nyuuzyou/wikis huggingface dataset:
NOR is not Associative

== Theorem ==
Let $\downarrow$ signify the NOR operation.
Then there exist propositions $p,q,r$ such that:
:$p \downarrow \paren {q \downarrow r} \not \vdash \paren {p \downarrow q} \downarrow r$
That is, NOR is not associative.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NAND is not Associative
== Sources ==
* : $\S 2.10$: Exercise $2.13$
Category:Logical NOR
Category:NOR is not Associative
NOR is not Associative/Proof by Truth Table

== Theorem ==
== Proof ==
Apply the Method of Truth Tables:
:$\begin{array}{|ccccc||ccccc|} \hline
p & \downarrow & (q & \downarrow & r) & (p & \downarrow & q) & \downarrow & r \\
\hline
\F & \F & \F & \T & \F & \F & \T & \F & \F & \F \\
\F & \T & \F & \F & \T & \F & \T & \F & \F & \T \\
\F & \T & \T & \F & \F & \F & \F & \T & \T & \F \\
\F & \T & \T & \F & \T & \F & \F & \T & \F & \T \\
\T & \F & \F & \T & \F & \T & \F & \F & \T & \F \\
\T & \F & \F & \F & \T & \T & \F & \F & \F & \T \\
\T & \F & \T & \F & \F & \T & \F & \T & \T & \F \\
\T & \F & \T & \F & \T & \T & \F & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives do not match for all boolean interpretations.
Category:NOR is not Associative
NOR with Equal Arguments

== Theorem ==
Let $\downarrow$ signify the NOR operation.
Then for any proposition $p$:
:$p \downarrow p \dashv \vdash \neg p$
That is, the NOR of a proposition with itself corresponds to the negation operator.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* NAND with Equal Arguments
== Sources ==
* : $\S 2.3.3$
Category:Logical NOR
Category:NOR with Equal Arguments
NOR with Equal Arguments/Proof by Truth Table

== Theorem ==
== Proof ==
Apply the Method of Truth Tables:
:$\begin {array} {|ccc||cc|} \hline
p & \downarrow & p & \neg & p \\
\hline
\F & \T & \F & \T & \F \\
\T & \F & \T & \F & \T \\
\hline
\end{array}$
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
Category:NOR with Equal Arguments
NP-Complete Problems are all Equally Hard

== Theorem ==
Let $\LL$ be a decision problem which is NP-complete.
Suppose $\LL$ can be shown to be solvable by an algorithm running in polynomial time.
Then, in theory, so should all decision problems which are NP-complete
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:NP-Complete
N (n + 1) (2n + 1) over 6 is Integer

== Theorem ==
Let $n \in \Z$ be an integer.
Then $\dfrac {n \paren {n + 1} \paren {2 n + 1} } 6$ is also an integer.
== Proof ==
This is equivalent to proving that $n \paren {n + 1} \paren {2 n + 1}$ is a multiple of $6$.
There are $6$ cases to consider:
:$(1): \quad n \equiv 0 \pmod 6$: we have $n = 6 k$
:$(2): \quad n \equiv 1 \pmod 6$: we have $n = 6 k + 1$
:$(3): \quad n \equiv 2 \pmod 6$: we have $n = 6 k + 2$
:$(4): \quad n \equiv 3 \pmod 6$: we have $n = 6 k + 3$
:$(5): \quad n \equiv 4 \pmod 6$: we have $n = 6 k + 4$
:$(6): \quad n \equiv 5 \pmod 6$: we have $n = 6 k + 5$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $4$
Category:Divisibility
N Choose Negative Number is Zero

== Theorem ==
Let $n \in \Z$ be an integer.
Let $k \in \Z_{<0}$ be a (strictly) negative integer.
Then:
:$\dbinom n k = 0$
== Proof ==
From Pascal's Rule we have:
:$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$
Thus it is sufficient to prove that:
:$\forall n \in \Z: \dbinom n {-1} = 0$
So:
Hence the result.
Category:Binomial Coefficients
N Choose k is not greater than n^k

