Find all real numbers x satisfying floor of (x+1/2) = floor xandFind all real numbers x satisfying ceiling of (x+1/2) = ceiling x
Suppose floor(x+1/2) = floor(x). Then noting floor(x) is an integer and using the definition of floor(x+1/2), we have floor(x) <= x+1/2 < floor(x) + 1. Now, since we always have floor(x) <= x < x+1/2, the above inequality is equivalent to x - floor(x) < 1/2. Finally, this inequality has as solution the union of [m, m+1/2) where m is any integer. All of the steps I made are trivially reversible (either by definition or I gave an explanation), so that union is also the solution set for floor(x+1/2) = floor(x). --- You can very essentially mimic the above proof to get one for the ceiling, but I'll let you do so.
Sorry, it's been a while since I've done floor & ceiling, but I believe that these functions take a number and either round it down to the integer (floor) or round it up. ie: floor would take 2.1, 2.4, 2.7, 2.9 and make them all equal to 2, and ceiling would make all of those numbers equal to 3. Only difference is if it is an even integer, than it stays that integer. The way I see it, there are NO numbers that satisfy the first one. This is because the floor function automatically lowers a number (takes 3.4 and makes it 3, or also takes -1.4 and makes it -2) so if you add .5 to it, it is automatically going to be higher. If x was 2.3, floor (x) = 2, but 2.3 +.5 = 2.8. These aren't equal For the Ceiling function, it is every integer that ends in .5. IE: say x=1.5. Ceiling (1.5) = 2, and 1.5 + .5 =2 - They are equal. Similarly, if x= - 2.5, Ceiling (-2.5)= -2, and -2.5 + .5 = -2. - they are equal. So I guess the way you'd say it is that the answer is i+.5 where i is the set of all integers. Something like that. Hope that helped...
