Floor of m+n-1 over n

Theorem

Let $m, n \in \Z$ such that $n > 0$.

Then:

$\floor {\dfrac {m + n - 1} n} = \ceiling {\dfrac m n}$

The identity does not necessarily apply for $n < 0$.

Example

Let $n \in \Z$.

Then:

$\floor {\dfrac {n + 2 - \floor {n / 25} } 3} = \floor {\dfrac {8 n + 24} {25} }$

Proof

First let $n > 0$ as stated.

Suppose $n \divides m$.

Then $m = k n$ for some $k \in \Z$.

It follows that:

$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 - \dfrac 1 n} = k$

and:

$\ceiling {\dfrac m n} = k$

Now suppose $n \nmid m$.

Since $n > 0$, we have $m = k n + r$ for some $k \in\Z$ and $r \in \N$, $0 < r < n$.

Therefore:

$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 + \dfrac {r - 1} n} = k + 1$

and:

$\ceiling {\dfrac m n} = k + 1$

$\Box$

Setting $m = 1, n = -2$ we have:

\(\ds \floor {\dfrac {m + n - 1} n}\)

\(=\)

\(\ds \floor {\dfrac {1 + \paren {-2} - 1} {\paren {-2} } }\)

\(\ds \)

\(=\)

\(\ds \ceiling 1\)

\(\ds \)

\(=\)

\(\ds 1\)

\(\ds \)

\(\ne\)

\(\ds 0\)

\(\ds \)

\(=\)

\(\ds \ceiling {\dfrac 1 {\paren {-2} } }\)

\(\ds \)

\(=\)

\(\ds \ceiling {\dfrac m n}\)

Thus, as stated, it is confirmed that the identity does not hold for $n < 0$.

It is noted that when $n = 0$ the expressions on either side are not defined.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $48 \ \text{(a)}$