Let k be some integer, and x be some real number. Then kx is a real number.

The max distance any real number is from the nearest integer is (1/2) .

∴ there is some integer M such that M-(1/2) ≤ kx ≤ M+(1/2).Case 1: M-(1/2) ≤ kx < M

Then $\displaystyle \ll kx\gg\,=M-kx \,.$

M-(1/2) ≤ kx < M → -M < -kx ≤ (1/2) - M → k - M < k - kx ≤ (1/2) + k - M

Since k - M is the closest integer to k - kx = k(1 - x), we have $\displaystyle \ll k(1-x)\gg\,=k-kx-(k-M)=M-kx \,.$

For Case 1: $\displaystyle \ll kx\gg\,=\ll k(1-x)\gg \,.$

Case 2: kx = M (This case should be easy.)

Case 3: M < kx ≤ M+(1/2) (This case proceeds much like Case 1.