nyuuzyou/wikis huggingface dataset:
Nicomachus's Theorem/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $1^3 = 1$.
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:
:$k^3 = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$
Then we need to show:
:$\paren {k + 1}^3 = \paren {\paren {k + 1}^2 - \paren {k + 1} + 1} + \paren {\paren {k + 1}^2 - \paren {k + 1} + 3} + \dotsb + \paren {\paren {k + 1}^2 + \paren {k + 1} - 1}$
=== Induction Step ===
Let $T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$.
We can express this as:
:$T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 - k + 2k - 1}$
We see that there are $k$ terms in $T_k$.
Let us consider the general term $\paren {\paren {k + 1}^2 - \paren {k + 1} + j}$ in $T_{k+1}$:
So, in $T_{k + 1}$, each of the terms is $2 k$ larger than the corresponding term for $T_k$.
So:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
 
Finally, note that the first term in the expansion for $\paren {n + 1}^3$ is $n^2 - n + 1 + 2 n = n^2 + n + 1$.
This is indeed two more than the last term in the expansion for $n^3$ .
== Sources ==
* : $\S 1.2.1$: Mathematical Induction: Exercise $8 \ \text{(a)}$
Category:Nicomachus's Theorem
Nicomachus's Theorem/Proof 2

== Theorem ==
== Proof ==
From the definition:
:$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
can be written:
:$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}$
Writing this in sum notation:
Category:Nicomachus's Theorem
Nilpotent Element is Contained in Prime Ideals

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\mathfrak p \subset R$ be a prime ideal of $R$.
Let $x \in R$ be a nilpotent element of $R$.
Then $x \in \mathfrak p$.
== Proof ==
Let $x$ be nilpotent in $R$ as asserted.
Then by definition:
:$\exists n \in \Z_{>0}: x^n = 0$
But $0 \in \mathfrak p$ so:
:$x^n \in \mathfrak p$
Category:Prime Ideals of Rings
Nilpotent Element is Zero Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Suppose further that $R$ is not the null ring.
Let $x \in R$ be a nilpotent element of $R$.
Then $x$ is a zero divisor in $R$.
== Proof ==
First note that when $R$ is the null ring the result is false.
This is because although $0_R$ is nilpotent element in the null ring, it is not actually a zero divisor.
Hence in this case $0_R$ is both nilpotent and a zero divisor.
So, let $R$ be a non-null ring.
By hypothesis, there exists $n \in \Z_{>0}$ such that $x^n = 0_R$.
If $n = 1$, then $x = 0_R$.
By hypothesis, $R$ is not the null ring, so we may choose $y \in R \setminus \set 0$.
By Ring Product with Zero:
:$y \circ x = y \circ 0_R = 0_R$
Therefore $x$ is a zero divisor in $R$.
If $n \ge 2$, define $y = x^{n - 1}$.
Then:
:$y \circ x = x^{n - 1} \circ x = x^n = 0_R$
so $x$ is a zero divisor in $R$.
Category:Nilpotent Ring Elements
Category:Zero Divisors
Nilpotent Elements of Commutative Ring form Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$ and whose unity is $1_R$.
The subset of nilpotent elements of $R$ form an ideal of $R$.
== Proof ==
Let $N$ be the subset of nilpotent elements.
Because $0_R$ is nilpotent, $0_R \in N$ and so $N$ is non-empty.
Let $x \in N$ and $a \in R$.
We have:
Let $x, y \in N$.
Let $x^n = 0$ and $y^m = 0$.
By the Binomial Theorem, $\paren {x - y}^p = 0$ for $p \ge n + m - 1$.
Thus $x - y \in N$.
Thus from Test for Ideal, $N$ is an ideal.
== Also see ==
* Definition:Nilradical of Ring
== Sources ==
* : Chapter $1$: Rings and Ideals: $\S$ Nilradical and Jacobson Radical
* : Chapter $9$: Rings: Exercise $11 \ \text {(ii)}$
Category:Nilpotent Ring Elements
Category:Commutative Rings
Category:Ideal Theory
Nilpotent Ring Element plus Unity is Unit

