nyuuzyou/wikis huggingface dataset:
Not Preceding implies There Exists Meet Irreducible Element Not Preceding

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Let $x, y \in S$ such that
:$y \npreceq x$
Then
:$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$
== Proof ==
By definition of continuous:
:$L$ satisfies the axiom of approximation
and
:$\forall x \in S: x^\ll$ is directed.
By Axiom of Approximation in Up-Complete Semilattice
:$\exists u \in S: u \ll y \land u \npreceq x$
By Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure:
:there exists way below open filter $F$ in $L$: $y \in F \subseteq u^\gg$
By Way Above Closure is Subset of Upper Closure of Element:
:$u^\gg \subseteq u^\succeq$
By definition of upper section:
:$x \notin F$
By definition of relative complement:
:$x \in \relcomp S F$
By Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement:
:$\exists m \in \relcomp S F: x \preceq m \land m = \max \relcomp S F$
Thus by Maximal Element of Complement of Filter is Meet Irreducible:
:$m$ is meet irreducible.
Thus
:$x \preceq m$
:$y \preceq m$
By definition of upper section:
:$m \in F$
This contradicts $m \in \relcomp S F$ by definition of greatest element.
== Sources ==
* 
* 
Category:Order Theory
Category:Meet Irreducible Elements
Not every Closed Set is G-Delta Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be a closed set of $T$.
Then it is not necessarily the case that $V$ is a $G_\delta$ set of $T$.
== Proof ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $V$ be a closed set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $G_\delta$:
:$V$ is not a $G_\delta$ set of $T$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:G-Delta Sets
Category:Closed Sets
Not every Non-Empty Subset of Natural Numbers has Greatest Element

== Theorem ==
Let $S \subseteq \N_{>0}$.
Then, despite Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element, it is not necessarily the case that $S$ has a greatest element.
== Proof ==
Let:
:$S = \set {x \in \N_{>0}: x > 1}$
Then $S \subseteq \N_{>0}$.
 $S$ has a greatest element.
Let the greatest element of $S$ be $k$.
But $\N_{>0}$ is an inductive set.
Therefore $k + 1 \in \N_{>0}$.
By definition of $S$:
:$k + 1 \in S$
Therefore $k$ cannot be the greatest element of $S$.
By Proof by Contradiction it follows that $S$ has no greatest element.
It is not possible to apply Non-Empty Subset of Initial Segment of Natural Numbers has Greatest Element, because $S$ is not a subset of an initial segment of $\N_{>0}$.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $4 \ \text{(b)}$
Category:Natural Numbers
Not every Open Set is F-Sigma Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $V$ be an open set of $T$.
Then it is not necessarily the case that $V$ is an $F_\sigma$ set of $T$.
== Proof ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an uncountable set $S$.
Let $U$ be an open set of $T$.
From Closed Set of Uncountable Finite Complement Topology is not $F_\sigma$:
:$U$ is not an $F_\sigma$ set of $T$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:F-Sigma Sets
Category:Open Sets
Nowhere Dense iff Complement of Closure is Everywhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Then $H$ is nowhere dense in $T$ $S \setminus H^-$ is everywhere dense in $T$, where $H^-$ denotes the closure of $H$.
=== Corollary ===
== Proof ==
Let:
:$H^\circ$ denote the interior of any $H \subseteq S$
:$H^-$ denote the closure of any $H \subseteq S$.
From the definition, $H$ is nowhere dense in $T$ $\paren {H^-}^\circ = \O$.
From the definition of interior, it follows that $\paren {H^-}^\circ = \O$ every open set of $T$ contains a point of $S \setminus \paren {H^-}$.
Thus $S \setminus \paren {H^-}$ is everywhere dense by definition.
== Warning ==
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.28$
Category:Set Closures
Category:Nowhere Dense iff Complement of Closure is Everywhere Dense
Category:Nowhere Denseness
Nowhere Dense iff Complement of Closure is Everywhere Dense/Corollary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $H$ be a closed set of $T$.
Then $H$ is nowhere dense in $T$ $S \setminus H$ is everywhere dense in $T$.
== Proof ==
From Closed Set equals its Closure, $H$ is closed in $T$ :
:$H = H^-$
where $H^-$ is the closure of $H$.
The result then follows directly from Nowhere Dense iff Complement of Closure is Everywhere Dense.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Corollary $3.7.29$
Category:Nowhere Dense iff Complement of Closure is Everywhere Dense
Nowhere Dense iff Complement of Closure is Everywhere Dense/Warning

