nyuuzyou/wikis huggingface dataset:
Fixed Point of Idempotent Mapping

== Theorem ==
Let $S$ be a set.
Let $f: S \to S$ be an idempotent mapping.
Let $\Img f$ be the image of $f$.
Let $x \in S$.
Then $x$ is a fixed point of $f$ $x \in \Img f$.
== Proof ==
=== Necessary Condition ===
Let $x$ be a fixed point of $f$.
Then:
:$\map f x = x$
and so by definition of image of mapping:
:$x \in \Img f$
=== Sufficient Condition ===
Let $x \in \Img f$.
Then by the definition of image:
:$\exists y \in S: \map f y = x$
Then:
Thus by definition $x$ is a fixed point of $f$.
Category:Idempotent Mappings
Fixed Point of Mappings is Fixed Point of Composition

== Theorem ==
Let $S$ be a set.
Let $f, g: S \to S$ be mappings.
Let $x \in S$ be a fixed point of both $f$ and $g$.
Then $x$ is also a fixed point of $f \circ g$, the composition of $f$ and $g$.
=== General Result ===
== Proof ==
Since $x$ is a fixed point of $g$:
: $\map g x = x$
Thus:
: $\map f {\map g x} = \map f x$
Since $x$ is a fixed point of $f$:
: $\map f x = x$
It follows that:
: $\map {\paren {f \circ g} } x = \map f {\map g x} = x$
Thus $x$ is a fixed point of $f \circ g$.
Category:Mapping Theory
Fixed Point of Mappings is Fixed Point of Composition/General Result

== Theorem ==
Let $S$ be a set.
Let $n \in \N$ be a strictly positive integer.
Let $\N_n$ be the initial segment of $n$ in $\N$.
That is, let $\N_n = \set {0, 1, \dots, n-1}$.
For each $i \in \N_n$, let $f_i: S \to S$ be a mapping.
Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_n$.
Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1}$ be the composition of all the $f_i$s.
Then $x$ is a fixed point of $g$.
== Proof ==
The proof proceeds by mathematical induction on $n$, the number of mappings.
=== Base Case ===
If $n = 1$, then $g = f_0$.
Since $x$ is a fixed point of $f_0$, it is also a fixed point of $g$.
=== Inductive Step ===
Suppose that the theorem holds for $n$. We will show that it holds for $n+1$.
Let $x \in S$ be a fixed point of $f_i$ for each $i \in \N_{n+1}$.
Let $g = f_0 \circ f_1 \circ \dots \circ f_{n-1} \circ f_n$ be the composition of all the $f_i$s.
Since the theorem holds for $n$, $x$ is a fixed point of $f_0 \circ f_1 \circ \dots \circ f_{n-1}$.
By Composition of Mappings is Associative:
: $g = \paren { f_0 \circ f_1 \circ \dots \circ f_{n-1} } \circ f_n$
Thus by Fixed Point of Mappings is Fixed Point of Composition (for two mappings), $x$ is a fixed point of $g$.
Category:Mapping Theory
Fixed Point of Permutation is Fixed Point of Power

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\sigma \in S_n$.
Let $i \in \Fix \sigma$, where $\Fix \sigma$ denotes the set of fixed elements of $\sigma$.
Then for all $m \in \Z$:
:$i \in \Fix {\sigma^m}$
== Proof ==
It follows from Integers form Ordered Integral Domain that for any integer $m$ either:
:$m = 0$
or
:$m > 0$
or:
:$m < 0$
=== Case 1 ===
By Element to the Power of Zero is Identity:
: $m = 0 \implies \sigma^m = e$
So:
:$\sigma^m \paren i = e \paren i = i$
=== Case 2 ===
Follows from Fixed Point of Mappings is Fixed Point of Composition: General Result.
=== Case 3 ===
For all $m < 0$, $m = -k$ for some (strictly) positive integer $k$.
Therefore:
:$\sigma^k \paren i = i \implies i = \sigma^{-k} \paren i$
Hence the result.
Category:Symmetric Groups
Category:Permutations
Fixed Point of Progressing Mapping on Minimally Closed Class is Greatest Element

