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1 != 0.99...

Name: Anonymous 2010-10-15 21:42

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.

Name: Anonymous 2010-10-31 7:11

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.

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