== Theorem ==
Let $A$ be a ring with unity.
Let $1 \in A$ be its unity.
Let $a \in A$ be nilpotent.
Then $1 + a$ is a unit of $A$.
== Proof ==
Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$.
By Sum of Geometric Sequence in Ring:
:$\paren {1 + a} \cdot \ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$
:$\paren {\ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$
where $\sum$ denotes summation.
By Negative of Nilpotent Ring Element:
:$\paren {-a}^n = 0$
Thus $1 + a$ is a unit.
== Also see ==
* Nilpotent Ring Element plus Unit is Unit
* Topologically Nilpotent Element plus Unity is Invertible
Category:Nilpotent Ring Elements
Nine Point Circle Theorem

== Theorem ==
Let $\triangle ABC$ be a triangle.
These $9$ points:
:the feet of the altitudes of $\triangle ABC$
:the midpoints of the sides of $\triangle ABC$
:the midpoints of the lines from the vertices of $\triangle ABC$ to the orthocenter $H$ of $\triangle ABC$
all lie on the circumference of a circle.
The center $M$ lies on the Euler line of $\triangle ABC$, at the midpoint between the orthocenter $H$ and the circumcenter $O$.
:
This circle is known as the Feuerbach circle, or the nine point circle.
== Proof ==
Let the altitudes of $\triangle ABC$ be $AD$, $BE$ and $CF$.
Let $H$ be the orthocenter of $\triangle ABC$.
Let $X$, $Y$ and $Z$ bisect $AH$, $BH$ and $CH$, respectively.
Let $A_m$, $B_m$, and $C_m$ bisect $BC$, $AC$, and $AB$, respectively.
:
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$\Box B_mZYC_m$ is a parallelogram.
By Parallelogram with One Right Angle is Rectangle:
:$\Box B_mZYC_m$ is a rectangle.
Draw the circle with $B_mY$ as diameter.
By Thales' Theorem/Converse:
:$Z$ and $C_m$ are also on the circle.
Since $ZC_m = Y B_m$:
:$ZC_m$ is a diameter of the circle.
mutatis mutandis:
:$XA_m$ is a diameter of the circle.
Using diameter $B_mY$, by Thales' Theorem/Converse:
:$E$ is on the circle.
Using diameter $C_mZ$, by Thales' Theorem/Converse:
:$F$ is on the circle.
Using diameter $A_mX$, by Thales' Theorem/Converse:
:$D$ is on the circle.
The result follows.
== Also known as ==
== Also see ==
* Feuerbach's Theorem
== Historical Note ==
== Sources ==
* : $9$
* : $9$
* 
* 
* 
Category:Nine Point Circle Theorem
Category:Triangles
Category:Circles
Category:Named Theorems
Nine Point Circle Theorem/Historical Note

== Historical Note on Nine Point Circle Theorem ==
The Nine Point Circle Theorem was proved in $1820$ by and .
In $1822$, proved that the nine point circle was tangent to the incircle and all $3$ excircles.
Hence this circle is often referred to as the Feuerbach circle.
== Sources ==
* : $9$
* : $9$
* 
* 
Category:Historical Notes
Nine Regular Polyhedra

== Theorem ==
There exist $9$ regular polyhedra.
== Proof ==
From Five Platonic Solids, there exist $5$ regular polyhedra which are convex:
:the regular tetrahedron
:the cube
:the regular octahedron
:the regular dodecahedron
:the regular icosahedron.
From Four Kepler-Poinsot Polyhedra:
All $4$ of the above are regular polyhedra which are non-convex.
making the total $9$.
== Sources ==
* : $9$
* : $9$
Category:Regular Polyhedra
Niven's Theorem

