Proof of a nearest-integer inequality

Question

Let $N(y)$ be the nearest-integer function and undefined on half-integers.

For all $r \in \mathbb R$ that are not half-integers, prove $$\forall{\ i \in \mathbb Z}:\left|N(r)-r\right|\leq\left|i-r\right|$$

Does this even need to be "proven" or perhaps only "demonstrated"? Does it follow directly from the typical definition of the nearest-integer function?

Answer 1

Giving a rigorous definition of the nearest-integer function, $$N(r)::=j\in\mathbb Z:\forall i\in\mathbb Z, |j-r|\le|i-r|,$$

it is indeed immediate that $$\forall i\in\mathbb Z, |N(r)-r|\le|i-r|.$$

That definition is not completely valid, as there could be several $j$ that realize the minimum distance (and this is indeed the case for half-integer $r$). But even if we arbitrarily choose for $N(r)$ any $j$ that achieves the minimum, the claimed property holds despite non-uniqueness.

Answer 2

If I had to define a nearest-integer function $N(r)$, one way to do it is this:

First, the distance to the nearest integer:

$D(r) =\min_{n \in \mathbb{Z}} |r-n| $

Then, the integer(s) whose distance is the same as the distance to the nearest integer:

$N(r) =\{n \mid |r-n| = D(r)\} $

Then you have to derive uniqueness, existence, and things like that.

Another way, if you have the integer-part function $I(r)$ defined as the integer $n$ such that $n \le r < n+1$, then you could define $N(r) =I(r)\ if\ r-I(r)< \frac12\ otherwise\ I(r)+1 $ or, alternatively but equivalently, $N(r) = I(r)+I(2(r-I(r))) $.