== Theorem ==
:$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$ 
where $\dbinom n k$ is a binomial coefficient.
Equality holds when $k = 0$ and $k = 1$.
== Proof 1 ==
== Proof 2 ==
Category:Binomial Coefficients
Category:N Choose k is not greater than n^k
N Choose k is not greater than n^k/Proof 1

== Theorem ==
== Proof ==
First let $k > 1$.
For $k > 1$, the product has at least two factors.
Hence at least one factor is strictly less than $n$.
Let $k < 0$.
By definition of binomial coefficient:
:$\dbinom n k = 0$
while $n^k > 0$.
Hence the result follows for $k < 0$.
Finally let $0 \le k \le 1$.
From Binomial Coefficient with Zero:
:$\dbinom n 0 = 1 = n^0$
From Binomial Coefficient with One:
:$\dbinom n 1 = n = n^1$
So equality holds when $k = 0$ or $k = 1$.
Hence the result has been shown to hold for all $k \in \Z$.
Category:N Choose k is not greater than n^k
N Choose k is not greater than n^k/Proof 2

== Theorem ==
== Proof ==
We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$.
Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$.
Let:
: $N = \set {1, \ldots, n}$
: $K = \set {1, \ldots, k}$
From Cardinality of Set of Strictly Increasing Mappings, $\dbinom n k$ is the number of strictly increasing mappings from $K$ to $N$.
From Cardinality of Set of All Mappings, $n^k$ is the number of all mappings from $K$ to $N$.
For $k = 1$ there is exactly 1 mapping from $K$ to $N$ which is trivially strictly increasing.
Otherwise we have that $K$ has more than one element, and so therefore does $n$.
The mapping $f: K \to N: \forall s \in K: \map f s = 1$ is clearly not a strictly increasing mapping.
So not all mappings from $K$ to $N$ are strictly increasing.
Hence a strict inequality holds, and so $\dbinom n k < n^k$.
Category:N Choose k is not greater than n^k
N over 2 times Reciprocal of 1 Plus n Squared x Squared to the Power of 3/2 Delta Sequence

== Theorem ==
<onlyinclude>
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\ds \map {\delta_n} x := \frac n 2 \frac 1 {\paren{1 + n^2 x^2}^{3 / 2} }$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof ==
By Definite Integral of Even Function:
Let $a,b \in \R$.
Then:
Suppose $0 < a < b$.
Then:
Analogously, suppose $a < b < 0$.
Then:
Let $\epsilon \in \R_{> 0}$.
Then:
$\epsilon$ is an arbitrary positive real number.
Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.
Suppose $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
:$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
Then $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.
But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.
Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.
== Also see ==
* N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence

== Theorem ==
Let $\sequence {\map {\delta_n} x}$ be a sequence such that:
:$\ds \map {\delta_n} x := \frac n \pi \frac 1 {1 + n^2 x^2}$
Then $\sequence {\map {\delta_n} x}_{n \mathop \in {\N_{>0} } }$ is a delta sequence.
That is, in the distributional sense it holds that:
:$\ds \lim_{n \mathop \to \infty} \map {\delta_n} x = \map \delta x$
or
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map {\delta_n} x \map \phi x \rd x = \map \delta \phi$
where $\phi \in \map \DD \R$ is a test function, $\delta$ is the Dirac delta distribution, and $\map \delta x$ is the abuse of notation, usually interpreted as an infinitely thin and tall spike with its area equal to $1$.
== Proof 1 ==
== Proof 2 ==
Category:N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence
Category:Examples of Delta Sequences
Category:Dirac Delta Distribution
N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence/Proof 1

== Theorem ==
== Proof ==
Furthermore:
Let $a,b \in \R$.
Then:
Suppose $0 < a < b$.
Then:
Analogously, suppose $a < b < 0$.
Then:
Let $\epsilon \in \R_{> 0}$.
Then:
$\epsilon$ is an arbitrary positive real number.
Hence, for every $\epsilon \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_+}$ and $\map \phi {\xi_-}$ vanish.
Suppose $\xi_\epsilon \ne 0$.
By Real Numbers are Densely Ordered:
:$\forall \epsilon \in \R_{> 0} : \exists \epsilon' \in \R_{> 0} : 0 < \epsilon' < \epsilon$
Then $\epsilon'$ we have that $\xi_\epsilon = \xi_{+'}$ or $\xi_\epsilon = \xi_{-'}$, where $\xi_{+'} \in \hointr {\epsilon'} \infty$ and $\xi_{-'} \in \hointl {-\infty} {-\epsilon'}$.
But from the result above, for every $\epsilon' \in \R_{> 0}$ contributions from expressions with $\map \phi {\xi_{+'}}$ and $\map \phi {\xi_{-'}}$ vanish.
Therefore, the only nonvanishing contribution can come from $\xi_\epsilon = 0$.
== Sources ==
: Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence
N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence/Proof 2

