nyuuzyou/wikis huggingface dataset:
Successive Solutions of Phi of n equals Phi of n + 2

== Theorem ==
Let $\phi$ denote the Euler $\phi$ function.
$7$ and $8$ are two successive integers which are solutions to the equation:
:$\map \phi n = \map \phi {n + 2}$
== Proof ==
From Euler Phi Function of Prime:
:$\map \phi 7 = 7 - 1 = 6$
From Euler Phi Function of Prime Power:
:$\map \phi 9 = \map \phi {3^2} = 2 \times 3^{2 - 1} = 6 = \map \phi 7$
From the corollary to Euler Phi Function of Prime Power:
:$\map \phi 8 = \map \phi {2^3} = 2^{3 - 1} = 4$
From Euler Phi Function of Integer:
:$\map \phi {10} = \map \phi {2 \times 5} = 10 \paren {1 - \dfrac 1 2} \paren {1 - \dfrac 1 5} = 10 \times \dfrac 1 2 \times \dfrac 4 5 = 4 = \map \phi 8$
== Sources ==
* : $7$
Category:Euler Phi Function
Successor Mapping is Progressing

== Theorem ==
Let $V$ be a basic universe.
Let $s: V \to V$ denote the successor mapping on $V$:
:$\forall x \in V: \map s x := x \cup \set x$
Then $s$ is a progressing mapping.
== Proof ==
Recall
By Set is Subset of Union:
:$x \subseteq x \cup \set x$
That is:
:$x \subseteq \map s x$
Thus $s$ is by definition a progressing mapping.
== Proof ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications
Category:Successor Mapping
Category:Progressing Mappings
Successor Mapping is Slowly Progressing

== Theorem ==
Let $V$ be a basic universe.
Let $s: V \to V$ denote the successor mapping on $V$:
:$\forall x \in V: \map s x := x \cup \set x$
Then $s$ is a slowly progressing mapping.
== Proof ==
From Successor Mapping is Progressing we have that $s$ is a progressing mapping.
Then we have that:
:$\set x \notin x$
Thus:
:$\card {x \cup \set x} = \card x + 1$
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 9$ Supplement -- optional
Category:Successor Mapping
Category:Slowly Progressing Mappings
Successor Mapping of Peano Structure has no Fixed Point

== Theorem ==
Let $\PP = \struct {P, s, 0}$ be a Peano structure.
Then:
:$\forall n \in P: \map s n \ne n$
That is, the successor mapping has no fixed points.
== Proof ==
Let $T$ be the set:
:$T = \set {n \in P: \map s n \ne n}$
We will use to prove that $T = P$.
=== Part $1$: $0 \in T$ ===
From , a fortiori:
:$\map s 0 \ne 0$
Hence $0 \in T$.
=== Part $2$: $n \in T \implies \map s n \in T$ ===
Let $n \in T$.
Let $m = \map s n$.
Then $n \ne m$, as $\map s n \ne n$.
 $m \notin T$.
That is:
:$\map s m = m$
Then that would mean:
:$\map s n = \map s m = m$
But from , $s$ is injective.
From this contradiction, it follows that $m \in T$.
Hence we have that:
:$n \in T \implies \map s n \in T$
=== Conclusion ===
By , we conclude that $T = P$.
From the definition of $T$, the result follows.
== Sources ==
* : Introduction $\S 4$: The natural numbers: Exercise $1$
* : $\S 12$: The Peano Axioms
Category:Peano's Axioms
Successor Mapping on Natural Numbers has no Fixed Element

== Theorem ==
Let $\N$ denote the set of natural numbers.
Then:
:$\forall n \in \N: n + 1 \ne n$
== Proof ==
Consider the set of natural numbers as defined by the von Neumann construction.
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\N$ is a minimally inductive class under the successor mapping.
Let $s: \N \to \N$ denote this successor mapping:
:$\forall x \in \N: \map s x := x + 1$
 $\exists n \in \N: n = n + 1$
From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $n$ is the greatest element of $\N$.
From Minimally Inductive Class under Progressing Mapping with Fixed Element is Finite it follows that $\N$ is a finite set.
This contradicts the fact that the natural numbers are by definition countably infinite.
== Sources ==
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(d)}$
Category:Natural Numbers
Successor Mapping on Natural Numbers is Progressing

