nyuuzyou/wikis huggingface dataset:
Unsigned Stirling Number of the First Kind of n with n-3

== Theorem ==
Let $n \in \Z_{\ge 3}$ be an integer greater than or equal to $3$.
Then:
:$\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$
where:
:$\ds {n \brack n - 3}$ denotes an unsigned Stirling number of the first kind
:$\dbinom n 6$ denotes a binomial coefficient.
== Proof ==
The proof proceeds by induction.
=== Basis for the Induction ===
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$\ds {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$
$\map P 3$ is the case:
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds {k \brack k - 3} = \binom k 6 + 8 \binom {k + 1} 6 + 6 \binom {k + 2} 6$
from which it is to be shown that:
:$\ds {k + 1 \brack k - 2} = \binom {k + 1} 6 + 8 \binom {k + 2} 6 + 6 \binom {k + 3} 6$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{\ge 3}: {n \brack n - 3} = \binom n 6 + 8 \binom {n + 1} 6 + 6 \binom {n + 2} 6$
== Also see ==
* Stirling Number of the Second Kind of n with n-3
* Particular Values of Unsigned Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(57)$
Category:Examples of Stirling Numbers of the First Kind
Unsymmetric Functional Equation for Riemann Zeta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Let $\Gamma$ be the gamma function.
Then for all $s \in \C$:
:$\map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\dfrac {\pi s} 2} \map \Gamma s \map \zeta s$
== Proof ==
For $s \notin \Z$, we have Euler's Reflection Formula:
:$\map \Gamma s \map \Gamma {1 - s} = \dfrac \pi {\map \sin {\pi s} }$
Replacing $s \mapsto \dfrac {1 + s} 2$ we deduce:
Also, we have Legendre's Duplication Formula for $z \notin -\dfrac 1 2 \N_0$:
:$\map \Gamma s \map \Gamma {s + \dfrac 1 2} = 2^{1 - 2 s} \sqrt \pi \map \Gamma {2 s}$
Replacing $s \mapsto s / 2$ this yields:
:$\map \Gamma {\dfrac s 2} \map \Gamma {\dfrac {1 + s } 2} = 2^{1 - s} \sqrt \pi \map \Gamma s$
Together these give:
:$(1): \quad \dfrac {\map \Gamma {s / 2} } {\map \Gamma {\paren {1 - s} / 2} } = 2^{1 - s} \pi^{-1/2} \map \Gamma s \map \cos {\pi s / 2}$
Now we take the Functional Equation for Riemann Zeta Function:
:$\pi^{-s/2} \map \zeta s \map \Gamma {s / 2} \map \Gamma {\dfrac {1 - s} 2}^{-1} = \pi^{\paren {s - 1} / 2} \map \zeta {1 - s}$
and substitute $(1)$ to give:
:$\pi^{\paren {s - 1} / 2} \map \zeta {1 - s} = \pi^{-\paren {s + 1} / 2} \map \zeta s 2^{1 - s} \map \Gamma s \map \cos {\pi s / 2}$
Multiplying by $\pi^{\paren {s - 1} / 2}$ this becomes:
:$\map \zeta {1 - s} = \pi^{-s} 2^{1 - s} \map \cos {\pi s / 2} \map \Gamma s \map \zeta s$
as desired.
Category:Riemann Zeta Function
Category:Reflection Formulas
Unwarranted Assumptions

== Fallacy ==
Let $p$ be a proposition whose truth value is to be proved (either true or false).
During the course of a logical argument it is permitted to make an assumption by use of the Rule of Assumption.
However, each of the assumptions being made has to be discharged before the proof of $p$ can be considered complete.
To fail to discharge all the assumptions is to commit the fallacy of unwarranted assumptions.
== Also see ==
* Begging the Question
== Sources ==
* : $\text{I}$: 'NOT' and 'IF': $\S 4$
Category:Fallacies and Mistakes
Up-Complete Lower Bounded Join Semilattice is Complete

