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...

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...?

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!

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?

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