== Theorem ==
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
Let $s: \omega \to \omega$ denote the successor mapping on $\omega$.
Then $s$ is a progressing mapping.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications
Category:Progressing Mappings
Category:Successor Mapping on Natural Numbers is Progressing
Successor Mapping on Natural Numbers is not Surjection

== Theorem ==
Let $f: \N \to \N$ be the successor mapping on the natural numbers $\N$:
:$\forall n \in \N: \map f n = n + 1$
Then $f$ is not a surjection.
== Proof ==
There exists no $n \in \N$ such that $n + 1 = 0$.
Thus $\map f 0$ has no preimage.
The result follows by definition of surjection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Example $5.2$
Category:Surjections
Successor Mapping on Ordinals is Strictly Progressing

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $s$ denote the successor mapping on $\On$:
:$\forall \alpha \in \On: \map s \alpha := \alpha \cup \set \alpha$
where $\alpha$ is an ordinal.
Then $s$ is a strictly progressing mapping.
== Proof ==
From Ordinal is Proper Subset of Successor:
:$\alpha \subsetneqq \alpha^+$
where $\alpha^+$ is identified as $\map s \alpha$.
The result follows by definition of strictly progressing mapping.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers
Category:Successor Mapping
Category:Strictly Progressing Mappings
Category:Ordinals
Successor Set of Ordinal is Ordinal

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $\alpha \in \On$ be an ordinal.
Then its successor set $\alpha^+ = \alpha \cup \set \alpha$ is also an ordinal.
== Proof 1 ==
== Proof 2 ==
Category:Successor Set of Ordinal is Ordinal
Category:Ordinals
Category:Successor Mapping
Successor Set of Ordinal is Ordinal/Proof 1

== Theorem ==
== Proof ==
We have the result that Class of All Ordinals is Minimally Superinductive over Successor Mapping.
Hence $\On$ is a superinductive class the successor mapping.
Hence, by definition of superinductive class:
:$\On$ is closed under the successor mapping.
That is:
:$\forall \alpha \in \On: \alpha^+ \in \On$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.2 \ (2)$
Category:Successor Set of Ordinal is Ordinal
Successor Set of Ordinal is Ordinal/Proof 2

== Theorem ==
== Proof ==
From Ordinal is Transitive, it follows by Successor Set of Transitive Set is Transitive that $\alpha^+$ is transitive.
We now have to show that $\alpha^+$ is strictly well-ordered by the epsilon restriction $\Epsilon {\restriction_{\alpha^+} }$.
So suppose that a subset $A \subseteq \alpha^+$ is non-empty.
Then:
We need to show that $A$ has a smallest element.
We first consider the case where $A \cap \alpha$ is empty.
By equation $\paren 1$, it follows that $A \cap \set \alpha$ is non-empty (because $A$ is non-empty).
Therefore:
:$\alpha \in A$
That is:
:$\set \alpha \subseteq A$
By Union with Empty Set and Intersection with Subset is Subset, equation $\paren 1$ implies that $A \subseteq \set \alpha$.
Therefore, $A = \set \alpha$ by the definition of set equality.
So $\alpha$ is the smallest element of $A$.
We now consider the case where $A \cap \alpha$ is non-empty.
By Intersection is Subset:
:$A \cap \alpha \subseteq \alpha$
By the definition of a well-ordered set, there exists a smallest element $x$ of $A \cap \alpha$.
Let $y \in A$.
If $y \in \alpha$, then $y \in A \cap \alpha$
Therefore, by the definition of the smallest element, either $x \in y$ or $x = y$.
Otherwise, $y = \alpha$, and so $x \in \alpha = y$.
That is, $x$ is the smallest element of $A$.
== Sources ==
* : $\S 7.24$
Category:Successor Set of Ordinal is Ordinal
Successor Set of Ordinary Transitive Set is Ordinary