== Theorem ==
Let $\struct {S, \preceq}$ be an up-complete lower bounded join semillattice.
Then $\struct {S, \preceq}$ is complete.
== Proof ==
Let $X$ be a subset of $S$.
In the case when $X = \O$:
by definition of lower bounded:
:$\exists L \in S: L$ is lower bound for $S$.
By definition of empty set:
:$L$ is upper bound for $X$.
By definition of lower bound:
:$\forall x \in S: x$ is upper bound for $X \implies L \preceq x$
Thus by definition
:$L$ is a supremum of $X$.
Thus:
:$X$ admint a supremum.
In the case when $X \ne \O$:
Define
:$Y := \set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$
where $\map {\operatorname {Fin} } X$ denotes the set of all finite subsets of $X$.
By Existence of Non-Empty Finite Suprema in Join Semilattice
:all suprema in $Y$ exist,
By definition of non-empty set:
:$Y$ is a non-empty set.
We will prove that
:$Y$ is directed
Let $x, y \in Y$.
By definition of $Y$:
:$\exists A \in \map {\operatorname {Fin} } X \setminus \set \O: x = \sup A$
and
:$\exists B \in \map {\operatorname {Fin} } X \setminus \set \O: y = \sup B$
By Finite Union of Finite Sets is Finite:
:$A \cup B$ is finite
:$A \cup B \ne \O$
By Union is Smallest Superset:
:$A \cup B \subseteq X$
By definition of $Y$:
:$\map \sup {A \cup B} \in Y$
By Set is Subset of Union:
:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
Thus by Supremum of Subset:
:$x \preceq \map \sup {A \cup B}$ and $y \preceq \map \sup {A \cup B}$
Thus by definition:
:$Y$ is directed.
By definition up-complete:
:$Y$ admits a supremum
By definition of supremum
:$\sup Y$ is upper bound for $Y$
We will prove that
:$X \subseteq Y$
Let $x \in X$.
By definitions of subset and singleton:
:$\set x \subseteq X$
:$\set x$ is finite
:$\set x \ne \O$
By definition of $Y$:
:$\sup {\set x} \in Y$
Thus by Supremum of Singleton:
:$x \in Y$
By Upper Bound is Upper Bound for Subset:
:$\sup Y$ is upper bound for $X$
We will prove that
:$\forall x \in S: x$ is upper bound for $X \implies \sup Y \preceq x$
Let $x \in S$ such that
:$x$ is upper bound for $X$
We will prove as sublemma that
:$x$ is upper bound for $Y$
Let $y \in Y$.
By definition of $Y$:
:$\exists A \in \map {\operatorname {Fin} } X \setminus \set \O: y = \sup A$
By definition of $\operatorname {Fin}$:
:$A \subseteq X$
By Upper Bound is Upper Bound for Subset
:$x$ is upper bound for $A$
Thus by definition of supremum:
:$y \preceq x$
Thus by definition
:$x$ is upper bound for $Y$
This ends the proof of sublemma.
Thus by definition of supremum:
:$\sup Y \preceq x$
This ends the proof of lemma.
By definition
:$\sup Y$ is a supremum of $X$
and thus:
:$X$ admits a supremum.
Thus result follows by Lattice is Complete iff it Admits All Suprema.
== Sources ==
* 
* 
Category:Complete Lattices
Up-Complete Product/Lemma 1

== Theorem ==
Let $X$ be a directed subset of $S$.
Let $Y$ be a directed subset of $T$.
Then $X \times Y$ is also a directed subset of $S \times T$.
== Proof ==
Let $\tuple {s_1, t_1}, \tuple {s_2, t_2} \in X \times Y$.
By definition of Cartesian product:
:$s_1, s_2 \in X$ and $t_1, t_2 \in Y$
By definition of directed subset:
:$\exists h_1 \in X: s_1 \preceq_1 h_1 \land s_2 \preceq_1 h_1$
and
:$\exists h_2 \in X: t_1 \preceq_2 h_2 \land t_2 \preceq_2 h_2$
By definition of simple order product:
:$\exists \tuple {h_1, h_2} \in X \times Y: \tuple {s_1, t_1} \preceq \tuple {h_1, h_2} \land \tuple {s_2, t_2} \preceq \tuple {h_1, h_2}$
Thus by definition:
:$X \times Y$ is a directed subset of $S \times T$.
== Sources ==
* 
* 
Category:Order Theory
Up-Complete Product/Lemma 2

