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.999 repeating = 1?

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Adam Russell

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Oct 30, 1998, 3:00:00 AM10/30/98
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It is equal.
1 - 0.999... = 0.000... = 0
or
(.999...)/3 = .333... = 1/3

Zero0Zero wrote in message <19981030235322...@ng37.aol.com>...
>Someone recently told me that .999 repeating is equal to 1, and I was shown
a
>few proofs, but then another came and said that they aren't, and I've met
as
>many who think they are equal as those who don't. Could anyone please
clarify
>this for me? And please give reasons either way. I'd appreciate that :-)

Zero0Zero

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Oct 31, 1998, 3:00:00 AM10/31/98
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Victor Ng

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Oct 31, 1998, 3:00:00 AM10/31/98
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Adam Russell wrote:
>
> It is equal.
> 1 - 0.999... = 0.000... = 0
> or
> (.999...)/3 = .333... = 1/3

I would say it is equal as well.

let x=0.9999....
then 10x=9.9999.....

10x - x= 9x (LHS)

9.9999.... - 0.9999... = 9.000.. (RHS)

9x=9 --> x=1

--

Victor Ng
v_...@alcor.concordia.ca
http://alcor.concordia.ca/~v_ng

torqu...@my-dejanews.com

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Oct 31, 1998, 3:00:00 AM10/31/98
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In article <19981030235322...@ng37.aol.com>,
zero...@aol.com (Zero0Zero) wrote:

If you've been given some proofs that .999... = 1 that's all you need.
Mathematics works by proofs. If you have a proof you don't need 'reasons'. If
the proof seems problematic you may need to learn some foundational
mathematics (eg. basic analysis) so you can make your own judgement. Without
such knowledge weighing up various people's 'reasons' is a fairly worthless
activity. Trust no- one else's opinion. -- http://travel.to/tanelorn

-----------== Posted via Deja News, The Discussion Network ==----------
http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own

graham...@hotmail.com

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Oct 31, 1998, 3:00:00 AM10/31/98
to
In article <363AB8...@alcor.concordia.ca>,

Victor Ng <v_...@alcor.concordia.ca> wrote:
> Adam Russell wrote:
> >
> > It is equal.
> > 1 - 0.999... = 0.000... = 0
> > or
> > (.999...)/3 = .333... = 1/3
>
> I would say it is equal as well.
>
> let x=0.9999....
> then 10x=9.9999.....
>
> 10x - x= 9x (LHS)
>
> 9.9999.... - 0.9999... = 9.000.. (RHS)
>
> 9x=9 --> x=1

Also, there is the summation series:

0.9999... = 9/10 + 9/100 + 9/1000 + 9/10000 + ...

which is clearly equal to 9 * Sum_over_n(1/10^n)

which is a simple geometric sum in the form Sum(1/K^n) which is known and
provably equal to 1/(K-1).

So, we get 9 * (1/(10-1)), which is 1.

- Graham

Russell Easterly

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Oct 31, 1998, 3:00:00 AM10/31/98
to

If you are talking about real numbers
then most people would say that .999... = 1.000...

But there differences between these two numbers.
1.000... is bounded above and below by 1.000....
It is a solution to this inequality:

1.000... >= x >= 1.000...

But .999... would not be a solution
because it doesn't have a rational
greatest lower bound.
(it does have a least upper bound, 1.000...)


Russell
-2 many 2 count


Jim Hunter

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Oct 31, 1998, 3:00:00 AM10/31/98
to

Russell Easterly wrote:

I don't get it.
Doesn't 1.000.. = 0.999... imply 1.000.... >= 0.999.... ?


---
Jim


Adam Russell

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Oct 31, 1998, 3:00:00 AM10/31/98
to
If you agree that 0.333... = 1/3, then multiplying both sides by 3 you see
that 0.999... = 1. If not, then I guess you must think that repeating
decimals are not rational, or perhaps even not real. If they are not
rational or not real then I guess you could say that they are undefined.

Russell Easterly wrote in message <71g7ak$t3t$1...@sparky.wolfe.net>...


>
>If you are talking about real numbers
>then most people would say that .999... = 1.000...
>
>But there differences between these two numbers.
>1.000... is bounded above and below by 1.000....
>It is a solution to this inequality:
>
>1.000... >= x >= 1.000...
>
>But .999... would not be a solution
>because it doesn't have a rational
>greatest lower bound.
>(it does have a least upper bound, 1.000...)
>
>

Pertti Lounesto

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Nov 1, 1998, 3:00:00 AM11/1/98
to
torqu...@my-dejanews.com wrote:
> If you've been given some proofs that .999... = 1 that's all you need.
> Mathematics works by proofs. If you have a proof you don't need 'reasons'.
> If the proof seems problematic you may need to learn some foundational
> mathematics (eg. basic analysis) so you can make your own judgement.
> Without such knowledge weighing up various people's 'reasons' is a fairly
> worthless activity. Trust no-one else's opinion.

