nyuuzyou/wikis huggingface dataset:
Characteristic Subgroup is Transitive

== Theorem ==
Let $G$ be a group.
Let $H$ be a characteristic subgroup of $G$.
Let $K$ be a characteristic subgroup of $H$.
Then $K$ is a characteristic subgroup of $G$.
== Proof ==
Let $\phi: G \to G$ be a group automorphism.
Since $H$ is a characteristic subgroup of $G$, we have:
:$\phi \sqbrk H = H$
Thus, from Group Homomorphism Preserves Subgroups, $\phi {\restriction_H}$, the restriction of $\phi$ to $H$, is an automorphism of $H$.
Now since $K$ is a characteristic subgroup of $H$, we have that:
:$\phi {\restriction_H} \sqbrk K = K$
but this immediately implies that:
:$\phi \sqbrk K = K$
by definition of the restriction $\phi {\restriction_H}$.
That is, $K$ is a characteristic subgroup of $G$ as well.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \epsilon$
Category:Characteristic Subgroups
Characteristic Subgroup of Normal Subgroup is Normal

== Theorem ==
Let $G$ be a group.
Let $N\leq G$ be normal.
Let $H\leq N$ be characteristic.
Then $H$ is normal in $G$.
== Proof ==
Let $g \in G$.
Because $N$ is normal, conjugation by $g$ is an automorphism of $N$.
Because $H$ is characteristic in $N$, $g H g^{-1} = H$.
Thus $H$ is normal in $G$.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 64 \epsilon$
Category:Characteristic Subgroups
Category:Normal Subgroups
Characteristic of Cayley Table of Left Operation

== Theorem ==
Let $S$ be a finite set.
Let $\leftarrow$ denote the left operation on $S$.
The Cayley table of the algebraic structure $\struct {S, \leftarrow}$ is characterised by the fact that each row contains just one distinct element.
== Proof ==
A row of a Cayley table headed by $x$ contains all those elements of the form $x \leftarrow y$.
By definition of the left operation:
:$x \leftarrow y = x$
Hence the result.
== Also see ==
* Characteristic of Cayley Table of Right Operation
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.9$
Category:Cayley Tables
Category:Left Operation
Characteristic of Cayley Table of Right Operation

== Theorem ==
Let $S$ be a finite set.
Let $\rightarrow$ denote the right operation on $S$.
The Cayley table of the algebraic structure $\struct {S, \rightarrow}$ is characterised by the fact that each column contains just one distinct element.
== Proof ==
A column of a Cayley table headed by $y$ contains all those elements of the form $x \rightarrow y$.
By definition of the right operation:
:$x \rightarrow y = y$
Hence the result.
== Also see ==
* Characteristic of Cayley Table of Left Operation
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.9$
Category:Cayley Tables
Category:Right Operation
Characteristic of Common Logarithm of Number Greater than 1

== Theorem ==
Let $x \in \R_{>1}$ be a (strictly) positive real number greater than $1$.
The characteristic of its common logarithm $\log_{10} x$ is equal to one less than the number of digits to the left of the decimal point of $x$.
== Proof ==
Let $x$ be expressed in scientific notation:
:$x = a \times 10^e$
where:
:$1 \le a < 10$
:$e \in \Z_{\ge 0}$
From Range of Common Logarithm of Number between 1 and 10:
:$0 \le \log_{10} a < 1$
The characteristic of $\log_{10} x$ equals $\map {\log_{10} } {10^e} = e$.
Thus the characteristic of $\log_{10} x$ is equal to the exponent of $x$.
By Multiplication by Power of 10 by Moving Decimal Point, multiplication by $10^e$ is the same as moving the decimal point $e$ places to the right.
Hence $a \times 10^e$ has $e$ more digits to the left of the decimal point than $a$ does.
That is: $e + 1$.
Hence the result.
== Sources ==
* : Chapter $1$: Logarithms
Category:Common Logarithms
Characteristic of Common Logarithm of Number Less than 1

