0.9999....(recurring) = 1?

George,Y · 2007-02-24 18:47:48

Also the plane worked bad when describing the world map. It worked pretty well in depicting a small ploughing field though.

If the concept of a plane can be decomponented into two concepts- flat and infinite, we will find which really counts. The concept of something flat applies to perhaps all the occassions when the concept of the plane is applied. And it is the concept that reflects reality well. The other sub-concept-infinite scale, however, is rarely applied.

Last edited by George,Y (2007-02-25 01:48:09)

Anthony.R.Brown · 2007-02-25 01:03:27

To luca-deltodesco

Quote:

" 0.99 + 0.001 = 0.991, not 1, if you can't even get that right..."

A.R.B

The Main thing is that you understood what I meant! Strange that! everything else I put forward you don't understand! I make a mistake with a Zero! and you understand it!?

Anthony.R.Brown · 2007-02-25 01:41:10

Lets have a look at why Infinite ( 0.9 ....0.1 ) and ( 0.99....0.01 ) etc. have the same Infinite Difference!

If we are talking about normal single Numbers! then sure 0.9 and 0.1 and 0.99 and 0.01 all have different values!

But because we are talking about Infinite Numbers! the Values are different! because the Decimal point is Infinite.

Example 1 ( 1 ) - ( 0.9 ) " Stage one/Decimal place " = 0.1

Example 2 ( 1 ) - ( 0.99 ) " Stage two/Decimal place's " = 0.01

Example 3 ( 1 ) - Infinite ( 0.9 ) " Stage Infinite/Decimal place's " = 0.001...etc.

For Example 1 above both Infinite ( 0.9 ) and the Infinite Difference ( 0.1 ) Shift one Decimal place! and so remain equally Different! they both will always have a Value!

For Example 2 above both Infinite ( 0.99 ) and the Infinite Difference ( 0.01 ) Shift two Decimal place's! and so remain equally Different! they both will always have a Value!

For Example 3 above both Infinite ( 0.9 ) Infinite Decimal place's) and the Infinite Difference
( 0.001...etc. ) Infinite Decimal place's) Shift Infinite Decimal place's! and so remain equally Different! they both will always have a Value!
Neither can ever equal Zero! because it would be a Contradiction to the Definition Infinite!

George,Y · 2007-02-25 01:54:28

All right, then, there is some difference between 0.999... and 1.
The difference shall be real-infinitesimal or one point if the concept of infinity is allowed.

luca-deltodesco · 2007-02-25 02:21:30

please, stop using 0.001.... that would be a number -> 0.001111111111 (infinate 1's).

you are implying 0.000....1, i.e. a number -> 0.0000000.... (infinate 0's)...... 1, the problem you seem to be having is equating such a number to 0.

as ive said earlier, that number can't exist by its definition, let me try and explain this to you anthony.

lets say, i have a number, 5.4444, and i put a 9 on the end of it, to make it the number 5.44449, the 9 is at the end of the 4's.

7.11, put a 9 on the end, and you get 7.119. you add the 9 to the end of the 1's.

if you have a number: 0. (one hundred 0's) 1, the 1 is at the end of the hundred zero's, you have one hundred zero's, then a one.

0.00000000000000001, you have after the decimal point, 16 zero's, then a one, you put it at the end of the list of zeros.

you seem to accept infinity, as having no end, infinity is unreachable, and has no end, it just carrys on indefinately, for example:

0.777.... (recuring 7) or in your words, infinate (0.7) there are an infinate amount of 7's after the decimal point, there aernt 100 7's, there aern't 1 million 7's, there aernt 10^1million 7's, there is no number to describe the number of 7's, because they never end, there is an infinate amount of them, you can never reach a number that would describe the number of 7's, there is no end to them.

if there is no end to the number of 7's, then you can't possibly put another number, lets say 3 after the 7's, becuase you would have to at some point, stop putting more 7's down, and put a 3, but no matter when you stop putting down 7's to put a 3, there would not be enough 7's, you'd have to take the 3 back off, and start adding 7's again, you would never be able to reach a state at which you could put the 3 down, because there is no end to the 7's for which to put the 3. in otherwords, there is no such number as 0.777....3, the 3 is irrelevant, it is 0.777....

apply the same process to 0.000....1, an infinate amount of 0's, followed by a 1, but to put a 1 after the 0's, you have to put a cap on the number of 0's at some point, for which to put the 1, but there is no end to the 0's for which to put a 1, its impossible, if you did that, there wouldn't be an infinate amount of 0's, 0.000....1 = 0.000...., because no matter what you 'want' to put after the zero's you can't, without making it not have infinate amount of 0's, and ofcourse 0.000... = 0.

