Is 1 = 0.9999......

[JupiterJones]

Originally posted by: SilentRunning

Originally posted by: silverpig

I read the link, I know the difference between giving a value and finding what it equals.

The value given to 0.999... is the limit of the sum, this limit equals 1.

So you get, I thought you were one of the ones who thought 1 = 0.99999...

They are not equal but the are given the same value. That is the concept that people here do not seem to grasp. That is why my tag line is what it is. Yes 1 and 0.9999... are the same value for mathematical purposes due to the approximation for 0.9999..., but they are not truely equal.

They thread asked only one question are the two numbers equal, not do we treat them as the same value.

WRONG! 1 = .99999... because there exist no number between the two values. They are EXACTLY equal - by definition of equality.

 

[HomeBrewerDude]

here for information on the proof

 

[SilentRunning]

Originally posted by: JupiterJones

Originally posted by: SilentRunning

Originally posted by: silverpig

I read the link, I know the difference between giving a value and finding what it equals.

The value given to 0.999... is the limit of the sum, this limit equals 1.

So you get, I thought you were one of the ones who thought 1 = 0.99999...

They are not equal but the are given the same value. That is the concept that people here do not seem to grasp. That is why my tag line is what it is. Yes 1 and 0.9999... are the same value for mathematical purposes due to the approximation for 0.9999..., but they are not truely equal.

They thread asked only one question are the two numbers equal, not do we treat them as the same value.

WRONG! 1 = .99999... because there exist no number between the two values. They are EXACTLY equal - by definition of equality.

You are presuming that you can do math operations on a nonterminating decimal number.

In spite of the apparent similarity, finite and infinite quantities are separated by a deep and wide gulf. Mathematicians overlooked this gulf until the 19th Century. Ignoring the dangers, they treated infinite objects in the same way as finite ones and sometimes obtained absurd results. In the 19th Century, the way for dealing with the infinite was discovered gradually and reliable bridges were erected across the separating gulf. Let us walk across one of these bridges.

Note that a terminating decimal fraction is in no way different from an ordinary fraction; the only difference is the notation. The fraction 0.33 has the numerator 33 and the denominator 100. But what is the numerator of the non-terminating fraction 0.333···? We do not yet know the answer to this question, which indicates clearly that a non-terminating decimal fraction does not have the meaning carried by a terminating one. N.N. Luzon used to say that drawing a symbol 0.333··· does not impart a meaning to this symbol. It remains but a pattern. However, we can give this symbol a meaning.

The sum 1 + 1/2 + 1/4 has a meaning, because we can calculate it by successive additions: 1 + 1/2 = 3/2, 3/2 + 1/4 = 7/4. But we cannot determine the infinite sum 1 + 1/2 + 1/4 + 1/8 + ··· by this method, because the process of consecutive additions will never terminate. And this is not a technicality, but a basic obstacle. It would be frivolous to hide behind the argument that successive additions of terms in 1 + 1/2 + 1/4 + 1/8 + ··· yield approximate values of the infinite sum. You cannot look for what does not exist! The meaning of infinite sums must first be defined; only afterwards can we speak of approximate values of such sums.




Relinked for JupiterJones Edumacation

 

[Haircut]

SilentRunning, from your link it says:

It would be frivolous to hide behind the argument that successive additions of terms in 1 + 1/2 + 1/4 + 1/8 + ··· yield approximate values of the infinite sum. You cannot look for what does not exist! The meaning of infinite sums must first be defined

The inifinite sum is defined as the limit of the sum to n as n -> infinity.
See here.
From this definition 0.999... = 1

I agree with you when you say that the two are the same value for mathematical purposes, this also must mean they are equal in value. They occupy the same point on the real line and so must be equal.

 

[SilentRunning]

Originally posted by: Haircut
SilentRunning, from your link it says:

It would be frivolous to hide behind the argument that successive additions of terms in 1 + 1/2 + 1/4 + 1/8 + ··· yield approximate values of the infinite sum. You cannot look for what does not exist! The meaning of infinite sums must first be defined

The inifinite sum is defined as the limit of the sum to n as n -> infinity.
See here.
From this definition 0.999... = 1

I agree with you when you say that the two are the same value for mathematical purposes, this also must mean they are equal in value. They occupy the same point on the real line and so must be equal.

