Proving nearest integer function identity

Question

While trying to express the nearest integer function in terms of the modulo or floor function to make some lunar ephemeris equations more compact, I noticed (using GeoGebra) that $$\lfloor x\rceil=1+\lfloor x-1/2\rfloor$$ Is it true? If so, how would one prove it?

Answer 1

There are four cases. (There may be fewer cases; I've made no attempt to determine if these cases can be handled together.)

• $x$ is an integer. Then $1+\lfloor x-1/2 \rfloor = 1 + (x-1) = x$, as expected.
• $x$ is one-half more than an integer, $n$. Then $1 + \lfloor (n+1/2) - 1/2 \rfloor = 1 + \lfloor n \rfloor = n+1$, which may or not be the behaviour you expect.
• $n$ is an integer and $x \in (n,n+1/2)$. Then \begin{align*} n &< x < n+1/2 \\ n - 1/2 &< x -1/2 < n \\ \lfloor n - 1/2 \rfloor &\leq \lfloor x -1/2 \rfloor < \lfloor n \rfloor \\ n-1 &\leq \lfloor x -1/2 \rfloor < n \\ 1 + (n-1) &\leq 1 + \lfloor x -1/2 \rfloor < 1 + n \\ n &\leq 1 + \lfloor x -1/2 \rfloor < n + 1 \text{,} \end{align*} and since the expression in the middle is an integer, it must be $n$, as expected.
• A similar argument for integer $n$ and $x \in (n-1/2, n)$: \begin{align*} n - 1/2 &< x < n \\ n - 1 &< x - 1/2 < n - 1/2 \\ \lfloor n - 1 \rfloor &\leq \lfloor x -1/2 \rfloor \leq \lfloor n - 1/2 \rfloor \\ n-1 &\leq \lfloor x -1/2 \rfloor \leq n-1 \\ 1 + (n-1) &\leq 1 + \lfloor x -1/2 \rfloor \leq 1 + (n-1) \\ n &\leq 1 + \lfloor x -1/2 \rfloor \leq n \text{,} \end{align*} so we obtain $n$, as expected.