As has been repeatedly stated, that proof is extremely problematic in that most only accept that 1/3 = .333... because they've been conditioned to do so. The equations don't actually explain how or why .000...=1 in any serious way.BiscuitTrouser said:Dont mock these people, this proof is correct. Let me write it differently for you.Sturmdolch said:Third last step, at 2a(a-b) = a(a-b). a = b, so dividing by (a-b) is impossible. Yet they still do. So no.Naheal said:Actually, it didn't :-/ That proof works.
The third last step is still valid, because a-b = 0, so both sides equal 0.
You could just as easily say
532531521215a(a-b) = a(a-b)
532531521215 = 1
which is equally as wrong.
You're doing it wrong, too. At step 4, 9x = 9, that is not true.havass said:If x = 0.999999...
Then 10x = 9.9999...
Therefore, 10x - x = 9
Which implies 9x = 9
Thus, x = 1
x also = 0.99999...
In conclusion, I have just proven 1 = 0.9999...
9x = 8.99999999
Edit: I'm not even a math major, but the amount of ignorant false intellectualism in this thread is about to make me cry.
EDIT: I see what you did there... your saying x = 1 because 0.999 rec = 1. So 9x SHOULD be 8.999
1/3 = 0.333 recurring. times 0.33 recurring by three. You know have 0.9999 recurring logically. However you have 3/3 if written as a fraction. This shows that 3/3 = 0.999 recuring AND 3/3 also = 1. Therefor 0.999 reccuring must also = 1.
No infinites there, just fractions of an irational nature. These are logical. These work. You cannot tell me that 3/3 does not = 1 but that 1/3 = 0.33333 recuring. These are both true. Leading to the logical conclusion that 0.99999 recurring = 1.
You need analytic proofs or proofs from the construction of the real numbers to really shed light on the question. I think every such well-known proof has already been posted and people should really be looking at those rather than trying to cling to these algebraic proofs.