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number such that $x < 1$.
The characteristic of its common logarithm $\log_{10} x$ is equal to one less than the number of zero digits to the immediate right of the decimal point of $x$.
== Proof ==
Let $x$ be expressed in scientific notation:
:$x = a \times 10^{-e}$
where:
:$1 \le a < 10$
:$e \in \Z_{>0}$
From Range of Common Logarithm of Number between 1 and 10:
:$0 \le \log_{10} a < 1$
The characteristic of $\log_{10} x$ equals $\map {\log_{10} } {10^{-e} } = \overline e$.
Thus the characteristic of $\log_{10} x$ is equal to the exponent of $x$.
When $e = -1$, we have that:
:$0 \cdotp 1 \le x < 1$
and so $x$ is in the form:
:$x = 0 \cdotp rstu \cdotp$
where $rstu \ldots$ are the digits of $x$ following the decimal point such that $r > 0$.
Thus when $e = -1$ there are no zero digits to the immediate right of the decimal point of $x$
That is $e$ is one less than the number of zero digits to the immediate right of the decimal point of $x$.
By Multiplication by Power of 10 by Moving Decimal Point, multiplication by $10^{-e}$ is the same as moving the decimal point $e$ places to the right.
Hence $a \times 10^{-e}$ has $e - 1$ more digits to the left of the decimal point than $a$ does.
That is: $e - 1$.
Hence the result.
== Sources ==
* : Chapter $1$: Logarithms
Category:Common Logarithms
Characteristic of Division Ring is Zero or Prime

== Theorem ==
Let $\struct {D, +, \circ}$ be a division ring.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number.
== Proof ==
By definition, a division ring has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
Thus it follows that $\Char D$ is $0$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm
Category:Division Rings
Characteristic of Envelope of Family of Curves

== Theorem ==
Let $\FF$ be a one-parameter family of curves defined by a parameter $m$.
Let elements $A$ and $B$ of $\FF$ have parameters which differ by a small amount $\delta m$.
If $\delta m$ is sufficiently small, $A$ and $B$ will intersect.
The locus of the points of intersection of elements of $\FF$ as $\delta m \to 0$ becomes the envelope of $\FF$.
== Proof ==
== Sources ==
* 
* 
Category:Envelopes
Characteristic of Extending Operation

== Theorem ==
Let $E$ be an extending operation.
Then there exists a mapping $F$ on the class of all ordinals $\On$ such that:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
where:
:$F \restriction \alpha$ denotes the restriction of $F$ to $\alpha$
:$\alpha^+$ denotes the successor ordinal of $\alpha$.
== Proof ==
Let $M$ be the class which is minimally superinductive under $E$.
Recall the Transfinite Recursion Theorem:
where:
:for an arbitrary ordinal $\alpha$, $M_\alpha$ denotes the $\alpha$th element of $M$ under the well-ordered class $\struct {M, \subseteq}$
:$K_{II}$ denotes the class of all limit ordinals.
We know that $M_0 = 0$, so $M_0$ is a $0$-sequence.
Let $M_\alpha$ be an arbitrary $\alpha$-sequence.
Then from $(2)$ above, $M_{\alpha^+}$ is an $\alpha^+$-sequence.
Let $\lambda$ be a limit ordinal.
Suppose that, for each $\alpha < \lambda$, $M_\alpha$ is an $\alpha$-sequence.
Then by Length of Union of Chain of Ordinal Sequences, $M_\lambda$ is a $\lambda$-sequence.
Hence by the Second Principle of Transfinite Induction:
:for every $\alpha \in \On$, $M_\alpha$ is an ordinal sequence of length $\alpha$.
Let $F = \bigcup M$ be the union of $M$.
Then by Length of Union of Chain of Ordinal Sequences again, $F$ is a mapping on $\On$.
Because each $M_\alpha \subseteq F$ and $\Dom {M_\alpha} = \alpha$, it follows that $M_\alpha = F \restriction \alpha$.
Hence from Union of Nest of Ordinal Sequences which is Proper Class:
:$\forall \alpha \in \On: F \restriction \alpha^+ = \map E {F \restriction \alpha}$
== Sources ==
* : Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Theorem $5.3$
Category:Extending Operations
Characteristic of Field by Annihilator