== Theorem ==
$g$ has no fixed point, unless possibly the greatest element, if there is one.
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.17 \ (3)$
Category:Minimally Closed Classes under Progressing Mapping
Fixed Point of Progressing Mapping on Minimally Closed Class is Greatest Element/Proof

== Theorem ==
== Proof ==
Suppose $g$ has a fixed point.
Let $x$ be an element of $M$ such that $\map g x = x$.
We have from Smallest Element of Minimally Closed Class under Progressing Mapping that:
:$b \subseteq x$
Suppose that $y \subseteq x$.
Then by Image of Proper Subset under Progressing Mapping on Minimally Closed Class:
:$\map g y \subseteq \map g x$
But we have that $\map g x = x$.
Thus:
:$\map g y \subseteq x$
So we have:
:$b \subseteq x$
and:
:$y \subseteq x \implies \map g y \subseteq x$
and the result follows by the Principle of General Induction for Minimally Closed Class.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Exercise $4.1$
Category:Minimally Closed Classes under Progressing Mapping
Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element

== Theorem ==
Let $M$ be a class which is minimally inductive under a progressing mapping $g$.
Let $x$ be a fixed point of $g$.
Then $x$ is the greatest element of $M$.
== Proof 1 ==
== Proof 2 ==
Category:Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element
Category:Minimally Inductive Classes
Category:Progressing Mappings
Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element/Proof 1

== Theorem ==
== Proof ==
Let $x$ be an element of $M$ such that $\map g x = x$.
From Empty Set is Subset of All Sets, we have that:
:$\O \subseteq x$
Suppose that $y \subseteq x$.
Then by Characteristics of Minimally Inductive Class under Progressing Mapping:
:$\map g y \subseteq \map g x$
But we have that $\map g x = x$.
Thus:
:$\map g y \subseteq x$
That is:
:$\O \subseteq x$
and:
:$y \subseteq x \implies \map g y \subseteq x$
and the result follows by the Principle of General Induction.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Theorem $4.14$
Category:Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element
Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element/Proof 2

== Theorem ==
== Proof ==
A minimally inductive class under $g$ is the same thing as a minimally closed class under $g$ with respect to $\O$.
The result then follows by a direct application of Fixed Point of Progressing Mapping on Minimally Closed Class is Greatest Element.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Exercise $4.1$
Category:Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element
Fixed Point of g-Tower is Greatest Element

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x = \map g x$.
Then $x$ is the greatest element (under the subset relation) of $M$.
== Proof ==
Let $x$ be an element of $M$ such that $\map g x = x$.
We will demonstrate by the Principle of Superinduction on $y$ that:
:$\forall y \in M: y \subseteq x$
The proof proceeds by superinduction.
For all $y \in M$, let $\map P y$ be the proposition:
:$y \subseteq x$
=== Basis for the Induction ===
From Empty Set is Subset of All Sets, we have that:
:$\O \subseteq x$
Thus $\map P \O$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P y$ is true, where $y \in M$, then it logically follows that $\map P {\map g y}$ is true.
So this is the induction hypothesis:
:$y \subseteq x$
from which it is to be shown that:
:$\map g y \subseteq x$
=== Induction Step ===
This is the induction step:
Suppose that $y \subseteq x$.
We note that by the 
Then by the Sandwich Principle for G-Towers: Corollary 2:
:$\map g y \subseteq \map g x$
But we have that $\map g x = x$.
Thus:
:$\map g y \subseteq x$
So $\map P y \implies \map P {\map g y}$.
=== Closure under Chain Unions ===
It remains to demonstrate closure under chain unions.
Let $\map P y$ be true for all $y \in C$, where $C$ is an arbitrary chain of elements of $M$.
That is:
:$\forall y \in C: y \subseteq x$
Then it follows by Union of Subsets is Subset that:
:$\ds \bigcup C \subseteq x$
Thus:
:$\forall C: \forall y \in C: \map P y \implies \map P {\ds \bigcup C}$
The result follows by the Principle of Superinduction.
Therefore:
:$\forall y \in M: y \subseteq x$
and so $x$ is the greatest element of $M$.
== Also see ==
* Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.5$
Category:G-Towers
Fixed Points of Idempotent Operator

