Uncle_Milton

Weird. I can definitely say he has a very good selling and buying rep on here. I'd keep trying to get a hold of him.

remember .333... is equal to 1/3 Or is it not, but functionally defined as such?

Not if I start writing my set first:poke:

Yes it does. There have already been about 4 or 5 proofs/demonstrations that show this. It seems your flaw of thinking is that with 0.9999.... the 9s end at some point. Well they don't. They keep on going for infinity. If they did end at some point, then yes, you could say "it gets close but does not equal 1." But since they never end, it is equal to 1. Look, no one ever said that it would be easy.... But there is an infinite number of real numbers between 1.0 and 0.9999...

Let me sum this up real nice and quick lol

X = {x|x e/e x}

roof 1.999 = 2;

Quality vs. Quantity

Quoted for fail. No, it's right. Multiplying by the empty set, an then comparing it to something is not legal in any of the fundamental logics.

This is how .999_ = 1 x = .999_ 10x = 9.999_ 10x - x = (9.999_ - .999_) 9x = 9 x = 1 .999_ = 1 To further prove this, look at 3/3. .333_ = 1/3, and 3/3 = 1. Therefore, .999 = 1. Now, it can be argued that 1=2. 0 x 1 = 0 0 x 2 = 0 So than this must be true: 0 x 1 = 0 x 2 Similarly: 0/0 x 1 = 0/0 x 2 Simplified, 1 = 2 Multiplying by zero is a no no in logic, and the first one is a mistake of significant decimals.

not anymore. I'm {censored}ing done as of last month. Double major in who gives a {censored}, with a minor in I don't care anymore. I GET A {censored}ING BREAK. in all honesty in eight months I'll be back at it for six + more years. {censored}ing PH.D. programs are long.