nyuuzyou/wikis huggingface dataset:
Odd Power Function is Surjective

== Theorem ==
Let $n \in \Z_{\ge 0}$ be an odd positive integer.
Let $f_n: \R \to \R$ be the real function defined as:
:$\map {f_n} x = x^n$
Then $f_n$ is a surjection.
== Proof ==
From Existence of Positive Root of Positive Real Number we have that:
:$\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$
From Power of Ring Negative:
:$\paren {-x}^n = -\paren {x^n}$
and so:
:$\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$
Thus:
:$\forall x \in \R: \exists y \in \R: y^n = x$
and so $f_n$ is a surjection.
Hence the result.
Category:Powers
Category:Surjections
Category:Real Functions
Odd Power of 2 is Congruent to 2 Modulo 3

== Theorem ==
Let $n \in \Z_{\ge 0}$ be an odd positive integer.
Then:
:$2^n \equiv 2 \pmod 3$
== Proof ==
From Congruence Modulo 3 of Power of 2:
:$2^n \equiv \paren {-1}^n \pmod 3$
We have that $n$ is odd.
Hence:
Category:Number Theory
Category:2
Category:3
Odd Square Modulo 8

== Theorem ==
Let $x \in \Z$ be an odd square.
Then $x \equiv 1 \pmod 8$.
== Proof ==
Let $x \in \Z$ be an odd square.
Then $x = n^2$ where $n$ is also odd.
Thus $n$ can be expressed as $2 k + 1$ for some $k \in \Z$.
Hence:
But $k$ and $k + 1$ are of opposite parity and can therefore be expressed as $2 r$ and $2 s + 1$ (either way round).
Hence the result.
== Examples ==
== Also see ==
* Square Modulo 8
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm
Category:Modulo Arithmetic
Category:Odd Squares
Odd Square is Eight Triangles Plus One

== Theorem ==
Let $n \in \Z$ be an odd integer.
Then $n$ is square $n = 8 m + 1$ where $m$ is triangular.
== Proof ==
Follows directly from the identity:
as follows:
Let $m$ be triangular.
Then from Closed Form for Triangular Numbers:
:$\exists k \in \Z: m = \dfrac {k \paren {k + 1} } 2$
From the above identity:
:$8 m + 1 = \paren {2 k + 1}^2$
which is an odd square.
Let $n$ be an odd square.
Then $n = r^2$ where $r$ is odd.
Let $r = 2 k + 1$, so that $n = \paren {2 k + 1}^2$.
From the above identity:
:$n = 8 \dfrac {k \paren {k + 1} } 2 + 1 = 8 m + 1$
where $m$ is triangular.
== Illustration ==
:
== Also see ==
* Odd Square Modulo 8
== Historical Note ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(b)}$
* : $8$
* : $15$
* : $8$
* : $15$
Category:8
Category:Triangular Numbers
Category:Odd Squares
Odd Square is Eight Triangles Plus One/Historical Note

== Historical Note on Odd Square is Eight Triangles Plus One ==
This result, according to in in $1986$, was known to .
However, in his of $1980$, attributes the result to , circa $100$ C.E.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $1 \ \text {(b)}$
* : $15$
* : $15$
Category:Historical Notes
Odd Squares 7 Less than Nearest Power of 2

== Theorem ==
There exist exactly $3$ odd squares which are $7$ less than the nearest power of $2$:
:$5^2 = 25 = 2^5 - 7$
:$11^2 = 121 = 2^7 - 7$
:$181^2 = 32 \, 761 = 2^{15} - 7$
Note that this sequence includes $1$ and $3$, being all the squares which are $7$ less than a power of $2$.
However, for $1$ and $3$, those powers ($8$ and $16$ respectively) are not the nearest power of $2$ ($1$ and $4$ respectively).
== Proof ==
From Solutions of Ramanujan-Nagell Equation, the only solutions to the equation:
:$x^2 + 7 = 2^n$
are $\tuple {x, n} =$:
:$\tuple {1, 3}, \tuple {3, 4}, \tuple {5, 5}, \tuple {11, 7}, \tuple {181, 15}$
so no more solutions exist.
== Sources ==
* 
* : $7$
Category:Number Theory
Odd Untouchable Numbers