== Theorem ==
== Proof ==
Let $\map g x = \map \phi x - \map \phi 0$.
Then:
:$\ds \int_{- \infty}^\infty \map \phi x \map {\delta_n} x \rd x = \map \phi 0 + \int_{- \infty}^\infty \map g x \map {\delta_n} x \rd x$
Let $A \in \R_{> 0}$.
Then:
Let:
:$\ds \max_{x \mathop \in \closedint {-A} A} \size {\map g x} := \map M A$
Then:
We have that $\map g 0 = 0$.
By definition, $\phi$ is a smooth real function on $\R$.
By Differentiable Function is Continuous, $\map g x$ is continuous at $x = 0$. 
Furthermore:
By definition of the limit of a real function:
:$\forall \epsilon' \in \R_{>0} : \exists \delta \in \R_{>0}: \forall A \in \R_{> 0}: 0 < A < \delta \implies \map M A < \epsilon'$
Let $\ds \epsilon' = \frac \epsilon 2$.
It follows that:
:$\ds \forall \epsilon \in \R_{> 0} : \exists A \in \R_{> 0} : I_3 \le \map M A < \frac \epsilon 2$
Suppose $A$ is such that the above inequality holds.
By definition, $\map \phi x$ is bounded.
Then:
It follows that:
:$\exists b \in \R_{> 0} : \forall x \in \R : \size {\map g x} < b$ 
Then:
With the number $A$ fixed: 
:$\ds \lim_{n \mathop \to \infty} \frac 2 \pi \map \arctan {n A} = 1$.
By Squeeze Theorem and for a fixed $A$ we have:
:$\ds \lim_{n \mathop \to \infty} \size {I_1 + I_2} = 0$
By definition of the limit of a real sequence: 
:$\ds \forall \overline \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} \le b \size {1 - \frac 2 \pi \map \arctan {n A} } < \overline \epsilon$.
Let $\ds \overline \epsilon = \frac \epsilon 2$.
Then:
:$\ds \forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \size {I_1 + I_2} < \frac \epsilon 2$.
Let $A$ and $N$ be such that the above inequalities for $I_3$ and $I_1 + I_2$ hold.
Then:
To sum up:
:$\ds \forall \epsilon \in \R_{>0} : \exists N \in \R_{>0} : \forall n \in \N_{>0} : \forall n > N \implies \size {\int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x} < \epsilon$
By definition of the limit of a real sequence:
:$\ds \lim_{n \mathop \to \infty} \int_{-\infty}^\infty \map g x \map {\delta_n} x \rd x = 0$
However:
== Sources ==
: Chapter $1$ Mathematical Preliminaries $1.11$ Dirac Delta Function
Category:N over Pi times Reciprocal of 1 Plus n Squared x Squared Delta Sequence
Nagata-Smirnov Metrization Theorem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is metrizable $T$ is regular and has a basis that is $\sigma$-locally finite.
== Proof ==
=== Necessary Condition ===
Let $T$ be metrizable.
=== Sufficient Condition ===
Let $T$ be regular.
Let $T$ have a basis that is $\sigma$-locally finite.
== Historical Note ==
== Sources ==
* : Chapter $4$: Embedding and Metrization, $\S$ Metrization, Theorem $18$
* : Chapter $7$: Metrizable Spaces: $\S23$: Metrization: Theorem $23.9$
* : Chapter $6$: Metrization Theorems and Paracompactness: $\S40$: The Nagata-Smirnov Metrization Theorem: Theorem: $40.3$
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Nagata-Smirnov Metrization Theorem
Category:Metrization Theorems
Category:Regular Spaces
Category:Metrizable Topologies
Nagata-Smirnov Metrization Theorem/Historical Note