== Theorem ==
Let $X$ be a directed subset of $S \times T$.
Then
:$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ are directed
where 
:$\pr_1$ denotes the first projection on $S \times T$
:$\pr_2$ denotes the second projection on $S \times T$
:$\map {\pr_1^\to} X$ denotes the image of $X$ under $\pr_1$
== Proof ==
Let $x, y \in \map {\pr_1^\to} X$.
By definitions of image of set and projections:
:$\exists x' \in T: \tuple {x, x'} \in X$
and
:$\exists y' \in T: \tuple {y, y'} \in X$
By definition of directed:
:$\exists \tuple {a, b} \in X: \tuple {x, x'} \preceq \tuple {a, b} \land \tuple {y, y'} \preceq \tuple {a, b}$
By definition of simple order product:
:$\exists a \in \map {\pr_1^\to} X: x \preceq_1 a \land y \preceq_1 a$
Thus by definition
:$\map {\pr_1^\to} X$ is directed.
By mutatis mutandis:
:$\map {\pr_2^\to} X$ is directed.
== Sources ==
* 
* 
Category:Order Theory
Upper Adjoint Preserves All Infima

== Theorem ==
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.
Let $g: S \to T$ be an upper adjoint of Galois connection.
Then $g$ preserves all infima.
== Proof ==
By definition of upper adjoint
:$\exists d: T \to S: \left({g, d}\right)$ is a Galois connection
Let $X$ be a subset of $S$ such that
:$X$ admits an infimum.
We will prove as lemma 1 that
:$\forall t \in T: t$ is lower bound for $g^\to\left({X}\right) \implies t \precsim g\left({\inf X}\right)$
Let $t \in T$ such that
:$t$ is lower bound for $g^\to\left({X}\right)$
We will prove as sublemma that
:$d\left({t}\right)$ is lower bound for $X$
Let $s \in X$.
By definition of image of set:
:$g\left({s})\right) \in g^\to\left({X}\right)$
By definition of lower bound:
:$t \precsim g\left({s}\right)$
Thus by definition of Galois connection:
:$d\left({t}\right) \preceq s$
This ends the proof of sublemma.
By definition of infimum:
:$d\left({t}\right) \preceq \inf X$
Thus by definition of Galois connection:
:$t \precsim g\left({\inf X}\right)$
This ends the proof of lemma 1.
We will prove as lemma 2 that
:$g\left({\inf X}\right)$ is lower bound for $g^\to\left({X}\right)$
Let $t \in g^\to\left({X}\right)$.
By definition of image of set:
:$\exists s \in S: s \in X \land g\left({s}\right) = t$
By definition of infimum:
:$\inf X$ is lower bound for $X$
By definition of lower bound:
:$\inf X \preceq s$
By definition of Galois connection:
:$g$ is increasing mapping.
Thus by definition of increasing mapping:
:$g\left({\inf X}\right) \precsim t$
This ends the proof of lemma 2.
Thus by definition of infimum:
:$g^\to\left({X}\right)$ admits an infimum
and
:$\inf\left({g^\to\left({X}\right)}\right) = g\left({\inf X}\right)$
Thus by definition:
:$g$ preserves infimum on $X$
Thus by definition:
:$g$ preserves all infima.
== Sources ==
* 
* 
Category:Galois Connections
Upper Adjoint of Galois Connection is Surjection implies Lower Adjoint at Element is Minimum of Preimage of Singleton of Element

== Theorem ==
Let $L = \struct {S, \preceq}, R = \paren {T, \precsim}$ be ordered sets.
Let $g: S \to T, d:T \to S$ be mappings such that:
:$\tuple {g, d}$ is a Galois connection
and
:$g$ is a surjection.
Then
:$\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {\set t} }$
== Proof ==
By definition of Galois connection:
:$g$ is an increasing mapping.
Let $t \in T$.
By definition of surjection:
:$\Img g = T$
By :
:$g \sqbrk {g^{-1} \sqbrk {t^\succeq} } = t^\succeq$
By Galois Connection is Expressed by Minimum:
:$\map d t = \min \set {g^{-1} \sqbrk {t^\succeq} }$
By definition of min operation:
:$\map d t = \inf \set {g^{-1} \sqbrk {t^\succeq} }$ and $\map d t \in g^{-1} \sqbrk {t^\succeq}$
By definition of image of set:
:$\map g {\map d t} \in g \sqbrk {g^{-1} \sqbrk {t^\succeq} }$
By definition of upper closure of element:
:$t \precsim \map g {\map d t}$
By definition of minimum element:
:$g^{-1} \sqbrk {t^\succeq}$ admits an infimum.
By definition of infimum:
:$\map d t$ is lower bound for $g^{-1} \sqbrk {t^\succeq}$
By definition of surjection:
:$\exists s \in S: t = \map g s$
By definition of singleton:
:$t \in \set t$
By Set is Subset of Upper Closure
:$\set t \subseteq \set t^\succeq$
By Upper Closure of Singleton:
:$\set t^\succeq = t^\succeq$
By definition of image of set:
:$s \in g^{-1} \sqbrk {t^\succeq}$
By definition of lower bound:
:$\map d t \preceq s$
By definition of increasing mapping:
:$\map g {\map d t} \precsim t$
By definition of antisymmetry:
:$\map g {\map d t} = t$
By definition of preimage of set:
:$\map d t \in g^{-1} \sqbrk {\set t}$
By Preimage of Subset is Subset of Preimage:
:$g^{-1} \sqbrk {\set t} \subseteq g^{-1} \sqbrk {t^\succeq}$
We will prove that
:$\map d t$ is an infimum of $g^{-1} \sqbrk {\set t}$
Thus by Lower Bound is Lower Bound for Subset:
:$\map d t$ is lower bound for $g^{-1} \sqbrk {\set t}$
Thus by definition:
:$\forall s \in S: s$ is lower bound for $g^{-1} \sqbrk {\set t} \implies s \preceq \map d t$
Thus by definition of min operation:
:$\map d t = \min \set {g^{-1} \sqbrk {\set t} }$
== Sources ==
* 
Category:Galois Connections
Upper Bound for Abscissa of Absolute Convergence of Product of Dirichlet Series

