Does 0.999... Equal 1?

[vlad321]

Menago KF said:

steven787 said:

vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

"n" is a theoretical number, so the proof is invalid.

It's a theoretical number, but one that exists in reality, that belongs to the so called "reality numbers".It can range from one to any postive number known that is not a fracture,or anything as strange (don't have the vocab to say what i want, but you understand i hope). His equation is valid.

 

You mean complex, as in it includes sqrt(-1)?

Anyways, there are a lot more theoretical things than will blow your mind than a simple sigma notation statement. Did you know that some infinities are greater than others? While the real numbers are infinite, the numbers which are not whole are more numerous than the reals?

The question of infinity was put to rest when Newton came along with his epsilon-delta definition.