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Then of the following two cases, exactly one applies:
=== Characteristic Zero ===
=== Prime Characteristic ===
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Characteristics of Fields
Category:Characteristic of Field by Annihilator
Characteristic of Field by Annihilator/Characteristic Zero

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Suppose that:
:$\map {\mathrm {Ann} } F = \set 0$
That is, the annihilator of $F$ consists of the zero only.
Then:
:$\Char F = 0$
That is, the characteristic of $F$ is zero.
== Proof ==
Let the zero of $F$ be $0_F$ and the unity of $F$ be $1_F$.
By definition of characteristic, $\Char F = 0$ :
:$\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$
That is, there exists no $n \in \Z, n > 0$ such that $n \cdot r = 0_F$ for all $r \in F$.
But note that $\forall r \in F: 0 \cdot r = 0_F$ by definition of integral multiple.
 there exists a non-zero element of $\map {\mathrm {Ann} } F$.
From Non-Trivial Annihilator Contains Positive Integer, $\map {\mathrm {Ann} } F$ must contain a (strictly) positive integer.
But this would contradict the statement that $\Char F = 0$.
So it follows that:
:$\map {\mathrm {Ann} } F = \set 0 \iff \Char F = 0$
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Characteristics of Fields
Characteristic of Field by Annihilator/Prime Characteristic

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Suppose that:
:$\exists n \in \map {\mathrm {Ann} } F: n \ne 0$
That is, there exists (at least one) non-zero integer in the annihilator of $F$.
If this is the case, then the characteristic of $F$ is non-zero:
:$\Char F = p \ne 0$
and the annihilator of $F$ consists of the set of integer multiples of $p$:
:$\map {\mathrm {Ann} } F = p \Z$
where $p$ is a prime number.
== Proof ==
Let $A := \map {\mathrm {Ann} } F$.
We are told that:
:$\exists n \in A: n \ne 0$
Consider the set $A^+ \set {n \in A: n > 0}$.
From Non-Trivial Annihilator Contains Positive Integer we have that $A^+ \ne \O$.
As $A^+ \subseteq \N$ it follows from the well-ordering principle that $A^+$ has a least value $p$, say.
 $p$ is not a prime number.
Then $p$ can be expressed as $p = a b$ where $1 < a, b < p$.
But then either $a \in A$ or $b \in A$, and so $p$ is not the minimal positive element of $A$ after all.
So from this contradiction it follows that $p$ is necessarily prime.
Next let $n \in \Z$.
Then:
So all multiples of $p$ are in $A$.
Finally, suppose $k \in A$.
By the Division Theorem $k = q p + r$ where $0 \le r < p$.
Then:
So $r \in A$ contradicting the stipulation that $p$ is the smallest positive element of $A$.
Hence all and only multiples of $p$ are in $A$.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties
Category:Characteristics of Fields
Characteristic of Field is Zero or Prime

== Theorem ==
Let $F$ be a field.
Then the characteristic of $F$ is either zero or a prime number.
== Proof ==
From the definition, a field is a ring with no zero divisors.
So by Characteristic of Finite Ring with No Zero Divisors, if $\Char F \ne 0$ then it is prime.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Theorem $30$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 89 \alpha$
Category:Characteristics of Fields
Characteristic of Finite Ring is Non-Zero

== Theorem ==
Let $\struct {R, +, \circ}$ be a finite ring with unity.
Then the characteristic of $R$ is not zero.
== Proof ==
We have that $\struct {R, +, \circ}$ is finite, so its additive group $\struct {R, +}$ is likewise finite.
The result follows by Element of Finite Group is of Finite Order and the definition of characteristic.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Theorem $29$
Category:Finite Rings
Characteristic of Finite Ring with No Zero Divisors

== Theorem ==
Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
:$(1): \quad n$ must be a prime number
:$(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the cyclic group of order $n$.
== Proof 1 ==
== Proof 2 ==
Category:Finite Rings
Category:Rings with Unity
Category:Characteristics of Rings
Category:Characteristic of Finite Ring with No Zero Divisors
Characteristic of Finite Ring with No Zero Divisors/Proof 2