== Theorem ==
Let $f, g: \N \to \C$ be arithmetic functions with Dirichlet convolution $h = f * g$.
Let $F, G, H$ be their Dirichlet series.
Let $\sigma_f, \sigma_g, \sigma_h$ be their abscissae of absolute convergence.
Then:
:$\sigma_h \le \max \set {\sigma_f, \sigma_g}$
== Proof ==
Follows from Dirichlet Series of Convolution of Arithmetic Functions
Category:Dirichlet Series
Upper Bound for Abundancy Index

== Theorem ==
Let $n \in \N_{>0}$ be a non-zero natural number.
Let $\map h n$ be the abundancy index of $n$.
Then:
:$\ds \map h n \le \prod_{p \mathop \divides n} \dfrac p {p - 1}$
such that equality holds $n = 1$.
== Proof ==
First we note that $\map h 1 = 1$ demonstrating that equality holds if $n = 1$.
Let $n$ be expressed as its prime decomposition:
where $p_1, p_2, \ldots, p_k$ are distinct primes.
By definition of abundancy index:
:$\map h n = \dfrac {\map {\sigma_1} n} n$
where $\map {\sigma_1} n$ denotes the divisor sum function of $n$.
We have:
It remains to be shown that:
:$\map h {p^a} < \dfrac p {p - 1}$
for prime $p$.
We have that:
Category:Abundancy
Upper Bound for Binomial Coefficient

== Theorem ==
Let $n, k \in \Z$ such that $n \ge k \ge 0$.
Then:
:$\dbinom n k \le \left({\dfrac {n e} k}\right)^k$
where $\dbinom n k$ denotes a binomial coefficient.
== Proof ==
From Lower and Upper Bound of Factorial, we have that:
:$\dfrac {k^k} {e^{k - 1} } \le k!$
so that:
:$(1): \quad \dfrac 1 {k!} \le \dfrac {e^{k - 1} } {k^k}$
Then:
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $67$
Category:Binomial Coefficients
Upper Bound for Harmonic Number

== Theorem ==
:$H_{2^m} \le 1 + m$
where $H_{2^m}$ denotes the $2^m$th harmonic number.
== Proof ==
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$ 
where $\ds s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$
From Ordering of Reciprocals:
:$\forall m, n \in \N_{>0}: m > n: \dfrac 1 m < \dfrac 1 n$
so each of the summands in a given $s_k$ is less than $\dfrac 1 {2^k}$.
The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:
:$s_k < \dfrac {2^k} {2^k} = 1$
Hence the harmonic sum $H_{2^m}$ satisfies the following inequality:
Hence the result.
== Sources ==
* : $\S 1.2.7$: Harmonic Numbers: Exercise $2$
Category:Harmonic Numbers
Upper Bound for Lowest Common Multiple

== Theorem ==
Let $a, b \in \Z$ be integers such that $a b \ne 0$.
Then:
:$\lcm \set {a, b} \le \size {a b}$
where:
:$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$
== Proof ==
By Product of GCD and LCM:
:$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
where:
:$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
By Existence of Greatest Common Divisor $\gcd \set {a, b}$ exists.
By definition of GCD, $\gcd \set {a, b} \in \Z_{>0}$.
Hence the result.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Lowest Common Multiple
Upper Bound for Lucas Number

