Floor
For any \( x \in \mathbb{ R } \), we define the floor of \( x \) denoted by \( \lfloor x \rfloor \) as \( \max \left( \left\{ m \in \mathbb{ Z } : m \le x \right\} \right) \). Note that \( \lfloor \cdot \rfloor : \mathbb{ R } \to \mathbb{ Z } \)

The max exists as it is a subset of \( \mathbb{ Z } \) which is bounded above, its explicit value can be by using the archimedian property to find the smallest \( n \) such that \( n \gt x \) then taking \( m = n - 1 \).

From here we can see that \( \lfloor \pi \rfloor = 3 \)

Fractional Part of Floor
Let \( x \in \mathbb{ R } \) then we define \[ \left\rfloor x \right\lfloor := x - \left\lfloor x \right\rfloor \]

In other words \( x = \left\lfloor x \right\rfloor + \left\rfloor x \right\lfloor \)

Fractional Part of Floor Bound
For any \( x \in \mathbb{ R } \) \[ \left\rfloor x \right\lfloor \in [0, 1) \]
Ceiling
For any \( x \in \mathbb{ R } \), we define the ceiling of \( x \) denoted by \( \lceil x \rceil \) as \( \min \left( \left\{ n \in \mathbb{ Z } : n \ge x \right\} \right) \), here \( \lceil \cdot \rceil : \mathbb{ R } \to \mathbb{ Z } \)

And that \( \lceil \pi \rceil = 4 \)

Floor of an Integer is Itself
Suppose \( x \in \mathbb{ Z } \), then \( \lfloor x \rfloor = x \)
The greatest integer less than or equal to \( x \) is \( x \).
Ceiling of an Integer is Itself
Suppose \( x \in \mathbb{ Z } \), then \( \lceil x \rceil = x \)
The smallest integer greater than or equal to \( x \) is \( x \).
Floor of a Non Integer is Smaller
Suppose that \( a \in \mathbb{ R } \setminus \mathbb{ Z } \), then \[ \lfloor a \rfloor \lt a \]
Ceiling of a Non Integer is Greater
Suppose that \( a \in \mathbb{ R } \setminus \mathbb{ Z } \) then \[ \lceil a \rceil \gt a \]
The floor of a Sum of a Real and an Integer
Suppose that \( x \in \mathbb{ R } \) and \( n \in \mathbb{ Z } \) then we have \[ \left\lfloor x + n \right\rfloor = \left\lfloor x \right\rfloor + n \]
A Number is a Perfect Square if its Square Root is an Integer
Let \( n \in \mathbb{ N } _ 1 \) then \( n \) is a perfect square if and only if \( \sqrt{ n } \in \mathbb{ N } _ 1 \)
Counting Squares with Floor
The number of squares in the set \( \left[ 1, \ldots , n \right] \) is given by \[ \left\lvert a \in \left[ 1, \ldots , n \right] : \left\lfloor \sqrt{ a } \right\rfloor = \sqrt{ a } \right\rvert \]
When Flooring Tells us Two Numbers Divide Eachother
Suppose \( a, b \in \mathbb{ Z } \) then \[ a \mid b \iff \left\lfloor \frac{a}{b} \right\rfloor = \frac{a}{b} \]
Counting Multiples with Floor
Let \( n, d \in \mathbb{ N } _ 1 \) then there are \( \left\lfloor \frac{n}{d} \right\rfloor \) multiples of \( d \) within the set \( \left[ 1, \ldots , n \right] \)
Largest Prime Power Dividing the Factorial
Suppose that \( n \in \mathbb{ N } _ 2 \) and that \( p \in \mathbb{ P } \) such that \( p \mid n! \) then \[ p ^ { \sum _ { i \in \mathbb{ N } _ 1 } \left\lfloor \frac{n}{p ^ i} \right\rfloor } \mid n! \] and \( \sum _ { i \in \mathbb{ N } _ 1 }\left\lfloor \frac{n}{p ^ i} \right\rfloor \) is the largest integer with this property
Prime Factorization of the Factorial
Suppose that \( n \in \mathbb{ N } _ 1 \) then \[ n! = \prod _ { p \in \mathbb{ P } } p ^ { \sum { i \in \mathbb{ N } _ 1 } \left\lfloor \frac{n}{p ^ i} \right\rfloor } \]
Double Sum of Divisors Equation
For any arithmetic \( f: \mathbb{ N } _ 1 \to C \) we have \[ \sum _ { k = 1 } ^ { n } \sum _ { d \mid k } f \left( d \right) = \sum _ { e = 1 } ^ { n } \left\lfloor \frac{n}{e} \right\rfloor f \left( e \right) \]