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Question:

Combinatorics-Floor and Ceiling Problem?

Combinatorics-Floor and Ceiling Problem?

Answer:

If x=2k, k∈ℤ, ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = ceiling[2k + 1/ 2] - ceiling[k + 1/4] + floor[k + 1/4] = = (2k+1) - (k+1) + k = 2k = x = floor(x) = ceiling(x). ---------- If x=2k+1, k∈ℤ, ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = ceiling[2k + 1] - ceiling[k + 1/2] + floor[k + 1/2] = = (2k+1) - (k+1) + k = 2k = x = floor(x) = ceiling(x). ---------- If 2k < x ≤ 2k+1/2, k∈ℤ, = 2k+1/2 < (2x+1)/2 = x+1/2 ≤ 2k+1 = ceiling[(2x+1)/2] = 2k+1 and k+1/4 < (2x+1)/4 = x/2+1/4 ≤ k+1/2 = ceiling[(2x+1)/4] = k+1 floor[(2x+1)/4] = k Then ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = (2k+1) - (k+1) + k = 2k = floor(x). ---------- If 2k+1/2 < x < 2k+1, k∈ℤ, = 2k+1 < (2x+1)/2 = x+1/2 < 2k+3/2 = ceiling[(2x+1)/2] = 2k+2 and k+1/2 < (2x+1)/4 = x/2+1/4 < k+3/4 = ceiling[(2x+1)/4] = k+1 floor[(2x+1)/4] = k Then ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = (2k+2) - (k+1) + k = 2k+1 = ceiling(x). ---------- If 2k+1 < x ≤ 2k+3/2, k∈ℤ, = 2k+3/2 < (2x+1)/2 = x+1/2 ≤ 2k+2 = ceiling[(2x+1)/2] = 2k+2 and k+3/2 < (2x+1)/4 = x/2+1/4 ≤ k+1 = ceiling[(2x+1)/4] = k+1 floor[(2x+1)/4] = k Then ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = (2k+2) - (k+1) + k = 2k+1 = floor(x). ---------- If 2k+3/2 < x < 2k+2, k∈ℤ, = 2k+2 < (2x+1)/2 = x+1/2 < 2k+5/2 = ceiling[(2x+1)/2] = 2k+3 and k+1 < (2x+1)/4 = x/2+1/4 < k+5/4 = ceiling[(2x+1)/4] = k+2 floor[(2x+1)/4] = k+1 Then ceiling[(2x + 1) / 2)] - ceiling[(2x + 1) / 4] + floor[(2x + 1) / 4] = = (2k+3) - (k+2) + (k+1) = 2k+2 = floor(x).

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