If 0.9999~ = 1... then what is ⌊0.9999~⌋ ? Doesnt the floor value equal 0, just as integer portion would?

Answer 1

Nope. 0.999999.... is exactly equal to 1, which means that its floor is the same as the floor of 1, namely 1 itself.

Answer 2

No, because the floor is not a continuous function at 1 or any other integer value.

Answer 3

0.9 reoccuring is equal to 1.
A proof of this is can be writen as:
Let x = 0.9 reoccuring......(1)
So 10x = 9.9 reoccuring....(2)
(2) - (1) gives:
9x = 9
So x = 1
So writing 0.9 reoccuring is just a covoluted way of writing 1 as they are exactly the same number. They do not differ by anything, not even a very small amount.
So the floor of 0.9 reoccuring is just the floor of 1 which is 1.

Answer 4

The thing to understand here is that you are dealing with an infinitely long number of 9's, there is no number small enough to add to this chain of 9's that wouldn't make the sum larger than 1. If that doesn't help, think about it like this, we have 0.999.. ( with the 9s repeating forever), and we want to add that last bit to make it one, so we will add1 dividided by infinity, but 1/infinity is 0, making 0.999... the same as 1.

Answer 5

Of course 0.9999 ~=1 but 0.9999 <1

So @0.9999 ~= @ but @0.9999 < @

Then the floor value is @ - 1

Answer 6

Because with 0.99999 you're simple rounding to a position so close it makes little difference in most mathematics.
The function of [x] is to bring the number to the first whole integer that is closer to the negatives (not zero). That means, since 0.99999 isn't REALLY 1, the only whole integer available is 0.