== Theorem ==
Consider the angles $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$.
The only values of $\theta$ such that both $\dfrac \theta \pi$ and $\sin \theta$ are rational are:
:$\theta = 0: \sin \theta = 0$
:$\theta = \dfrac \pi 6: \sin \theta = \dfrac 1 2$
:$\theta = \dfrac \pi 2: \sin \theta = 1$
== Proof ==
We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then:
:$\theta \in \set{ {0, \dfrac \pi 3, \dfrac \pi 2} }$
=== Lemma ===
Suppose that $\dfrac \theta \pi$ is rational, meaning:
:$\theta = \dfrac {2 \pi k} n$
where $k, n \in \Z$ and $n \ge 1$.
Suppose also that $\cos \theta \in \Q$.
Denoting $c = 2 \cos \theta \in \Q$, we get:
:$\map {F_n} c = \map {F_n} {2 \cos \dfrac {2 \pi k} n} = 2 \map \cos {2 \pi k} = 2$
So $c$ is a rational root of $\map {F_n} x - 2$, which is a monic polynomial with integer coefficients.
By Rational Root Theorem, $c$ must be an integer.
But:
:$\size c = \size {2 \cos \theta} \le 2$
so:
:$c \in \set {-2, -1, 0, 1, 2}$
Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that:
:$\theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$
Thus for any $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are rational, then:
:$\theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$
Instead of the above, suppose that:
:$0 \le \alpha \le \dfrac \pi 2$
and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are rational.
Then we can denote $\theta = \dfrac \pi 2 - \alpha$ and get that:
:$0 \le \theta \le \dfrac \pi 2$
:$\dfrac \theta \pi \in Q$
:$\cos \theta \in \Q$
So:
:$\dfrac \pi 2 - \alpha = \theta \in \set {0, \dfrac \pi 3, \dfrac \pi 2}$
therefore:
:$\alpha \in \set {0, \dfrac \pi 6, \dfrac \pi 2}$
It is suspected that this result is considerably older, and may date back as far as .
Category:Trigonometry
Category:Niven's Theorem
Niven's Theorem/Lemma

== Theorem ==
For any integer $n \ge 1$, there exists a polynomial $\map {F_n} x$ such that:
:$\map {F_n} {2 \cos t} = 2 \cos n t$
In addition:
:$\deg F_n = n$
and $F_n$ is a monic polynomial with integer coefficients.
== Proof ==
The proof proceeds by induction.
For $n = 1$, it is seen that:
:$\map {F_1} x = x$
fulfils the propositions.
For $n = 2$:
:$\map {F_2} x = x^2 - 2$
For $n > 2$:
so:
: $\map {F_n} x = x \map {F_{n - 1}} x - \map {F_{n - 2}} x \in \Z \sqbrk x$
will fulfil:
: $\map {F_n} {2 \cos t} = 2 \cos n t$
Because $\deg F_{n - 1} = n - 1$ and $\deg F_{n - 2} = n - 2$, we can conclude that:
:$\deg F_n = \deg \paren {x \map {F_{n - 1}} x - \map {F_{n - 2}} x} = n$
In addition, the leading coefficient of $F_n$ is equal to the leading coefficient of $F_{n - 1}$, which is $1$.
Hence the result.
Category:Proofs by Induction
Category:Niven's Theorem
No 4 Fibonacci Numbers can be in Arithmetic Sequence

== Theorem ==
Let $a, b, c, d$ be distinct Fibonacci numbers.
Then, except for the trivial case:
:$a = 0, b = 1, c = 2, d = 3$
it is not possible that $a, b, c, d$ are in arithmetic sequence.
== Proof ==
Let:
:$a = F_i, b = F_j, c = F_k, d = F_l$
where $F_n$ denotes the $n$th Fibonacci number.
, further suppose that;
:$a < b < c < d$
or equivalently:
:$i < j < k < l$
Since $i, j, k, l$ are integers, the inequality could be written as:
:$i \le j - 1 \le k - 2 \le l - 3$
Now consider:
For $a, b, c, d$ be in arithmetic sequence:
:$d - c = b - a$
This means that the all the inequalities above must be equalities:
:$F_l - F_k = F_l - F_{l - 1}$
:$F_{l - 2} = F_j$
:$F_j = F_j - F_i$
So:
:$F_i = 0$
and:
:$F_k = F_{l - 1}$
:$F_j = F_{l - 2}$ 
The only Fibonacci numbers having different index but have the same value is $F_1 = F_2 = 1$.
So one of the following is true:
:$F_k = F_{l - 1} = 1$
:$F_j = F_{l - 2} = 1$ 
:$j - 1 = k - 2 = l - 3$
Suppose the third statement is true.
Write $k = j + 1$, $l = j + 2$.
Then:
The only zero term of the Fibonacci numbers is $F_0$.
This gives $j = 2$.
Therefore the only arithmetic sequence among Fibonacci numbers satisfying the condition above is:
:$F_0, F_2, F_3, F_4$
which corresponds to:
:$0, 1, 2, 3$
Now suppose $F_j = 1$.
Since $F_i, F_j, F_k, F_l$ form an arithmetic sequence:
:$F_k = F_j + \paren {F_j - F_i} = 2$ 
:$F_l = F_k + \paren {F_j - F_i} = 3$ 
Which again gives the arithmetic sequence $0, 1, 2, 3$.
Finally suppose $F_k = 1$. 
Since $F_i, F_j, F_k$ form an arithmetic sequence:
:$F_j = \dfrac 1 2 \paren {F_i + F_k} = \dfrac 1 2$
which is not an integer.
So $F_k \ne 1$.
All cases have been accounted for, and the only arithmetic sequence that can be formed is $0, 1, 2, 3$.
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Arithmetic Sequences
No 4 Fibonacci Numbers can be in Arithmetic Sequence/Mistake