== Theorem ==
Let $L_n$ denote the $n$th Lucas number.
Then:
:$L_n < \paren {\dfrac 7 4}^n$
== Proof ==
The proof proceeds by complete induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$L_n < \paren {\dfrac 7 4}^n$
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
=== Basis for the Induction ===
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
This is the induction hypothesis:
:$L_k < \paren {\dfrac 7 4}^k$
from which it is to be shown that:
:$L_{k + 1} < \paren {\dfrac 7 4}^{k + 1}$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: L_n < \paren {\dfrac 7 4}^n$
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Example $1 \text{-} 1$
Category:Lucas Numbers
Category:Proofs by Induction
Upper Bound for Number of Grains of Sand to fill Universe

== Theorem ==
The number of grains of sand that would fill the Universe, to the extent it was believed to exist in the days of , is bounded above by $10^{63}$.
== Proof ==
== Historical Note ==
== Sources ==
* : $10^{51}$
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* : $10^{51}$
Category:Number Theory
Upper Bound for Number of Grains of Sand to fill Universe/Historical Note

== Historical Note on Upper Bound for Number of Grains of Sand to fill Universe ==
As an exercise in his system of notation for denoting very large numbers, determines an Upper Bound for Number of Grains of Sand to fill Universe in his book .
It needs to be emphasised that the actual size of the universe was little understood in those days.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $10^{51}$
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* : $10^{51}$
Category:Historical Notes
Upper Bound for Number of Grains of Sand to fill Universe/Mistake

== Source Work ==
:
:The Dictionary
::$10^{51}$
:
:The Dictionary
::$10^{51}$
== Mistake ==
:... the number of grains of sand required to fill the universe turns out to be, in our notation, less than $10^{51}$.
Other sources report it as $10^{63}$. It is not clear from where sources his information.
== Sources ==
* : $10^{51}$
* : $10^{51}$
Category:Mistakes/Curious and Interesting Numbers
Upper Bound for Subset

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $U$ be an upper bound for $S$.
Let $\left({T, \preceq}\right)$ be a subset of $\left({S, \preceq}\right)$.
Then $U$ is an upper bound for $T$.
== Proof ==
By definition of upper bound:
:$\forall x \in S: x \preceq U$
But as $\forall y \in T: y \in S$ by definition of subset, it follows that:
:$\forall y \in T: y \preceq U$.
Hence the result, again by definition of upper bound.
== Also see ==
* Lower Bound for Subset
Category:Order Theory
Upper Bound is Dual to Lower Bound

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
:$a$ is an upper bound for $T$
:$a$ is a lower bound for $T$
== Proof ==
By definition, $a$ is an upper bound for $T$ :
:$\forall t \in T: t \preceq a$
The dual of this statement is:
:$\forall t \in T: a \preceq t$
by Dual Pairs (Order Theory).
By definition, this means $a$ is a lower bound for $T$.
The converse follows from Dual of Dual Statement (Order Theory).
== Also see ==
* Duality Principle (Order Theory)
Category:Dual Pairs (Order Theory)
Upper Bound is Lower Bound for Inverse Ordering

== Definition ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$.
Let $M$ be an upper bound for $\left({T, \preceq}\right)$.
Let $\succeq$ be the dual ordering of $\preceq$.
Then $M$ is a lower bound for $\left({T, \succeq}\right)$.
== Proof ==
Let $M$ be an upper bound for $\left({T, \preceq}\right)$.
That is:
:$\forall a \in T: a \preceq M$
By definition of dual ordering, it follows that:
:$\forall a \in T: M \succeq a$
That is, $M$ is a lower bound for $\left({T, \succeq}\right)$. 
== Also see ==
* Lower Bound is Upper Bound for Inverse Ordering
Category:Order Theory
Upper Bound is Upper Bound for Subset

== Theorem ==
Let $\left({S, \preceq}\right)$ be a preordered set.
Let $A, B$ be subsets of $S$ such that
:$B \subseteq A$
Let $U$ be an upper bound for $A$.
Then $U$ is an upper bound for $B$.
== Proof ==
Assume that:
: $U$ is upper bound for $A$.
By definition of upper bound:
:$\forall x \in A: x \preceq U$
By definition of subset:
:$\forall x \in B: x \in A$
Hence:
:$\forall x \in B: x \preceq U$
Thus by definition
: $U$ is sn upper bound for $B$.
== Sources ==
* 
Category:Preorder Theory
Upper Bound of Bloch's Constant