== Unproven Hypothesis ==
It is highly likely that $5$ is the only odd untouchable number.
== Progress ==
This would follow from a slightly stronger form of Goldbach's Conjecture:
:Every even integer $n > 6$ is the sum of two distinct primes.
Let $2 n + 1$ be an odd integer number greater than $7$.
Then by the conjecture:
:$2 n = p + q$
The aliquot parts of $p q$ are $1$, $p$ and $q$.
Then:
:$1 + p + q = 2 n + 1$
and so $2 n + 1$ is not untouchable. 
Then we have that $1$, $3$ and $7$ are not untouchable, as they are the aliquot sums of $2$, $4$ and $8$ respectively.
That leaves $5$ as the only odd untouchable number.
== Sources ==
* : $5$
* : $5$
* 
Category:Untouchable Numbers
Category:Unproven Hypotheses
Category:5
Odd Vertex of Graph/Examples/Graph with All Odd Vertices

== Examples of Odd Vertices ==
An example of a simple graph whose vertices are all odd includes the complete graph of order $4$:
:
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $7 \ \text{(b)}$
Category:Examples of Odd Vertices
Odd Vertex of Graph/Examples/Graphs of Order p with n Odd Vertices

== Examples of Odd Vertices ==
Examples of simple graphs of order $p$ with $n$ odd vertices for $0 \le n < p$ for various $p$ and $n$ are as follows:
:
Note that by the Handshake Lemma $n$ is always even.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $8$
Category:Examples of Odd Vertices
Odd Vertices Determines Edge-Disjoint Trails

== Theorem ==
Let $G$ be a loop-multigraph with $2 n$ odd vertices, $n > 0$.
Then $G$ has $n$ edge-disjoint trails such that every edge of $G$ is contained in one of these trails.
Each of these trails starts and ends on an odd vertex.
== Proof ==
We prove this result by induction:
For all $n \in \N_{>0}$, let $\map P N$ be the proposition:
:if $G$ has $2 n$ odd vertices, it consists entirely of $n$ edge-disjoint trails, each starting and ending on an odd vertex.
First note that from the corollary to the Handshake Lemma, no graph can have an odd number of odd vertices.
=== Basis for the Induction ===
$\map P 1$ is true, as this is the Characteristics of Traversable Graph.
This is our basis for the induction.
Note that, from the same result, the trail starts and ends on an odd vertex.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:If $G$ has $2 k$ odd vertices, it consists entirely of $k$ edge-disjoint trails, each starting and ending on an odd vertex.
Then we need to show:
:If $G$ has $2 \paren {k + 1}$ odd vertices, it consists entirely of $k + 1$ edge-disjoint trails, each starting and ending on an odd vertex.
=== Induction Step ===
This is our induction step:
Let $G_{k + 1}$ be a graph with $2 \paren {k + 1}$ odd vertices.
Let $u, v$ be any pair of odd vertices in $G_{k + 1}$.
We construct the graph $G'_{k + 1}$ which consists of $G_{k + 1}$ with the edge $u v$ added.
It can be seen that as $u, v$ are now even vertices, $G'_{k + 1}$ has $2 k$ odd vertices.
By our induction hypothesis, we know that $G'_{k + 1}$ consists entirely of $k$ edge-disjoint trails.
So, let us construct these trails: let them be $T_1, T_2, \ldots, T_k$.
The edge $u v$ must be somewhere in one of the above trails.
As the $k$ trails are all edge-disjoint, it will be in exactly one.
Suppose it is in $T_j$.
As $T_j$ must start and end on an odd vertex, $u v$ must be neither at the start nor the end of such a trail.
Suppose $T_j = \tuple {t_1, t_2, \ldots, u, v, \ldots, t_r}$.
Then it follows that $t_1$ and $t_r$ are odd vertices.
Now we remove $u v$ from $G'_{k + 1}$ so as to return to $G_{k + 1}$.
We see that $G_{k + 1}$ consists entirely of the edge-disjoint trails:
:$T_1, T_2, \ldots, T_{j - 1}$
:$T_{j + 1}, T_{j + 2}, \ldots, T_k$
:The two trails: $\tuple {t_1, t_2, \ldots, u}$ and $\tuple {v, \ldots, t_r}$.
That makes $k + 1$ edge-disjoint trails.
Also note that as $u$ and $v$ are odd vertices, both of $\tuple {t_1, t_2, \ldots, u}$ and $\tuple {v, \ldots, t_r}$ start and end on odd vertices.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Historical Note ==
== Sources ==
* : Chapter $3$: Transportation Problems: $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Problem $5$
Category:Degrees of Vertices
Category:Proofs by Induction
Odd and Even Permutations of Set are Equivalent