== Theorem ==
Let $x$ be a transitive set which is also ordinary.
Let $x^+$ denote the successor set of $x$:
:$x^+ = x \cup \set x$
Then $x^+$ is also an ordinary set.
== Proof ==
By definition of ordinary set:
:$x \notin x$
 $x^+ \in x^+$.
Then we have:
;Let $x^+ \in x$
We have that $x$ is transitive.
Hence:
:$x^+ \subseteq x$
;Let $x^+ = x$
We have that:
$x^+ = x \cup \set x$
Hence by definition of union of set of sets:
:$x^+ \subseteq x$
Hence by Proof by Cases:
:$x^+ \subseteq x$
But we have:
:$x \in x^+$
and because $x^+ \subseteq x$ it follows that:
:$x \in x$
But this contradicts our supposition that $x \notin x$.
Hence by Proof by Counterexample:
:$x^+ \notin x^+$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.2$
Category:Ordinary Sets
Category:Transitive Classes
Category:Successor Mapping
Successor Set of Transitive Set is Transitive

== Theorem ==
Let $S$ be a transitive set.
Then its successor set $S^+ = S \cup \set S$ is also transitive.
== Proof ==
Recall that $S$ is transitive :
:$x \in S \implies x \subseteq S$
Hence:
Then:
Thus we have:
:$x \in S \implies x \subseteq S$
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.1$
Category:Successor Mapping
Category:Transitive Classes
Successor Sets of Linearly Ordered Set Induced by Convex Component Partition

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\relcomp S {A^* \cup B^*}$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
:$M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Then each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$, the set of strict upper bounds for $A_\alpha$.
Similarly for $B_\beta$.
That immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
== Proof ==
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then $A_\alpha \cap {S_\alpha}^-$ contains exactly $1$ point, say $p$.
This belongs to the complement in $S$ of the closed set $\paren {B^*}^-$.
Hence there exists a neighborhood $\openint x y$ of $p$ which is disjoint from $\paren {B^*}^-$.
Then:
:$\openint x y \cap S_\alpha \ne \O$
and so:
:$\openint p y \ne \O$
But $\openint p y$ is disjoint from both $A^*$ and $B^*$.
Thus there must exist some $C_\gamma$ which contains $\openint p y$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $5$
Category:Linearly Ordered Spaces
Successor is Less than Successor/Sufficient Condition/Proof 1

== Theorem ==
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$
== Proof ==
Suppose $y^+ \in x^+$.
By the definition of successor, $y^+ \in x \lor y^+ = x$.
Suppose $y^+ = x$.
By Ordinal is Less than Successor, $y \in x$.
Suppose $y^+ \in x$.
By Ordinal is Less than Successor, $y \in y^+$.
By Ordinal is Transitive, $y \in x$.
Category:Ordinals
Successor is Less than Successor/Sufficient Condition/Proof 2

== Theorem ==
Let $x$ and $y$ be ordinals and let $x^+$ denote the successor set of $x$.
Let $x^+ \in y^+$.
Then:
: $x \in y$
== Proof ==
First note that by Successor Set of Ordinal is Ordinal, $x^+$ and $y^+$ are ordinals.
Let $x^+ \in y^+$.
Then since $y^+$ is transitive, $x^+ \subseteq y^+$.
Thus $x \in y$ or $x = y$.
If $x = y$ then $x^+ \in x^+$, contradicting Ordinal is not Element of Itself.
Thus $x \in y$.
Category:Ordinals
Successor is Supremum