== Historical Note on Bing's Metrization Theorem ==
The Nagata-Smirnov Metrization Theorem was discovered by ($1950$) and ($1951$) independently of Bing's Metrization Theorem discovered in $1951$ by .
The two theorems Bing's Metrization Theorem and Nagata-Smirnov Metrization Theorem are often merged as the Bing-Nagata-Smirnov Metrization Theorem.
Category:Historical Notes
Nagata-Smirnov Metrization Theorem/Lemma 1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB = \ds \bigcup_{n \mathop \in \N} \BB_n$ be a $\sigma$-locally finite basis of $T$ where $\BB_n$ is locally finite set of subsets for each $n \in \N$.
Let $I = \set {\tuple {B, n} : B \in \BB, B \in \BB_n}$.
For each $\tuple {B, n} \in I$, let $f_{\tuple {B, n} }: S \to \closedint 0 1$:
:$B = \set {x \in S : \map {f_{\tuple {B, n} }} x \ne 0}$
Then:
:$\forall x \in S$ and $n \in \N$:
::the generalized sum $\ds \sum_{B \mathop \in \BB_n} \map {f_{\tuple {B, n} }^2} x$ converges
== Proof ==
Let $s \in S$ and $m \in \N$.
By definition of locally finite set of subsets:
:$\exists U \in \tau : s \in U : \set {B \in \BB_m : B \cap U \ne \O}$ is finite
Hence:
:$\set {B \in \BB_m : s \in B}$ is finite
It follows that:
:$\set {\tuple {B, m} \in I : \map {f_{\tuple {B, m} } } s \ne 0}$ is finite
From Generalized Sum with Finite Non-zero Summands:
:the generalized sum $\ds \sum_{B \mathop \in \BB_m} \map {f_{\tuple {B, m} }^2} s$ converges
The result follows.
Category:Nagata-Smirnov Metrization Theorem
Nagata-Smirnov Metrization Theorem/Lemma 2

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB = \ds \bigcup_{n \mathop \in \N} \BB_n$ be a $\sigma$-locally finite basis of $T$ where $\BB_n$ is locally finite set of subsets for each $n \in \N$.
Let $I = \set{\tuple{B, n} : B \in \BB, B \in \BB_n}$.
For each $\tuple{B, n} \in I$, let $f_{\tuple{B, n}}:S \to \closedint 0 1$:
:$B = \set{x \in S : \map {f_{\tuple{B, n}}} x \ne 0}$
Let $g_n : S \to \closedint 0 1$ be the mapping defined by:
:$\map {g_n} x$ is the limit of the generalized sum $\ds \sum_{B \in \BB_n} \map {f_{\tuple{B, n}}^2} x$
For all $n \in \N$, let:
:$I_n = \set{\tuple{B, k} \in I : k \ge n}$
Then:
:for all $n \in \N$ and $x \in S$:
::the generalized sum $\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2$ converges 
and:
:$\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2 \le \sum_{m \mathop = n}^\infty \dfrac 1 {2^k}$
== Proof ==
Let $n \in \N$.
Let $x \in S$.
Let $\FF$ denote the set of finite subsets of $I_n$.
Let $F \in \FF$.
Hence:
:$\set{k \in \N : \exists B \in \BB_k : \tuple{B, k} \in F}$ is finite.
Let $\set{n_1, n_2, \ldots, n_m} = \set{k \in \N : \exists B \in \BB_k : \tuple{B, k} \in F}$.
We have:
Since $F$ was arbitrary, it follows that:
:$\forall F \in \FF : \ds \sum_{\tuple{B, k} \mathop \in F} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{b, k} } } x} {\sqrt {1 + \map {g_k} x} } }^2 \le \sum_{k \mathop = n}^\infty \dfrac 1 {2^k}$
From Bounded Generalized Sum is Absolutely Convergent:
:the generalized sum $\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2$ converges
and:
:$\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2 \le \sum_{k \mathop = n}^\infty \dfrac 1 {2^k}$
Category:Nagata-Smirnov Metrization Theorem
Nagata-Smirnov Metrization Theorem/Necessary Condition