== Theorem ==
Let $n \in \N_{> 0}$ be a natural number greater than $0$.
Let $S$ be a set of cardinality $n$.
Let $S_n$ denote the symmetric group on $S$ of order $n$.
Let $R_e$ and $R_o$ denote the subsets of $S_n$ consisting of even permutations and odd permutations respectively.
Then $R_e$ and $R_o$ are equivalent.
== Proof ==
From Symmetric Group on n Letters is Isomorphic to Symmetric Group, it is sufficient to investigate the symmetric group on n letters.
Let $\tau$ be a transposition.
By definition of sign:
:$\map \sgn \rho = -1$
By definition of odd permutation:
:$\tau \in R_o$
Moreover, also by definition of sign, for any $\rho \in R_e$ we have:
:$\rho \circ \tau \in R_o$
Define the mapping $\phi$:
:$\forall \rho \in R_e: \map \phi \sigma = \rho \circ \tau$
We claim that $\phi$ is injective.
Suppose:
:$\map \phi {\rho_1} = \map \phi {\rho_2}$
Then:
:$\rho_1 \circ \tau = \rho_2 \circ \tau$
Since each Transposition is Self-Inverse, it follows that:
:$\rho_1 = \rho_1 \circ \tau \circ \tau = \rho_2 \circ \tau \circ \tau = \rho_2$
Thus $\phi$ is indeed injective.
We claim that $\phi$ is also surjective.
Let $\sigma \in R_o$. 
We need to show that there exists an element $\rho\in R_e$ such that:
:$\map \phi \rho = \sigma$
By definition of sign:
:$\sigma \circ \tau \in R_e$
Since each Transposition is Self-Inverse, it follows that:
:$\map \phi {\sigma \circ \tau} = \sigma \circ \tau \circ \tau = \sigma$
Thus $\phi$ is surjective.
Since $\phi$ is injective and surjective, $\phi$ is the desired bijection.
Hence the result by definition of set equivalence.
== Sources ==
* : Appendix: Elementary set and number theory
Category:Even Permutations
Category:Odd Permutations
Category:Set Equivalence
Odds Ratio/Examples/2x2 Contingency Table

== Example of Empirical Odds Ratio ==
In general, as empirical odds ratio can be evaluated from a contingency table:
:$\begin{array}{r|cc|c}
& & & \\
\hline
 & a & b & a + b \\
 & c & d & c + d \\
\hline
\text {Totals} & a + c & b + d & 
\end{array}$
The empirical odds ratio is then the ratio $a d : b c$.
Let the expected numbers in the $4$ cells be $m_{11}$, $m_{12}$, $m_{21}$ and $m_{22}$.
Then:
:$\theta = \dfrac {m_{11} m_{22} } {m_{12} m_{21} }$
When $\theta = 1$ this implies no association.
In such a $2 \times 2$ contingency table:
:$0 \le \theta^* \le \infty$
== Sources ==
* 
* 
Category:Examples of Odds Ratios
Odds Ratio/Examples/Smoking Disease

== Example of Empirical Odds Ratio ==
Let $P$ be a population of $250$ people.
Let $P_1$ be the set consisting of the $150$ members of $P$ who are smokers.
Let $P_2$ be the set consisting of the $100$ members of $P$ who do not smoke.
Let $Q$ be the property of the members of $P$ defined as:
:$\forall x \in P: \map Q x$ means that $x$ has a certain disease supposed to be caused by smoking.
Let:
:the number of members of $P_1$ with $Q$ be $12$
:the number of members of $P_2$ with $Q$ be $2$.
The empirical odds ratio of having $Q$ relative to smoking or non-smoking is then:
:$\theta^* = 4 \cdotp 26$
It is common to use the empirical odds ratio as the basis of a hypothesis test.
The null hypothesis in such a case is whether $P_1 = P_2$.
This is equivalent to testing whether $\theta^*$ is significantly different from $\theta = 1$.
The above can be placed into a contingency table:
:$\begin{array}{r|cc|c}
& \text {Diseased} & \text {Not diseased} & \text {totals} \\
\hline
\text {Smokers} & 12 & 138 & 150 \\
\text {Nonsmokers} & 2 & 98 & 100 \\
\hline
\text {Totals} & 14 & 236 & 250
\end{array}$
== Proof ==
By definition of empirical odds ratio:
== Sources ==
* 
* 
Category:Examples of Odds Ratios
Oesterlé-Masser Conjecture