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $x, y \in S$.
Then:
:$x \preceq y$ $y = \map \sup {x, y}$
where $\map \sup {x, y}$ is the supremum of $x$ and $y$
== Proof ==
=== Necessary Condition ===
Let $x \preceq y$.
By :
:$y \preceq y$
Hence $y$ is an upper bound of $\set {x, y}$ by definition.
Let $z$ be an upper bound of $\set {x, y}$.
By definition of upper bound:
:$y \preceq z$
Hence $y$ is the supremum of $x$ and $y$ by definition.
=== Sufficient Condition ===
Let $y$ be the supremum of $x$ and $y$.
Then $x \preceq y$ by definition of a supremum.
Category:Orderings
Category:Suprema
Successor of Omega

== Theorem ==
:$\omega + 1 = \set {0, 1, 2, \ldots; \omega}$
where $\omega$ is the minimally inductive set and $\omega + 1$ is the successor of $\omega$.
Note the use of the semicolon; this is the notation for multipart infinite sets.
== Proof ==
== Comment ==
It is customary to use $\omega + 1$ rather than $\omega^+$ for transfinite arithmetic. 
However, it needs to be borne in mind that this is not conventional natural number addition. 
For example, $\omega + 1 \ne 1 + \omega$.
== Sources ==
* p. $205$
* 
Category:Transfinite Arithmetic
Successor of Ordinal Smaller than Limit Ordinal is also Smaller

== Theorem ==
Let $\On$ denote the class of all ordinals.
Let $\lambda \in \On$ be a limit ordinal.
Then:
:$\forall \alpha \in \On: \alpha < \lambda \implies \alpha^+ < \lambda$
== Proof 1 ==
== Proof 2 ==
Category:Successor of Ordinal Smaller than Limit Ordinal is also Smaller
Category:Ordinals
Category:Limit Ordinals
Category:Successor Mapping
Successor of Ordinal Smaller than Limit Ordinal is also Smaller/Proof 1

== Theorem ==
== Proof ==
Let $\lambda$ be a limit ordinal such that $\alpha < \lambda$.
From Successor of Element of Ordinal is Subset
Then as $\alpha^+$ is the successor set of $\alpha$ it follows that:
:$\alpha^+ \le \lambda$
But as $\lambda$ is not a successor ordinal:
:$\alpha^+ \ne \lambda$
Hence:
:$\alpha^+ < \lambda$
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.19$
Category:Successor of Ordinal Smaller than Limit Ordinal is also Smaller
Successor of Ordinal Smaller than Limit Ordinal is also Smaller/Proof 2

== Theorem ==
== Proof ==
Because $\lambda$ is a limit ordinal:
:$\lambda \ne \alpha^+$
Moreover, by Successor of Element of Ordinal is Subset:
:$\alpha \in \lambda \implies \alpha^+ \subseteq \lambda$
Therefore by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$\alpha^+ \subset \lambda$ and $\alpha^+ \in \lambda$
Category:Successor of Ordinal Smaller than Limit Ordinal is also Smaller
Successor to Natural Number

== Theorem ==
Let $\N_{> 0}$ be the 1-based natural numbers:
:$\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$
Let $<$ be the ordering on $\N_{> 0}$:
:$\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$
Let $a \in \N_{>0}$.
Then there exists no natural number $n$ such that $a < n < a + 1$.
== Proof ==
Using the following axioms:
Suppose that $\exists n \in \N_{>0}: a < n < a + 1$.
Then by the definition of ordering on natural numbers:
By Axiom $D$, either:
:$y = 1$
or:
:$y = t + 1$ for some $t \in \N_{>0}$
Then either:
:$x + 1 = 1$ when $y = 1$
or:
:$x + \left({t + 1}\right) = \left({x + t}\right) + 1 = 1$ when $y = t + 1$
Both of these conclusions violate Natural Number is Not Equal to Successor.
Hence the result.
== Sources ==
* : $2.2$: Theorem $2.15$
Category:Natural Numbers/1-Based
Sufficient Condition for 5 to divide n^2+1

