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Science Forum Index » Logic Forum » Squares of 0.999... tend away from 1
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| Garry Denke |
 Posted: Thu Jan 08, 2004 1:34 pm |
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Squares of 0.999... tend away from 1
.9^2 = .81
.9^4 = .6561
.99^2 = .9801
.99^4 = .96059601
.999^2 = .998001
.999^4 = .996005996001
.9999^2 = .99980001
.9999^4 = .9996000599960001
.99999^2 = .9999800001
.99999^4 = .99996000059999600001
.999999^2 = .999998000001
.999999^4 = .999996000005999996000001
.9999999^2 = .99999980000001
.9999999^4 = .9999996000000599999960000001
.99999999^2 = .9999999800000001
.99999999^4 = .99999998000000059999999600000001
.999999999^2 = .999999998000000001
.999999999^4 = .999999996000000005999999996000000001
.9999999999^2 = .99999999980000000001
.9999999999^4 = .9999999996000000000599999999960000000001
.99999999999^2 = .9999999999800000000001
.99999999999^4 = .99999999996000000000059999999999600000000001
.999999999999^2 = .999999999998000000000001
..999999999999^4 = .999999999996000000000005999999999996000000000001
[snip rest for brevity]
Challenge: Prove squares of 0.999... tend toward 1
Thank you
Garry Denke, Geologist
Denoco Inc. of Texas |
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| karl malbrain |
| Posted: Thu Jan 08, 2004 2:27 pm |
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*** post for FREE via your newsreader at post.newsfeed.com ***
"Garry Denke" <garrydenke@starmail.com> wrote in message
news:210a83f4.0401081034.24b607ce@posting.google.com...
Quote: Squares of 0.999... tend away from 1
.9^2 = .81
.9^4 = .6561
(...)
I think you mean that each Square from your list is adding digits away from
its trailing one, which is illustrated by your example. BTW, you still
haven't said what 0.999... means. karl m
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| Ignacio Larrosa Cañestro |
| Posted: Thu Jan 08, 2004 3:04 pm |
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En el mensaje:210a83f4.0401081034.24b607ce@posting.google.com,
Garry Denke <garrydenke@starmail.com> escribió:
Quote: Squares of 0.999... tend away from 1
[snip rest for brevity]
Quote:
Challenge: Prove squares of 0.999... tend toward 1
You meant prove that (1 - 10^k)^2 --> 1, as k ---> inf ?
--
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosaQUITARMAYUSCULAS@mundo-r.com |
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| The World Wide Wade |
| Posted: Thu Jan 08, 2004 3:32 pm |
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In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Quote: Squares of 0.999... tend away from 1
..99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
Assuming your post is not a troll, start with the fact that the sequence
..9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2
= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2,
.... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the
latter sequence tends to 0. Therefore the former sequence tends to 0, which
is the same as saying .9^2, .99^2, .999^2, ... tends to 1. |
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| karl malbrain |
| Posted: Thu Jan 08, 2004 3:59 pm |
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"The World Wide Wade" <waderameyxiii@comcast.remove13.net> wrote in message
news:waderameyxiii-4AC08C.12324508012004@news.supernews.com...
Quote: In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Squares of 0.999... tend away from 1
.99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
Oh, he started that already in another thread. He means squares from a list
drawn from a breakdown of 0.999... .
Quote: You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
In the past, he has meant "the Last One", but not as a limit value of the
sequence.
Hope this helps, karl m
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| Leonard Blackburn |
| Posted: Thu Jan 08, 2004 7:18 pm |
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garrydenke@starmail.com (Garry Denke) wrote in message news:<210a83f4.0401081034.24b607ce@posting.google.com>...
Quote: Squares of 0.999... tend away from 1
There is only one square of 0.999...; perhaps you can tell me why you
used the plural?
But your numerical evidence below supports that the squares of
0.9, 0.99, 0.999, ... tend TOWARD 1, not AWAY. To see this, all you
need to do is arithmetic.
I don't see anything in your post tending AWAY from 1.