== Conjecture ==
Let $\epsilon \in \R$ be a strictly positive real number.
=== Formulation 1 ===
=== Formulation 2 ===
=== Formulation 3 ===
== Also known as ==
== Also see ==
* Equivalence of Formulations of Oesterlé-Masser Conjecture
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Oesterlé-Masser Conjecture
Category:Number Theory
Category:Unproven Hypotheses
Oesterlé-Masser Conjecture/Formulation 1

== Theorem ==
Let $\epsilon \in \R$ be a strictly positive real number.
There exists only a finite number of triples of (strictly) positive integers $\tuple {a, b, c}$ with the conditions:
:$a + b = c$
:$a$, $b$ and $c$ are pairwise coprime
such that:
:$c > \map \Rad {a b c}^{1 + \epsilon}$
where $\Rad$ denotes the radical of an integer.
== Also see ==
* Equivalence of Formulations of Oesterlé-Masser Conjecture
Category:Oesterlé-Masser Conjecture
Oesterlé-Masser Conjecture/Formulation 2

== Theorem ==
Let $\epsilon \in \R$ be a strictly positive real number.
There exists a constant $K_\epsilon$ such that for all triples of (strictly) positive integers $\tuple {a, b, c}$ with the conditions:
:$a + b = c$
:$a$, $b$ and $c$ are pairwise coprime
such that:
:$c < K_\epsilon \map \Rad {a b c}^{1 + \epsilon}$
where $\Rad$ denotes the radical of an integer.
== Also see ==
* Equivalence of Formulations of Oesterlé-Masser Conjecture
Category:Oesterlé-Masser Conjecture
Oesterlé-Masser Conjecture/Formulation 3

== Theorem ==
Let $\epsilon \in \R$ be a strictly positive real number.
There exists only a finite number of triples of (strictly) positive integers $\tuple {a, b, c}$ with the conditions:
:$a + b = c$
:$a$, $b$ and $c$ are pairwise coprime
such that:
:$\map q {a, b, c} > 1 + \epsilon$
where $\map q {a, b, c}$ denotes the quality of $\tuple {a, b, c}$.
== Also see ==
* Equivalence of Formulations of Oesterlé-Masser Conjecture
Category:Oesterlé-Masser Conjecture
Oesterlé-Masser Conjecture/Historical Note

== Historical Note on Oesterlé-Masser Conjecture ==
The Oesterlé-Masser Conjecture was first proposed by in $1985$ and in $1988$.
In $2012$, published a series of papers which claimed to have proved it.
However, there is perceived to be a gap in this proof.
Despite the misgivings of a number of mathematicians, it was announced on $3$rd April $2020$ that 's proof would be published in a journal of which himself is the chief editor.
In the face of all this, it is still generally recognised in the mathematical community that the Oesterlé-Masser Conjecture remains unproven.
== Sources ==
* 
* 
Category:Historical Notes
Omega-Accumulation Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1

== Theorem ==
Let $\sequence {a_n}$ denote the sequence defined as:
Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $S$ denote the set of terms of $\sequence {a_n}$ considered as a subset of $\struct {\R, \tau_d}$.
Then $0$ is an $\omega$-accumulation point of $S$.
== Proof ==
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a countably infinite number of terms of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of $\omega$-accumulation point of $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $3$
Category:Sequences
Category:Omega-Accumulation Points
Omega Constant is Transcendental

== Theorem ==
The omega constant is transcendental.
== Proof ==
From the definition of omega constant, it is the real number $\Omega$ such that:
:$\Omega \, e^\Omega = 1$
where $e$ denotes Euler's number.
 $\Omega$ is not transcendental.
Hence, by definition, $\Omega$ is algebraic.
Then $e^\Omega$ is also algebraic, because:
:$e^\Omega = \dfrac 1 \Omega$
However, by the weaker Hermite-Lindemann-Weierstrass theorem, $e^\Omega$ is transcendental.
From this contradiction it follows that $\Omega$ is transcendental.
Category:Omega Constant
Category:Lambert W Function
Category:Transcendental Number Theory
Omega as Limit Point of Intervals of Uncountable Closed Ordinal Space

