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Question:

FLoor and ceiling function equivalence?

Define functions H and K from R to R by the following formulas:For all x in R,H(x) = [floor](1) + 1 and K(x) = [ceiling](x)Does H = K? Explain.I hate to ask for homework help, but I just do not get this. At all. Help?

Answer:

Have you tried some values of x? I would start there. For instance, what is H(0) and K(0)?
think floor(x+a million/2) = floor(x). Then noting floor(x) is an integer and using the definition of floor(x+a million/2), we've floor(x) <= x+a million/2 < floor(x) + a million. Now, considering that we continuously have floor(x) <= x < x+a million/2, the above inequality is such as x - floor(x) < a million/2. finally, this inequality has as answer the union of [m, m+a million/2) the place m is any integer. all the stairs I made are trivially reversible (the two by using definition or I gave a proof), so as that union is likewise the answer set for floor(x+a million/2) = floor(x). --- you are able to very in fact mimic the above evidence to get one for the ceiling, yet i will make it easier to gain this.
the floor of a number x is the largest integer not exceeding x. the ceiling of a number x is the least integer x is not greater than. if x is NOT an integer, then floor(x) + 1 = ceiling(x). but what if x IS an integer. if x = 1, floor(x) = 1, because 1 isn't bigger than 1, so floor(1) + 1 = 2 but ceiling(1) = 1, so in this case, H(x) is not the same as K(x). you might want to look at the graphs here: en.wikipedia.org/wiki/Floor_and_c...

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