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Let x be a real number. Show that floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3)

User Jeff
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Final answer:

To show that floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3), we consider three cases for the value of x and show that the equation holds for each case.

Step-by-step explanation:

To show that floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3), we need to apply the property that the floor function rounds a real number down to the nearest integer. Let's consider the value of x in three cases.

Case 1: If x is an integer, then the floors of x, x + 1/3, and x + 2/3 are equal to x. Therefore, floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3) holds.

Case 2: If x is between two consecutive integers, then the floors of x, x + 1/3, and x + 2/3 are equal to the smaller integer. Therefore, floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3) holds.

Case 3: If x is greater than an integer and less than that integer plus 1, then the floors of x, x + 1/3, and x + 2/3 are equal to that integer. Therefore, floor(3x) = floor(x)+floor(x + 1/3) +floor(x + 2/3) holds.

User Vicky S
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