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\Omega$ is a limit point of the set $\openint a \Omega$, but not the limit of any sequence of points in $\openint a \Omega$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $2$
Category:Ordinal Spaces
Category:Limit Points
Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set

== Theorem ==
Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\set \Omega$ is a closed set of $\closedint 0 \Omega$ but not a $G_\delta$ set.
== Proof ==
The complement relative to $\closedint 0 \Omega$ of $\set \Omega$ is $\hointr 0 \Omega$, which is open in $\closedint 0 \Omega$.
Hence, by definition, $\set \Omega$ is a closed set of $\closedint 0 \Omega$.
Let $G_i$ be a countable set of open sets of $\closedint 0 \Omega$ which contain $\Omega$.
Then we can find a set of basis elements of $\closedint 0 \Omega$ of the form $\hointl {\alpha_i} \Omega \subseteq G_i$ for each $i$.
Each $\hointr 0 {\alpha_i}$ is countable.
We also have Countable Union of Countable Sets is Countable.
Thus the infimum of the $\alpha_i$ is an ordinal $\gamma$ which is less than $\Omega$.
Therefore:
:$\ds \bigcap G_i \supseteq \hointl \gamma \Omega \ne \set \Omega$
That is, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $\sqbrk {0, \Omega}$: $1$
Category:Ordinal Spaces
Category:G-Delta Sets
One-Digit Number is Harshad

== Theorem ==
Let $n$ be a $1$-digit positive integer.
Then $n$ is a harshad number.
== Proof ==
By definition, a harshad number is divisible by the sum of its digits base $10$.
Let $n$ be a $1$-digit positive integer.
The sum of the digits of $n$ is trivially $n$.
The result follows from Integer Divides Itself.
Category:Harshad Numbers
One-Dimensional Random Walk/Examples/Gambling Game

== Example of One-Dimensional Random Walk ==
Consider the gambling game in which a player with initial capital $k$ wins $1$ unit with probability $p$ and loses $1$ unit with probability $1 - p$.
This is a one-dimensional random walk with absorbing state $k = 0$.
== Sources ==
* 
* 
Category:Examples of One-Dimensional Random Walks
One-Parameter Family of Curves/Examples/Circles of Equal Radius with Centers along x-Axis

== Example of One-Parameter Family of Curves ==
Consider the equation:
:$(1): \quad \paren {x - h}^2 + y^2 = a^2$
where $a$ is constant.
$(1)$ defines a one-parameter family of circles of constant radius $a$ whose centers are on the $x$-axis of a Cartesian plane at $\tuple {h, 0}$ determined by values of the parameter $h$.
== Sources ==
* 
* 
Category:Examples of One-Parameter Families of Curves
One-Parameter Family of Curves/Examples/Concentric Circles

== Example of One-Parameter Family of Curves ==
The equation:
:$x^2 + y^2 = r^2$
is a one-parameter family of concentric circles whose centers are at the origin of a Cartesian plane and whose radii are the values of the parameter $r$.
== Sources ==
* 
* 
Category:Examples of One-Parameter Families of Curves
One-Parameter Family of Curves for First Order ODE

== Theorem ==
Every one-parameter family of curves is the general solution of some first order ordinary differential equation.
Conversely, every first order ordinary differential equation has as its general solution some one-parameter family of curves.
== Proof ==
From Picard's Existence Theorem, every point in a given rectangle is passed through by some curve which is the solution of a given integral curve of a differential equation.
The equation of this family can be written as:
:$y = \map y {x, c}$
where different values of $c$ give different curves.
The integral curve which passes through $\tuple {x_0, y_0}$ corresponds to the value of $c$ such that:
:$y_0 = \map y {x_0, c}$
Conversely, consider the one-parameter family of curves described by:
:$\map f {x, y, c} = 0$
Differentiate $f$ $x$ to get a relation in the form:
:$\map g {x, y, \dfrac {\d y} {\d x}, c} = 0$
Then eliminate $c$ between these expressions for $f$ and $g$ and get:
:$\map F {x, y, \dfrac {\d y} {\d x} } = 0$
which is a first order ordinary differential equation.
== Sources ==
* : Chapter $\text {I}$: Equations of the First Order and Degree: $1$. Definitions: $(1.3)$
* : $1$: The Nature of Differential Equations: $\S 3$: Families of Curves. Orthogonal Trajectories
Category:First Order ODEs
Category:One-Parameter Families
One-Parameter Family of Surfaces/Examples/Concentric Spheres