== Theorem ==
Let $T = \struct {S, \tau}$ be a metrizable topological space.
Then:
:$T$ is regular and has a basis that is $\sigma$-locally finite.
== Proof ==
==== $T$ is Regular ====
We show that $T$ is regular.
By definition of metrizable topological space:
:there exists a metric $d: S \times S \to \R_{\gt 0}$ on $S$ such that the topology induced by $d$ is $\tau$
From Metric Space is Fully Normal, Fully Normal Space is Normal Space and Normal Space is Regular Space:
:$T$ is a regular space
==== Construction of Basis $\VV$ ====
We construct a $\sigma$-locally finite basis $\VV$.
For each $n \in \N$, let:
:$\UU_n = \set{\map {B_{1 / 2^n}} x : x \in S}$
That is, $\UU_n$ is the set of all open balls of radius $\dfrac 1 {2^n}$.
From Open Balls of Same Radius form Open Cover:
:$\forall n \in \N: \UU_n$ is an open cover of $T$
From Metric Space is Paracompact:
:$T$ is a paracompact space
By definition of paracompact space:
:$\forall n \in \N : \exists$ a open refinement $\VV_n$ of $\UU_n$ which is locally finite
Let $\VV = \ds \bigcup_{n \in \N} \VV_n$.
By definition, $\VV$ is $\sigma$-locally finite.
==== $\VV$ is Basis ====
We show that $\VV$ is a basis.
Let $U \in \tau$.
Let $x \in U$.
From Sequence of Powers of Number less than One:
:the sequence $\sequence{\dfrac 1 {2^n}}$ is a null sequence
From Null Sequence induces Local Basis in Metric Space:
:$\exists n \in N : \map {B_{1 / 2^n}} x \subseteq U$
By definition of open refinement:
:$\VV_{n + 1}$ is an open cover
By definition of open cover:
:$\exists V \in \VV_{n + 1} : x \in V$
By definition of open refinement:
:$\exists U \in \UU_{n + 1} : V \subseteq U$
By definition of $\UU_{n + 1}$:
:$\exists y \in S : U = \map {B_{1 / 2^{n + 1}}} y$
We have:
:$x \in V \subseteq \map {B_{1 / 2^{n + 1}}} y$
From Open Ball Contains Open Ball Less Than Half Its Radius:
:$\map {B_{1 / {2^{n + 1}}}} y \subseteq \map {B_{1 / 2^n}} x$
Hence:
:$x \in V \subseteq \map {B_{1 / {2^{n + 1}}}} y \subseteq \map {B_{1 / 2^n}} x$
By definition $\VV$ is a basis.
Hence $\VV$ is a basis that is $\sigma$-locally finite.
Category:Nagata-Smirnov Metrization Theorem
Nakayama's Lemma

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module. 
Let $\map {\operatorname{Jac} } A$ be the Jacobson radical of $A$.
Let $\mathfrak a \subseteq \map {\operatorname{Jac} } A$ be an ideal of $A$.
Suppose $\mathfrak a M = M$.
Then:
:$M = 0$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Cayley-Hamilton Theorem
Category:Commutative Algebra
Nakayama's Lemma/Corollary 1

== Corollary to Nakayama's Lemma ==
Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module. 
Let there exist a submodule $N \subseteq M$ such that:
:$M = N + \operatorname{Jac} \left({A}\right) M$
Then $M = N$.
== Proof ==
If $M = N + \operatorname{Jac} \left({A}\right) M$ then:
:$\operatorname{Jac} \left({A}\right) \left({M / N}\right) = M/N$
so by Nakayama's Lemma:
:$M/N = 0$
and so:
:$M = N$
Category:Commutative Algebra
Nakayama's Lemma/Corollary 2

== Corollary to Nakayama's Lemma ==
Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module. 
Let:
:$m_1 + \operatorname{Jac} \left({A}\right) M, \dotsc, m_n + \operatorname{Jac} \left({A}\right) M$
generate $M / \operatorname{Jac} \left({A}\right) M$ over $A / \operatorname{Jac} \left({A}\right)$.
Then $m_1,\ldots, m_n$ generate $M$ over $A$.
== Proof ==
Let $N$ be the submodule of $M$ generated by $m_1, \dotsc, m_n$.
Then:
:$M = N + \operatorname{Jac} \left({A}\right) M$
Hence by Corollary 1:
:$M = N$
Category:Commutative Algebra
Nakayama's Lemma/Proof 1

