0.99999........

Michaelrsh

0.9999999999999... to infinity.

Should this be considered as 1 or should it be considered as forever heading towards 1.

Podge_irl

It is one. They're the exact same thing.

LeixlipRed

To say they're the exact same thing is not exactly correct as they're not equal. It's the closest to 1 you can get without ever reaching it

gamblitis

It's equivilent to but not equal to 1.

Michael Collins

LeixlipRed wrote: »

To say they're the exact same thing is not exactly correct as they're not equal. It's the closest to 1 you can get without ever reaching it

Is it not actually considered just simply equal to one?

The classic proof:

x = 0.999...
10x = 9.99999

10x-x = 9
9x = 9
x=1

I'm not suggesting this is a rigorous proof, but i don't see anything wrong with it!

Here's what wiki says

Clinker

That Wikipedia article does a very thorough job showing that 0.999... = 1 in lots of different ways!

It struck me to think about what 1 - 0.9999... would be: clearly

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

Longboard

Michael Collins wrote: »

Is it not actually considered just simply equal to one?

The classic proof:

x = 0.999...
10x = 9.99999

10x-x = 9
9x = 9
x=1

I'm not suggesting this is a rigorous proof, but i don't see anything wrong with it!

Here's what wiki says

The devil is in the detail.

10x-x = 8.999........9991
9x = 8.999........9991
x = .999........99999

Clinker

Longboard wrote: »

The devil is in the detail.

10x-x = 8.999........9991
9x = 8.999........9991
x = .999........99999

Your argument only works if the sequence of 9s is taken to be finite. But the whole point is that it's infinite!

Michael Collins

Longboard wrote: »

The devil is in the detail.

10x-x = 8.999........9991
9x = 8.999........9991
x = .999........99999

Sorry was in a bit of a rush there in my last post let me show it more clearly:

[latex]\displaystyle x = 0.999\overline{9}[/latex]
[latex]\displaystyle 10x = 9.999\overline{9}[/latex]
[latex]\displaystyle 10x- x = 9x = 9.999\overline{9} - 0.999\overline{9} = 9 [/latex]
[latex]\displaystyle x = 1[/latex]

Where the bar above a digit means that digit recurs ad infinitum (sometimes done using a period either e.g. [latex] 9.999\dot{9} [/latex])

(Incidentally is there any way to use more advanced equation enviroments such as "eqnarray" with LateX on boards?)

Longboard

I'm aware of the concept that .9999999~ is infinty "long".

However, Imagine 1 as being the peak of a mountain. moving down one side goes to .9 and moving down the other goes to 1.1. No matter how small a distance you move, (inifintesmally small) you will be off the "true" point (1). What you are arguing is [1 + (1/~)] = 1 = [1 - (1/~)]. I am arguing that when moving away from 1, there has to be a first point where that value is not equal to 1. This point is 0.9999~9 or 1.000~1 .

If it is not at this point, then can you define the first number ether side not equal to 1?

bnt

Well:

• 1/3 = 0.33333333 ...
• 2/3 = 0.66666666 ...
• 3/3 = 0.99999999 ... and 3/3 = 1.

The Wikipedia page has the following analytical proof:

[latex]\displaystyle
0.999\ldots = \lim_{n\to\infty}0.\underbrace{ 99\ldots9 }_{n} = \lim_{n\to\infty}\sum_{k = 1}^n\frac{9}{10^k} = \lim_{n\to\infty}\left(1-\frac{1}{10^n}\right) = 1-\lim_{n\to\infty}\frac{1}{10^n} = 1.[/latex]

LeixlipRed

You learn something new everyday, was never aware of the proof they were equal. Can't believe I've never come across that or considered it before *looks shamed*

red_fox

Longboard wrote: »

I'm aware of the concept that .9999999~ is infinty "long".

However, Imagine 1 as being the peak of a mountain. moving down one side goes to .9 and moving down the other goes to 1.1. No matter how small a distance you move, (inifintesmally small) you will be off the "true" point (1). What you are arguing is [1 + (1/~)] = 1 = [1 - (1/~)]. I am arguing that when moving away from 1, there has to be a first point where that value is not equal to 1. This point is 0.9999~9 or 1.000~1 .

If it is not at this point, then can you define the first number ether side not equal to 1?

If you assume that there's a number just before 1 (call it l, for less than) then you have a pair of real numbers such that the open interval (l,1) is empty, but since l<1 and R is dense this can't happen, for example the average, (l+1)/2, is in there. (and in fact there's a bijection between (l,1) and R for any l<1).

Grudaire

Another way to think of it might be that the Real Numbers form a dense set. ie between any two real numbers there are infinitly many more real numbers.

However 1, and .99999~ don't have this property...

Fringe

Longboard wrote: »

I'm aware of the concept that .9999999~ is infinty "long".