== Example of One-Parameter Family of Surfaces ==
The equation:
:$x^2 + y^2 + z^2 = r^2$
is a one-parameter family of concentric spheres whose centers are at the origin of a Cartesian $3$-space and whose radii are the values of the parameter $r$.
== Sources ==
* 
* 
Category:Examples of One-Parameter Families of Surfaces
One-Sided Continuity/Examples/Reciprocal of 1 + e to the Reciprocal of x

== Example of One-Sided Continuity ==
Consider the real function $f$ defined as:
:$f := \dfrac 1 {1 + e^{1 / x} }$
Then from One-Sided Limit of Real Function: Examples: $\dfrac 1 {1 + e^{1 / x} }$:
Hence however $\map f 0$ is defined, $f$ cannot be made to be continuous at $x = 0$.
However, let us define $g$ as:
:$g := \begin {cases} \dfrac 1 {1 + e^{1 / x} } & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
Then $g$ is right-continuous.
Similarly, let us define $h$ as:
:$h := \begin {cases} \dfrac 1 {1 + e^{1 / x} } & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Then $h$ is left-continuous.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity
Category:Examples of One-Sided Continuity
One-Sided Derivative/Examples/Absolute Value Function at Zero

== Examples of One-Sided Derivatives == 
Let $f$ be the real function defined as:
:$\map f x = \size x$
where $\size x$ denotes the absolute value function.
Then:
while the derivative of $f$ at $x = 0$ does not exist.
== Proof ==
Demonstrated in Derivative of Absolute Value Function.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.2$ Derivatives
* 
Category:Examples of One-Sided Derivatives
Category:Absolute Value Function
One-Sided Invertible Hermitian Element of Unital *-Algebra is Invertible

== Theorem ==
Let $\tuple {A, \ast}$ be a unital $\ast$-algebra over $\C$. 
Let $a \in A$ be Hermitian such that either:
:$a b = {\mathbf 1}_A$ for some $b \in A$
or:
:$b a = {\mathbf 1}_A$ for some $b \in A$.
Then $a$ is invertible with inverse $b$. 
Further, $b$ is Hermitian.
== Proof ==
Suppose first that:
:$a b = {\mathbf 1}_A$ for some $b \in A$.
From the definition of an involution and Identity Element in Unital *-Algebra is Hermitian, we have:
:$b^\ast a = {\mathbf 1}_A$
For Left Inverse and Right Inverse is Inverse, we have that:
:$a$ is invertible with inverse $b$.
Further Left Inverse and Right Inverse is Inverse prescribes uniqueness and so:
:$b = b^\ast$
So $b$ is Hermitian.
Suppose now that:
:$b a = {\mathbf 1}_A$ for some $b \in A$. 
From the definition of an involution and Identity Element in Unital *-Algebra is Hermitian, we have:
:$a b^\ast = {\mathbf 1}_A$
For Left Inverse and Right Inverse is Inverse, we again have that:
:$a$ is invertible with inverse $b$ with $b$ Hermitian.
Category:Hermitian Elements of *-Algebras
One-Sided Limit of Real Function/Examples

== Examples of One-Sided Limits of Real Functions ==
=== Example: $\dfrac 1 {1 + e^{1 / x} }$ ===
=== Example: $e^{-1 / x}$ to $0$ from Right ===
</onlyinclude>
=== Example: $e^{-1 / x}$ to $0$ from Left ===
</onlyinclude>
Category:Examples of Limits of Real Functions
One-Sided Limit of Real Function/Examples/Floor of 1 minus Size x

== Example of One-Sided Limit of Real Functions ==
:
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = \floor {1 - \size x}$
Then:
:$\ds \lim_{x \mathop \to 0} \map f x = 0$
but:
:$\map f 0 = 1$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $2$
Category:Examples of Limits of Real Functions