5(1-0.1(bar))=4.5555555555555555...(4.5(bar))
6=(54/9)=6(1-0.1(bar))=5.4444444444444444...(5.4(bar))
7=(63/9)=7(1-0.1(bar))=6.3333333333333333...(6.3(bar))
8=(72/9)=8(1-0.1(bar))=7.2222222222222222...(7.2(bar))
9=(81/9)=9(1-0.1(bar))=9-0.9(bar)=9-1=8.0000000000(8.0(bar))
10=(90/9)=10(1-0.1(bar))=(10-1.1(bar))=8.88888888888888(8.8(bar))
99=(891/9)=99(1-0.1(bar))=99-11.1(bar)=87.8888888888888888(87.8(bar))
If this is true, this screws up math.
USE THIS POWER WISELY
choube...@gmail.com
> Sooo,
> 1=(9/9)=[1-0.1(bar)] = 0.9999999999999999...(0.9(bar))
> 2=(18/9)=2(1-0.1(bar))=1.8888888888888888...(1.8(bar))
and 2(1-0.1(bar))= 1.7777777777777777...(1.7(bar))
If you meant 2*0.9(bar)
then 2*0.9(bar)= 1.9999999999999999...(1.9(bar)) and not 1.8(bar)
2*0.9 = 1.8,
2*0.99 = 1.98,
2*0.999 = 1.998,
2*0.9999 = 1.9998,
2*0.99999 = 1.99998 and so on.
> 99=(891/9)=99(1-0.1(bar))=99-11.1(bar)=87.8888888888888888(87.8(bar))
>
99*0.1(bar) = 10.9(bar) and not 11.1(bar)
similarly 99*0.9(bar) = 98.9999999999999999... = 98.9(bar)
> If this is true, this screws up math.
> USE THIS POWER WISELY