However, Imagine 1 as being the peak of a mountain. moving down one side goes to .9 and moving down the other goes to 1.1. No matter how small a distance you move, (inifintesmally small) you will be off the "true" point (1). What you are arguing is [1 + (1/~)] = 1 = [1 - (1/~)]. I am arguing that when moving away from 1, there has to be a first point where that value is not equal to 1. This point is 0.9999~9 or 1.000~1 .

If it is not at this point, then can you define the first number ether side not equal to 1?

Infinitesimals don't exist. If you suppose that we have this number x which is the closest point to 1 but not 1 and we take the average ie (x + 1)/2, it's clear then that x < (x + 1)/2 < 1 which is a contradiction.

Longboard

Infintesmal (1/~) don't exist therefore "round off" to 0 or 1. [1/~ = 0 or (1 - (1/~)) = 1 or 0.99~]

x < (x + 1)/2 < 1 when x = (1-(1/~)) = .999~

(1- (1/~)) < ((1-(1/~)+1)/2) < 1
(1-(0)) < ((1-(0)+1)/2) < 1

1 < 1 < 1

Fremen

In the standard construction of the real numbers, 0.9 recurring is equal to one. This is not debatable. If you disagree, then you haven't properly understood the definition of a limit.

However, there are other formalisms in which this may not hold. In a branch of mathematics known as non-standard analysis, one can construct a system of real numbers for which 0.9 recurring is not equal to one.
The wiki page is quite technical, so here's a slightly easier read:
http://u.cs.biu.ac.il/~katzmik/999.html

Eliot Rosewater

I would agree with the others Longboard in that your notion of limit and/or infinity is off.

Its something Ive noticed a bit in my Maths course: people have a general misconception of infinity as just a "really really big number". They then make loads of mistakes by treating it as a finite entity.

I blame this on the Leaving Cert. Every time you use infinity in Secondary School its to get the limit of something as some variable tends to infinity. The problem is that you always get a concrete finite number as an answer, giving the impression that infinity is some concrete and finite quantity.

kev_ofla

Michael Collins wrote: »

Sorry was in a bit of a rush there in my last post let me show it more clearly:

[latex]\displaystyle x = 0.999\overline{9}[/latex]
[latex]\displaystyle 10x = 9.999\overline{9}[/latex]
[latex]\displaystyle 10x- x = 9x = 9.999\overline{9} - 0.999\overline{9} = 9 [/latex]
[latex]\displaystyle x = 1[/latex]

Where the bar above a digit means that digit recurs ad infinitum (sometimes done using a period either e.g. [latex] 9.999\dot{9} [/latex])

(Incidentally is there any way to use more advanced equation enviroments such as "eqnarray" with LateX on boards?)

im just wondering would 10x not be

[latex]\displaystyle 10x = 9.999\overline{9}[/latex]
[latex]\displaystyle 10x- x = 9x = 9.999\overline{9}0 - 0.999\overline{9} = 9.0\overline{0}9 [/latex]
[latex]\displaystyle x = 1.0\overline{0}9 [/latex]


i dont see (even though it does have an infinite amount of 9's) why multiplying it by 10 would mean you would create an extra 9.

Victor

Surely it depends on what you are using it for?

xoxyx

Surely 1/3 = .33333333333333333333~

.33333333333333333333~ x 3 = .99999999999999999999~

But, obviously, if 1/3 = .33333333333333333333~, then .33333333333333333333~ x 3 = 1.

Ergo .99999999999999999999~ = 1.

Davidius

kev_ofla wrote: »

im just wondering would 10x not be

[latex]\displaystyle 10x = 9.999\overline{9}[/latex]
[latex]\displaystyle 10x- x = 9x = 9.999\overline{9}0 - 0.999\overline{9} = 9.0\overline{0}9 [/latex]
[latex]\displaystyle x = 1.0\overline{0}9 [/latex]


i dont see (even though it does have an infinite amount of 9's) why multiplying it by 10 would mean you would create an extra 9.

Well that'd be the thing, it's an unending/infinite sequence of 9s. By having
[latex]\displaystyle 10x = 9.99\overline{9}0[/latex] you'd be basically saying that the sequence will end at some point, hence that would mean there'd have to be a finite numbers of 9s.

As far as I see it 10x doesn't have 'more' 9s than x does seeing as 9 has already has 'infinite' 9, and you can't really have more than an 'infinite' amount seeing as it's a not a real number. Hence there's no extra 9 added in there.

Well that's my understanding of it and I'm always open to correction.

Iwasfrozen

The way I see it is quite simple:

If:
1/3 = 0.3333333333...
2/3 = 0.6666666666...

Then:
3/3 = 0.9999999999... = 1

Thus:
0.9999999999... = 1

Jippo

Longboard wrote: »

The devil is in the detail.

10x-x = 8.999........9991
9x = 8.999........9991
x = .999........99999

Oh dear