nyuuzyou/wikis huggingface dataset:
Unique Code for State of URM Program

== Theorem ==
Every state of a URM program can be assigned a unique code number.
This code number is called the state code (or situation code).
== Proof ==
The state of a URM program at a particular point in time is defined as:
:the value of the instruction pointer
:the value, at that point, of each of the registers that are used by the program.
Let $P$ be a URM program.
Suppose that, at a given stage of computation:
:the value of the instruction pointer is $a$;
:the value of register $R_k$ is $r_k$.
Let $b = \map \rho P$ be the number of registers used by $P$.
Then we can define the state code $s$ as:
:$s = p_1^a p_2^{r_1} p_3^{r_2} \cdots p_{b + 1}^{r_b}$
where $p_j = \map p j$ is defined as the $j$th prime number.
Hence it follows from the Fundamental Theorem of Arithmetic that $s$ is uniquely specified for any given state.
== Comment ==
So, if we know the state code at a given stage of computation, we can determine the number of the instruction to be performed, and also the contents of the registers at that stage.
This is done by using the prime exponent function.
The important cases are as follows:
Let $s$ be a state code.
Then $\tuple s_1$ is the index number of the instruction about to be carried out.
If $P$ has terminated, then $\tuple s_2$ is the value held by $R_1$, which is the output.
Category:URM Programs
Unique Code for URM Instruction

== Theorem ==
Each basic instruction $I$ in a URM Program can be identified with a unique code number $\beta \left({I}\right)$.
We also define the following sets:
* $\operatorname{Zinstr}$ is the set of codes of all the Zero instructions
* $\operatorname{Sinstr}$ is the set of codes of all the Successor instructions
* $\operatorname{Cinstr}$ is the set of codes of all the Copy instructions
* $\operatorname{Jinstr}$ is the set of codes of all the Jump instructions.
Then we define $\operatorname{Instr}$ to be the set of codes of all basic URM instructions.
That is:
:$\operatorname{Instr} = \operatorname{Zinstr} \cup \operatorname{Sinstr} \cup \operatorname{Cinstr} \cup \operatorname{Jinstr}$.
== Proof ==
Each basic URM instruction is of one of the following forms:
Zero 	$Z \left({n}\right)$
Successor	$S \left({n}\right)$
Copy 	$C \left({m, n}\right)$
Jump 	$J \left({m, n, q}\right)$
Let $\Bbb I$ be the set of all basic URM instructions.
We define the mapping $\beta: \Bbb I \to \N$ as follows:
* $\beta \left({Z \left({n}\right)}\right) = 6 n - 3$
* $\beta \left({S \left({n}\right)}\right) = 6 n$
* $\beta \left({C \left({m, n}\right)}\right) = 2^m 3^n + 1$
* $\beta \left({J \left({m, n, q}\right)}\right) = 2^m 3^n 5^q + 2$
We note that:
* $\beta \left({Z \left({n}\right)}\right) \equiv 3 \pmod 6$
* $\beta \left({S \left({n}\right)}\right) \equiv 0 \pmod 6$
* $\beta \left({C \left({m, n}\right)}\right) \equiv 1 \pmod 3$
* $\beta \left({J \left({m, n, q}\right)}\right) \equiv 2 \pmod 3$
So, if $\beta \left({I_1}\right) = \beta \left({I_2}\right)$, then both instructions must be of the same type.
Hence it follows from the Fundamental Theorem of Arithmetic that $\beta$ is uniquely specified for any given basic URM instruction.
Thus $\beta$ is an injection.
== Note ==
The coding scheme for $\Bbb I$ (i.e., the mapping $\beta$) is not unique.
This particular scheme lends itself especially to number-theoretical analysis techniques.
Category:URM Programs
Unique Code for URM Program