JaneFairfax · 2007-02-25 02:42:20

Maybe its just a flaw in the notation used to write down real numbers? The flaw is that any real number with a terminating decimal (except 0) can be written in two different ways in the notation: e.g. 1 = 1.0000 = 0.9999, 1.234 = 1.2340000 = 1.2339999, etc. Its all too easy to be fooled into thinking that two different ways of representing something must mean two different things.

Maybe we should switch to a better notation for real numbers, one in which every real number would have its own unique representation. (Continued fractions, perhaps?) Then the mind would not be fooled into looking for nonexistent infiinitesimal differences between numbers from the way they appear when written down.

Just my tuppenceworth.

Last edited by JaneFairfax (2007-02-25 02:49:12)

John E. Franklin · 2007-02-25 04:26:52

lucadd, are you saying these numbers are okay?
1e1
1e2
1e3
1e4
1e0
1e-1
1e-2
1e-3
Then why not
1e-9999999999999999999999999999999999999999999999999
And of course the unspeakable:
1e-∞

luca-deltodesco · 2007-02-25 04:28:21

do you mean e as in the programming notation.

1e4 = 1×10^4
1.53e-4 = 1.53×10^-4

?

John E. Franklin · 2007-02-25 08:28:32

Yes lucadd. "e notation" seems to be mentioned in maple program and fortran maybe.

luca-deltodesco · 2007-02-25 08:56:13

its in all languages ive ever come across

and well yes, ofcourse 1e-999999999999999999 is fine, its still a defined number.

1e-∞ would be the infintesimal, which ive explained, is simply equal to 0

or rather, i agree with jane, theres no such thing as 0.999... or 0.000...1 they are just misrepresentations of 1 and 0

Last edited by luca-deltodesco (2007-02-25 08:57:00)

George,Y · 2007-02-25 15:26:19

luca-deltodesco wrote:

you seem to accept infinity, as having no end, infinity is unreachable, and has no end, it just carrys on indefinately, for example:
0.777.... (recuring 7) or in your words, infinate (0.7) there are an infinate amount of 7's after the decimal point, there aernt 100 7's, there aern't 1 million 7's, there aernt 10^1million 7's, there is no number to describe the number of 7's, because they never end, there is an infinate amount of them, you can never reach a number that would describe the number of 7's, there is no end to them.

In a word, Growing.
You still treat an infinite digits as a growing variable, luca. A number cannot carry on, only can a variable.

George,Y · 2007-02-25 15:31:14

Ricky has admitted that whenever he states 0.999...=1, he means {0.9, 0.99, 0.999...} getting closer and closer to 1 in this thread

Ricky · 2007-02-25 16:12:25

Ricky has admitted that whenever he states 0.999...=1, he means {0.9, 0.99, 0.999...} getting closer and closer to 1 in this thread

What? No, I don't mean a set of numbers with increasing amount of 9's. I mean a limit, in exactly the same way as every other infinite decimal expansion is interpreted as a limit, as I had define oh so many posts ago.

And there you go again with your loaded words. "Admitted"? You make it sound like I confessed to murder. I haven't change what I've been saying for this entire time.

George,Y · 2007-02-25 18:22:15

Well, I am sorry not having repeated your words one by one, however I have given the thread's whereabout, haven't I? Readers have their own judgement on whether I have translated your words correctly.

But I have to translate the terminology "limit" because it may be unfamiliar to some readers. "Closer and closer" is just a simple translation of "arbitarily close on condition that n is large enough", and the verb "get" clearly describes the nature of the variable (in this case S[sub]n[/sub])- varying.

George,Y · 2007-02-25 18:26:21

Ricky wrote:

You make it sound like I confessed to murder.

No you haven't, but you have done something like defining holding a gun as a murder. (This is also illustrated in that thread)

luca-deltodesco · 2007-02-25 18:44:15

george, im not talking about a variable that is increasing, im saying going on and on indefinately when you try and write it down, or even, just imaganing it, if you are writing down the number 0.777....4, youd be constantly writing down 7's, and no matter what, you could never write the 4 digit, even if it were somehow possible lets say, to write an infinate amount of digits in a lifetime, you still wouldn't be able to write the 4 if you know what i mean (i know thats a bit of a contradiction, but you should be able to get at what im saying here) the 4 digit can never be placed, it can never be part of the number, its a falacy to write 0.777...4 because that number can't exist, the 4 can never exist

luca-deltodesco · 2007-02-25 18:46:14

it's like saying, i think ill eat donuts until the end of time, and afterwards, for all infinitum, but then ill take a drink.

thats a lie, because no matter what, you can never take a drink, the only truth, would be that you are eating donuts for infinitum.