== Source Work ==
:
:The Dictionary
:: $5$
:
:The Dictionary
:: $5$
== Mistake ==
:(Incidentally, no four terms of the Fibonacci sequence can be in arithmetic progression.)
The trivial case:
::$F_0 = 0, F_1 = 1, F_3 = 2, F_4 = 3$
has been forgotten.
== Sources ==
* : $5$
* : $5$
Category:Mistakes/Curious and Interesting Numbers
No Arithmetic Sequence of 4 Primes with Common Difference 2

== Definition ==
There exist no $n \in \Z_{>0}$ such that $n, n + 2, n + 4, n + 6$ are all prime.
=== Corollary ===
== Proof ==
 $S$ is a set of $4$ prime numbers of the form $n, n + 2, n + 4, n + 6$.
$S$ must contain as a subset a set of primes $\set {n, n + 2, n + 4}$
From Unique Sequence of Consecutive Odd Numbers which are Prime, the only one of these is $\set {3, 5, 7}$.
The only sets of the form $\set {n, n + 2, n + 4, n + 6}$ containing $\set {3, 5, 7}$ are:
:$(1): \quad \set {1, 3, 5, 7}$: as $1$ is by convention not a prime, then this is not $S$.
:$(2): \quad \set {3, 5, 7, 9}$: as $9 = 3 \times 3$ is not a prime, then this is not $S$.
There are no more possible $\set {n, n + 2, n + 4, n + 6}$ all prime.
Hence, by Proof by Contradiction, $S$ does not exist.
Category:Prime Numbers
Category:Arithmetic Sequences
No Arithmetic Sequence of 4 Primes with Common Difference 2/Corollary

== Definition ==
Let $n \in \Z$.
Let $\map {S_k} n = \set {n, n + 2, n + 4, \ldots, n + 2 k}$ where $k > 2$.
Then it can not be the case that all elements of $S$ are primes.
== Proof ==
From Unique Sequence of Consecutive Odd Numbers which are Prime the set $\map {S_2} 3 = \set {3, 5, 7}$ consists only of primes.
From No Arithmetic Sequence of 4 Primes with Common Difference 2, the set $\map {S_3} n$ contains at least one non-prime.
As $\map {S_{k - 1} } n \subset \map {S_k} n$ it follows that for $k \ge 4$, the set $\map {S_k} n$ also contains at least one non-prime.
Category:Prime Numbers
Category:Arithmetic Sequences
No Bijection between Finite Set and Proper Subset

== Theorem ==
A finite set can not be in one-to-one correspondence with one of its proper subsets.
That is, a finite set is not Dedekind-infinite.
== Proof 1 ==
== Proof 2 ==
== Note ==
Some sources use this result as the property which defines a finite set.
== Sources ==
* : Chapter $1$: Mappings: $\S 15 \theta$
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.3$
Category:No Bijection between Finite Set and Proper Subset
Category:Proper Subsets
Category:Finite Sets
No Bijection between Finite Set and Proper Subset/Proof 1

== Theorem ==
== Proof ==
Let $S$ be a finite set.
Let $T$ be a proper subset of $S$.
Let $f: T \to S$ be an injection.
By Cardinality of Image of Injection and Cardinality of Subset of Finite Set:
:$\card {\Img f} = \card T < \card S$
where $\Img f$ denotes the image of $f$.
Thus $\Img f \ne S$, and so $f$ is not a bijection.
Category:No Bijection between Finite Set and Proper Subset
No Bijection from Set to its Power Set

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
There is no bijection $f: S \to \powerset S$.
== Proof ==
A bijection is by its definition also a surjection.
By Cantor's Theorem there is no surjection from $S$ to $\powerset S$.
Hence the result.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
* : $\S 2.5$: Theorem $6$
* : Chapter $1$: Mappings: $\S 14$
Category:Power Set
Category:Bijections
No Boolean Interpretation Models a WFF and its Negation