torquemada is right: knowledge of the math topic is needed to
evaluate validity of math proofs and no-one's math proofs should
be trusted, without making your own judgement. I have falsified
several math proofs with counterexamples which satisfy all the
assumptions of the theorems without the conclusion being valid.
To evaluate possible validity of my counterexamples, one needs
to know the topic in question. Those interested in my counter-
examples falsifying proofs in recent mathematical literature
can point their browsers to

http://gemini.tntech.edu/~plounesto/counterexamples.htm

Pertti Lounesto

Haoyong Zhang

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Nov 1, 1998, 3:00:00 AM11/1/98
to
In article <19981030235322...@ng37.aol.com>,

Zero0Zero <zero...@aol.com> wrote:
>Someone recently told me that .999 repeating is equal to 1, and I was shown a
>few proofs, but then another came and said that they aren't, and I've met as
>many who think they are equal as those who don't. Could anyone please clarify
>this for me? And please give reasons either way. I'd appreciate that :-)

The short answer to your question is .999... = 1 because that is the
definition.

The long answer is the following:

Well, the questions are "how is real number defined?" and
"what is decimal expansion?"

- Natural Numbers:

If you take a look at some set theory books, they would give you
definitions of natural numbers (or non-negative integers). For
example, Paul R. Halmos's Naive Set Theory is a great book.

- Integers:

Assume we have numbers 0,1,2,3,..., then we can define
Integer using equivalent class of pairs of natural number.
For example, every integer can be represent as (a,b) where
both a and b are natural numbers.

1 can be written in the form (a+1,a), i.e. (2,1) or (6,5)...
0 can be written in the form (a,a), i.e. (0,0) or (10,10)...
-1 can be written in the form (a,a+1), i.e. (1,2) or (99,100)...
-10 can be written in the form (a,a+10), i.e. (100,110)...

and so on. Of couse, we usually use a short form of representation,
i.e. with +/- signs in front of the number.

- Rational Numbers:

Similiarly, rational number is defined as equivalent class of
pairs, we usually written it in the form of m/n where m,n are integers.

1 can be written as 1/1, 2/2, 3/3, 10/10, ...
0.5 can be written as 1/2, 2/4, 3/6, ...

Once we have rational number, we can then define real number.

- Real Numbers:

I believe the standard definition of a real number is the
equivalent class of Cauchy sequences that converge to that number.
And to get decimal expansion of a real number r, we simply take the set
Cauchy sequences converges to r with the form
Sum(n=1 to infinity) {a_n*10^-n} where 0 <= a_n <= 9, and we list all of
a_n's.

In fact, any rational number ended with repeating 0's will have
an alternative representation.

For example,
0.5 can be written as 0.49999... or 0.500000...
0.123456 can be written as 0.123455999999999... or 0.12345600000...

So, in this sense, 0.999... is SAME as 1.0000... they are just two
way to write the same number 1.

Actually, the fact that the decimal expansion is not unique often
yields problems when one has to use it to prove properties of real number,
but that should be easy to eliminate. (i.e. Have a rule, always choose
repeating 9's over repeating 0's, or the other way, whenever is convenient.)

Hope that will help,

Haoyong


Benne de Weger

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Nov 2, 1998, 3:00:00 AM11/2/98
to Zero0Zero
Here's an argument that I like:

If two real numbers are not the same, then there is a number in between them,
e.g. their average. So if a < b (say) then a < (a+b)/2 < b. Note that the average
is
not equal to both a or b.
Now try to find a number in between 0.999... and 1.000..., e.g. compute their
average.

Benne de Weger
--------------------------------

ry...@edge.net

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Nov 2, 1998, 3:00:00 AM11/2/98
to
In article <71g7ak$t3t$1...@sparky.wolfe.net>,

"Russell Easterly" <logi...@wolfenet.com> wrote:
>
> If you are talking about real numbers
> then most people would say that .999... = 1.000...
>
> But there differences between these two numbers.

Hold on a minute! If they are equal, there is no difference. If there are
differences, then they are not equal.

Matthew H. Burch

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Nov 2, 1998, 3:00:00 AM11/2/98
to
The meaning can be the same, even if the terms are different :

For example, 3*7 = 2*10 + 1

Each side of the equation is different, BUT they are equal!

1 = .99999...

Because the three little dots mean "continued to infinity"

Therefore they mean the same thing, even though they are arrived at in a
different manner.


Matthew H. Burch
ap...@pipeline.com


ry...@edge.net wrote in message <71ktkr$894$1...@nnrp1.dejanews.com>...

Haoyong Zhang

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Nov 3, 1998, 3:00:00 AM11/3/98
to
That's right. The difference between .999... and 1 is the same
as the difference betwenn 2/4 and 1/2. (If you call that "difference").

Haoyong

In article <71l3j5$mdo$1...@camel19.mindspring.com>,

Mike McCarty

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Nov 3, 1998, 3:00:00 AM11/3/98
to
In article <F1tqB...@undergrad.math.uwaterloo.ca>,
Haoyong Zhang <h2z...@neumann.uwaterloo.ca> wrote:
)That's right. The difference between .999... and 1 is the same
)as the difference betwenn 2/4 and 1/2. (If you call that "difference").
)
)Haoyong

Well, 2/4 and 1/2 are not the same (as fractions). But they are both
elements (representatives) of the same rational number.

I.E.

2/4 =/= 1/2
but
[2/4] = [1/2]

Mike
--
----
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