-Leonard (email defunct)
Quote:
.9^2 = .81
.9^4 = .6561
.99^2 = .9801
.99^4 = .96059601
.999^2 = .998001
.999^4 = .996005996001
.9999^2 = .99980001
.9999^4 = .9996000599960001
.99999^2 = .9999800001
.99999^4 = .99996000059999600001
.999999^2 = .999998000001
.999999^4 = .999996000005999996000001
.9999999^2 = .99999980000001
.9999999^4 = .9999996000000599999960000001
.99999999^2 = .9999999800000001
.99999999^4 = .99999998000000059999999600000001
.999999999^2 = .999999998000000001
.999999999^4 = .999999996000000005999999996000000001
.9999999999^2 = .99999999980000000001
.9999999999^4 = .9999999996000000000599999999960000000001
.99999999999^2 = .9999999999800000000001
.99999999999^4 = .99999999996000000000059999999999600000000001
.999999999999^2 = .999999999998000000000001
.999999999999^4 = .999999999996000000000005999999999996000000000001
[snip rest for brevity]
Challenge: Prove squares of 0.999... tend toward 1
Thank you
Garry Denke, Geologist
Denoco Inc. of Texas |
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| Garry Denke |
| Posted: Thu Jan 08, 2004 9:45 pm |
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Quote: I think you mean that each Square from your list is adding digits away from
its trailing one, which is illustrated by your example.
Wrong. I don't. The squares of 0.999... tend toward 0, not toward 1
.9^2 = .81
.9^3 = .729
.9^4 = .6561
.99^2 = .9801
.99^3 = .970299
.99^4 = .96059601
.999^2 = .998001
.999^3 = .997002999
.999^4 = .996005996001
.9999^2 = .99980001
.9999^3 = .999700029999
.9999^4 = .9996000599960001
.99999^2 = .9999800001
.99999^3 = .999970000299999
.99999^4 = .99996000059999600001
.999999^2 = .999998000001
.999999^3 = .999997000002999999
.999999^4 = .999996000005999996000001
.9999999^2 = .99999980000001
.9999999^3 = .999999700000029999999
.9999999^4 = .9999996000000599999960000001
.99999999^2 = .9999999800000001
.99999999^3 = .999999970000000299999999
.99999999^4 = .99999998000000059999999600000001
.999999999^2 = .999999998000000001
.999999999^3 = .999999997000000002999999999
.999999999^4 = .999999996000000005999999996000000001
.9999999999^2 = .99999999980000000001
.9999999999^3 = .999999999700000000029999999999
.9999999999^4 = .9999999996000000000599999999960000000001
.99999999999^2 = .9999999999800000000001
.99999999999^3 = .999999999970000000000299999999999
.99999999999^4 = .99999999996000000000059999999999600000000001
.999999999999^2 = .999999999998000000000001
.999999999999^3 = .999999999997000000000002999999999999
..999999999999^4 = .999999999996000000000005999999999996000000000001
[snip rest for brevity]
Prove the squares of 0.999... tend toward 1
or please stop claiming that 0.999... = 1
Thank you
Garry Denke, Geologist
Denoco Inc. of Texas |
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| Garry Denke |
| Posted: Fri Jan 09, 2004 1:13 am |
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The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-4AC08C.12324508012004@news.supernews.com>...
Quote: In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Squares of 0.999... tend away from 1
.99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
What would be a more accurate term for a starter?
Quote:
You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
I am interested in the sequence .9^2, .9^3, .9^4, ...
and the sequence .99^2, .99^3, .99^4 ...
and the sequence .999^2, .999^3, .999^4 ...
etc., etc., etc.
Quote: Assuming your post is not a troll, start with the fact that the sequence
.9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2
= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2,
... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the
latter sequence tends to 0. Therefore the former sequence tends to 0, which
is the same as saying .9^2, .99^2, .999^2, ... tends to 1.
Do you agree the sequence .9^2, .9^3, .9^4, ... tends to 0?
Do you agree the sequence .99^2, .99^3, .99^4 ... tends to 0?
Do you agree the sequence .999^2, .999^3, .999^4 ... tends to 0?
etc., etc., etc.
If so, do you agree that 0.999... = 0.999... and is not equal to 1?
Thank you.
Garry Denke, Geologist
Denoco Inc. of Texas |
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| Virgil |
| Posted: Fri Jan 09, 2004 2:48 am |
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In article <210a83f4.0401082213.1c57d6a4@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Quote: The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message
news:<waderameyxiii-4AC08C.12324508012004@news.supernews.com>...
In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Squares of 0.999... tend away from 1
.99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
What would be a more accurate term for a starter?
Saying what you mean would help.
