Is 1 really equal to 0.9999...?

[Doron Shadmi]

... = infinitely many.

Fact A:
-------

Let n = 0.9999...
10n = 9.9999...
10n - n = 9.9999... - 0.9999...
9n = 9
n = 1 = 0.9999...


Fact B:
-------

not = ~

A nonempty R member(= set's content exists) never reach
0(={}= set's content does not exit) by definition, (0__A nonempty R member).

It means that 0.0000...001 ~= 0

0.0000...0009999... = 0.0000...001 (by fact A)

Therefore, there must be a quantum-leap path between any
nonempty R member to 0.

But then 0.9999... cannot exist.

Therefore, 1 = 0.999...999 + 0.000...001



Please tell me what do you think.

Thank you.
Yours,

Doron

 

[selfAdjoint]

Doron, what I really think is that you don't understand either set theory or real numbers. Your version of set theory is full of predjudices that aren't in the real thing. Likewise your denial of the real line. What you imagine to be logical proof isn't.

 

[Integral]

Originally posted by Doron Shadmi
... = infinitely many.

Fact A:
-------

Let n = 0.9999...
10n = 9.9999...
10n - n = 9.9999... - 0.9999...
9n = 9
n = 1 = 0.9999...

I consider this a demonstration, not a proof. Though since your algebra is correct, the result is correct.

Fact B:
-------

not = ~

A nonempty R member(= set's content exists) never reach
0(={}= set's content does not exit) by definition, (0__A nonempty R member).

This sort of poorly defined,poorly elucidated comment makes me think of a homeless person wandering aimlessly down the street muttering to himself. His own thoughts, his own language, no one else can make any sense of it.

It means that 0.0000...001 ~= 0

Since the use of an ellipsis in the middle of a number indicates that a FINITE number of digits have been omitted, .
Edit: Since you are using ~ as not, this is true.

0.0000...0009999... = 0.0000...001 (by fact A)

This works but is a bit dangerous, as the ellipsis must represent the correct number of digits.

Therefore, there must be a quantum-leap path between any
nonempty R member to 0.

This is more muttering, where did this conclusion come from? What does it mean?

But then 0.9999... cannot exist.

OK, now, when used at the end of a number, the ellipsis represents an infinite number of digits. Seems to me your statement is self contradictory. If .999... does not exist how can you represent it and talk about as a number? Your very reference to it gives it meaning therefore existence.

Therefore, 1 = 0.999...999 + 0.000...001

Once again, this is a rather trivial, since it is a complicated way of saying 1 = .9999 + .0001 , where I have chosen to represent all of the digits.

Please tell me what do you think.

Thank you.
Yours,

Doron

Are you are trying to use the ellipsis to represent and infinite number of digits omitted from the "middle" of a number? If so, sorry, it simply does not work. Remember that every digit must be assigned a position integer, that is, every digit in a number MUST have a unique power of 10 (or 2 or 3 or what ever your base is) assigned to it. So When your write .000...0001, that 1 has a integer power of the base assigned, since it is an integer it is finite, by definition.

 

[HallsofIvy]

I agree with both adjoint and integral: as long as you quote mathematical facts, you do okay. As soon as you assert your own "facts", you seem to lose yourself in a morass of undefined terms, non-sequiturs, and just plain bad grammar.

By the way, integral is correct when he asserts that "..." MUST represent a finite number of terms. At the beginning of your post you write "... = infinitely many". The reason that makes no sense is that such things as 0.00...0001, with "..." meaning "infinitely many" is impossible.
The sequence 0.1, 0.01, 0.001, 0.0001, etc. converges to 0. That's exactly the conclusion you are trying to avoid. You avoid it by thinking of this as an "infinitely long" sequence of digits ENDING in 1. That's impossible. An infinitely long sequence doesn't END in anything! You keep talking about things that are demonstrably NOT real numbers as if they were. How do YOU define "real numbers"?

 

[Inquiring_Mike]

0.99999999...=1

I'm just a newbie, but wouldn't this prove 0.9999... = 1...
1/3 + 2/3 = 3/3 = 1

1/3 = 0.333... + 2/3= 0.666... = 0.999... which according to the above equals 1.

 

[Integral]

Originally posted by Inquiring_Mike
I'm just a newbie, but wouldn't this prove 0.9999... = 1...
1/3 + 2/3 = 3/3 = 1

1/3 = 0.333... + 2/3= 0.666... = 0.999... which according to the above equals 1.