== Theorem ==
Any URM program can be assigned a unique code number.
== Proof ==
Let $\mathbf P$ be the set of all URM programs.
Let $P \in \mathbf P$ be a URM program with $k$ basic instructions:
Line 		Command 	
$1$ 		$I_1$ 	
$2$ 		$I_2$ 	
$\vdots$		$\vdots$	
$k$ 		$I_k$ 	
We define the mapping $\gamma: \mathbf P \to \N$ as follows:
:$\ds \map \gamma P = \prod_{i \mathop = 1}^k p_i^{\map \beta {I_i} }$
where:
* $p_i$ is the $i$th prime number;
* $\map \beta {I_i}$ is the unique code for instruction $i$.
Hence it follows from the Fundamental Theorem of Arithmetic that $\gamma$ is uniquely specified for any given URM program.
Thus $\gamma$ is an injection.
For a given $P$, the number $\map \gamma P$ is referred to as the code number of $P$.
== Does Not Code ==
Not every $e \in \N$ is the code number of a URM program.
If $e$ is not the code of any URM program, we say that $e$ does not code a URM program.
== Note ==
The coding scheme for $\mathbf P$ is not unique.
This particular scheme lends itself especially to number-theoretical analysis techniques.
Category:URM Programs
Unique Composition of Group Element whose Order is Product of Coprime Integers

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $g \in G$ be an element of $g$.
Let:
:$\order g = m n$
where:
:$\order g$ denotes the order of $g$ in $G$
:$m$ and $n$ are coprime integers.
Then $g$ can be expressed uniquely as the product of two commuting elements $a$ and $b$ of order $m$ and $n$ respectively.
== Proof ==
Let $g_1 = g^n$ and $g_2 = g^m$.
By Powers of Group Element Commute, $g_1$ and $g_2$ commute.
We have:
:${g_1}^m = g^{n m} = e$
:${g_2}^n = g^{m n} = e$
It follows that:
:the order of $g_1$ is $m$
:the order of $g_2$ is $n$
otherwise if either $g_1$ or $g_2$ were of a smaller order then $g$ would also be of a smaller order.
By Bézout's Identity:
:$\exists u, v \in \Z: u n + v m = 1$
as $m \perp n$.
Thus:
:$g = g^{u n + v m} = \paren {g^n}^u \paren {g^m}^v = {g_1}^u {g_2}^v$
Also by Bézout's Identity:
:$u \perp m$
and:
:$v \perp n$
Thus by Order of Group Element equals Order of Coprime Power:
:$\order { {g_1}^u} = m$
and:
:$\order { {g_2}^v} = n$
We have that $g_1$ and $g_2$ commute.
So by Commutativity of Powers in Group, ${g_1}^u$ and ${g_2}^v$ also commute.
Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions.
It remains to prove uniqueness.
Suppose that:
:$(1): \quad g = r_1 r_2 = s_1 s_2$
where:
:$r_1$ and $r_2$ commute
:$s_1$ and $s_2$ commute
:$\order {r_1} = \order {s_1} = m$
:$\order {r_2} = \order {s_2} = n$
Raising $(1)$ to the $n u$th power:
:$g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$
and so:
It follows directly from $(1)$ that $r_2 = s_2$.
Hence the result.
== Also see ==
* Order of Product of Commuting Group Elements of Coprime Order is Product of Orders, a converse of this
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 8$: The Order (Period) of an Element: Theorem $4$
* : $0$: Some Conventions and some Basic Facts: Exercise $7$
Category:Order of Group Elements
Unique Constant in Category of Monoids

== Theorem ==
Let $\mathbf{Mon}$ be the category of monoids.
Then every object $M$ of $\mathbf{Mon}$ has precisely one constant.
=== Corollary ===
The category of monoids $\mathbf{Mon}$ does not have enough constants.
== Proof ==
From Trivial Monoid is Terminal Element, we obtain that a constant of $M$ is a morphism $f: \left\{{e}\right\} \to M$.
By Trivial Monoid is Initial Element, there is precisely one such morphism.
Hence $M$ has one constant.
== Sources ==
* : $\S 2.3$: Example $2.12$: $2$
Category:Category of Monoids
Unique Epimorphism from Quotient Group to Quotient Group