== Theorem ==
Let:
where $\nmid$ denotes non-divisibility.
Then:
:$5 \divides n^2 + 1$
where $\divides$ denotes divisibility.
== Proof ==
We have that:
So either:
:$n \equiv 2 \pmod 5$
or:
:$n \equiv 3 \pmod 5$
and so:
or:
== Sources ==
* : $\text {3-2}$ Fermat's Little Theorem: Exercise $6$
Category:Modulo Arithmetic
Sufficient Condition for C* Identity

== Theorem ==
Let $\tuple {A, \norm {\, \cdot \,} }$ be a Banach algebra over $\C$.
Let $\ast$ be an involution on $A$ such that:
:$\norm a^2 \le \norm {a a^\ast}$
Then $\tuple {A, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
== Proof ==
Let $a \in A$.
From Zero Vector in *-Algebra is Hermitian and , we have:
:$\norm a^2 = \norm {a a^\ast}$
for $a = {\mathbf 0}_A$.
Now take $a \ne {\mathbf 0}_A$ so that $\norm a \ne 0$. 
From the definition of an algebra norm:
:$\norm {a a^\ast} \le \norm a \norm {a^\ast}$
Hence:
:$\norm a^2 \le \norm a \norm {a^\ast}$ 
Dividing by $\norm a$ we have:
:$\norm a \le \norm {a^\ast}$
for each $a \in A \setminus \set { {\mathbf 0}_A}$.
Replacing $a$ with $a^\ast$ we get:
:$\norm {a^\ast} \le \norm {a^{\ast \ast} } = \norm a$
Hence we have:
:$\norm a = \norm {a^\ast}$
Hence from:
:$\norm {a a^\ast} \le \norm a \norm {a^\ast}$
we have:
:$\norm {a a^\ast} \le \norm a^2$
Then, we have:
:$\norm a^2 \le \norm {a a^\ast} \le \norm a^2$
giving that:
:$\norm a^2 = \norm {a a^\ast}$
Hence the $C^\ast$ identity holds, so $\tuple {A, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
Category:C*-Algebras
Sufficient Condition for Quaternion Multiplication to Commute

== Theorem ==
In general, quaternion multiplication does not commute.
But, for $\mathbf x,\mathbf y \in \H$, $\mathbf x \times \mathbf y = \mathbf y \times \mathbf x$ if any one of the following conditions hold:
== Proof ==
=== Proof of $\paren 1$ ===
It follows directly from Complex Numbers form Subfield of Quaternions and Complex Multiplication is Commutative.
=== Proof of $\paren 2$ ===
Let $\mathbf x \in \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a\in \R}$.
Let $\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k: e, f, g, h \in \R$.
Then:
The above is the proof of $\paren {2a}$, and the proof of $\paren {2b}$ is similar.
=== Proof of $\paren 3$ ===
Category:Quaternions
Sufficient Condition for Quotient Group by Intersection to be Abelian

== Theorem ==
Let $G$ be a group.
Let $N$ and $K$ be normal subgroups of $G$.
Let the quotient groups $G / N$ and $G / K$ be abelian.
Then the quotient group $G / \paren {N \cap K}$ is also abelian.
== Proof ==
From Intersection of Normal Subgroups is Normal, we have that $N \cap K$ is normal in $G$.
We are given that $G / N$ and $G / K$ are abelian.
Hence:
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $16$
Category:Quotient Groups
Sufficient Condition for Square of Product to be Triangular

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $2 n^2 \pm 1 = m^2$ be a square number.
Then $\paren {m n}^2$ is a triangular number.
== Proof ==
That is, either:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2 - 1} \paren {2 n^2} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2 - 1}$
or:
:$\paren {m n}^2 = \dfrac {\paren {2 n^2} \paren {2 n^2 + 1} } 2$
and so:
:$\paren {m n}^2 = T_{2 n^2}$
Hence the result.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $6 \ \text {(a)}$
Category:Triangular Numbers
Category:Square Numbers
Sufficient Condition for Stationary Point to be Local Maximum