Again this is a demonstration, but not a proof.

The trouble with both of these demonstrations (yours and Doron's) is the implied operations at infinity. A true proof involves operations only on well defined, finite, digits. I have a proof on my hard drive, of which I am not entirely happy with my explanations, which I'll dig out, proof read and post RSN.
Edit:
http://home.comcast.net/~rossgr1/Math/one.PDF a link to a fundamental proof that 1=.999...

(and some spelling)

 

[bogdan]

1 is the limit of 0.999... as the number of 9 goes to infinity...
1/3 is the limit of 0.33333... as the number of 3 goes to infinity...

 

[Doron Shadmi]

Hi Integral, selfadjoint and HallsofIvy

I will be glad to see a proof that shows how a non-empty R member = 0 = {}.

For me it is a simple axiomatic state, that a non-empty set can never be an empty set,
and it can only be done by a phase transition (a quantum leap) between these states.

As I see it, no infinitely many points can do it, because any set's content means content exists, and "set's content exists"(= a non-empty set) can never be "set's content does not exist"(= an empty set).

Integral, 9n = 9 = 1 because 0.9999... - 0.9999... = 0, therefore 0.9999... has no influence on the result from the beginning.

Instead of writing:

Let n = 0.9999...
10n = 9.9999...
10n - n = 9.9999... - 0.9999...
9n = 9
n = 1 = 0.9999...

We can simply write:

Let n = 1.0000...
10n = 10.0000...
10n - n = 10.0000... - 1.0000...
9n = 9
n = 1.0000... and we don’t need all this n = 0.9999... mambo jambo.
So, from this point of view, 0.9999... does not exist.

Are you are trying to use the ellipsis to represent and infinite number of digits omitted from the "middle" of a number? If so, sorry, it simply does not work. Remember that every digit must be assigned a position integer, that is, every digit in a number MUST have a unique power of 10 (or 2 or 3 or what ever your base is) assigned to it. So When your write .000...0001, that 1 has a integer power of the base assigned, since it is an integer it is finite, by definition.

(... = infinitely many)

Integral and what if I write: 0.place1,place2,place3...place n-1,place n ?

A non-empty 0.place1,place2,place3...place n-1,place n+[oo] information's cell can never be an empty set.

Therefore, any non-empty fraction's information-cell, is the first and last cell through the "eyes" of the empty set.

This is, however, the deep reason, why n/0 is meaningless, because any non-empty set = 1 (= content exists) through the "eyes" of the empty set (any n = 1 = content exists).

The vice versa holds too, because through any non-empty set, the empty set looks the same(=0=content does not exist).

 

[Integral]

Both 1 and .999... are fixed numbers, .999... does not change therefore it cannot "approach" anything. It is true that 1 is the limit of the set of sequences
S(N)= 9∑ .1ⁿ (n=1 to N) as N-> infinity.

But this is not what we are talking about. We are talking about the FIXED number which consists of 9s for each digit after the decimal. There is no limit here, only nines.

 

[Integral]

Originally posted by Doron Shadmi
Hi Integral, selfadjoint and HallsofIvy

I will be glad to see a proof that shows how a non-empty R member = 0 = {}.

Talking to yourself again. What does that mean? Why should I make any effort to learn your notation when it clearly leads you to false conclusions, and since you refuse to learn the standard notation.

For me it is a simple axiomatic state, that a non-empty set can never be an empty set,
and it can only be done by a phase transition (a quantum leap) between these states.

All good and fine, as far as I know, a non empty set is not empty. How does this connect to anything else you are saying.

As I see it, no infinitely many points can do it, because any set's content means content exists, and "set's content exists"(= a non-empty set) can never be "set's content does not exist"(= an empty set).

You really need to do something to tie your set ramblings to the Real numbers. I do not have a clue how what you have written relates to the Real numbers.

Integral, 9n = 9 = 1 because 0.9999... - 0.9999... = 0, therefore 0.9999... has no influence on the result from the beginning.

Instead of writing:

Let n = 0.9999...
10n = 9.9999...
10n - n = 9.9999... - 0.9999...
9n = 9
n = 1 = 0.9999...

We can simply write:

Let n = 1.0000...
10n = 10.0000...
10n - n = 10.0000... - 1.0000...
9n = 9
n = 1.0000... and we don't need all this n = 0.9999... mambo jambo.
So, from this point of view, 0.9999... does not exist.