== Nowhere Dense iff Complement of Closure is Everywhere Dense: Warning ==
Note that in the result:
:$H$ is nowhere dense in $T$ the relative complement of its closure is everywhere dense in $T$
must be applied to the closure of $H$.
Otherwise, consider the real numbers $\R$ and the rational numbers $\Q$.
We have that:
:$\R \setminus \Q$ is the set of irrational numbers.
We have that Irrationals are Everywhere Dense in Reals.
But we also have from Rationals are Everywhere Dense in Reals:
:$\map \cl \Q = \R$
and so:
:$\paren {\map \cl \Q}^\circ = \R$
So it is not the case that $\R \setminus \Q$ is nowhere dense in $\R$.
However:
:$\R \setminus \map \cl \Q = \O$
which is indeed nowhere dense in $\R$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Example $3.7.30$
Category:Nowhere Dense iff Complement of Closure is Everywhere Dense
Nth Derivative of Exponential of Minus One over x

== Theorem ==
Let $n \in \N$ be a natural number.
Let $\map {P_n} x$ be a real polynomial of degree $n$.
Then:
:$\ds \dfrac {\d^n} {\d x^n} \map \exp {- \frac 1 x} = \frac {\map {P_n} x}{x^{2n}} \map \exp {- \frac 1 x}$
where $\map {P_n} x$ satisfies the following recurrence relation:
:$\map {P_{n + 1}} x = x^2 \dfrac \d {\d x} \map {P_n} x - \paren {2 n x - 1} \map {P_n} x$
and $\map {P_0} x = 1$.
== Proof ==
Proof by induction:
=== Basis for the Induction ===
We have that:
:$\ds \map \exp {- \frac 1 x} = \frac {\map {P_0} x} {x^{2 \cdot 0}} \map \exp {- \frac 1 x}$
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\ds \dfrac {\d^n} {\d x^n} \map \exp {- \frac 1 x} = \dfrac {\map {P_n} x}{x^{2n}} \map \exp {- \frac 1 x}$
Now we need to show true for $n + 1$:
:$\ds \dfrac {\d^{n + 1}} {\d x^{n + 1}} \map \exp {- \frac 1 x} = \dfrac {\map {P_{n + 1}} x}{x^{2\paren {n + 1}}} \map \exp {- \frac 1 x}$
=== Induction Step ===
This is our induction step:
The result follows by induction.
For all $n \in \N$ $\map {P_n} x$ is a real polynomial because we have $\map {P_0} x = 1$, and every step involves only addition, subtraction and multiplication of polynomials and their derivatives.
Category:Proofs by Induction
Nth Derivative of General Harmonic Number Order One

== Theorem ==
Let $n \in \Z_{>0}$.
The $n$th derivative of $\harm 1 x$ $x$ is:
:$\dfrac {\d^n} {\d x^n} \harm 1 x = \paren {-1}^{n + 1} n! \paren {\map \zeta {n + 1} - \harm {n + 1} x}$
where:
:$\harm n x$ denotes the general harmonic number of order $n$ evaluated at $x$
:$\map \zeta n$ is the Riemann zeta function
:$x \in \C$ with $x \notin \Z_{<0}$
== Proof ==
Category:Derivatives
Category:General Harmonic Numbers
Nth Derivative of Mth Power

== Theorem ==
Let $m \in \Z$ be an integer such that $m \ge 0$.
The $n$th derivative of $x^m$ $x$ is:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$
where $m^{\underline n}$ denotes the falling factorial.
=== Corollary ===
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\dfrac {\d} {\d x} x^m = m x^{m - 1}$
This follows by Power Rule for Derivatives, which also includes the case:
:$\dfrac {\d} {\d x} x^0 = 0$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\dfrac {\d^k} {\d x^k} x^m = \begin{cases} m^{\underline k} x^{m - k} & : k \le m \\ \\ 0 & : k > m \end{cases}$
Then we need to show:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \begin{cases} m^{\underline {k + 1} } x^{m - \paren {k + 1} } & : k + 1 \le m \\ \\ 0 & : k + 1 > m \end{cases}$
=== Induction Step ===
This is our induction step:
First, let $k < m$. Then we have:
At this stage, as $k < m$, we have that $k+1 \le m$. So far so good.
Now suppose that $k = m$.
Then by the induction hypothesis:
:$\dfrac {\d^k} {\d x^k} x^m = m^{\underline k} x^0 = m!$
Then by Derivative of Constant:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } x^m = \dfrac \d {\d x} m! = 0$
At this stage, as $k = m$, we have that $k + 1 > m$. Again, so far so good.
Finally, suppose that $k > m$.
Then by the induction hypothesis:
:$\dfrac {\d^k} {\d x^k} x^m = 0$
Then by Derivative of Constant:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } = 0$
At this stage, as $k > m$, we have that $k + 1 > m$.
This is all that is needed.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\dfrac {\d^n} {\d x^n} x^m = \begin{cases} m^{\underline n} x^{m - n} & : n \le m \\ \\ 0 & : n > m \end{cases}$
Category:Derivatives
Nth Derivative of Natural Logarithm