== Theorem ==
== Proof ==
Suppose $\Char R = n$ where $n$ is composite.
Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
First note that:
Then:
as $R$ has no proper zero divisors.
But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$.
So if $\Char R = n$ it follows that $n$ must be prime.
Now let $x \in R^*$.
Then by Characteristic times Ring Element is Ring Zero, $n \cdot x = 0_R$.
It follows from Element to Power of Multiple of Order is Identity that:
:$\order x \divides n$
Since $n$ is prime, either $\order x = 1$ or $\order x = n$.
It cannot be $1$, from Null Ring iff Characteristic is One, so the result follows.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Theorem $30$
* : $\S 61.1$ Characteristic of an integral domain or field
Category:Characteristic of Finite Ring with No Zero Divisors
Characteristic of Galois Field is Prime

== Theorem ==
Let $\GF$ be a Galois field. 
Then the characteristic of $\GF$ is a prime number.
== Proof ==
By Characteristic of Field is Zero or Prime, it follows that $\Char \GF$ is $0$ or a prime number. 
By Finite Field has Non-Zero Characteristic:
:$\Char \GF \ne 0$
Thus $\Char \GF$ is a prime number.
Category:Galois Fields
Characteristic of Integral Domain is Zero or Prime

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D$ is either $0$ or a prime number.
== Proof ==
By definition, an integral domain has no proper zero divisors.
If $\struct {D, +, \circ}$ is finite, then from Characteristic of Finite Ring with No Zero Divisors, $\Char D$ is prime.
On the other hand, suppose $\struct {D, +, \circ}$ is not finite.
Then there are no $x, y \in D, x \ne 0 \ne y$ such that $x + y = 0$.
Thus it follows that $\Char D$ is $0$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm
* 
* 
Category:Integral Domains
Characteristic of Interior Point of Circle whose Center is Origin

== Theorem ==
Let $\CC$ be a circle of radius $r$ whose center is at the origin $O$ of a Cartesian plane.
Let $P = \tuple {x, y}$ be a point in the plane of $\CC$.
Then $P$ is in the interior of $\CC$ :
:$x^2 + y^2 - r^2 < 0$
== Proof ==
Let $d$ be the distance of $P$ from $O$.
Then by definition of interior of $\CC$:
:$P$ is in the interior of $\CC$ $d^2 < r^2$
and the result follows.
== Sources ==
* : Chapter $\text {III}$. The Circle: $10$.
Category:Circles
Characteristic of Ordered Integral Domain is Zero

== Theorem ==
Let $\struct {D, +, \circ}$ be an ordered integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\Char D$ be the characteristic of $D$.
Then $\Char D = 0$.
Let $g: \Z \to D$ be the mapping defined as:
:$\forall n \in \Z: \map g n = n \cdot 1_D$
where $n \cdot 1_D$ is defined as the $n$th power of $1_D$.
Then $g$ is the only monomorphism from the ordered ring $\Z$ onto the ordered ring $D$.
== Proof ==
By Properties of Ordered Ring $(5)$:
:$\forall n \in \Z_{>0}: n \cdot 1_D > 0$
Thus:
:$\forall p > 0: \Char D \ne p$
Hence:
:$\Char D = 0$
and so $g$ is a monomorphism from $\Z$ into $D$.
Also, if $m < p$, then $p - m \in \Z_+$, so:
:$p \cdot 1_D - m \cdot 1_D > 0_D$
Hence:
:$\map g m < \map g p$
Thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.11$
Category:Ordered Integral Domains
Characteristic of Quadratic Equation that Represents Circle