== Theorem ==
Let $v$ be a boolean interpretation.
Let $\mathbf A$ be a WFF of propositional logic.
Then $v$ can not model both $\mathbf A$ and $\neg \mathbf A$.
== Proof ==
Suppose that $v$ models $\mathbf A$:
:$v \models \mathbf A$
Then $v \left({\mathbf A}\right) = T$ by definition of models.
By definition of boolean interpretation, $v \left({\neg \mathbf A}\right) = F$.
In particular, $v (\neg \mathbf A) \ne T$, so that:
:$v \not\models \neg \mathbf A$
Hence the result.
Category:Propositional Logic
No Group has Two Order 2 Elements

== Theorem ==
A group can not contain exactly two elements of order $2$.
== Proof ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Suppose:
: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
That is. they are self-inverse:
:$s^2 = e = t^2$
The identity is of order $1$.
Hence $s$ nor $t$ is the identity
Hence, as $s \ne t$, then $s \circ t \in G$ is distinct from both $s$ and $t$.
Also $s \circ t \ne e$ because $s \ne t^{-1}$.
Suppose $s$ and $t$ commute.
Then $\paren {s \circ t}^2 = e$ from Self-Inverse Elements Commute iff Product is Self-Inverse.
Thus there is a third element (at least) in $G$ which is of order $2$.
Now suppose $s$ and $t$ do not commute.
Then from Commutation Property in Group, $s \circ t \circ s^{-1}$ is another element of $G$ distinct from both $s$ and $t$.
But from Order of Conjugate Element equals Order of Element:
:$\order {s \circ t \circ s^{-1} } = \order t$
and thus $s \circ t \circ s^{-1}$ is another element of order $2$.
Thus there is a third element (at least) in $G$ which is of order $2$.
== Sources ==
* : Chapter $6$: An Introduction to Groups: Exercise $12 \ \text {(ii)}$
Category:Order of Group Elements
No Infinitely Descending Membership Chains

== Theorem ==
Let $\omega$ denote the minimally inductive set.
Let $F$ be a mapping whose domain is $\omega$.
Then:
:$\exists n \in \omega: \map F {n^+} \notin \map F n$
=== Corollary ===
== Proof ==
Let $F$ be a mapping whose domain is $\omega$.
By the axiom of replacement, the image of $F$ is a set.
Let the image of $F$ be denoted $\map \WW F$.
Then:
But:
:$\map F {n^+} \in \map \WW F$
So:
:$\map F {n^+} \notin \map F n$
== Sources ==
* : $\S 7.34$
Category:Set Theory
No Infinitely Descending Membership Chains/Corollary

== Theorem ==
There cannot exist a sequence $\sequence {x_n}$ whose domain is $\N_{\gt 0}$ such that:
:$\forall n \in \N_{>0}: x_{n + 1} \in x_n$
== Proof ==
 there is a sequence like that.
From the definition of a sequence, let $f$ be the the mapping that is defined by $\sequence {x_n}$.
Let $\omega$ denote the minimally inductive set.
Let $g: \omega \to \N_{>0}$ be defined as:
:$\map g \alpha = \alpha + 1$
Then the composition $f \circ g$ is a mapping whose domain is $\omega$ such that:
:$\forall n \in \omega: \map {\paren {f \circ g} } {n^+} \in \map {\paren {f \circ g} } n$
But this contradicts No Infinitely Descending Membership Chains.
Therefore by Proof by Contradiction there cannot exist such a sequence.
Hence the result.
Category:Axiom of Foundation
No Injection from Power Set to Set/Lemma

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then there does not exist a set $B$ such that there is an injection from $B$ into $S$ and a surjection from $B$ onto $\powerset S$.
== Proof ==
 there exists such a $B$.
Let $i: B \to S$ be an injection.
Let $f: B \to \powerset S$ be a surjection.
Let $i^\gets: \powerset S \to \powerset B$ be the inverse image mapping of $i$.
By Mapping Induced by Inverse of Injection is Surjection, $i^\gets$ is a surjection.
Let $f^\to: \powerset B \to \powerset {\powerset S}$ be the direct image mapping of $f$.
By Direct Image Mapping of Surjection is Surjection, $f^\to$ is a surjection:
We have that $i^\gets: \powerset S \to \powerset B$ and $f^\to: \powerset B \to \powerset {\powerset S}$ are surjective.
By Composite of Surjections is Surjection, their composition $f^\to \circ i^\gets: \powerset S \to \powerset {\powerset S}$ is a surjection by Composite of Surjections is Surjection.
But this violates Cantor's Theorem, contradicting the assumption that such a $B$ exists.
Category:No Injection from Power Set to Set
No Injection from Power Set to Set/Proof 1