== Theorem ==
Let $\struct {G_1, \circledcirc}$ and $\struct {G_2, \circledast}$ be groups.
Let $\phi: \struct {G_1, \circledcirc} \to \struct {G_2, \circledast}$ be a group epimorphism.
Let $K$ be the kernel of $\phi$.
Let $H_1$ be a normal subgroup of $\struct {G_1, \circledcirc}$.
Let $H_2 := \phi \sqbrk {H_1}$ be the image of $H_1$ under $\phi$.
Then there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:
:$\psi \circ q_{H_1} = q_{H_2} \circ \phi$
where:
:$G_1 / H_1$ and $G_2 / H_2$ are quotient groups of $G_1$ by $H_1$ and of $G_2$ by $H_2$ respectively
:$q_{H_1}: G_1 \to G_1 / H_1$ and $q_{H_2}: G_2 \to G_2 / H_2$ are their respective quotient mappings.
The kernel of $\psi$ is given by:
:$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$
== Proof ==
First we note that by Group Epimorphism Preserves Normal Subgroups, $H_2$ is indeed a normal subgroup of $\struct {G_2, \circledast}$.
Hence $G_2 / H_2$ and $q_{H_2}: G_2 \to G_2 / H_2$ are appropriately defined.
Let $\RR$ be the congruence relation induced on $\struct {G_1, \circledcirc}$ by $H$.
Let $q_\RR$ be the quotient mapping induced by $\RR$.
Then by definition $q_\RR = q_{H_1}$.
We are given that $\phi$ is a group epimorphism.
We also have from Quotient Group Epimorphism is Epimorphism that $q_{H_2}$ is also a group epimorphism.
Hence from Composite of Group Epimorphisms is Epimorphism, $q_{H_2} \circ \phi$ is also a group epimorphism.
So from Quotient Theorem for Epimorphisms, there exists a unique epimorphism $\psi$ from $G_1 / H_1$ to $G_2 / H_2$ such that:
:$\psi \circ q_{H_1} = q_{H_2} \circ \phi$
It remains to be shown that:
:$\map \ker \psi = \dfrac {H_1 \circledcirc K} {H_1}$
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.20 \ \text {(c)}$
Category:Group Epimorphisms
Category:Quotient Groups
Unique Factorization Domain is Integrally Closed

== Theorem ==
Let $A$ be a unique factorization domain (UFD).
Then $A$ is integrally closed.
== Proof ==
Let $K$ be the field of quotients of $A$.
Let $x \in K$ be integral over $A$.
Let:
$x = a / b$
for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
This makes sense because a UFD is GCD Domain.
There is an equation:
:$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$
with $a_i \in A$, $i = 0, \dotsc, n - 1$.
Multiplying by $b^n$, we obtain:
:$a^n + b c = 0$
with $c \in A$.
Therefore:
:$b \divides a^n$
 $b$ is not a unit.
Then:
:$\gcd \set {a, b} \notin A^\times$
which is a contradiction.
So $b$ is a unit, and:
:$x = a b^{-1} \in A$
By definition of integrally closed integral domain:
:$A$ is integrally closed it is integrally closed in its field of fractions $K$.
Since $x$ is an arbitrary element in the integral closure of $A$ in $K$:
:$A$ is integrally closed.
== Sources ==
* : Chapter $2$ Ring of Integers: Integral elements: Proposition $2.9$
Category:Algebraic Number Theory
Category:Factorization
Category:Unique Factorization Domains
Unique Factorization Theorem for Gaussian Integers

== Theorem ==
The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
forms a unique factorization domain.
That is, every Gaussian integer can be expressed as the product of one or more Gaussian primes, uniquely up to the order in which they appear.
== Proof ==
We have that:
:Gaussian Integers form Euclidean Domain
:Euclidean Domain is Principal Ideal Domain
:Principal Ideal Domain is Unique Factorization Domain.
So $\struct {\Z \sqbrk i, +, \times}$ forms a unique factorization domain.
Hence the result, by definition of unique factorization domain.
== Historical Note ==
== Sources ==
* 
* 
Category:Gaussian Integers
Category:Unique Factorization Domain‎s
Category:Gaussian Primes
Category:Algebraic Number Theory
Category:Named Theorems
Unique Factorization Theorem for Gaussian Integers/Historical Note

== Historical Note on Unique Factorization Theorem for Gaussian Integers ==
The Unique Factorization Theorem for Gaussian Integers was proved by .
This result, and the ideas associated with it, ushered in the field of algebraic number theory.
== Sources ==
* : Chapter $\text {A}.25$: Gauss ()
Category:Historical Notes
Unique Integer Close to Rational in Valuation Ring of P-adic Norm