Do you mean that the sequence .9^2, .99^2, .999^2, ..., does not
converge to 1?
If you do, you are wrong, as any fool (except you) can plainly see.
Quote:
You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
I am interested in the sequence .9^2, .9^3, .9^4, ...
and the sequence .99^2, .99^3, .99^4 ...
and the sequence .999^2, .999^3, .999^4 ...
etc., etc., etc.
OK, you have, essentially, f(m,n) = (1 - 1/10^m)^n, where m is the
number of nines in the decimal number, and n is the power it is being
raised to.
Now for any fixed positive integer, n, lim_{m -> oo} f(m,n) = 1
and for any fixed positive integer, m, lim_{n -> oo} f(m,n) = 0.
You seem to think that the second of these limits disproves the first,
but you are wrong. Again.
Quote: Assuming your post is not a troll, start with the fact that the sequence
.9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2
= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2,
... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the
latter sequence tends to 0. Therefore the former sequence tends to 0, which
is the same as saying .9^2, .99^2, .999^2, ... tends to 1.
Do you agree the sequence .9^2, .9^3, .9^4, ... tends to 0?
Do you agree the sequence .99^2, .99^3, .99^4 ... tends to 0?
Do you agree the sequence .999^2, .999^3, .999^4 ... tends to 0?
etc., etc., etc.
See above. That lim_{n -> oo} (1 - 1/10^m)^n = 0 has no effect on the
fact than lim_{m -> oo} (1 - 1/10^m)^n = 1.
You are conflating the effect of changing one variable with the effect
of changing the other. It don't work that way.
The quality of your mathematics leads me to ask, hesitantly, what sort
of dowsing rods do you use to find your oil
Quote:
If so, do you agree that 0.999... = 0.999... and is not equal to 1?
I never agree to nonsense, except possibly on April first.
Quote: Thank you.
Garry Denke, Geologist
And non-mathematician
> Denoco Inc. of Texas |
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| Virgil |
| Posted: Fri Jan 09, 2004 2:58 am |
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In article <210a83f4.0401081845.59aecb83@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Quote: I think you mean that each Square from your list is adding digits away from
its trailing one, which is illustrated by your example.
Wrong. I don't. The squares of 0.999... tend toward 0, not toward 1
Since 0.999... is just one number, it only has one square, which does
not tend to be any different than its one value.
What you might possibly mean is that the sequence
0.9^2, 0.99^2, 0.999^2, ..., tends towards zero, not one.
But you would then be wrong.
If what you might mean above were true, then f(x) = x^2 would have to be
discontinuous at x = 1, which means that polynomials of degree greater
than one need not be continuous or differentiable or integrable, and all
of calculus would collapse. The mathematical and engineering worlds
would then burn you in effigy. Or perhaps not in effigy. |
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| David McAnally |
| Posted: Fri Jan 09, 2004 5:16 am |
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garrydenke@starmail.com (Garry Denke) writes:
Quote: I think you mean that each Square from your list is adding digits away from
its trailing one, which is illustrated by your example.
Wrong. I don't. The squares of 0.999... tend toward 0, not toward 1
By your own calculations, .9^2, .99^2, .999^2, is an *increasing*
sequence, getting closer to 1. Simple observation shows that,
Quote: .9^2 = .81
.99^2 = .9801
.999^2 = .998001
.9999^2 = .99980001
.99999^2 = .9999800001
.999999^2 = .999998000001
.9999999^2 = .99999980000001
.99999999^2 = .9999999800000001
.999999999^2 = .999999998000000001
.9999999999^2 = .99999999980000000001
.99999999999^2 = .9999999999800000000001
.999999999999^2 = .999999999998000000000001
David McAnally
"Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills." |
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| David McAnally |
| Posted: Fri Jan 09, 2004 5:20 am |
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garrydenke@starmail.com (Garry Denke) writes:
Quote: The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-4AC08C.12324508012004@news.supernews.com>...
In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Squares of 0.999... tend away from 1
.99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
What would be a more accurate term for a starter?
You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
I am interested in the sequence .9^2, .9^3, .9^4, ...
and the sequence .99^2, .99^3, .99^4 ...
and the sequence .999^2, .999^3, .999^4 ...
etc., etc., etc.
This would only be a problem if limits commute. They need not commute.
There is no reason to expect that
lim_{x -> a} lim_{y -> b} f(x,y) = lim_{y -> b} lim_{x -> a} f(x,y).