Actually 0.99999... does not occupy a point on the number line. What we are trying to do is define a point which approximates the infinite sum. That point is 1.

Cantor's Continuity Axiom If an infinite sequence of segments is given on a straight line such that
1. each next segment is nested within the preceding one
2. the length of the segments tends to zero, then there exists a unique point which belongs to all the segments.



So for 0.99999... we have the nested segments:

[0.9;1.0] , [0.99;1.00], [0.999;1.000], [0.9999;1.0000] ......

We see that 1 belongs to all of the segments, so we give it a value of 1.

 

[Haircut]

What we are trying to do is define a point which approximates the infinite sum. That point is 1.

No, from the definition above, the point 1 is the point that is equal to the infinite sum

Cantor's Continuity Axiom If an infinite sequence of segments is given on a straight line such that
1. each next segment is nested within the preceding one
2. the length of the segments tends to zero, then there exists a unique point which belongs to all the segments.

I fail to see how that is applicable here.
Even if we take [0.9,1) , [0.99,1), [0.999,1) ... then as these are open intervals we cannot find a single point which belongs to all of the intervals.

 

[SilentRunning]

Originally posted by: Haircut

What we are trying to do is define a point which approximates the infinite sum. That point is 1.

No, from the definition above, the point 1 is the point that is equal to the infinite sum

Cantor's Continuity Axiom If an infinite sequence of segments is given on a straight line such that
1. each next segment is nested within the preceding one
2. the length of the segments tends to zero, then there exists a unique point which belongs to all the segments.

I fail to see how that is applicable here.
Even if we take [0.9,1) , [0.99,1), [0.999,1) ... then as these are open intervals we cannot find a single point which belongs to all of the intervals.

So how do you find a limit if you don't want to use Cantor's Continuity Axiom? Don't tell me something is defined as the limit of something else. Tell me how you find the limit.

 

[Haircut]

OK, I think I was trying to get something different out of that. I agree with that if we use a closed interval then the limit is 1.

I still don't see how this shows that 0.999... doesn't occupy a point on the real line though, in any of the intervals [0.9,1] , [0.99,1] etc. the lowest number in the interval is always < 0.999...

 

[silverpig]

Originally posted by: SilentRunning

Originally posted by: silverpig

I read the link, I know the difference between giving a value and finding what it equals.

The value given to 0.999... is the limit of the sum, this limit equals 1.

So you get, I thought you were one of the ones who thought 1 = 0.99999...

They are not equal but the are given the same value. That is the concept that people here do not seem to grasp. That is why my tag line is what it is. Yes 1 and 0.9999... are the same value for mathematical purposes due to the approximation for 0.9999..., but they are not truely equal.

They thread asked only one question are the two numbers equal, not do we treat them as the same value.

1 = 0.999...

simple as that really.

 

[silverpig]

Originally posted by: SilentRunning

Actually 0.99999... does not occupy a point on the number line. What we are trying to do is define a point which approximates the infinite sum. That point is 1.

No. What some people are trying to do is come up with a sum that approximates the description of the point occupied by 0.999...

0.999... is not an approximation to anything. It's just a number.

That's like saying that 2.000... is an approximation. It's not. It's just a number. S = Sum (i=0..n) 1/2^i is a sum which approximates the value 2 for any finite n, but that's getting us sidetracked.

 

[josphII]

You are presuming that you can do math operations on a nonterminating decimal number.

thats not true at all. the proof that shows that 0.999... = 1 does not do any math operations on 0.999...

They are not equal but the are given the same value.

that makes no sense. if 0.999... is ever encountered in a problem, you can simply substitue 1 in its place, becasue 0.999... equals 1

 

[josphII]

]Actually 0.99999... does not occupy a point on the number line. What we are trying to do is define a point which approximates the infinite sum. That point is 1.

what infinite sum?? 0.999... is a number and most definetely exists on a number line - as 1.

and if your referring to the following infinite sum:
S=Sum 9/10^n, for n = 1 to inf

that sum equals 1. 1 is not an approximation in any way.

