floor and ceiling

Floor

For any

x \in R

, we define the floor of $x$ denoted by

⌊ x ⌋

max ({m \in Z : m \leq x})

. Note that

⌊ \cdot ⌋ : R \to Z

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

From here we can see that $⌊ π ⌋ = 3$

Fractional Part of Floor

Let

x \in R

then we define

⌋ x ⌊ : = x - ⌊ x ⌋

In other words $x = ⌊ x ⌋ + ⌋ x ⌊$

Fractional Part of Floor Bound

For any

x \in R

⌋ x ⌊ \in [0, 1)

Ceiling

For any

x \in R

, we define the ceiling of $x$ denoted by

⌈ x ⌉

min ({n \in Z : n \geq x})

, here

⌈ \cdot ⌉ : R \to Z

And that $⌈ π ⌉ = 4$

Floor of an Integer is Itself

Suppose

x \in Z

, then

⌊ x ⌋ = x

The greatest integer less than or equal to

x

x

Ceiling of an Integer is Itself

Suppose

x \in Z

, then

⌈ x ⌉ = x

The smallest integer greater than or equal to

x

x

Floor of a Non Integer is Smaller

Suppose that

a \in R ⧵ Z

, then

⌊ a ⌋ < a

Ceiling of a Non Integer is Greater

Suppose that

a \in R ⧵ Z

then

⌈ a ⌉ > a

The floor of a Sum of a Real and an Integer

Suppose that

x \in R

and

n \in Z

then we have

⌊ x + n ⌋ = ⌊ x ⌋ + n

A Number is a Perfect Square if its Square Root is an Integer

Let

n \in N_{1}

then

n

is a perfect square if and only if

\sqrt{n} \in N_{1}

Counting Squares with Floor

The number of squares in the set

[1, \dots, n]

is given by

| a \in [1, \dots, n] : ⌊ \sqrt{a} ⌋ = \sqrt{a} |

When Flooring Tells us Two Numbers Divide Eachother

Suppose

a, b \in Z

then

a ∣ b ⟺ ⌊ \frac{a}{b} ⌋ = \frac{a}{b}

Counting Multiples with Floor

Let

n, d \in N_{1}

then there are

⌊ \frac{n}{d} ⌋

multiples of

d

within the set

[1, \dots, n]

Largest Prime Power Dividing the Factorial

Suppose that

n \in N_{2}

and that

p \in P

such that

p ∣ n!

then

p^{\sum_{i \in N_{1}} ⌊ \frac{n}{p^{i}} ⌋} ∣ n!

and

\sum_{i \in N_{1}} ⌊ \frac{n}{p^{i}} ⌋

is the largest integer with this property

Prime Factorization of the Factorial

Suppose that

n \in N_{1}

then

n! = \prod_{p \in P} p^{\sum i \in N_{1} ⌊ \frac{n}{p^{i}} ⌋}

Double Sum of Divisors Equation

For any arithmetic

f : N_{1} \to C

we have

\sum_{k = 1}^{n} \sum_{d ∣ k} f (d) = \sum_{e = 1}^{n} ⌊ \frac{n}{e} ⌋ f (e)

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)