avewright/proofwiki-math huggingface dataset:
Theorem: Translation of Intersection of Subsets of Vector Space

Let $K$ be a field. 
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $x \in X$.
Then:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x = \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
where $E_\alpha + x$ denotes the translation of $E_\alpha$ by $x$.

Proof:
Let $v \in X$.
We have:
:$\ds v \in \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x$
{{iff}}:
:$\ds v - x \in \bigcap_{\alpha \mathop \in I} E_\alpha$
{{iff}}:
:$v - x \in E_\alpha$ for each $\alpha \in I$
{{iff}}:
:$v \in E_\alpha + x$ for each $\alpha \in I$
{{iff}}:
:$\ds v \in \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
That is:
:$\ds \paren {\bigcap_{\alpha \mathop \in I} E_\alpha} + x = \bigcap_{\alpha \mathop \in I} \paren {E_\alpha + x}$
{{qed}}
Category:Translation of Subsets of Vector Spaces
Category:Set Intersection
4jsm82san7t8xatwyvecx8p5k1l037r
Theorem: Reduced Gyromagnetic Ratio of Proton

The reduced gyromagnetic ratio of the proton is given by:
{{begin-eqn}}
{{eqn | l = \gamma_{\mathrm p} / 2 \pi
      | o = \approx
      | r = 4 \cdotp 25774 \, 785 \times 10^7 \, \mathrm {Hz \, T^{-1} }
      | c = 
}}
{{eqn | o = \approx
      | r = 4 \cdotp 25774 \, 785 \times 10^3 \, \mathrm {Hz \, Gs^{-1} }
      | c = 
}}
{{end-eqn}}

Proof:
By definition, the reduced gyromagnetic ratio of a proton is given as:
:$\dfrac {\gamma_{\mathrm p} } {2 \pi}$
where $\gamma_{\mathrm p}$ denotes the gyromagnetic ratio of the proton.
Then we have:
{{begin-eqn}}
{{eqn | l = \gamma_{\mathrm p}
      | o = \approx
      | r = 2 \cdotp 67522 \, 18744 \, (11) \times 10^8 \, \mathrm {rad \, s^{-1} \, T^{-1} }
      | c = Gyromagnetic Ratio of Proton
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\gamma_{\mathrm p} } {2 \pi}
      | o = \approx
      | r = \dfrac {2 \cdotp 67522 \, 18744 \, (11) \times 10^{-15} } {2 \pi} \, \mathrm {Hz \, T^{-1} }
      | c = 
}}
{{eqn | o = \approx
      | r = 4 \cdotp 25774 \, 785 \times 10^7 \, \mathrm {Hz \, T^{-1} }
      | c = by calculation
}}
{{eqn | o = \approx
      | r = 4 \cdotp 25774 \, 785 \times 10^3 \, \mathrm {Hz \, Gs^{-1} }
      | c = by calculation
}}
{{end-eqn}}
{{qed}}
Theorem: Ratio of Proton Moment to Nuclear Magneton

The ratio of the proton moment to the nuclear magneton is given by:
{{begin-eqn}}
{{eqn | l = \dfrac {\mu_{\mathrm p} } {\mu_{\mathrm N} }
      | r = 2 \cdotp 79284 \, 73446 \, 3(82)
      | c = 
}}
{{end-eqn}}

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \mu_{\mathrm N}
      | r = 5 \cdotp 05078 \, 3699 \, (31) \times 10^{-27} \, \mathrm {J \, T^{-1} }
      | c = {{Defof|Nuclear Magneton|subdef = Value}}
}}
{{eqn | l = \mu_{\mathrm p}
      | r = 1 \cdotp 41060 \, 67973 \, 6 \, (60) \times 10^{-26} \, \mathrm {J \, T^{-1} }
      | c = {{Defof|Proton Moment|subdef = Value}}
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\mu_{\mathrm p} } {\mu_{\mathrm N} }
      | r = 2 \cdotp 79284 \, 73446 \, 3 \, (82)
      | c = by calculation
}}
{{end-eqn}}
{{qed}}
Theorem: Finite Generalized Sum Converges to Summation

Let $G$ be a commutative topological semigroup with identity $0_G$.
Let $\set{i_0, i_1, \ldots, i_n}$ be a finite enumeration of a finite set $I$. 
Let $\family{g_i}_{i \in I}$ be an indexed family of elements of $G$.
Then:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ 
converges to:
:the summation over finite index $\ds \sum_{i \mathop \in I} g_i$

