Replicative Function of x minus Floor of x is Replicative/Lemma

Theorem

Let $x \in \R$.

Suppose $x - \floor x < \dfrac 1 n$.

Then:

$\floor {x + \dfrac k n} = \dfrac {\floor {n x} } n$

for any $0 \le k \le n - 1$.

Proof

We have $n x < n \floor x + 1$.

By Number less than Integer iff Floor less than Integer:

$\floor {n x} < n \floor x + 1$

Thus $\floor {n x} \le n \floor x$.

From definition of floor function:

$n x \ge n \floor x$

By Number not less than Integer iff Floor not less than Integer:

$\floor {n x} \ge n \floor x$

Therefore $\floor {n x} = n \floor x$ and thus $\floor x = \dfrac {\floor {n x} } n$.

Now we show $\floor {x + \dfrac k n} = \floor x$ for $0 \le k \le n - 1$.

We have:

$\floor x \le \floor {x + \dfrac k n} \le \floor {x + \dfrac {n - 1} n}$

We also have:

$x + \dfrac {n - 1} n < \floor x + \dfrac 1 n + \dfrac {n - 1} n = \floor x + 1$

By Number less than Integer iff Floor less than Integer:

$\floor {x + \dfrac {n - 1} n} < \floor x + 1$

This gives:

$\floor {x + \dfrac {n - 1} n} \le \floor x$

Combining the above:

$\floor {x + \dfrac k n} = \floor x$

Therefore $\floor {x + \dfrac k n} = \floor x = \dfrac {\floor {n x} } n$.

$\blacksquare$