== Theorem ==
== Proof ==
We induct on the number of generators of $M$. 
;Basis for the Induction
Let $M$ have a single generator $m_1 \in M$.
Then $\mathfrak a m_1 = M$.
So:
:$m_1 \in \mathfrak a m_1$
That is:
:$m_1 = a m_1$
for some $a \in \mathfrak a$.
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
So:
:$\paren {1 - a}^{-1} \paren {1 - a} m = 0$
thus $m = 0$.
;Inductive Step
Suppose that $M$ is generated by $n$ elements:
:$M = A m_1 + \dotsb + A m_n$
for some $m_1, \dotsc, m_n \in M$.
Then we have:
:$M = \mathfrak a M = \mathfrak a m_1 + \dotsb + \mathfrak a m_n $
Thus for some $a_1, \dotsc, a_n \in \mathfrak a$:
:$ m_1 = a_1 m_1 + \dotsb + a_n m_n $
Then:
:$\paren {1 - a_1} m_1 = a_2 m_2 + \dotsb + a_n m_n$
By Characterisation of Jacobson Radical, $1 - a$ is a unit in $A$.
Multiplying both sides by $\paren {1 - a_1}^{-1}$ gives:
:$m_1 = \paren {1 - a_1}^{-1} a_2 m_2 + \dotsb + \paren {1 - a_1}^{-1} a_n m_n$
so we have:
:$m_1 \in A m_2 + \cdots + A m_n$
Therefore $M$ has $n - 1$ generators $m_2, \dotsc, m_n$.
By the induction hypothesis:
:$\mathfrak a M = M \implies M = 0$
Category:Commutative Algebra
Nakayama's Lemma/Proof 2

== Theorem ==
== Proof ==
Let $\phi : M \to M$ be the identity mapping on $M$, i.e.:
:$\forall x \in M : \map \phi x = x$
Since $\mathfrak a M = M$ , $\phi$ is an endomorphism of $M$ such that:
:$\map \phi M \subseteq a M$
By Cayley-Hamilton Theorem, there exist $a_0, \ldots , a_{n-1} \in \mathfrak a$ such that:
:$(1):\quad \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
Evaluating $(1)$ at $x=1$, especially, we have:
:$(2):\quad 1 + a= 0$
where:
:$a := a_{n - 1} + \cdots + a_1 + a_0$
Since:
:$a \in \mathfrak a \subseteq \map {\operatorname {Jac} } A$
by Characterisation of Jacobson Radical, $1 + a$ is a unit in $A$.
Thus for each $x \in M$, we have:
That is:
:$M = 0$
Category:Commutative Algebra
Napier's Analogies

== Theorem ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
=== Tangent of Half Sum of Sides ===
=== Tangent of Half Difference of Sides ===
=== Tangent of Half Sum of Angles ===
=== Tangent of Half Difference of Angles ===
== Also presented as ==
== Also see ==
* Delambre's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Napier's Analogies
Category:Spherical Trigonometry
Napier's Analogies/Tangent of Half Difference of Angles

== Napier's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac {A - B} 2 = \dfrac {\sin \frac {a - b} 2} {\sin \frac {a + b} 2} \cot \dfrac C 2$
== Proof ==
== Also see ==
* Delambre's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Napier's Analogies
Napier's Analogies/Tangent of Half Difference of Sides

== Napier's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac {a - b} 2 = \dfrac {\sin \frac {A - B} 2} {\sin \frac {A + B} 2} \tan \dfrac c 2$
== Proof ==
== Also see ==
* Delambre's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Napier's Analogies
Napier's Analogies/Tangent of Half Sum of Angles

== Napier's Analogies ==
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\tan \dfrac {A + B} 2 = \dfrac {\cos \frac {a - b} 2} {\cos \frac {a + b} 2} \cot \dfrac C 2$
== Proof ==
== Also see ==
* Delambre's Analogies
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $16$. Delambre's and Napier's analogies.
* 
* 
Category:Napier's Analogies