# Floor function proof

## Homework Statement

Note: [x] denotes the floor of x.

Prove that [3x] = [x] + [x + 1/3] + [x + 2/3]

## The Attempt at a Solution

Let x = n + E, where n is an integer and 0≤ E < 1. We have three cases:

Case 1: 0 ≤ E < 1/3
3x = 3n + 3E and [3x] = 3n, since 0 ≤ 3E < 1.
[x + 1/3} = n, since x + 1/3 = n + (1/3 + E) and 0 ≤ 1/3 + E < 1.
[x + 2/3} = n, since x + 2/3 = n + (2/3 + E) and 0 ≤ 1/3 + E < 1.
Thus, [3x] = 3n, and [x] + [x + 1/3] + [x + 2/3] = n + n + n = 3n.

Case 2: 1/3 ≤ E < 2/3
3x = 3n = 3E = (3n + 1) + (3E -1) and [3x] = 3n + 1, since 0 ≤ 3E - 1 < 1.
[x + 1/3] = n, since x + 1/3 = n + (1/3 + E) and 0 ≤ 1/3 + E < 1.
[x + 2/3] = [n + (2/3 + E)] = [n + 1 + (E - 1/3)] = n + 1, since 0 ≤ E - 1/3 < 1.
Thus, [3x] = 3n + 1, and [x] + [x + 1/3] + [x + 2/3] = n + n + (n + 1) = 3n + 1.

Case 3: 2/3 ≤ E < 1
3x = 3n = 3E = (3n + 2) + (3E -2) and [3x] = 3n + 2, since 0 ≤ 3E - 2 < 1.
[x + 1/3] = [n + (1/3 + E)] = [n + 1 + (E - 2/3)] = n + 1, since 0 ≤ E - 2/3 < 1.
[x + 2/3] = [n + (2/3 + E)] = [n + 1 + (E - 1/3)] = n + 1, since 0 ≤ E - 1/3 < 1.
Thus, [3x] = 3n + 2, and [x] + [x + 1/3] + [x + 2/3] = n + (n + 1) + (n + 1) = 3n + 2.

How does this look? I just want to make sure I'm not over looking something.

Thanks.

Dick