[Dersan00]

The proof I've always seen associated with this involves fractions.

(1/3) = .3333...
3 x (1/3) = (3/3) = 1
3 x .3333... = .9999...

(3/3) = .9999....

Who the hell invented math anyways? I mean subtraction is the addition of negative numbers, division is multiplication of fractions, and all math can be done with just 1's and 0's.

CRAZY.

[juanrga]

let c = 0.99999...

10c = 9.9999...

10c - c = 9.9999... - 0.99999...

9c = 9

c = 1

0.99999... = 1

Fastest proof

A=1; B=0.99999...

A - B = 0.00000...

[Benzene Martini]

In theory, no.

In reality, it depends.

It is all relative.

[SirReal]

You can't make that argument since the series of 9's after the decimal point is infinite.

[Benzene Martini]

You can't make that argument since the series of 9's after the decimal point is infinite.

Yes you can. You just have to know how to present the data.

[SirReal]

But 1/3 is = .33333, it is the infinite sum of:

Σ(n=1 ∞) (3/10^n)  right?

So doesn't that mean that the limit of this series will indeed approach 1/3, and then the math is valid?  I think the problem is that the series is being treated as a number.  However, I am not a mathematician.  I will have to consult my calculus 2 professor from several terms ago, she is quite literally a genius who does research on algebraic topology.  Perhaps she will be able to shed light on this dilemma