HUH? This is YOUR argument,not mine. Whats your point? Once again, your words are self contradictory, if .999... does not exist then what are we talking about? The mere fact that I can think about the number gives it existence. Since I can think of it YOU must deal with it. It cannot be swept under the rug simply because we do not need it.

Integral and what if I write: 0.place1,place2,place3...place n-1,place n [/B]

What if you did write that? So What? That implies that you left out n-4 digits, as I was saying an ellipsis in the middle of a number represents a finite number of missing digits, n-4 in your last example. Not sure what your point is.

Did you take a moment to look at my proof?

 

[Doron Shadmi]

Hi Integral,

Before I'll give my point of view on the proof, please read again my previous post.

I added new things at the end of it.

Sorry, and thank you,

Doron

 

[Integral]

Doron,
Sorry it didn't help me any. Still not sure what you are trying to say. Why do you keep talking about the empty set? I cannot see where that has anything to do with this disscussion. Zero is not the empty set, nor is .999... or any of its digits. So why are you so concerned about something being, or not being, in the empty set?

Sorry, time for me to call it a night.

 

[Doron Shadmi]

Integral, thanks for your answers, and have a good night.

What if you did write that? So What? That implies that you left out n-4 digits, as I was saying an ellipsis in the middle of a number represents a finite number of missing digits, n-4 in your last example. Not sure what your point is.

(Integral and what if I write: 0.place1,place2,place3...place n-1,place n ?)

By ... I mean infinitely many places.

Zero is not the empty set

By Von Neumann Hierarchy we have:

0 = { }

1 = {{ }} = {0}

2 = {{ },{{ }}} = {0,1}

3 = {{ },{{ }},{{ },{{ }}}} = {0,1,2}

4 = {{ },{{ }},{{ },{{ }}},{{ },{{ }},{{ },{{ }}}}} = {0,1,2,3}

and so on ...

If you mean that I do not distinguish between { } to {{ }}, so let me tell you that I do distinguish between them.

HUH? This is YOUR argument, not mine. Whats your point? Once again, your words are self contradictory, if .999... does not exist then what are we talking about? The mere fact that I can think about the number gives it existence. Since I can think of it YOU must deal with it. It cannot be swept under the rug simply because we do not need it.

I will try to explain myself step by step.

... = infinitely many

{} = empty set

~{} = non-empty set

By writing any fractional representation of some number by 0.x1,x2,x3,... we ignore the fact that any transition from {} to ~{} cannot be but a phase transition (and vice versa), and this leap has nothing in between (like in electron's quantum leap).

From {} point of view, any ~{} = 1(set's content exists).

From ~{} point of view, any {} = 0{set's content does not exist).

If we are not ignoring the {} to ~{} phase transition, when we are using the representation of some number, than we can get the closest fractional representation to 0,
which is: 0.000...01 that can be ordered as 0.x1,x2,x3,...,xn-1,xn

If we try to move another fractional step to the right, we have a phase transition to 0(={}).

It means that the representation of 0.000...01 as 0.000...00999... , does not hold.

Therefore any representation of some number by 0.x1,x2,x3,... does not hold, because we can't ignore the phase transition from ~{} to {} (and vice versa).

So instead of 0.999... we have 0.999...99 ,
and we get 1 = 0.999...99 + 0.000...01

(... = infinitely many)

 

[HallsofIvy]

I'll say this one more time and then quit: you can't just make up new meanings for mathematical symbols and think that you are talking mathematics. No one can answer any of your questions because what you are saying does not make sense.

1.0 and 0.9999... (infinitely repeating) are two different notations for the same number in the system that mathematicians mean when they talk about real numbers. I know several ways of defining that system (Dedekind Cut, equivalence classes of increasing sequences, equivalence classses of Cauchy sequences, etc.) and they all lead to the result that 1.0= 0.99999... If you are using a different definition, then you will have to tell us what you are talking about.
There is no such thing as "0.000...0001 with ... meaning infinitely many". If you are using a system in which there is, you will have to define it for us.

If you are going to talk about the step from 0 to 1 or from the empty set to a non-empty set (things that are commonplace in discrete mathematics) as a "phase transition" then you are going to have to tell us what in the world you mean by "phase transition".

Perhaps if you do that people will not suspect that you are not defining your terms because YOU have no idea what you are doing.