The proof for the property ⌊−x⌋ = −⌈x⌉ is established through the definitions of the floor and ceiling functions. By applying the relationship x – 1 < ⌊x⌋ ≤ x, it follows that −x + 1 > −⌊x⌋ ≥ −x, leading to the conclusion that −⌊x⌋ = ⌈−x⌉. The discussion highlights the importance of considering both integer and non-integer cases, ultimately confirming that only one integer can exist in the range defined by the ceiling function, thus validating the proof.
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Yes, that's a valid relationship regardless of the sign of x.twoski said:Can i use the property ⌊x⌋ = n, x = n + m where 0 <= m < 1
That wouldn't be very satisfactory. You really need to prove it for all m in the range, not just special values.I think there would be 2 cases to this proof.
Case 1: m = 0, the lower bound.
Case 2: m = 1/2.
m < 1, so just two cases (as you wrote). But it probably isn't necessary to break it into separate cases for m at all. Separate cases for x +ve/-ve might be useful.Case 3: m = 1, the upper bound.
I hope you mean "and one where it is not a whole number"; otherwise you're leaving out the irrationals. Anyway, as I said, I doubt it's useful to break it up that way. Go ahead and try it, but do the non-integer case first; I expect you'll find it handles both.twoski said:wouldn't there be 2 cases, one where x is a whole number and one where it is a rational number?
The above is correct, but the last step could be made clearer:twoski said:Proof for ⌊−x⌋ = −⌈x⌉
By the definition of the floor function, x – 1 < ⌊x⌋ ≤ x
It follows that −x + 1 > −⌊x⌋ ≥ −x
Next, let ⌈x⌉ = n where x ≤ n < x + 1
We note that -x ≤ −⌊x⌋ < -x + 1
Clearly, −⌊x⌋ = ⌈-x⌉
It is the second part of the OP.the result of the proof isn't what we set out to prove, the signs are misplaced.