== Theorem ==
Let $X$ be a vector space.
Let $T : X \to X$ be an idempotent operator.
Then the set of fixed points of $T$ is precisely the range $\Rng T$. 
== Proof ==
If $x$ is a fixed point of $T$, then:
:$T x = x$
and we clearly have $x \in \Rng T$. 
Conversely, if $y \in \Rng T$ then there exists $x \in X$ such that $y = T x$. 
Then we have:
So $y$ is a fixed point of $T$.
Category:Idempotent Operators
Fixed Points of Orthogonal Projection

== Theorem ==
Let $\struct {H, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\norm \cdot$ be the inner product norm of $H$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Let $h \in H$. 
Then: 
:$\map {P_K} h = h$
 $h \in K$. 
== Proof ==
Let $d$ be the metric induced by $\norm \cdot$. 
Let $h \in H$.
By the definition of orthogonal projection, we have: 
:$\map d {h, \map {P_K} h} = \map d {h, K}$
Note that by the definition of a metric, we have that: 
:$\map d {h, \map {P_K} h} = 0$
:
:$h = \map {P_K} h$
So, we have: 
:$h = \map {P_K} h$
:
:$\map d {h, K} = 0$
Since $K$ is closed, from Subset of Metric Space is Closed iff contains all Zero Distance Points, this is the case : 
:$h \in K$
Category:Orthogonal Projections
Fixed Points of Projection in Plane

== Theorem ==
Let $M$ and $N$ be distinct lines in the plane.
:
Let $\pr_{M, N}$ be the projection on $M$ along $N$:
:$\forall x \in \R^2: \map {\pr_{M, N} } x =$ the intersection of $M$ with the line through $x$ parallel to $N$.
Then $M$ is the set of fixed points of $\pr_{M, N}$ in the sense that:
:$x \in M$
:
:$\map {\pr_{M, N} } x = x$
== Proof ==
=== Sufficient Condition ===
Let $x \in M$.
Let $\LL$ be the straight line through $x$ which is parallel to $N$.
As $x \in M$ it follows that $x$ is on the intersection of $M$ with $\LL$.
Hence by definition:
:$\map {\pr_{M, N} } x = x$
=== Necessary Condition ===
Again, let $\LL$ be the straight line through $x$ which is parallel to $N$.
Let $\map {\pr_{M, N} } x = x$.
Then by definition $x$ is on the intersection of $M$ with $\LL$.
Hence by definition of intersection:
:$x \in M$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.5$
Category:Geometric Projections
Flip Bifurcation/Examples/Arbitrary Example 1

== Example of Flip Bifurcation ==
Consider the mapping:
:$\forall x \in \R: \map {T_\lambda} x = x^2 - \paren {1 + \lambda}$
$T_\lambda$ has a flip bifurcation at $\lambda = 0$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Flip Bifurcations
Floating-Point Representation/Examples/105.7/Mistake

== Source Work ==
:
:floating-point representation
== Mistake ==
:For example, $105.7 = 0.1057 \times 10^{-3}$.
== Correction ==
In fact:
:$105.7 = 0.1057 \times 10^3$
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Floor Function/Examples

== Examples of Use of Floor Function ==
=== Floor of $0 \cdotp 99999$ ===
=== Floor of $1 \cdotp 1$ ===
=== Floor of $-1 \cdotp 1$ ===
=== Floor of $\sqrt 2$ ===
=== Floor of $\sqrt 5$ ===
=== Floor of $\sqrt {10}$ ===
=== Floor of $\dfrac 1 2$ ===
=== Floor of $- \dfrac 1 2$ ===
=== Floor of $\dfrac 5 2$ ===
=== Floor of $4 \cdotp 35$ ===
=== Floor of $-\dfrac 5 2$ ===
=== Floor of $3$ ===
=== Floor of $14$ ===
=== Floor of $\lg 35$ ===
Category:Examples of Floor Function
Floor Function/Examples/Floor of -1.1

