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1=/ .9999999..... proof
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Posted by: damsell
The Australian philosopher colin leslie proves .99999[BAR]=/1
mathematicians say .999999 [BAR] = 1
http://mathforum.org/dr.math/faq/faq.0.9999.html
Dean shows you it does not
let
x=.33333[BAR]
3x = .99999999 [BAR]
but you say
.99999999 [BAR] = 1
then
3x = 1
now 3x-x = 1- .333333[BAR]
2x = .6666666........7
(ie 0-3=7
so when we subtract the last .3 at infinity from the last 0 at infinity we get .6666666........7)
therefore
x= (.6666666........7)/2
therefore
x= .333333.....35
3x=1.000000...05
therefore we have now
.999999[bar] = 3x= 1.0.......5
we have one proof proving .99999[BAR] = 1
and another proving .99999[BAR] =/1
that demonstrating colin leslie deans claim that mathematics ends in meaninglessness ie self contradiction
Posted by: Jason Cawley
"the last .3 at infinity"
Meaningless...
Posted by: damsell
"the last .3 at infinity"
Meaningless...
no more meaningless as at infinity there are some infinities with different sizes
if there is no last .3 at infinity
then 1.0000[R] - .33333[R] must be indeterminate as with no final number you cant have any subtraction
normaly you would say
1.0000[R] -.33333[R] = .6666666[R]
BUT WITH NO FINAL 0 OR .3 AS YOU SAY
you cant perform the subtraction
Posted by: Jason Cawley
I think you need a basic review of the definition of an equivalence class...
Posted by: damsell
you say
I think you need a basic review of the definition of an equivalence class.
for support of deans claim
from sci logic
http://www.talkaboutscience.com/gro...ges/199419.html
Actually its legal to talk about a 7 at the end
of the infinity line. Namely because we are
dealing with ordinals and not with cardinals.
A sequence has more to do with an ordinal
than wie a cardinal, as it orders its elements.
Omega denotes the standard ordinal, which
we normally connect with infinite decimal
numbers. What elsiemelsi does, is simple he
puts forward omega+1.
But there are much much more orders and ordinals,
we could even have omega+omega, or omega+mirror(omega),
omega+omega+1, omega^2:
12... 12... = omega + omega
12... ...12 = omega + mirror(omega)
12... 12... 1 = omega + omega + 1
12...
2....
..... = omega^2
.....
.....
Nothing special. For example Z, the positive and
negative numbers, are easily seen as mirror(omega)+
omega, namely:
... -2 -1 0 1 2 ...
But mirror(omega)+omega is not healthy for numbers,
because the get big!
... 3333.3333 ....
Quite a big number, isn't it?
http://www.talkaboutscience.com/gro...ges/199487.html
> Actually its legal to talk about a 7 at the end
> of the infinity line. Namely because we are
> dealing with ordinals and not with cardinals.
OK. Now how does one handle
..333.... x .333... ?
In this New Arithmetic(tm) we should get something like
..111....11088....889
<i> <w+i> <2w>
where 'w' is omega.
Since 1/3 = .333.... the product 1/3 x 1/3 is clearly
not represntable by a fraction (1/9 = .111..... does not
extend past <w>), and therefore the rational numbers is
not closed under this new system.
This is a minor problem, to be sure...but it *is* a problem.
I'm assuming you have a fix.
Posted by: Enexseenge
I had a similar idea. Or perhaps what i am currently discerning by skimming over your comments simply seems familiar, irrationally..
As for Jason, i wish i was rich enough to hire someone to make comments like that (as well as explanations) full time.
The idea that i had pertained to irrational numbers and the idea of not actually being able to do operations like subtraction and addition with them. Something about how the digit sequence is not complete.
It feels like we must assume a level of completeness within an irrational number before we can operate on it, as if we have to make a choice to "stop" enumerating it's digit sequence and use it as a discrete element in an operation.
Hummm? what ever.
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