From: The Ghost In The Machine on
In sci.logic, RLG
<Junk(a)Goldolfo.com>
wrote
on Sat, 24 Mar 2007 18:59:26 -0800
<Gfednb222eDiSpjbnZ2dnUVZ_ragnZ2d(a)comcast.com>:
>
> "The Ghost In The Machine" <ewill(a)sirius.tg00suus7038.net> wrote in message
> news:v8tgd4-1qv.ln1(a)sirius.tg00suus7038.net...
>> In sci.physics, RLG
>> <Junk(a)Goldolfo.com>
>> wrote
>> on Sat, 24 Mar 2007 16:40:04 -0800
>> <nM6dnRdlcfJKK5jbnZ2dnUVZ_rWnnZ2d(a)comcast.com>:
>>>
>>> "Tony Orlow" <tony(a)lightlink.com> wrote in message
>>> news:460520e5(a)news2.lightlink.com...
>>>>
>>>> 1=0.999...?
>>>
>>>
>>> 1 = 0.999...
>>>
>>> Multiply both sides of this equation by 10 to get:
>>>
>>> 10 = 9.999...
>>>
>>> Substract the original equation from this one to get:
>>>
>>> 9 = 9.
>>
>> A little more complicated than that, as one could either use that logic,
>> resulting in 9.0000..., or assume an infinite borrow, resulting in
>> 8.9999... . The proper method of handling this would be using limits,
>> which you implicitly do below:
>
> Just curious, how would an infinite borrow be done? The method I used had
> 10-1=9 and:
>
> 9.99999...
> 0.99999...
> -----------
> 9.00000...
>
>
> R
>

Well, that's just it...there's no last digit. However,
were there a last digit one might run into either

.....999999999
.....999999999

which has no borrow, or

.....999999990
.....999999999

which will need one. After all, we multiplied by 10...

Of course that's why limits need to be used anyway; my
strawman logic verges on the ridiculous. :-)

But a few paradoxes get messy without them:

-1 = 1 + 2 + 4 + 8 + ...
since if x = 1 + 2 + 4 + 8 + ... then 2x = x-1
? = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ...
could be 0, 1, 1/2.
log 2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ....
= (1 + 1/3 + 1/5 + ...) - (1/2 + 1/4 + 1/6 + ...)
= infinity - infinity
1 = 1/2 + 1/4 + 1/8 + 1/16 + ... (Zeno's Paradox)

and of course

1.00000... = 0.99999...

Makes life interesting. :-)

--
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From: Bob Kolker on
The Ghost In The Machine wrote:
> and of course
>
> 1.00000... = 0.99999...
>
> Makes life interesting. :-)

You can see what a service that Bishop Berkeley did for mathematics by
savaging the illogic of Newton and Leibniz and ultimately forcing
mathematicians to define limits and convergence properly. That is the
only way of beating Zeno.

Bob Kolker

>
From: RLG on

"The Ghost In The Machine" <ewill(a)sirius.tg00suus7038.net> wrote in message
news:slchd4-8v1.ln1(a)sirius.tg00suus7038.net...
>
>
> Well, that's just it...there's no last digit. However,
> were there a last digit one might run into either
>
> ....999999999
> ....999999999
>
> which has no borrow, or
>
> ....999999990
> ....999999999
>
> which will need one. After all, we multiplied by 10...
>
> Of course that's why limits need to be used anyway; my
> strawman logic verges on the ridiculous. :-)

Yes, on the standard real number line there are no infinitessimals.
Non-standard number lines have infinitessimals, like the surreal number
line, but they are not given decimal representations.

> But a few paradoxes get messy without them:
>
> -1 = 1 + 2 + 4 + 8 + ...
> since if x = 1 + 2 + 4 + 8 + ... then 2x = x-1
> ? = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ...
> could be 0, 1, 1/2.

Yes, divergent series can be arranged to converge to any number one likes.


R



From: RLG on

"The Ghost In The Machine" <ewill(a)sirius.tg00suus7038.net> wrote in message
news:slchd4-8v1.ln1(a)sirius.tg00suus7038.net...
>
>
> Well, that's just it...there's no last digit. However,
> were there a last digit one might run into either
>
> ....999999999
> ....999999999
>
> which has no borrow, or
>
> ....999999990
> ....999999999
>
> which will need one. After all, we multiplied by 10...
>
> Of course that's why limits need to be used anyway; my
> strawman logic verges on the ridiculous. :-)

Yes, on the standard real number line there are no infinitessimals.
Non-standard number lines have infinitessimals, like the surreal number
line, but they are not given decimal representations.

> But a few paradoxes get messy without them:
>
> -1 = 1 + 2 + 4 + 8 + ...
> since if x = 1 + 2 + 4 + 8 + ... then 2x = x-1
> ? = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ...
> could be 0, 1, 1/2.

Yes, divergent series can be arranged to converge to any number one likes.


R


From: Don Stockbauer on
The Definition of Points


"Widdle bitty itsy bitsy guys what ain't got no derminshuns in n e
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(Endless Discussions) cornserning what the heck they is phil-o-soffet-
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"An dats de truff."

- Edith Ann