== Theorem ==
:$\floor {-1 \cdotp 1} = -2$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$-2 \le -1 \cdotp 1 < -1$
Hence $-2$ is the floor of $-1 \cdotp 1$ by definition.
== Also see ==
* Floor of $1 \cdotp 1$: $\floor {1 \cdotp 1} = 1$
* Ceiling of $-1 \cdotp 1$: $\ceiling {-1 \cdotp 1} = -1$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $1$
Category:Examples of Floor Function
Floor Function/Examples/Floor of 0.99999

== Theorem ==
:$\floor {0 \cdotp 99999} = 0$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$0 \le 0 \cdotp 99999 < 1$
Hence $0$ is the floor of $0 \cdotp 99999$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $1$
Category:Examples of Floor Function
Floor Function/Examples/Floor of 1.1

== Theorem ==
:$\floor {1 \cdotp 1} = 1$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$1 \le 1 \cdotp 1 < 2$
Hence $1$ is the floor of $1 \cdotp 1$ by definition.
== Also see ==
* Floor of $-1\cdotp 1$: $\floor {-1 \cdotp 1} = -2$
* Ceiling of $-1\cdotp 1$: $\ceiling {-1 \cdotp 1} = -1$
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $1$
Category:Examples of Floor Function
Floor Function/Examples/Floor of 14

== Theorem ==
:$\floor {14} = 14$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that $14$ is an integer.
Thus this is a specific example of Real Number is Integer iff equals Floor:
$\floor x = x \iff x \in \Z$
== Sources ==
* : $\S 10$: The well-ordering principle
Category:Examples of Floor Function
Floor Function/Examples/Floor of 3

== Theorem ==
:$\floor 3 = 3$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that $3$ is an integer.
Thus this is a specific example of Real Number is Integer iff equals Floor:
$\floor x = x \iff x \in \Z$
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercises $3$
Category:Examples of Floor Function
Floor Function/Examples/Floor of 4.35

== Theorem ==
:$\floor {4 \cdotp 35} = 4$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$4 \le 4 \cdotp 35 < 5$
Hence $4$ is the floor of $4 \cdotp 35$ by definition.
== Also see ==
* Ceiling of $4 \cdotp 35$: $\ceiling {4 \cdotp 35} = 5$
== Sources ==
* 
* 
Category:Examples of Floor Function
Floor Function/Examples/Floor of 5 over 2

== Theorem ==
:$\floor {\dfrac 5 2} = 2$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
Hence $2$ is the floor of $\dfrac 5 2$ by definition.
== Sources ==
* : $\S 10$: The well-ordering principle
Category:Examples of Floor Function
Floor Function/Examples/Floor of Binary Logarithm of 35

== Theorem ==
:$\floor {\lg 35} = 5$
where:
:$\lg x$ denotes the binary logarithm ($\log_2$) of $x$
:$\floor x$ denotes the floor of $x$.
== Proof ==
Hence $5$ is the floor of $\lg 35$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $1$
Category:Examples of Floor Function
Floor Function/Examples/Floor of Minus 5 over 2

== Theorem ==
:$\floor {-\dfrac 5 2} = -3$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
Hence $-3$ is the floor of $-\dfrac 5 2$ by definition.
== Sources ==
* : $\S 10$: The well-ordering principle
Category:Examples of Floor Function
Floor Function/Examples/Floor of Minus One Half

== Theorem ==
:$\floor {-\dfrac 1 2} = -1$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$-1 = -\dfrac 2 2 \le -\dfrac 1 2$
and:
:$0 > -\dfrac 1 2$
Hence $-1$ is the floor of $-\dfrac 1 2$ by definition.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $3$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Examples of Floor Function
Floor Function/Examples/Floor of One Half

== Theorem ==
:$\floor {\dfrac 1 2} = 0$
where $\floor x$ denotes the floor of $x$.
== Proof ==
We have that:
:$0 \le \dfrac 1 2$
and:
:$1 = \dfrac 2 2 > \dfrac 1 2$
Hence $0$ is the floor of $\dfrac 1 2$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Examples of Floor Function
Floor Function/Examples/Floor of Root 2

== Theorem ==
:$\floor {\sqrt 2} = 1$
where $\floor x$ denotes the floor of $x$.
== Proof ==
The decimal expansion of $\sqrt 2$ is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 \le \sqrt 2 < 2$
Hence $1$ is the floor of $\sqrt 2$ by definition.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Examples of Floor Function
Floor Function/Examples/Floor of Root 5