== Theorem ==
== Proof ==
 $f: \powerset S \to S$ is an injection.
By Injection has Surjective Left Inverse Mapping, there is a surjection $g: S \to \powerset S$.
But this contradicts Cantor's Theorem.
Thus there can be no such injection. 
Category:No Injection from Power Set to Set
No Injection from Power Set to Set/Proof 2

== Theorem ==
== Proof ==
=== Lemma ===
The identity mapping $I_{\powerset S}: \powerset S \to \powerset S$ is a surjection by Identity Mapping is Surjection.
Thus, by the lemma, there can be no injection from $\powerset S$ into $S$.
Category:No Injection from Power Set to Set
No Largest Ordinal

== Theorem ==
Let $a$ be a set of ordinals.
Then:
:$\forall x \in a: x \prec \paren {\bigcup a}^+$
== Proof ==
For this proof, we shall use $\prec$, $\in$, and $\subset$ interchangeably.
We are justified in doing this because of Ordering on Ordinal is Subset Relation and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
== Remark ==
This theorem allows us to create an ordinal strictly greater than any ordinal in the set. Thus, this is another means of proving the Burali-Forti Paradox. If the ordinals are a set, then we may construct an ordinal greater than the set of all ordinals, a contradiction.
== Sources ==
* : $\S 7.26$
Category:Ordinals
No Membership Loops

== Theorem ==
For any proper classes or sets $A_1, A_2, \ldots, A_n$:
:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
== Proof ==
Either $A_1, A_2, \ldots, A_n$ are all sets, or there exists a proper class $A_m$ such that $1 \le m \le n$.
Suppose there exists a proper class $A_m$.
Then, by the definition of a proper class, $\neg A_m \in A_{m+1}$, since it is not a member of any class.
The result then follows directly.
Otherwise it follows that all $A_1, A_2, \ldots, A_n$ are sets.
Then, by the fact that Epsilon Relation is Strictly Well-Founded and a Strictly Well-Founded Relation has no Relational Loops, it follows that: 
:$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$
== Also see ==
* Definition:Strictly Well-Founded Relation
* Strictly Well-Founded Relation has no Relational Loops
* Epsilon Relation is Strictly Well-Founded
* Axiom:Axiom of Foundation
== Sources ==
* : $5.20$
Category:Class Theory
Category:Axiom of Foundation
No Natural Number between Number and Successor

== Theorem ==
Let $\N$ be the natural numbers.
Let $n \in \N$.
Then no natural number $m$ exists strictly between $n$ and its successor $n + 1$:
:$\neg \exists m \in \N: \paren {n < m < n + 1}$
That is:
:If $n \le m \le n + 1$, then $m = n$ or $m = n + 1$.
== Proof using Minimally Inductive Set ==
== Proof using Von Neumann Construction ==
Category:Ordering on Natural Numbers
Category:No Natural Number between Number and Successor
No Natural Number between Number and Successor/Proof using Minimally Inductive Set

== Theorem ==
== Proof ==
We use the model that $\N \cong \omega$ where $\omega$ is the minimally inductive set.
 such an ordinal $m$ exists.
Then, by Ordering on Ordinal is Subset Relation:
:$n \in m$
and from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$m \in n^+$
Applying the definition of a successor set, we have:
:$n \in m \lor n = m$
But this creates a membership loop, in that:
:$m \in n \in m \lor m \in m$
By No Membership Loops, we have created a contradiction.
The result follows from Proof by Contradiction.
== Sources ==
* : $\S 7.25$
Category:No Natural Number between Number and Successor
Category:Ordinals
No Natural Number between Number and Successor/Proof using Von Neumann Construction

== Theorem ==
== Proof ==
Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
The result follows from the Sandwich Principle:
:$\forall m, n \in \omega: m \subseteq n \subseteq m^+ \implies m = n \lor n = m^+$
where $m^+$ is identified as $m + 1$.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.3 \ (1)$
Category:No Natural Number between Number and Successor