== Theorem ==
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime number $p$.
Let $x \in \Q$ such that $\norm{x}_p \le 1$.
Then for all $i \in \N$ there exists a unique $\alpha \in \Z$ such that:
:$(1): \quad \norm {x - \alpha}_p \le p^{-i}$
:$(2): \quad 0 \le \alpha \le p^i - 1$
== Proof ==
Let $i \in \N$.
From Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm:
:$\exists \mathop {\alpha'} \in \Z: \norm{x - \alpha'}_p \le p^{-i}$
By Integer is Congruent to Integer less than Modulus, then there exists $\alpha \in \Z$:
:$\alpha \equiv \alpha' \pmod {p^i}$.
:$0 \le \alpha \le p^i - 1$
Then $\norm {\alpha' - \alpha}_p \le p^{-i}$
Hence:
Now suppose $\beta \in \Z$ satisfies:
:$(\text a): \quad 0 \le \beta \le p^i - 1$
:$(\text b): \quad \norm {x -\beta}_p \le p^{-i}$
Then:
Hence $p^i \divides \alpha - \beta$, or equivalently, $\alpha \equiv \beta \pmod {p^i}$
By Initial Segment of Natural Numbers forms Complete Residue System then $\alpha = \beta$.
The result follows.
== Also see ==
* Integer Arbitrarily Close to Rational in Valuation Ring of P-adic Norm
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Lemma $1.29$
Category:P-adic Number Theory
Unique Isomorphism between Equivalent Finite Totally Ordered Sets

== Theorem ==
Let $S$ and $T$ be finite sets of the same cardinality.
That is:
:$\card S = \card T$
Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be totally ordered sets.
Then there is exactly one order isomorphism from $\struct {S, \preceq}$ to $\struct {T, \preccurlyeq}$.
== Proof ==
It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$.
Let $A$ be the set of all $n \in \N$ such that if:
:$(1): \quad S$ is any set such that $\card S = n$, and
:$(2): \quad \preceq$ is any total ordering on $S$
then there is exactly one isomorphism from $\struct {S, \preceq}$ to $\struct {\N_n, \le}$.
$\O \in A$ from Empty Mapping is Mapping and Equivalence of Mappings between Finite Sets of Same Cardinality.
Now let $n \in A$.
Let $\struct {S, \preceq}$ be a totally ordered set with $n + 1$ elements.
By Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $S$ has a greatest element $b$.
Then $\card {S \setminus \set b} = n$ by Cardinality Less One.
The total ordering on $S \setminus \set b$ is the one induced from that on $S$.
So as $n \in A$, there exists a unique order isomorphism $f: S \setminus \set b \to \struct {\N_n, \le}$.
Let us define the mapping $g: S \to \N_{n + 1}$ as follows:
:$\forall x \in S: \map g x = \begin{cases}
\map f x: & x \in S \setminus \set b \\
n: & x = b
\end{cases}$
This is the desired order isomorphism from $\struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$.
Now let $h: \struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$ be an order isomorphism.
Then $\map h b = n$ (it has to be).
So the restriction of $h$ to $S \setminus \set b$ is an isomorphism from $S \setminus \set b$ to $\struct {\N_n, \le}$, and hence $h = f$ (as $n \in A$, any such order isomorphism is unique).
Thus:
:$\forall x \in S \setminus \set b: \map h x = \map f x = \map g x$
and:
:$\map h b = n = \map g b$
thus $h = g$.
Therefore $n + 1 \in A$.
The result follows from the Principle of Mathematical Induction.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.10$
Category:Total Orderings
Category:Order Isomorphisms
Unique Isomorphism between Ordinal Subset and Unique Ordinal

== Theorem ==
Let $\On$ be the class of all ordinals.
Let $S \subset \On$ where $S$ is a set.
Then there exists a unique mapping $\phi$ and a unique ordinal $x$ such that $\phi : x \to S$ is an order isomorphism.
== Proof ==
Since $S \subset \On$, $\struct {S, \in}$ is a strict well-ordering.
The result follows directly from Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping.
== Sources ==
* : $\S 7.52$
Category:Ordinals
Category:Order Isomorphisms
Unique Isomorphism from Quotient Mapping to Epimorphism Domain

