$\lceil m\mathrm{/}n\rceil$ for naturals $m$ and $n\ge 1$: the least $q\in \mathbb{N}$ with $m\le nq$

Let $m,n\in \mathbb{N}$ with $n\ge 1$ (The natural numbers $\mathbb{N}$ (von Neumann), Order on the natural numbers, On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), and put

$Q(m,n):=\{q\in \mathbb{N}:m\le nq\},$

the multiplication being that of $\mathbb{N}$ (Multiplication of natural numbers).

$Q(m,n)$ is nonempty, so the definition below has something to pick from. Since $n\mathrm{\ne }0$, Every nonzero natural number is a successor gives $n=\sigma (d)$ for some $d\in \mathbb{N}$, and then $n\cdot m=\sigma (d)\cdot m=d\cdot m+m=m+d\cdot m$ by the successor-left law of Distributivity and the successor law for multiplication and the commutativity of addition (Addition is commutative), so $m\le n\cdot m$ by the definition of the order (Order on the natural numbers), which asks for a natural $j$ with $m+j=nm$ and is met by $j=d\cdot m$. Hence $m\in Q(m,n)$.

Definition. $\lceil m\mathrm{/}n\rceil$ is the least element of $Q(m,n)$, which exists by the well-ordering principle (The well-ordering principle) applied to the nonempty subset $Q(m,n)$ of $\mathbb{N}$. It is a natural number, and it is defined for $n\ge 1$ only.

Four clauses, recorded here because they are what the notation is used for.

(a) $m\le n\lceil m\mathrm{/}n\rceil$. This is membership of $\lceil m\mathrm{/}n\rceil$ in $Q(m,n)$.

(b) Minimality. If $q\in \mathbb{N}$ satisfies $m\le nq$ then $\lceil m\mathrm{/}n\rceil \le q$; equivalently, every $q$ with $q<\lceil m\mathrm{/}n\rceil$ satisfies $nq<m$, by trichotomy (Trichotomy of the order on $\mathbb{N}$).

(c) Two values read off directly. $\lceil 0\mathrm{/}n\rceil =0$, since $0\le n\cdot 0$ makes $0$ the least element of $\mathbb{N}$ that qualifies; and $\lceil m\mathrm{/}1\rceil =m$, since $m\le 1\cdot m=m$ while $m\le 1\cdot q$ forces $m\le q$ (Zero and one under multiplication, Multiplication is commutative).

(d) The reading in $\mathbb{R}$. With $\iota :\mathbb{N}\to \mathbb{R}$ the canonical natural (The canonical natural $\iota (n)=n\cdot 1_F$ of a field), clause (a) gives $\iota (m)\le \iota (n)\iota (\lceil m\mathrm{/}n\rceil )$ by the multiplicativity of $\iota$ (clause 0 of Laws of finite sums and products in $\mathbb{N}$, and $\iota \left({\sum }_{k<n}{a}_k\right)={\sum }_{k<n}\iota (a_k)$) and its strict monotonicity (clause 7); and $\iota (n)>0$ because $n\ge 1$. Since $\mathbb{R}$ is an ordered field (Ordered field, Field), dividing by $\iota (n)$ gives

$\frac{\iota (m)}{\iota (n)}\le \iota (\lceil m\mathrm{/}n\rceil )\mathrm{.}$

• This is not a floor and it is not a ceiling function. It is defined for a pair of natural numbers with $n\ge 1$, its value is a natural number, and it is fixed by one order property and one minimality property. It is not defined for a real argument, it does not extend to negative numbers, and it carries no division: the symbol $m\mathrm{/}n$ inside the brackets is part of the notation and not an operation performed anywhere above.

• Why it is introduced at all. The strong form of the pigeonhole principle says that some fibre has at least "the average, rounded up" elements, and that phrase needs a name for the rounding. The least $q$ with $m\le nq$ is exactly what the proof produces, and the well-ordering principle is exactly what makes it exist, so nothing stronger is required.

• Nothing among this page's declared prerequisites supplies a division with remainder, and the definition above deliberately does not attempt one: no $r$ with $m=nq+r$ and $r<n$ is produced or claimed here.