 

[MadRat]

Here we go again with voodoo to make a 1-infinity, 2-infinity, 3-infinity... LOL

No, you cannot move the decimal place (division by 10) without affecting the last "9" of the value, a last "9" that doesn't exist on .999...

 

[bleeb]

Lets try to get this thread to reach 1000 posts!~

 

[RossGr]

Originally posted by: MadRat
Here we go again with voodoo to make a 1-infinity, 2-infinity, 3-infinity... LOL

No, you cannot move the decimal place (division by 10) without affecting the last "9" of the value, a last "9" that doesn't exist on .999...

Ah... more wisdom from Nadrat,

To repeat what has been said before, the algebraic manipultions are NOT proof that .999...= 1, that is merely a demonstration. Formal proofs of this type do not use such methods, the method is to show, formally, that the difference of 1- .999... =0.

I have already posted such a proof, but you proved incapable of understanding, or appriciating it. I will consider posting it again.

.999... does indeed constitute a infinite sum, but there is NO LIMIT INVOLVED it is 9 times the sum n= 1 to infinity of .1^n, the upper limit is not approaching infinity it IS infinity.

 

[TuxDave]

Originally posted by: bleeb
Lets try to get this thread to reach 1000 posts!~

you mean 999.9999.... posts

 

[NovaTone]

Originally posted by: TuxDave

Originally posted by: bleeb
Lets try to get this thread to reach 1000 posts!~

you mean 999.9999.... posts

lol!

 

[Pastore]

The best argument I found in the thread was that they must be equal because there is no number between 0.99..... and 1.

 

[bleeb]

Thug LIfe, Outlawww.com, Wesyde! The rate at which people are posting to this thread is slowing and I still don't think the issue has been resolved beyond a doubt yet.

 

[GroundZero]

depends on how you look at it.
0.999... is as close to 1 as you can get mathmatically, but it will still be short of 1 by the smallest amount imaginable.

 

[RossGr]

Why does the phrase "Casting Pearls before swine" come to mind?

 

[silverpig]

Originally posted by: GroundZero
depends on how you look at it.
0.999... is as close to 1 as you can get mathmatically, but it will still be short of 1 by the smallest amount imaginable.

What is this amount?

 

[SilentRunning]

Originally posted by: silverpig

Originally posted by: GroundZero
depends on how you look at it.
0.999... is as close to 1 as you can get mathmatically, but it will still be short of 1 by the smallest amount imaginable.

What is this amount?

For any n:

1 = (Sum(i = 1..n)[9/10^i]) + 10^(-n)

so for n=3:

1 = (0.9+0.09+.0.009) + 10^ (-3)
1 = 0.999 + 0.001
1 = 1

Let n --> inf

1 = (Sum(i = 1..inf)[9/10^i]) + 10^(-n) (as n--> inf)

But

0.999... = Sum(i = 1..inf)[9/10^i]

Therefore

1-0.999... = [(Sum(i = 1..inf)[9/10^i]) + 10^(-n)] -[Sum(i = 1..inf)[9/10^i]] (as n --> inf)
1-0.999... = [(Sum(i = 1..inf)[9/10^i]) - [Sum(i = 1..inf)[9/10^i]] + 10^(-n) (as n --> inf)
1-0.999... = 10^(-n) (as n --> inf)



Edit fixed this line 1 = (0.9+0.09+.0.009) + 10^ (-3)

 

[RossGr]

1-0.999... = 10^(-n) (as n --> inf)

Ok, now that you have taken this far you need to complete it.

1-0.999... = 10^(-n) (as n --> inf)= 0

Therefore 1= .999....

 

[SilentRunning]

Originally posted by: RossGr

1-0.999... = 10^(-n) (as n --> inf)

Ok, now that you have taken this far you need to complete it.

1-0.999... = 10^(-n) (as n --> inf)= 0

Therefore 1= .999....

Nope it doesn't = 0, it tends toward zero

If you examine the equations the difference is equal to the remainder that one must ignore when dividing 1 by itself to result in the answer 0.999...


Do you understand that infinite is the opposite of finite? As n --> infinity you will never reach it, it will never have a finite value. That is why we have limits.