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Oct 30, 1998, 3:00:00 AM10/30/98

to

It is equal.

1 - 0.999... = 0.000... = 0

or

(.999...)/3 = .333... = 1/3

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 :-)

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

to

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

to

Adam Russell wrote:

>

> It is equal.

> 1 - 0.999... = 0.000... = 0

> or

> (.999...)/3 = .333... = 1/3

>

> 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

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

to

In article <19981030235322...@ng37.aol.com>,

zero...@aol.com (Zero0Zero) wrote:

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

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

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

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

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

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.

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...)

>

>

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.

> 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

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

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

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

--------------------------------

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.

"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.

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>...

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").

as the difference betwenn 2/4 and 1/2. (If you call that "difference").

Haoyong

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

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

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

--

----

char *p="char *p=%c%s%c;main(){printf(p,34,p,34);}";main(){printf(p,34,p,34);}

This message made from 100% recycled bits.

I don't speak for Alcatel <- They make me say that.

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