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**Anthony.R.Brown****Banned**- Registered: 2006-11-16
- Posts: 516

A.R.B

The Masses! cant Resist!...Fame is Good for the Soul!.................................................................................................................

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Is there anything wrong with this proof?

1/9=.111...

2/9=.222...

3/9=.333...

4/9=.444...

5/9=.555...

6/9=.666...

7/9=.777...

8/9=.888...

9/9=.999...

See the pattern? x/9=x((1/10)+(1/100)+(1/1000)+...). 9/9=1, because x/x=1, but 9/9, as I show above, is .999... Therefore, using substitution, I can conclude that .999...=1.

Is there a flaw? I think so...:/

Linux FTW

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**Kargoneth****Member**- Registered: 2007-08-11
- Posts: 33

simron wrote:

Is there a flaw? I think so...:/

I don't think there is a flaw. I use this to prove that .999... = 1

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

There certainly isn't a flaw in that, but I'd say it's more of a demonstration than a proof.

For example, how do you know that 1/9 = 0.111...?

Why did the vector cross the road?

It wanted to be normal.

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Also:

1-.9=.1

1-.99=.01

1-.999=.001

limit as x goes to infinity (1-x(.1+.01+.001+.0001...))=**0**

limit as x goes to infinity (x(.1+.01+.001+.0001...))=**1***SO THE LIMIT OF .99999... IS 1*

Q.E.D.

I think I know where Anthony is coming from. Infinity is kind of a hard concept for everyone (well... me for sure. It sure took me a while for me to figure it out).

*Last edited by simron (2008-04-03 12:32:49)*

Linux FTW

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**RickIsAnIdiot****Member**- Registered: 2009-06-13
- Posts: 5

For simron

Quote:

Also:

1-.9=.1

1-.99=.01

1-.999=.001

limit as x goes to infinity (1-x(.1+.01+.001+.0001...))=0

limit as x goes to infinity (x(.1+.01+.001+.0001...))=1

SO THE LIMIT OF .99999... IS 1

Q.E.D.

I think I know where Anthony is coming from. Infinity is kind of a hard concept for everyone (well... me for sure. It sure took me a while for me to figure it out).

RickIsAnIdiot

There is NO LIMIT OF .99999... AS 1 ( To Limit .9999... is to Stop the .9's being continuous!...as soon as you make a Limit it will be

contradictory!

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Ack, another ARB sock puppet.

(At least, I think so...)

Linux FTW

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

simron wrote:

Ack, another ARB sock puppet.

(At least, I think so...)

Moderation has been discussing this and our (hmm, perhaps my) best guess is no, it is not.

To be fair, RickIsAnIdiot has a point. To say "0.999..." has a limit is wrong. The increasing sequence of adding 9's:

0.9, 0.99, 0.999, 0.9999, ...

This *sequence* has a limit. And of course, this is what you meant (as context in your post shows).

And I have no idea what you were trying to do here:

limit as x goes to infinity (1-x(.1+.01+.001+.0001...))=0

limit as x goes to infinity (x(.1+.01+.001+.0001...))=1

The limit as x goes to infinity of 1 - x * 0.111... = 0?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**simron****Real Member**- Registered: 2006-10-07
- Posts: 237

Well then, sorry, RickIsAnIdiot, no offense meant then.

I probably should have written that up in LaTeX, I'll try sometime.

Linux FTW

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**RickyIsAnIdiot****Banned**- Registered: 2009-06-15
- Posts: 18

For simron

Quote:

Well then, sorry, RickIsAnIdiot, no offense meant then.

I probably should have written that up in LaTeX, I'll try sometime.

RickyIsAnIdiot

No Problem! so do you now agree your counter Math example is Flawed!...

As I stated... There is NO LIMIT OF .99999... AS 1

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