== Theorem ==
The $n$th derivative of $\map \ln x$ for $n \ge 1$ is:
:$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says:
:$\dfrac \d {\d x} \ln x = \dfrac 1 x$
This follows by Derivative of Natural Logarithm Function.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\dfrac {\d^k} {\d x^k} \ln x = \dfrac {\paren {k - 1}! \paren {-1}^{k - 1} } {x^k}$
Then we need to show:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \ln x = \dfrac {k! \paren {-1}^k} {x^{k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\dfrac {\d^n} {\d x^n} \ln x = \dfrac {\paren {n - 1}! \paren {-1}^{n - 1} } {x^n}$
Category:Derivative of Natural Logarithm Function
Category:Natural Logarithms
Category:Proofs by Induction
Nth Derivative of Natural Logarithm by Reciprocal

== Theorem ==
:$\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
where $H_n$ denotes the $n$th harmonic number:
:$H_n = \ds \sum_{r \mathop = 1}^n \dfrac 1 r = 1 + \dfrac 1 2 + \dfrac 1 3 + \cdots + \dfrac 1 r$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
:$\dfrac \d {\d x} \dfrac {\ln x} x = \paren {-1}^n n! \dfrac {H_n - \ln x} {x^{n + 1} }$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\dfrac {\d^k} {\d x^k} \dfrac {\ln x} x = \paren {-1}^{k + 1} k! \dfrac {H_k - \ln x} {x^{k + 1} }$
from which it is to be shown that:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac {\ln x} x = \paren {-1}^{k + 2} \paren {k + 1}! \dfrac {H_{k + 1} - \ln x} {x^{k + 2} }$
=== 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:
:$\forall n \in \Z_{>0}: \dfrac {\d^n} {\d x^n} \dfrac {\ln x} x = \paren {-1}^{n + 1} n! \dfrac {H_n - \ln x} {x^{n + 1} }$
Category:Derivatives involving Logarithm Functions
Category:Natural Logarithms
Nth Derivative of Nth Power

== Corollary to Nth Derivative of Mth Power ==
Let $n \in \Z$ be an integer such that $n \ge 0$.
The $n$th derivative of $x^n$ $x$ is:
:$\dfrac {\d^n} {\d x^n} x^n = n!$
where $n!$ denotes $n$ factorial.
== Proof ==
From Nth Derivative of Mth Power, we have:
:$\dfrac {\d^n} {\d x^n} x^m = \begin {cases}
m^\underline n \, x^{m - n} & : n \le m \\
0 & : n > m
\end {cases}$
where $m^\underline n$ denotes the falling factorial.
Putting $m = n$:
:$\dfrac {\d^n} {\d x^n} x^n = n^\underline n$
where from the definition of the falling factorial:
:$n^\underline n = n!$
Hence the result.
Category:Derivatives
Nth Derivative of Reciprocal of Mth Power

== Theorem ==
Let $m \in \Z$ be an integer such that $m > 0$.
The $n$th derivative of $\dfrac 1 {x^m}$ $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$
where $m^{\overline n}$ denotes the rising factorial.
=== Corollary ===
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 {x^m} = \dfrac {\paren {-1}^n m^{\overline n}} {x^{m + n}}$
=== Basis for the Induction ===
$P(1)$ is true, as this is the case:
which matches the proposition as $m^{\overline 1} = m$ from the definition of rising factorial.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\dfrac {\d^k} {\d x^k} \dfrac 1 {x^m} = \dfrac {\paren {-1}^k m^{\overline k} } {x^{m + k} }$
Then we need to show:
:$\dfrac {\d^{k + 1} } {\d x^{k + 1} } \dfrac 1 {x^m} = \dfrac {\paren {-1}^{k + 1} m^{\overline {k + 1} } } {x^{m + k + 1} }$
=== Induction Step ===
This is our induction step:
First, let $k < m$. Then we have:
Category:Derivatives
Category:Reciprocals
Nth Derivative of Reciprocal of Mth Power/Corollary