== Theorem ==
:$\floor {\sqrt 5} = 2$
where $\floor x$ denotes the floor of $x$.
== Proof ==
The decimal expansion of $\sqrt 5$ is:
:$\sqrt 5 \approx 2 \cdotp 23606 \, 79774 \, 99789 \, 6964 \ldots$
Thus:
:$2 \le \sqrt 5 < 3$
Hence $2$ is the floor of $\sqrt 5$ by definition.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $3$
Category:Examples of Floor Function
Floor Function is Continuous on Right at Integer

== Theorem ==
Let $f$ be the real function defined as:
:$\map f x = \floor x$
where $\floor{\, \cdot \,}$ denotes the floor function.
Let $n \in \Z$ be an integer.
Then $\map f x$ is continuous on the right at $n$.
== Proof ==
From Real Number is Integer iff equals Floor:
:$\floor n = n$
By definition $\floor x$ is the unique integer such that:
:$\floor x \le x < \floor x + 1$
Consider the open real interval:
:$\II = \openint n {n + 1}$
By definition of $\floor x$ we have that:
:$\forall x \in \II: \floor x = n$
That is:
:$\forall x \in \II: \floor x - n = 0$
Let $\epsilon \in \R_{>0}$.
Let $\delta \in \openint n {n + 1}$.
Then:
:$\forall x \in \II: n < x < n + \delta: \size {\floor x - n} < \epsilon$
That is:
:$\ds \lim_{x \mathop \to n^+} \floor x = \floor n$
Hence, by definition, $\map f x$ is continuous on the right at $n$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $2$
Category:Floor Function
Category:Examples of One-Sided Continuity
Floor Function is Idempotent

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$.
Then:
:$\floor {\floor x} = \floor x$
That is, the floor function is idempotent.
== Proof ==
Let $y = \floor x$.
By Floor Function is Integer, $y$ is an integer.
Then from Real Number is Integer iff equals Floor:
:$\floor y = y$
So:
:$\floor {\floor x} = \floor x$
== Also see ==
* Ceiling Function is Idempotent
* Floor of Ceiling is Ceiling
* Ceiling of Floor is Floor
Category:Floor Function
Floor Function is Not Continuous on Left at Integer

== Theorem ==
Let $f$ be the real function defined as:
:$\map f x = \floor x$
where $\floor{\, \cdot \,}$ denotes the floor function.
Let $n \in \Z$ be an integer.
Then $\map f x$ is not continuous on the left at $n$.
== Proof ==
From Real Number is Integer iff equals Floor:
:$\floor n = n$
By definition $\floor x$ is the unique integer such that:
:$\floor x \le x < \floor x + 1$
Consider the open real interval:
:$\II = \openint {n - 1} n$
By definition of $\floor x$ we have that:
:$\forall x \in \II: \floor x = n - 1$
That is:
:$\forall x \in \II: \floor x - n = -1$
Let $\epsilon \in \R_{>0}$.
Let $\delta \in \openint {n - 1} n$.
Then:
:$\forall x \in \II: \paren n - \delta < x < n: \size {\floor x - \paren {n - 1} } < \epsilon$
That is:
:$\ds \lim_{x \mathop \to n^-} \floor x = n - 1$
But we have that:
:$\map f n = n$
That is:
:$\ds \lim_{x \mathop \to n^-} \map f x \ne \map f n$
Hence, by definition, $\map f x$ is not continuous on the left at $n$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $2$
Category:Floor Function
Category:Examples of One-Sided Continuity
Floor Function is Replicative

== Theorem ==
The floor function is a replicative function in the sense that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 0}^{n - 1} \floor {x + \frac k n} = \floor {n x}$
== Proof ==
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $38$
Category:Floor Function
Category:Replicative Functions
Floor defines Equivalence Relation