Proof:
Let $\FF$ be the set of finite subsets of $I$.
Let $h = \ds \sum_{i \mathop \in I} g_i$ be the summation over finite index $I$.
Let $U$ be an open subset of $G$ such that $h \in U$.
From Set is Subset of Itself:
:$I \in \FF$
Let: 
:$J \in \FF : I \subseteq J$
Let $h'= \ds \sum_{j \mathop \in J} g_j$ be the summation over finite index $J$.
By definition of $\FF$:
:$J \subseteq I$
By definition of set equality:
:$I = J$
Hence:
:$h'= h \in U$
Since $U$ was arbitrary, by definition of convergance:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ 
converges to:
:the summation over finite index $\ds \sum_{i \mathop \in I} g_i$
{{qed}}
Category:Generalized Sums
ha76k600b0ls6hj5kcpodqxxzw30i99
Theorem: Inequality of Height of Proper Ideal

Let $A$ be a commutative ring with unity.
Let $I$ be a proper ideal in $A$.
Then:
:$\map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } I \le \map {\operatorname {dim_{Krull} } } A$
where:
:$A/I$ is the quotient ring of $A$ by $I$
:$\operatorname {dim_{Krull} }$ denotes the Krull dimension
:$\map {\operatorname {ht} } I$ is the height of $I$

Proof:
Let:
:$n := \map {\operatorname {dim_{Krull} } } {A / I}$
Then there are $\mathfrak q_0, \ldots, \mathfrak q_n \in \Spec {A / I}$ such that:
:$\mathfrak q_0 \subsetneqq \cdots \subsetneqq \mathfrak q_n$
Let $\pi : A \to A / I$ be the quotient epimorphism.
Let:
:$\tilde {\mathfrak q_i} := \pi^{-1} \sqbrk {\mathfrak q_i}$
for $i=0, \ldots, n$.
Then:
:$\tilde {\mathfrak q_0}, \ldots, \tilde {\mathfrak q_n} \in \Spec A$
and:
:$(1):\quad I \subseteqq \tilde {\mathfrak q_0} \subsetneqq \cdots \subsetneqq \tilde {\mathfrak q_n}$
{{improve|More details}}
On the other hand, let:
:$m := \map {\operatorname {ht} } {\tilde {\mathfrak q_0} }$
Then there exist $\mathfrak p_0, \ldots, \mathfrak p_m \in \Spec A$ such that:
:$(2):\quad \mathfrak p_0 \subsetneqq \cdots \subsetneqq \mathfrak p_m = \tilde {\mathfrak q_0}$
Hence:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } I
      | o = \le
      | r = \map {\operatorname {dim_{Krull} } } {A / I} + \map {\operatorname {ht} } {\tilde {\mathfrak q_0} }
      | c = as $I \subseteqq \tilde {\mathfrak q_0}$
}}
{{eqn | r = n + m
}}
{{eqn | o = \le
      | r = \map {\operatorname {dim_{Krull} } } A
      | c = by $(1)$ and $(2)$
}}
{{end-eqn}}
{{qed}}
Theorem: Generalized Hilbert Sequence Space is Metric Space

Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $H^\alpha$ be the generalized Hilbert sequence space of weight $\alpha$.
Then:
:$H^\alpha$ is a metric space.

Proof:
Recall $H^\alpha$ is the structure $\struct{A, d_2}$ where:
:$A$ is the set of all real-valued functions $x : I \to \R$ such that:
::$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
::$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
:$d_2: A \times A \to \R$ is the real-valued function defined as:
::$\ds \forall x = \family {x_i}, y = \family {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$
Theorem: Equivalence of Definitions of Associated Prime of Module

Let $A$ be a commutative ring with unity.
Let $M$ be a module over $A$.
Let $\mathfrak p$ be a prime ideal in $A$.
{{TFAE|def = Associated Prime of Module}}

Proof:
=== Definition 1 implies Definition 2 ===
Suppose that $x \in M$ satisfies:
:$\map {\operatorname {Ann}_A} x = \mathfrak p$
Define a submodule of $M$ by:
:$N := \set { a x : a \in A }$
Define a module homomorphism $\phi : A \to N$ by:
:$a \mapsto a x$
Then the kernel of $\phi$ is:
:$\map \ker \phi = \map {\operatorname {Ann}_A} x = \mathfrak p$
By First Isomorphism Theorem:
:$N \cong A / \mathfrak p$
{{qed|lemma}}
Theorem: Summation over Finite Subset is Well-Defined

Let $\struct{G, +}$ be a commutative monoid.
Let $F \subseteq G$ be a finite subset of $G$.
Then the summation $\ds \sum_{g \mathop \in F} g$ is well-defined.