== Theorem ==
The $n$th derivative of $\dfrac 1 x$ $x$ is:
:$\dfrac {\d^n} {\d x^n} \dfrac 1 x = \dfrac {\paren {-1}^n n!} {x^{n + 1} }$
where $n!$ denotes $n$ factorial.
== Proof ==
Follows directly from Nth Derivative of Reciprocal of Mth Power by putting $m = 1$.
Category:Derivatives
Category:Reciprocals
Nth Forward Difference is Constant iff Function is Polynomial of Degree N

== Theorem ==
Let $y = \map f x$ be a real function.
Let $\Delta y$ denote the forward difference operator on $y$.
Then:
:$\Delta^n y$ is constant for all $y$
:$f$ is a polynomial function of degree $n$.
== Proof ==
== Sources ==
* 
* 
Category:Forward Difference Operator
Category:Polynomial Functions
Nth Root Test

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$.
Let the sequence $\sequence {a_n}$ be such that the limit superior $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
Then:
:If $l > 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
=== Weak Form ===
If we restrict our domain of $\sequence {a_n}$ to the strictly positive, we can use a weaker form:
== Proof ==
=== Absolute Convergence ===
Let $l < 1$.
Then let us choose $\epsilon > 0$ such that $l + \epsilon < 1$.
Consider the real sequence $\sequence {b_n}$ defined by $\sequence {b_n} = \sequence {\size {a_n} }$.
Here, $\size {a_n}$ denotes either the absolute value of $a_n$, or the complex modulus of $a_n$.
Then:
:$\ds l = \limsup_{n \mathop \to \infty} {b_n}^{1/n}$
It follows from Terms of Bounded Sequence Within Bounds that for sufficiently large $n$,:
:$b_n < \paren {l + \epsilon}^n$
By Sum of Infinite Geometric Sequence, the series $\ds \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ converges.
By the comparison test, $\ds \sum_{n \mathop = 1}^\infty b_n$ converges.
Hence $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely by the definition of absolute convergence.
=== Divergence ===
Let $l > 1$.
Then we choose $\epsilon > 0$ such that $l - \epsilon > 1$.
 that there exist an upper bound for the set:
:$S := \set {n \in \N: \size {a_n}^{1/n} > l - \epsilon}$
Then for all sufficiently large $n$:
:$\size {a_n}^{1/n} \le l - \epsilon$
However, this implies that:
:$\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} \le l - \epsilon$
which is false by the definition of $l$.
The set $S$, then, is not bounded.
This means that there exist arbitrarily large $n$ such that:
:$\size {a_n} > \paren {l - \epsilon}^n$
Thus:
:$\ds \lim_{n \mathop \to \infty} \size {a_n} \ne 0$
and so $\ds \lim_{n \mathop \to \infty} a_n \ne 0$.
Hence from Terms in Convergent Series Converge to Zero, $\ds \sum_{n \mathop = 1}^\infty a_n$ must be divergent.
== Warning ==
== Also known as ==
== Sources ==
* : $\S 6.18$
* : Appendix: $\S 18.5$: Tests for convergence of series
* 
* 
Category:Nth Root Test
Category:Convergence Tests
Category:Named Theorems
Nth Root Test/Also known as

== Nth Root Test: Also known as ==
The Nth Root Test is also seen presented just as the Root Test.
Some sources use:
* the Cauchy Convergence Test
* Cauchy's Convergence Test
* Cauchy's Test
for , but these are imprecise enough for to prefer Nth Root Test
== Sources ==
* 
* 
* 
Category:Nth Root Test
Nth Root Test/Warning

== Nth Root Test: Warning ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of real numbers $\R$ or complex numbers $\C$.
Let the sequence $\sequence {a_n}$ be such that the limit superior $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
If $l = 1$, the Nth Root Test provides no information on whether $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely, converges conditionally, or diverges.
If $\ds \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = \infty$, then of course $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
== Sources ==
* : $\S 6.18$
Category:Nth Root Test
Nth Root Test/Weak Form