== Theorem ==
A quadratic equation in $2$ variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
describes a circle embedded in the Cartesian plane :
:$(1): \quad h = 0$
:$(2): \quad a = b \ne 0$
== Proof ==
=== Necessary Condition ===
Consider the equation of a circle:
:$A \paren {x^2 + y^2} + B x + C y + D = 0$
which is the equation of a circle with radius $R$ and center $\tuple {a, b}$, where:
:$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
:$\tuple {a, b} = \tuple {\dfrac {-B} {2 A}, \dfrac {-C} {2 A} }$
provided:
:$A > 0$
:$B^2 + C^2 \ge 4 A D$
Thus it is seen that the coefficient for $x^2$ is the same as that for $y^2$, and there is no $x y$ term.
As can be seen, this matches the form as described: $a = b$ and $h = 0$.
=== Sufficient Condition ===
Consider the general quadratic equation in $2$ variables:
:$a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
Let us set $a = b$ and $h = 0$.
We obtain:
:$a x^2 + a y^2 + 2 g x + 2 f y + c = 0$
Assuming $a \ne 0$, we get:
It is seen that $(1)$ is Equation of Circle in Cartesian Plane, where:
:the represents the distance from $\tuple {x, y}$ to the point $\tuple {-\dfrac g a, -\dfrac f a}$ which is its center
:the represents the radius $\sqrt {\dfrac {g^2 + f^2 - a c} {a^2} }$.
If $a$, $g$, $f$ and $c$ are all real numbers, then the center is always a point in the cartesian plane.
The radius, however, is real only if $g^2 + f^2 > a c$.
If $g^2 + f^2 < a c$, then no real $x$ and $y$ can satisfy $(1)$, as $\paren {x + \dfrac g a}^2 + \paren {y + \dfrac f a}^2 + \dfrac {a c - g^2 + f^2} {a^2}$ is always strictly positive.
Hence the locus of $(1)$ is a virtual circle.
If $g^2 + f^2 = a c$, then $(1)$ degenerates to:
:$\paren {x + \dfrac g a}^2 + \paren {y + \dfrac f a}^2 = 0$
which is satisfied for $\tuple {x, y} = \tuple {-\dfrac g a, -\dfrac f a}$ only.
Hence the radius is $0$ and the locus of $(1)$ is a point-circle.
If $a = 0$ then the general quadratic equation degenerates to:
:$2 g x + 2 f y + c = 0$
which is the equation of a straight line.
== Sources ==
* : Chapter $\text {III}$. The Circle: $14$. To find the equation of the circle whose centre is $\tuple {\alpha, \beta}$ and radius $r$
Category:Circles
Characteristic of Quadratic Equation that Represents Two Straight Lines

== Theorem ==
Consider the quadratic equation in $2$ variables:
:$(1): \quad a x^2 + b y^2 + 2 h x y + 2 g x + 2 f y + c = 0$
where $x$ and $y$ are independent variables.
Then $(1)$ represents $2$ straight lines its discriminant equals zero:
:$a b c + 2 f g h - a f^2 - b g^2 - c h^2 = 0$
This can also be expressed in the form of a determinant:
:$\begin {vmatrix} a & h & g \\ h & b & f \\ g & f & c \end {vmatrix} = 0$
== Proof ==
Suppose that $a \ne 0$.
We have:
In order that the second part is a perfect square in $y$, it is necessary that:
Conversely, if $(3)$ is true, then $(2)$ can be expressed in the form of a Difference of Two Squares:
Hence $(2)$ has $2$ factors, which can be seen to be the equations of straight lines.
Let $a = 0$ but $b \ne 0$.
Then: 
In order that the second part is a perfect square in $x$, it is necessary that:
it is noted that $(4)$ is the same as $(3)$ but with $a = 0$.
Suppose $a = 0$ and $b = 0$ but $h \ne 0$.
Then:
and it is seen that in order for $(1)$ to be divisible into the $2$ required factors:
:$2 \paren {h x + f} \paren {h y + g} = 0$
it is necessary for $c h = 2 f g$.
This is again the same as $(3)$ when you set $a = 0$ and $b = 0$.
If $a = 0$ and $b = 0$ and $h = 0$, then $(1)$ is not a quadratic equation.
All cases have been covered.
Finally we see that:
== Sources ==
* : Chapter $\text {II}$. The Straight Line: $14$. Condition that the general quadratic equation of the second degree should represent two straight lines
Category:Quadratic Equations
Category:Straight Lines
Characteristic of Ring of Integers Modulo Prime