Ricky · 2007-02-25 19:05:39

Well, I am sorry not having repeated your words one by one, however I have given the thread's whereabout, haven't I? Readers have their own judgement on whether I have translated your words correctly.

Sorry George, but you aren't going to hide behind the "that's not what others may think". You keep trying to describe a limit as "getting closer and closer". While this is the concept behind a limit, a limit is a very fixed thing. The limit of your sum 0.9 + 0.99 + 0.999 + ... is 1. It's not that the limit is getting closer and closer, the limit is 1. It's not approaching 1, it is 1.

Just like 0.999... is 1.

George,Y · 2007-02-25 19:34:42

Ricky wrote:

Well, I am sorry not having repeated your words one by one, however I have given the thread's whereabout, haven't I? Readers have their own judgement on whether I have translated your words correctly.
Sorry George, but you aren't going to hide behind the "that's not what others may think". You keep trying to describe a limit as "getting closer and closer". While this is the concept behind a limit, a limit is a very fixed thing. The limit of your sum 0.9 + 0.99 + 0.999 + ... is 1. It's not that the limit is getting closer and closer, the limit is 1. It's not approaching 1, it is 1.
Just like 0.999... is 1.

word puzzle again, Ricky. Or you have mistranslated my words
I am not saying the limit is not 1. I am saying that the variable is not 1. Limit is a terminology to describe at least two variables, typically y and x, a[sub]n[/sub] and n, or S[sub]n[/sub] and n, don't you know? When the limit of the dependent variable is A, the dependent variable -whether it be y, a[sub]n[/sub], or S[sub]n[/sub]- approaches A, or literally be within a small interval around A. However, the limit concept does not guarantee the dependent variable being A, does it?

While you concentrate on the limit being 1, I also have the right to concentrate on the dependent variable -in this case the series{0.9, 0.99, 0.999, ...}or

- approaching 1, or getting close to 1. What's Wrong with My Focus?!

Remind you, limit only makes sense when it is related to at least two variables, one independent, one dependent. If you have any questions on this, start from Chapter 1 on most Calculus book.

Last edited by George,Y (2007-02-25 19:38:27)

George,Y · 2007-02-25 19:45:50

To luca:
If you aren't able to write out or imagine 0.00..01, neither are you able to write out or imagine 0.111.. with infinite digits. You may argue you can write or imagine 1's one by one, but you cannot get to the infinith 1 - 0.00..001 as the first clause, hence you lack the ability to have infinite digits.

Zhylliolom · 2007-02-25 19:51:00

I can see why someone might argue that 0.9 + 0.09 + 0.009 + ... only approaches 1. But with basic point set topology, we can easily prove that it actually is 1.

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1 (I have not seen anyone dispute that 0.9 + 0.09 + 0.009 + ... approaches 1, and this is basically what this sentence says). But S is closed, and thus contains all of its accumulation points. Then 1 ∈ S.

It should be clear that 0.999... is the element of S that is equal to 1.

Last edited by Zhylliolom (2007-02-25 21:31:08)

George,Y · 2007-02-25 20:03:51

Why is S closed? Is it based on the definition of Reals?
Frankly I abhor defining a number as an infinite number set. Making things more complex or more puzzling doesn't change the essence. 1 is 1, the basic, 2=1+1, not "all" the rational numbers before it. Cantor has defined something like {..., 0.9, ..., 0.99, ..., 0.999, ...}=1, quite equivalent to defining {0.9, 0.99, 0.999,...}=1 or 0.999...=1. Why does everyone have to accept this definition?

Zhylliolom · 2007-02-25 20:27:20

S is closed because its complement is an open set. An open set is a set whose points are all interior points. An interior point to a set A is a point in A such that the all the points in the open ball centered at that point are also in A. In the one-dimensional case we are considering, an open ball is simply an open interval. Also in this one-dimensional case, all open intervals are open sets. There is a theorem which states that the union of any collection of open sets is open. This is why S is closed in my proof.