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let $f$ have a stationary point at $\xi \in \openint a b$.
Let the second derivative of $f$ at $\xi$ be (strictly) negative.
Then $\xi$ is a local maximum.
== Proof ==
== Also see ==
* Sufficient Condition for Stationary Point to be Local Minimum
* Necessary Condition for Stationary Point to be Local Maximum
* Necessary Condition for Stationary Point to be Local Minimum
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: $3.4.2 \ (1)$
* 
* 
Category:Stationary Points
Sufficient Condition for Stationary Point to be Local Minimum

== Theorem ==
Let $f$ be a real function which is twice differentiable on the open interval $\openint a b$.
Let $f$ have a stationary point at $\xi \in \openint a b$.
Let the second derivative of $f$ at $\xi$ be (strictly) positive.
Then $\xi$ is a local minimum.
== Proof ==
== Also see ==
* Sufficient Condition for Stationary Point to be Local Maximum
* Necessary Condition for Stationary Point to be Local Minimum
* Necessary Condition for Stationary Point to be Local Maximum
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Maximum, Minimum and Point of Inflection
* : $3$: Elementary Analytic Methods: $3.4$ Limits, Maxima and Minima: $3.4.2 \ (1)$
* 
* 
Category:Stationary Points
Sufficient Condition for Twice Differentiable Functional to have Minimum

== Theorem ==
Let $J$ be a twice differentiable functional.
Let $J$ have an extremum for $y=\hat y$.
Let the second variation $\delta^2 J \sqbrk {\hat y; h}$ be strongly positive $h$.
Then $J$ acquires the minimum for $y = \hat y$ .
== Proof ==
By assumption, $J$ has an extremum for $y = \hat y$:
:$\delta J \sqbrk {\hat y; h} = 0$
The increment is expressible then as:
:$\Delta J \sqbrk {\hat y; h} = \delta^2 J \sqbrk {\hat y; h} + \epsilon \size h^2$
where $\epsilon \to 0$ as $\size h \to 0$.
By assumption, the second variation is strongly positive:
:$\delta^2 J \sqbrk {\hat y; h} \ge k \size h^2, \quad k \in \R_{>0}$
Hence:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2$
What remains to be shown is that there exists a set of $h$ such that $\epsilon$ is small enough so that is always positive.
Since $\epsilon \to 0$ as $\size h \to 0$, there exist $c \in \R_{>0}$, such that:
:$\size h < c \implies \size \epsilon < \dfrac 1 2 k$
Choose $h$ such that this inequality holds.
Then
Therefore:
:$\Delta J \sqbrk {\hat y; h} \ge \paren {k + \epsilon} \size h^2 > \dfrac 1 2 k \size h^2 $
For $k \in \R_{>0}$ and $\size h \ne 0$ is always positive.
Thus, there exists a neighbourhood around $y = \hat y$ where the increment is always positive: 
:$\exists c \in \R_{>0}: \size h < c \implies \Delta J \sqbrk {\hat y; h} > 0$ 
and $J$ has a minimum for $y = \hat y$.
== Sources ==
* : $\S 5.24$: Quadratic Functionals. The Second Variation of a Functional 
Category:Calculus of Variations
Sufficient Condition for Vector Equals Inverse iff Zero

== Theorem ==
Let $\struct {\mathbf V, +, \circ}_\GF$ be a vector space over a field $\GF$, as defined by the vector space axioms.
Let $\GF$ be infinite.
Then:
:$\forall \mathbf v, -\mathbf v \in \mathbf V: \mathbf v = - \mathbf v \iff \mathbf v = \mathbf 0$
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Utilizing the vector space axioms:
We have that $\GF$ is infinite.
By Characteristic of Division Ring is Zero or Prime:
:$\Char \GF = 0$
so $-2 \cdot 1_\GF \ne 0$.
Thus from Vector Product is Zero only if Factor is Zero, $\mathbf v = \mathbf 0$.
Category:Vector Algebra