Prove that the floor(2x)=floor(x) floor(x 1/2)i need the proof with cases.
Suppose that we split up x into its integer part y and its decimal part z (for example, we can split 3.6 up into 3 and 0.6). Then, we have: x = y z. Then: floor(2x) = floor(2y 2z), floor(x) = floor(y z), and floor(x 1/2) = floor(y z 1/2). For 0 < z < 1, floor(y z) = y for all integers y (y is defined to be an integer, so this is fine). As for floor(2y 2z) and floor(y z 1/2), we need to take cases. Now, if 0 < z < 0.5, then 0 < 2z < 1 and 0 < z 1/2 < 1 and so: 2y < 2y 2z < 2y 1 and y < y z 1/2 < y 1, which implies that 2x is some number between 2y and 2y 1 and that x 1/2 is some number between y and y 1, which implies: floor(2x) = x and floor(x 1/2) = x, so floor(2x) = floor(x) floor(x 1/2) when x is some integer plus a decimal between 0 and 0.5. As for the case 0.5 < z < 1, 1 < 2z < 2 and 1 < z 1/2 < 1.5, so: 2y 1 < 2y 2z < 2y 2 and y 1 < y z 1/2 < y 1.5, yielding floor(2y 2z) = 2y 1, floor(y z 1/2) = y 1, and floor(y z) = y, and so: floor(2y 2z) = floor(y z 1/2) floor(y z) == floor(2x) = floor(x) floor(x 1/2). Combining these two cases proves floor(2x) = floor(x) floor(x 1/2) when x is not an integer. To prove the case when x is an integer, notice that: floor(2x) = 2x, floor(x) = x, and floor(x 1/2) = x, when x is an integer, which shows that floor(2x) = floor(x) floor(x 1/2) when x is an integer. Combining these three cases yields floor(2x) = floor(x) floor(x 1/2) for all x (as required). I hope this helps!
Using the notation [x] for the floor(x) and correcting the equation, you need to show [2x] = [x] + [x+1/2]. By the definition of the floor function, we always have [x] ≤ x < [x] + 1 so our two cases are made just by splitting this inequality into two: Case 1: [x] ≤ x < [x] + 1/2 Then we have 2[x] ≤ 2x < 2[x] + 1 [x] ≤ [x] + 1/2 ≤ x + 1/2 < [x] + 1. In other words, this shows that for this case, [2x] = 2[x] = [x] + [x] = [x] + [x + 1/2] Case 2: [x] + 1/2 ≤ x < [x] + 1 Then we have 2[x] + 1 ≤ 2x < 2[x] + 2 [x] + 1 ≤ x + 1/2 < [x] + 3/2 < [x] + 2. In other words, this shows for this case, [2x] = 2[x] + 1 = [x] + ([x] + 1) = [x] + [x + 1/2] Since in both cases, we can derive the same equation, the equation must be true: [2x] = [x] + [x + 1/2]