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor function of $x$.
Let $\RR$ be the relation defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \floor x = \floor y$
Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointr n {n + 1}$.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
:$\forall x \in \R: \floor x = \floor x$
Thus the floor function is reflexive.
=== Symmetry ===
:$\forall x, y \in \R: \floor x = \floor y \implies \floor y = \floor x$
Thus the floor function is symmetric.
=== Transitivity ===
Let $\floor x = \floor y$ and $\floor y = \floor z$.
Let $n = \floor x = \floor y = \floor z$, which follows from transitivity of $=$.
Thus from Real Number is Floor plus Difference‎:
:$x = n + t_x, y = n + t_y, z = n + t_z: t_x, t_y, t_z \in \hointr 0 1$
Thus:
:$x = n + t_x, z = n + t_z$
and:
:$\floor x = \floor z$
Thus the floor function is transitive.
Thus we have shown that $\RR$ is an equivalence relation.
Now we show that the $\RR$-class of $n$ is the interval $\hointr n {n + 1}$.
Defining $\RR$ as above, with $n \in \Z$:
== Also see ==
* Ceiling defines Equivalence Relation
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $2$
* : $\S 1.4$: Decomposition of a set into classes. Equivalence relations: Example $3$
* : Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $1$
Category:Floor Function
Category:Examples of Equivalence Relations
Floor equals Ceiling iff Integer

== Theorem ==
Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor of $x$, and $\ceiling x$ denote the ceiling of $x$.
Then:
:$\floor x = \begin {cases}
\ceiling x & : x \in \Z \\
\ceiling x - 1 & : x \notin \Z \\
\end {cases}$
or equivalently:
:$\ceiling x = \begin {cases}
\floor x & : x \in \Z \\
\floor x + 1 & : x \notin \Z \\
\end {cases}$
where $\Z$ is the set of integers.
== Proof ==
From Real Number is Integer iff equals Floor:
:$x \in \Z \implies x = \floor x$
From Real Number is Integer iff equals Ceiling:
:$x \in \Z \implies x = \ceiling x$
So:
:$x \in \Z \implies \floor x = \ceiling x$
Now let $x \notin \Z$.
From the definition of the floor function:
:$\floor x = \map \sup {\set {m \in \Z: m \le x} }$
From the definition of the ceiling function:
:$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
Thus:
:$\floor x < x < \ceiling x$
Hence the result, from the definition of $\inf$ and $\sup$.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
* 
* 
Category:Floor Function
Category:Ceiling Function
Floor is between Number and One Less

== Theorem ==
:$x - 1 < \floor x \le x$
where $\floor x$ denotes the floor of $x$.
== Proof ==
By definition of floor function:
:$\floor x \le x < \floor x + 1$
Thus by subtracting $1$:
:$x - 1 < \paren {\floor x + 1} - 1 = \floor x$
So:
:$\floor x \le x$
and:
:$x - 1 < \floor x$
as required.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
Category:Floor Function
Floor of Ceiling is Ceiling

== Theorem ==
Let $x \in \R$ be a real number.
Let $\left \lfloor {x}\right \rfloor$ be the floor of $x$, and $\left \lceil {x}\right \rceil$ be the ceiling of $x$.
Then:
: $\left \lfloor {\left \lceil {x}\right \rceil}\right \rfloor = \left \lceil {x}\right \rceil$
That is, the floor of the ceiling is the ceiling.
== Proof ==
Let $y = \left \lceil{x}\right \rceil$.
By Ceiling Function is Integer, we have that $y \in \Z$.
From Real Number is Integer iff equals Floor, we have:
:$\left \lfloor{y} \right \rfloor = y$
So:
:$\left \lfloor {\left \lceil {x}\right \rceil}\right \rfloor = \left \lceil {x}\right \rceil$
== Also see ==
* Ceiling of Floor is Floor
* Floor Function is Idempotent
* Ceiling Function is Idempotent
Category:Floor Function
Category:Ceiling Function
Floor of Half of n+m plus Floor of Half of n-m+1

== Theorem ==
Let $n, m \in \Z$ be integers.
:$\floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2} = n$
where $\floor x$ denotes the floor of $x$.
== Proof ==
Either $n + m$ or $n - m + 1$ is even.
Thus:
:$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$
and so:
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $33 \ \text{(a)}$
Category:Floor Function