1 != 0.99...

Consider the floor function. It ignores the decimal part (everything right of the decimal point) of a number.

Thus, floor(0.999...) = 0, and if 0.999... = 1, floor(1) = 0.
However, floor(1) = 1, therefore we must conclude that either 1 = 0, or the hypothesis was wrong.

What you're doing is saying this:
floor(x)=x-frac{x}

But what is frac{0.999...}? it's 0.

Thus no contradiction, and what you end up doing is just putting off the question of whether 1=0.999.. by one step.

>>1
You don't need to go through any clever nonsense.  All you have to do is remember that the stardard definitions used by mathemeticians now make it so that ununderstandable infinites are inherent in how they define their idea of numbers.  We get bogged in notational bullshit that doesn't make sense if we try to imagine it in the world of our experience of apples and kittens, which is not the same thing as the wonderland of conceived math.  So, fuck it.

>>1
Your mistake is identifying the reals with decimal representations while allowing infinite sequences of 9's. If you allow infinite sequences of 9's, decimal representations become non-unique. In particular, 0.\bar{9} and 1 represent the same real. This also makes the intuitive mechanism of some functions, like the floor function, break down.

>>3
There are no excessively abstract ideas involved in constructing the reals, and after learning any such construction one understands why 0.\bar{9} = 1. Considering all modern engineering is underpinned by calculus, which is rife with ``notational bullshit that doesn't make sense,'' it seems that perhaps it does make sense after all.

JACKSON 5 GET

>>4
the idea of infinity is the pivotal concept.  it's pretty fuckin' abstract, albeit almost boringly familiar to an experienced student, enough to have long left behind the discomfort of not really knowing what the fuck and continuing on the merry way regardless

>>6
So very true...
Set theory anyone? Absolutely crazy, no one cares.

1/3 + 2/3 = 3/3 = 1

0.333... + 0.666... = 0.999... = 1

same shit

>>8
no it isn't.  I challenge you to add those decimals by hand, and don't skip any steps either you cheater

>>9
mfw I try then realise after 426,748 pieces of paper that I'll never complete the addition

It's been 5 yrs, and you guys still get trolled by this?

>>10
Indeed, it relies upon an abstraction that removes itself from the world of experience, retreated to the world of imagined truths, that abstraction being the idea of the infinite and its paradoxical containment within the finite.

It's very simple. You need to evaluate the expression first, then apply the floor function to it.

0.999... = lim n->infinity of (sum from i=1 to n of (9*10^(-i))) = 1

Thus:
floor(0.999...) = floor(lim n->infinity of (sum from i=1 to n of (9*10^(-i)))) = floor(1) = 1

How do you know that floor(1) = 1? Look at the fucking definition of the floor function:

floor(x) = max{integer z where z <= x}
What is the largest integer that is less than or equal to 1? IT HAPPENS TO BE FUCKING 1!

No need for your philosophical bullshit.

>>13
you used the idea of limit, geoius.  you can't hide from the mind beast we call infinity.  and all intellectual disciplines can be and should be examined for their underlying construction, otherwise you get too wedded to the idea that logical conclusions from assumed statements are the same thing as universal (as if godly) facts

>>14
>implying limits and the concept of infinity aren't well understood by now.

>>13
While your conclusion is correct, the reasoning is not. The question is how to evaluate the expression. You did not show lim n->inf (sum(9*10^-n, n, 1, n)) = 1, which is the essence of the entire proof. I'm going to assume you do not understand how to rigorously use limits.

>>15
no one understands them, stupid

>>16

Let me propose that: sum from i=1 to n of (9*10^(-i)) = 1 - 10^(-n) for all natural n

Show for n=1:
sum from i=1 to 1 of (9*10^(-i)) = 9*10^(-1) = 9/10 =
1 - 10^(-1) = 1 - 1/10 = 9/10

Assume true for n=k, show for k+1 etc:
sum from i=1 to (k+1) of (9*10^(-i))
= ( sum from i=1 to k of (9*10^(-i)) ) + 9*10^(-(k+1))
= 1-10^(-k) + 9*10^(-(k+1))
= 1-10^(-k) + (10-1)*10^(-(k+1))
= 1-10^(-k) + 10^(-k) - 10^(-(k+1))
= 1-10^(-(k+1))

Proof'd that sum from i=1 to n of (9*10^(-i)) = 1-10^(-n)

lim (1-10^(-n)) as n -> infinity
= lim (1-1/10^n) as n -> infinity
= 1 - lim(1/10^n) as n -> infinity
= 1 - 0
= fucking 1

Proved that you are a moron.