== Theorem ==
Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induced by $\RR_\phi$.
Let $\struct {S / \RR_\phi, \oplus_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.
Then there is one and only one isomorphism:
:$\psi: \struct {S / \RR_\phi, \oplus_{\RR_\phi} } \to \struct {T, *}$
which satisfies:
:$\psi \circ q_{\RR_\phi} = \phi$
where $\circ$ denotes composition of mappings.
== Proof ==
From the Quotient Theorem for Surjections, there is a unique bijection from $S / \RR_\phi$ onto $T$ satisfying $\psi \circ q_{\RR_\phi} = \phi$.
Also:
Therefore $\psi$ is an isomorphism.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.5$
Category:Epimorphisms (Abstract Algebra)
Category:Isomorphisms (Abstract Algebra)
Category:Quotient Mappings
Unique Linear Transformation Between Modules

== Theorem ==
Let $\struct {G, +_G, \times_G}_R$ and $\struct {H, +_H, \times_H}_R$ be unitary $R$-modules.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $G$.
Let $\sequence {b_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of $n$ elements of $H$.
Then there exists a unique linear transformation $\phi: G \to H$ satisfying:
:$\forall k \in \closedint 1 n: \map \phi {a_k} = b_k$
=== Corollary ===
== Proof ==
By Unique Representation by Ordered Basis, the mapping $\phi: G \to H$ defined as:
:$(1): \quad \ds \map \phi {\sum_{j \mathop = 1}^n \lambda_j \times_G a_j} = \sum_{j \mathop = 1}^n \lambda_j \times_H b_j$
is well-defined.
To verify that $\phi$ is a linear transformation, we need to show the following:
:$(1): \quad \forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$
:$(2): \quad \forall x \in G: \forall \lambda \in R: \map \phi {\lambda \times_G x} = \lambda \times_H \map \phi x$
Let $x, y \in G$ be arbitrary, such that:
:$\ds x = \sum_{k \mathop = 1}^n \lambda_k \times_G a_k$
:$\ds y = \sum_{k \mathop = 1}^n \mu_k \times_G a_k$
where:
:$a_k$ are elements of $\sequence {a_n}$
:$\lambda_k$ and $\mu_k$ are elements of $R$.
So:
and:
By Linear Transformation of Generated Module, $\phi$ is the only linear transformation whose value at $a_k$ is $b_k$ for all $k \in \closedint 1 n$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.4$
Category:Linear Transformations
Unique Linear Transformation Between Vector Spaces

== Corollary to Unique Linear Transformation Between Modules ==
Let $G$ be a finite-dimensional $K$-vector space.
Let $H$ be a $K$-vector space (not necessarily finite-dimensional).
Let $\sequence {a_n}$ be a linearly independent sequence of vectors of $G$.
Let $\sequence {b_n}$ be a sequence of vectors of $H$.
Then there is a unique linear transformation $\phi: G \to H$ satisfying:
:$\forall k \in \closedint 1 n: \map \phi {a_k} = b_k$
== Proof ==
From Generator of Vector Space Contains Basis, $\set {a_1, \ldots, a_m}$ is contained in a basis for $G$.
The result then follows from Unique Linear Transformation Between Modules.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.4$: Corollary
Category:Linear Transformations
Category:Vector Spaces
Unique Minimal Element may not be Smallest