== Theorem ==
Bloch's constant has an upper bound as follows:
:$B \le \sqrt {\dfrac {\sqrt 3 -1} 2} \times \dfrac {\map \Gamma {\frac 1 3} \map \Gamma {\frac {11} {12} } } {\map \Gamma {\frac 1 4} }$
== Proof ==
== Historical Note ==
== Sources ==
* 
Category:Bloch's Theorem
Upper Bound of Hamming Distance for Length of K-Error-Correcting Code

== Theorem ==
Let $C$ be an error-correcting code whose Hamming distance is no less than $2 k + 1$ for some $k \in \N$.
Then $C$ is $k$-error-correcting.
== Proof ==
== Sources ==
* 
Category:Hamming Distance
Category:Error-Correcting Codes
Upper Bound of Hermite Constant

== Theorem ==
Let $\gamma_n$ be the Hermite constant of dimension $n$.
Then:
:$\gamma_n \le \dfrac {\paren {1 + \epsilon_n} n} {\pi e}$
where $e_n \to 0$
== Proof ==
== Sources ==
* 
* : $1,33333 33333 33 \ldots$
Category:Hermite Constants
Upper Bound of Natural Logarithm

== Theorem ==
Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.
Then:
:$\ln x \le x - 1$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Bounds of Natural Logarithm
Category:Natural Logarithms
Category:Inequalities
Category:Upper Bound of Natural Logarithm
Upper Bound of Natural Logarithm/Corollary

== Theorem ==
Let $\ln y$ be the natural logarithm of $y$ where $y \in \R_{>0}$.
Then:
:$\forall s \in \R_{>0}: \ln x \le \dfrac {x^s} s$
== Proof ==
The result follows by dividing both sides by $s$.
== Also see ==
* Bounds of Natural Logarithm
== Sources ==
* : $\S 14.3 \ (2)$
Category:Upper Bound of Natural Logarithm
Upper Bound of Natural Logarithm/Proof 1

== Theorem ==
== Proof ==
From Logarithm is Strictly Concave:
:$\ln$ is (strictly) concave.
From Mean Value of Concave Real Function:
:$\ln x - \ln 1 \le \paren {\dfrac \d {\d x} \ln 1} \paren {x - 1}$
From Derivative of Natural Logarithm:
:$\dfrac \d {\d x} \ln 1 = \dfrac 1 1 = 1$
So:
:$\ln x - \ln 1 \le \paren {x - 1}$
But from Logarithm of 1 is 0:
:$\ln 1 = 0$
Hence the result.
== Sources ==
* : $\S 14.3 \ (2)$
Category:Upper Bound of Natural Logarithm
Upper Bound of Natural Logarithm/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {f_n}$ denote the sequence of mappings $f_n: \R_{>0} \to \R$ defined as:
:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
Fix $x \in \R_{>0}$.
We first show that
$\forall n \in \N : n \paren {\sqrt [n] x - 1} < x - 1 $
=== Case 1: $0 < x < 1$ ===
Suppose $0 < x < 1$.
Then:
 
=== Case 2: $x = 1$ ===
Suppose $x = 1$.
Then:
=== Case 3: $x > 1$ ===
Suppose $x > 1$.
Then:
Thus:
:$\forall n \in \N: n \paren {\sqrt [n] x - 1} \le x - 1$
by Proof by Cases.
Thus:
:$\ds \lim_{n \mathop \to \infty} \paren {\sqrt [n] x - 1 } \le x - 1$
from Limit of Bounded Convergent Sequence is Bounded.
Hence the result, from the definition of $\ln$.
Category:Upper Bound of Natural Logarithm
Upper Bound of Order of Non-Abelian Finite Simple Group

== Theorem ==
Let $G$ be a non-abelian finite simple group.
Let $t \in G$ be a self-inverse element of $G$.
Let $\map {C_G} t$ denote the centralizer of $t$ in $G$.
Let $m = \order {\map {C_G} t}$ be the order of $\map {C_G} t$.
Then:
:$\order G \le \paren {\dfrac {m \paren {m + 1} } 2}!$
=== Corollary ===
== Proof ==
== Sources ==
* 
* : $1$: Introduction to Finite Group Theory: $1.14$
Category:Centralizers
Category:Simple Groups
Category:Self-Inverse Elements
Category:Finite Groups