== Theorem ==
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a series of (strictly) positive real numbers $\R$.
Let the sequence $\sequence {a_n}$ be such that the limit $\ds \lim_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
Then:
:If $l > 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ diverges.
:If $l < 1$, the series $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
== Proof ==
== Sources ==
* 
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Nth Root of *-Algebra Homomorphism at Positive Element of C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ and $\struct {B, \dagger, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Let $a \in A$ be positive. 
Then we have:
:$\paren {\map \phi a}^{1/n} = \map \phi {a^{1/n} }$
for each $n \in \N$, where $\cdot^{1/n}$ denotes the positive $n$th root.
== Proof ==
From *-Algebra Homomorphism preserves Canonical Preordering of C*-Algebra, $\map \phi a$ is positive.
Since $a^{1/n}$ is positive by definition, we also obtain from *-Algebra Homomorphism preserves Canonical Preordering of C*-Algebra that $\map \phi {a^{1/n} }$ is positive.
Since $\phi$ is an algebra homomorphism, we have:
:$\paren {\map \phi {a^{1/n} } }^n = \map \phi {\paren {a^{1/n} }^n} = \map \phi a$
from the definition of positive $n$th root.
From Existence and Uniqueness of Positive Nth Root of Positive Element of C*-Algebra, $\map \phi a^{1/n}$ is the unique positive element such that $\paren {\paren {\map \phi a}^{1/n} }^n = \map \phi a$.
Hence we have:
:$\paren {\map \phi a}^{1/n} = \map \phi {a^{1/n} }$
Category:Positive Elements of C*-Algebras
Nth Root of 1 plus x not greater than 1 plus x over n

== Theorem ==
Let $x \in \R_{>0}$ be a (strictly) positive real number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\sqrt [n] {1 + x} \le 1 + \dfrac x n$
== Proof ==
From Bernoulli's Inequality:
:$\paren {1 + y}^n \ge 1 + n y$
which holds for:
:$y \in \R$ where $y > -1$
:$n \in \Z_{\ge 0}$
Thus it holds for $y \in \R_{> 0}$ and $n \in \Z_{> 0}$.
So:
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $13$
Category:Inequalities
Nth Root of Integer is Integer or Irrational

== Theorem ==
Let $n$ be a natural number.
Let $x$ be an integer.
If the $n$th root of $x$ is not an integer, it must be irrational.
== Proof ==
We prove the contrapositive: if the $n$th root of $x$ is rational, it must be an integer.
By Existence of Canonical Form of Rational Number, there exist an integer $a$ and a natural number $b$ which are coprime such that:
Since $a$ and $b$ are coprime, $a^n$ and $b^n$ are coprime by Powers of Coprime Numbers are Coprime.
Hence $\dfrac {a^n} {b^n}$ is by definition in canonical form.
Suppose $b \ne 1$.
As the denominator of $\dfrac {a^n} {b^n}$ is not $1$, $x = \dfrac {a^n} {b^n}$ is not an integer.
This is a contradiction.
Thus $b = 1$, and thus:
:$x^{1/n} = a$
which is an integer.
== Historical Note ==
== Sources ==
* : $\S 2.2$: Divisibility and factorization in $\mathbf Z$: Exercise $6$
Category:Irrationality Proofs
Category:Integers
Nth Root of Integer is Integer or Irrational/Historical Note

== Historical Note on Nth Root of Integer is Integer or Irrational ==
The fact that the Square Root of 2 is Irrational was known to .
 proved that the square roots of the natural numbers from $3$ to $17$, except for $4$, $9$ and $16$, are irrational.
He clearly did not have a general proof of this phenomenon.
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Historical Notes
Nu of Prime Number is 1

== Theorem ==
Let $p$ be a prime number.
Then:
:$\map \nu p = 1$
where $\nu$ denotes the $\nu$ function: the number of types of group of a given order.
== Proof ==
Let $G_1$ and $G_2$ be groups of order $p$.
From Prime Group is Cyclic, $G_1$ and $G_2$ are both cyclic groups.
From Cyclic Groups of Same Order are Isomorphic, $G_1$ and $G_2$ are isomorphic.
Thus by definition, $G_1$ and $G_2$ are of the same type.
Hence the result.
== Sources ==
* : $1$: Introduction to Finite Group Theory: $1.3$
Category:Prime Groups
Category:Nu Function
Nuisance Parameter/Examples/Variance of Normal Distribution

== Example of Nuisance Parameter ==
Consider the formation of a confidence interval for the expectation of a population with a normal distribution whose variance $\sigma^2$ is unknown.
Then $\sigma^2$ is an instance of a nuisance parameter.
This difficulty can be overcome by basing the confidence interval on the $t$-distribution, which does not involve $\sigma^2$.
== Sources ==
* 
* 
Category:Examples of Nuisance Parameters
Null Function/Examples/Example 1

== Example of Null Function ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} 1 & : t = 1/2 \\ -1 & : t = 1 \\ 0 & : \text {otherwise} \end {cases}$
Then $f$ is a null function.
== Proof ==
Let $x > 1$.
The cases where $x < \dfrac 1 2$ and $\dfrac 1 2 \le x \le 1$ are treated similarly.
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Special Functions: $\text {IX}$. Null functions: Example
Category:Examples of Null Functions