== Theorem ==
Let $\struct {\Z_p, +, \times}$ be the ring of integers modulo $p$, where $p$ is a prime number.
The characteristic of $\struct {\Z_p, +, \times}$ is $p$.
== Proof ==
From Ring of Integers Modulo Prime is Field we have that $\struct {\Z_p, +, \times}$ is a field.
So Characteristic of Finite Ring with No Zero Divisors applies, and so the characteristic of $\struct {\Z_p, +, \times}$ is prime.
The result follows.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Example $23$
Category:Ring of Integers Modulo m
Category:Finite Rings
Characteristic of Subfield of Complex Numbers is Zero

== Theorem ==
The characteristic of any subfield of the field of complex numbers is $0$.
== Proof ==
 to the contrary.
Let $K$ be a subfield of $\C$ such that $\Char K = n$ where $n \in \N, n > 0$.
Then from Characteristic times Ring Element is Ring Zero:
:$\forall a \in K: n \cdot a = 0$
But as $K$ is a subfield of $\C$ it follows that $K \subseteq \C$ which means:
:$\exists a \in \C: n \cdot a = 0$
Thus, by definition of characteristic:
:$0 < \Char \C \le n$
But $\C$ is infinite and so $\Char \C = 0$.
From that contradiction follows the result.
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Example $24$
Category:Subfields
Category:Characteristics of Fields
Category:Complex Numbers
Characteristic times Ring Element is Ring Zero

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
Let the characteristic of $R$ be $n$.
Then:
:$\forall a \in R: n \cdot a = 0_R$
== Proof ==
If $a = 0_R$ then $n \cdot a = 0_R$ is immediate.
So let $a \in R: a \ne 0_R$.
Then:
== Sources ==
* : Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Theorem $28$
Category:Characteristics of Rings
Characteristics of Birkhoff-James Orthogonality

== Theorem ==
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed linear space.
Let $x, y \in V$.
Then $x$ and $y$ are Birkhoff-James orthogonal either:
:$(1): \quad x = 0$
or:
:$(2): \quad$ there exists a continuous functional $ f$ on $\struct {V, \norm {\,\cdot\,} }$ such that:
::::$\norm f = 1$
::::$\map f x = \norm x$
::::$\map f y = 0$
== Proof ==
=== Necessary Condition ===
Let $x \perp_B y$.
Let $V' \subset V$ be the subspace spanned by $x$ and $y$.
Define $\overline f$ on $V'$ as:
:$\map {\overline f} {a x + b y} = a \norm x$
for $a$ and $b$ scalars.
Clearly, $\overline f$ is linear and:
:$\map {\overline f} x = \norm x$
:$\map {\overline f} y = 0$
Further:
proving that $\overline f$ is a bounded functional of norm $1$.
Now by Hahn-Banach Theorem, $\overline f$ can be extended to a functional $f$ on $V$ such that $\norm f = \norm {\overline f} = 1$
This proves the necessity.
=== Sufficient Condition ===
Let such a functional $f$ on $V$ exist.
Then for any scalar $\lambda$:
:$\norm {x + \lambda y} \ge \size {\map f {x + \lambda y} } = \norm x$
establishing the sufficiency.
== Sources ==
* 
* 
Category:Normed Vector Spaces
Characteristics of Cycle Graph

== Theorem ==
Let $G = \struct {V, E}$ be an (undirected) graph whose order is greater than $2$.
Then $G$ is a cycle graph :
:$G$ is connected
:every vertex of $G$ is adjacent to $2$ other vertices
:every edge of $G$ is adjacent to $2$ other edges.
== Proof ==
Recall that a cycle is a closed walk with the properties:
:all its edges are distinct
:all its vertices (except for the start and end) are distinct.
From Cycle Graph is Connected, $G$ is connected.
From Cycle Graph is 2-Regular, $G$ is $2$-regular.
Hence every vertex of $G$ is incident with $2$ edges.
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $23$
Category:Cycle Graphs