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Message: [QUOTE="Lie, post: 3534375"] It's not round up, why? In the usual sense -- when not comparing against round-half-even -- the number range we're talking is from x to lim((x+1)-y)[y -> 0 from the positive side], e.g. 1 to nearly 2 (thus the integer 2 itself is not included in the number range we're talking since it belongs to the next set), then we choose a uniformly distributed samples, i.e. 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, and 1.9. From these samples, we chose which are rounded down and which are rounded up, it happens that 1.0 is rounded down, while 1.5 is rounded up. IF for the sake of argument, you choose the number range to be lim(x + y)[y -> 0 from the positive side] to x + 1, e.g. barely above 1 to 2, then the number sample you use is 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, and 2.0, in this second number range, there are 5 round downs (1.1, 1.2, 1.3, 1.4, 1.5) and 5 round ups (1.6, 1.7, 1.8, 1.9, 2.0), but how logical is it to choose this number range (barely above 1 to 2) against choosing the more natural range (1 to nearly 2): in the correctly chosen number range (1 to nearly 2) all numbers in the form of 1.x (where x is any positive number and 0) is contained within it and there is nothing else in it, but in the second number range (barely above 1 to 2) the number 1.0 is not included while the number 2.0 is contained in it, clearly not a clean separation of numbers in the form of y.x where y is pre-determined and x is variable from other possible values of y. In short, choosing that x.0 is rounded down and x.5 is rounded up is arbitrary but not without a reason. [/QUOTE]

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