You seem to be demanding that limits commute in spite of the fact that
such an expectation is wrong.
In other words, your objections are based on a false premise.
David McAnally
"Despite anything you may have heard to the contrary,
the rain in Spain stays almost invariably in the hills." |
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| G. Frege |
| Posted: Fri Jan 09, 2004 6:01 am |
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On 8 Jan 2004 10:34:18 -0800, garrydenke@starmail.com (Garry Denke)
wrote:
Quote:
Challenge: Prove squares of 0.999..9 tend toward 1
<yawn
Use a pocket calculator:
Quote: .9^2 = .81
.9^2 - 1 = -0.19 ~ -0.2
.99^2 = .9801
.99^2 - 1 = -0.0199 ~ -0.02
.999^2 = .998001
.999^2 - 1 = -0.001999 ~ -0.002
.9999^2 = .99980001
.9999^2 - 1 = -0.00019999 ~ -0.0002
.99999^2 = .9999800001
etc. (ad infinitum)
The "trend" seems to be rather obvious. No?
.9999..9^2 - 1 tends to 0.
With other words,
.9999..9^2 tends to 1.
F.
P.S.
Quote:
[snip rest for brevity]
Nonsense. It's snipped because it's IMPOSSIBLE to write down a l l
lines. (Or did you try to be funny?) |
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| Guest |
| Posted: Fri Jan 09, 2004 9:13 am |
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garrydenke@starmail.com (Garry Denke) writes:
Quote:
Wrong. I don't. The squares of 0.999... tend toward 0, not toward 1
Sigh. Powers of any number strictly between zero and one converge to
zero. This fact is well known. You have hardly proven that the *sum*
of these terms converges to zero. Worse, the sum of powers is not
equal to the power of the sum, so your computation is also irrelevant.
In particular, any series whose sum is 1, and whose terms are
positive, will exhibit the property that powers of the terms converge
to zero. The powers of the sum, however, are all uniformly 1.
--Len |
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| Garry Denke |
| Posted: Fri Jan 09, 2004 9:14 am |
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The World Wide Wade <waderameyxiii@comcast.remove13.net> wrote in message news:<waderameyxiii-4AC08C.12324508012004@news.supernews.com>...
Quote: In article <210a83f4.0401081034.24b607ce@posting.google.com>,
garrydenke@starmail.com (Garry Denke) wrote:
Squares of 0.999... tend away from 1
.99999.... is a single number, so there is only one square of it. Talking
about "Squares of 0.999..." is a bit of a non-starter.
You seem to be interested in the sequence .9^2, .99^2, .999^2, ... Clearly
these numbers increase, so how can they "tend away from 1"? Your own
computations show they get closer to 1.
Assuming your post is not a troll, start with the fact that the sequence
.9, .99, .999, ... tends to 1. Next observe that if 0 < x < 1, then 1 - x^2
= (1+x)(1-x) < 2(1-x). Therefore the numbers 1-.9^2, 1-.99^2, 1-.999^2,
... are, respectively, less than 2(1-.9), 2(1-.99), 2(1-.999), ... and the
latter sequence tends to 0. Therefore the former sequence tends to 0, which
is the same as saying .9^2, .99^2, .999^2, ... tends to 1.
Sequence .9^2, .9^3, .9^4, ...
Sequence .99^2, .99^3, .99^4, ...
Sequence .999^2, .999^3, .999^4, ...
Sequence .9999^2, .9999^3, .9999^4, ...
Sequence .99999^2, .99999^3, .99999^4, ...
Sequence .999999^2, .999999^3, .999999^4, ...
Sequence .9999999^2, .9999999^3, .9999999^4, ...
Sequence .99999999^2, .99999999^3, .99999999^4, ...
Sequence .999999999^2, .999999999^3, .999999999^4, ...
Sequence .9999999999^2, .9999999999^3, .9999999999^4, ...
Sequence .99999999999^2, .99999999999^3, .99999999999^4, ...
Sequence .999999999999^2, .999999999999^3, .999999999999^4, ...
Sequence .9999999999999^2, .9999999999999^3, .9999999999999^4, ...
etc., etc., etc.,
tend to 0.
Does sequence .999...^2, .999...^3, .999...^4, ... tend to 0 or to 1?
Thank you.
Garry Denke, Geologist
Denoco Inc. of Texas |
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