If you want a headache at 12am

[MechaPiano]

ok, my last attempt would be to ask you to draw y=1/1-x function and see for yourself that x can never be 1, but it can be 0.999...

if that doesn't help, then I don't know what will

[chris-sharpe999]

A perfectly valid point, but doesn't prove a thing. A vertical asymptote is different from a horizontal one (well, depending on which axis you define as domain and which as range, usually x:domain y:range). In the case of y=1/1-x you place your own restriction on the domain so you don't end up dividing by zero. It is still fine to say that as x->1, y->∞. In the case of y = 0.9(1- 0.1^[x-1])/0.9 = 1- 0.1^[x-1] (adapting the geometric series formula to work on a graph as y in terms of x) you don't have to place any limit on the domain, but as x->∞, 0.1^[x-1] ->0, y->1. So the limit is 0.999... AND 1. And your definately right that we should stop now, we're just going in circles (probably won't happen though).

[dudeqwerty]

graphical methods rely on rounding of infinatly long numbers, what is gonna help you understand that 0.99.. = 1?

[phires]

are you insane? how on Earth did you get 0.999... by multiplying 0.333... with 3? (and please don't tell me you used a regular calculator to do it)

I would have to agree with Mech here.

0.333... * 3 does not equal 0.999...

If you multiple 0.333... * 3 there will ALWAYS still be an infinite amount of 3's that have yet to be multiplied by the 3. 0.999... is an infinte number of 9/(10^x) while 0.333... is an infintie number of 3/(10^x) in the process of being multiplied by 3. They're not the same and never will be.

Saying 0.333... * 3 = 0.999... means you've put a limit on 0.333... & 0.999...

0.333... * 3 = 0.999... will never be true unless at some point you stop multiplying 0.333... * 3 and state that they are the same at THAT POINT.

[chris-sharpe999]

You can multiply an infinite representation. You don't multiply each part as a seperate action, in which case you wouldn't ever have multiplied every term, but the multiplication is one action applied to every term. It is only one action, so you don't have an infinite amount left to multiply because all terms go at once. This is a case where using a fractional representation IS a good way of showing it. Using a different example, so we don't bring the whole 0.999...=1 thing into it:
1/9 = 0.111...
3 * 1/9 = 3 * 0.111...
3/9 = 1/3 = 0.333...
Therefore 3 * 0.111... = 0.333... and that is a perfectly acceptable conclusion.

[yasunobu13]

I would have to agree with Mech here.

0.333... * 3 does not equal 0.999...

If you multiple 0.333... * 3 there will ALWAYS still be an infinite amount of 3's that have yet to be multiplied by the 3. 0.999... is an infinte number of 9/(10^x) while 0.333... is an infintie number of 3/(10^x) in the process of being multiplied by 3. They're not the same and never will be.

Saying 0.333... * 3 = 0.999... means you've put a limit on 0.333... & 0.999...

0.333... * 3 = 0.999... will never be true unless at some point you stop multiplying 0.333... * 3 and state that they are the same at THAT POINT.

How do you people multiply?

Are you saying that, in order to multiply 3 * 0.333... I have to have intermediate steps?

At some point it becomes 0.9333...
Then it becomes 0.99333... and so on?
As though time must pass for any multiplication to occur.

This is ridiculous, what's the highest level math that you and Mecha have had?

I've proved 1 != 0.999... at least once already,

Actually, you failed to prove (or disprove) anything. What you described was a sum of an infinite geometric sequence. A well known example in fact. An example who's sum is, you guessed it, 1.

http://mathworld.wolfram.com/GeometricSeries.html
1/2 + 1/4 + 1/8 + ...

So r = 1/2. The sum of r^1 + r^2 + r^3 + ... is equal to
r / (1 - r) = (1/2) / (1 - (1/2)) = (1/2) / (1/2) = 1


But, my geuss is that you've had very little experience in that area seeing how you recently denied the definition of multiplication.

I'm done here