I did not define any number as an infinite set, but instead I defined a number as an element of an infinite set. Is there something wrong with this? Every real number is an element of an infinite set... Every integer is an element of an infinite set... How is this any different?

Last edited by Zhylliolom (2007-02-25 21:30:53)

George,Y · 2007-02-25 21:01:07

Zhylliolom wrote:

Also, 1 is an accumulation point (limit point) of S...But S is closed, and thus contains all of its accumulation points. Then 1 ∈ S.

Why? Is it based on the continum of the Reals? or Dedekind's cut?
I guess it's from a theorem stating "Any close number set includes its limit point, either its sup or its inf". But this theorem is a part of Real Analysis, and a natural derivative of Cantor's definition of a real Number as the Set of all the rationals Before it. So basically if one rejects Cantor's definition and accept only rationals, this proof can also be rejected.

Last edited by George,Y (2007-02-25 21:34:35)

Zhylliolom · 2007-02-25 21:27:49

Yes, it can be considered an effect of the completeness property of the reals: every nonempty set of real numbers with an upper bound has a supremum in R. The set S I used is bounded above since every element is less than or equal to 1 (don't think I am implying that 1 must be in the set here, although I did prove it, this is just the definition of an upper bound: for any s in S, s ≤ 1, the ≤ is in the definition. In general the upper bound is not necessarily an element of the set), and in fact the supremum is 1. A property of the supremum is that for any r > 0, there is an s[sub]r[/sub] ∈ S such that sup S - r < s[sub]r[/sub]. Since sup S is 1 in our case, we have that 1 - r < s[sub]r[/sub] ≤ 1. This means s[sub]r[/sub] ∈ B(1, r) = (1 - r, 1 + r). In case you need this defined, an accumulation point (also called a limit point) of a set A is a point x such that every open ball B(x, r) contains a point in A distinct from x for any radius r. We see from the argument several sentences ago that 1 is by definition an accumulation point of S. There is another theorem of point set topology that states a closed set contains all of its accumulation points. This is why it follows that 1 ∈ S. If you need any of the theorems I am referencing explicitly stated and proven, just ask.

Last edited by Zhylliolom (2007-02-25 21:30:17)

Math Is Fun Forum

#401 2007-02-24 18:47:48

Re: 0.9999....(recurring) = 1?

#402 2007-02-25 01:03:27

Re: 0.9999....(recurring) = 1?

#403 2007-02-25 01:41:10

Re: 0.9999....(recurring) = 1?

#404 2007-02-25 01:54:28

Re: 0.9999....(recurring) = 1?

#405 2007-02-25 02:21:30

Re: 0.9999....(recurring) = 1?

#406 2007-02-25 02:42:20

Re: 0.9999....(recurring) = 1?

#407 2007-02-25 04:26:52

Re: 0.9999....(recurring) = 1?

#408 2007-02-25 04:28:21

Re: 0.9999....(recurring) = 1?

#409 2007-02-25 08:28:32

Re: 0.9999....(recurring) = 1?

#410 2007-02-25 08:56:13

Re: 0.9999....(recurring) = 1?

#411 2007-02-25 15:26:19

Re: 0.9999....(recurring) = 1?

#412 2007-02-25 15:31:14

Re: 0.9999....(recurring) = 1?

#413 2007-02-25 16:12:25

Re: 0.9999....(recurring) = 1?

#414 2007-02-25 18:22:15

Re: 0.9999....(recurring) = 1?

#415 2007-02-25 18:26:21

Re: 0.9999....(recurring) = 1?

#416 2007-02-25 18:44:15

Re: 0.9999....(recurring) = 1?

#417 2007-02-25 18:46:14

Re: 0.9999....(recurring) = 1?

#418 2007-02-25 19:05:39

Re: 0.9999....(recurring) = 1?

#419 2007-02-25 19:34:42

Re: 0.9999....(recurring) = 1?

#420 2007-02-25 19:45:50

Re: 0.9999....(recurring) = 1?

#421 2007-02-25 19:51:00

Re: 0.9999....(recurring) = 1?

#422 2007-02-25 20:03:51

Re: 0.9999....(recurring) = 1?

#423 2007-02-25 20:27:20

Re: 0.9999....(recurring) = 1?

#424 2007-02-25 21:01:07

Re: 0.9999....(recurring) = 1?

#425 2007-02-25 21:27:49

Re: 0.9999....(recurring) = 1?

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