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $\struct {S, \preccurlyeq}$ have a unique minimal element.
Then it is not necessarily the case that $\struct {S, \preccurlyeq}$ has a smallest element.
== Proof ==
Let $S$ be the set defined as:
:$S = \Z \cup \set m$
where:
:$\Z$ denotes the set of integers
:$m$ is an arbitrary object such that $m \ne \Z$.
Let $\preccurlyeq$ be the relation on $\Z$ defined as:
:$\forall a, b \in S: a \preccurlyeq b \iff \begin {cases} a \le b & : a, b \in \Z \\ a = m = b & : a, b \notin \Z \end{cases}$
where $\le$ is the usual ordering on $\Z$.
Note that for all $x \in \Z$, we have that $x$ and $m$ are non-comparable.
Then $\struct {S, \preccurlyeq}$ is an ordered set such that:
:$m$ is the only minimal element of $\struct {S, \preccurlyeq}$
:$\struct {S, \preccurlyeq}$ has no smallest element.
This is shown as follows:
Let $x \in S$ such that $x \preccurlyeq m$.
Then by definition $x = m$.
Hence $m$ is a minimal element of $\struct {S, \preccurlyeq}$.
 $y \in \Z$ is another minimal element of $\struct {S, \preccurlyeq}$.
Let $z = y - 1$.
By definition of $\Z$, we have that $z \in \Z$.
But $z \le y$ and so $z \preccurlyeq y$.
Hence for all $y \in \Z$, it is not the case that $y$ is a minimal element of $\struct {S, \preccurlyeq}$.
Hence by Proof by Contradiction it follows that the minimal element $m$ of $\struct {S, \preccurlyeq}$ is unique.
 $x \in S$ is the smallest element of $\struct {S, \preccurlyeq}$.
Suppose $x \in \Z$.
Then because $x$ and $m$ are non-comparable, it is not the case that $x \preccurlyeq m$.
Hence $x$ cannot be the smallest element.
Suppose $x = m$.
Let $a \in S: a \ne m$.
Then $x$ and $a$ are again non-comparable.
Hence it is not the case that $x \preccurlyeq a$.
Hence $x$ again cannot be the smallest element of $\struct {S, \preccurlyeq}$.
So by Proof by Cases $x$ cannot be the smallest element of $\struct {S, \preccurlyeq}$.
It follows by Proof by Contradiction that there can be no smallest element of $\struct {S, \preccurlyeq}$.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(c)}$
Category:Minimal Elements
Category:Smallest Elements
Unique Ordinal Exponentiation Inequality

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x > 1$ and $y > 0$.
Then there exists a unique ordinal $z$ such that:
:$x^z \le y$ and $y < x^{z^+}$
== Proof ==
=== Existence of $z$ ===
By Lower Bound for Ordinal Exponentiation, $y \le x^y$ so $y < x^{y^+}$.
Therefore, $y$ is bounded above by $x^w$ for some $w$, so there is a smallest $w$ such that:
:$y < x^w$ by Subset of Ordinals has Minimal Element.
Moreover, if $w$ is a limit ordinal then there is some $v ∈ w$ such that $y < x^v$, contradicting the fact that $w$ is minimal.
Therefore, $w$ is not a limit ordinal and therefore must be the successor of some ordinal.
So $w = z^+$ for some $z$.
Since $z < z^+$, it follows that $x^z ≤ y$ and $y < x^{z^+}$.
=== Uniqueness of $z$ ===
Assume that $x^z \le y$ and $y < x^{z^+}$ for some $z = z_1$ and $z = z_2$.
Then $x^{z_1} < x^{z_2^+}$ and $x^{z_2} < x^{z_1^+}$.
By Membership is Left Compatible with Ordinal Exponentiation $z_1 < z_2^+$ and $z_1 \le z_2$.
Similarly $z_2 \le z_1$.
By definition of set equality:
:$z_1 = z_2$
== Sources ==
* : $\S 8.38$
Category:Ordinal Arithmetic
Unique Period Prime/Examples

== Examples of Unique Period Primes ==
In base $10$, the prime numbers $3$, $11$, $37$, and $101$ are the only ones with periods $1$, $2$, $3$ and $4$ respectively.
Hence they are unique period primes.
But:
:$41$ and $271$ both have period $5$
:$7$ and $13$ both have period $6$
:$239$ and $4649$ both have period $7$
:$73$ and $137$ both have period $8$.
Thus these primes are not unique period primes.
=== Various Number Bases ===
Category:Unique Period Primes
Unique Period Prime/Sequence

== Sequence ==
The sequence of unique period primes begins:
:$3$, $11$, $37$, $101$, $9091$, $9901$, $333 \, 667$, $909 \, 091$, $99 \, 990 \, 001$, $999 \, 999 \, 000 \, 001$, $9 \, 999 \, 999 \, 900 \, 000 \, 001$, $909 \, 090 \, 909 \, 090 \, 909 \, 091$, $1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$, $11 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$, $900 \, 900 \, 900 \, 900 \, 990 \, 990 \, 990 \, 991$, $909 \, 090 \, 909 \, 090 \, 909 \, 090 \, 909 \, 090 \, 909 \, 091$, $\ldots$
Category:Unique Period Primes