Proof:
To show that summation over $F$ is well-defined it needs to be shown:
:$(1) \quad \exists$ a finite enumeration of $F$
:$(2) \quad \forall$ finite enumerations $e$ and $d$ of $F : \ds \sum_{i \mathop = 1}^n e_i = \sum_{i \mathop = 1}^n d_i$
Theorem: Summation over Finite Index is Well-Defined

Let $\struct{G, +}$ be a commutative monoid.
Let $\family{g }_{i \mathop \in I}$ be an indexed subset of $G$ where the indexing set $I$ is finite.
Then the summation $\ds \sum_{i \mathop \in I} g_i$ is well-defined.

Proof:
To show that summation over $I$ is well-defined it needs to be shown:
:$(1) \quad \exists$ a finite enumeration of $I$
:$(2) \quad \forall$ finite enumerations $e$ and $d$ of $I : \ds \sum_{k \mathop = 1}^n g_{e_k} = \sum_{k \mathop = 1}^n g_{d_k}$
Theorem: Maximal Annihilator of Module is Associated Prime

Let $A$ be a commutative ring with unity.
Let $M$ be a module over $A$.
Let $\mathbf p$ be a maximal element of the set:
:$\set { \map {\operatorname {Ann}_A} x : x \in M , x \ne 0 }$
with respect to the subset relation.
Then $\mathfrak p$ is an associated prime of $M$.

Proof:
{{ProofWanted}}
Category:Commutative Algebra
kgdimxn2fu957dbfshwgkioxttr07qw
Theorem: Commutators are Identity iff Group is Abelian

Let $\struct {G, \circ}$ be a group whose identity is $e$.
For $g, h \in G$, let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\struct {G, \circ}$ is abelian {{iff}}:
:$\forall g, h \in G: \sqbrk {g, h} = e$

Proof:
=== Necessary Condition ===
Let $\struct {G, \circ}$ be such that:
:$\forall g, h \in G: \sqbrk {g, h} = e$
From Commutator is Identity iff Elements Commute:
:$\forall g, h \in G: g \circ h = h \circ g$
Hence $\struct {G, \circ}$ is abelian by definition.
{{qed|lemma}}
Theorem: Derived Subgroup is Normal

Let $G$ be a group whose identity is $e$.
Let $\sqbrk {G, G}$ denote the derived subgroup of $G$.
Then $\sqbrk {G, G}$ is a normal subgroup of $G$.

Proof:
Recall the definition of $\sqbrk {G, G}$:
:$\sqbrk {G, G}$ is the subgroup of $G$ generated by all its commutators.
Recall also the definition of the commutator of $g, h \in G$:
:$\sqbrk {g, h} = g^{-1} h^{-1} g h$
From Derived Subgroup is Subgroup we note that $\sqbrk {G, G}$ is indeed a subgroup of $G$.
Let $g, h \in G$ be arbitrary.
Let $x \in \sqbrk {G, G}$ be arbitrary.
By definition of generated subgroup:
:$\sqbrk {g, h} x \in \sqbrk {G, G}$
and:
:$\sqbrk {h, g} x \in \sqbrk {G, G}$
Hence also by definition of generated subgroup:
:$\sqbrk {g, h} x \sqbrk {h, g} \in \sqbrk {G, G}$
and:
:$\sqbrk {h, g} x \sqbrk {g, h} \in \sqbrk {G, G}$
But from Inverse of Group Commutator:
:$\sqbrk {g, h}^{-1} = \sqbrk {h, g}$
So:
:$\sqbrk {g, h} x \sqbrk {g, h}^{-1} \in \sqbrk {G, G}$
and:
:$\sqbrk {g, h}^{-1} x \sqbrk {g, h} \in \sqbrk {G, G}$
As $x \in \sqbrk {G, G}$ is arbitrary, by definition of coset:
:$\forall \sqbrk {g, h} \in G: \sqbrk {g, h} \sqbrk {G, G} \sqbrk {g, h}^{-1} \subseteq \sqbrk {G, G}$
and:
:$\forall \sqbrk {g, h} \in G: \sqbrk {g, h}^{-1} \sqbrk {G, G} \sqbrk {g, h} \subseteq \sqbrk {G, G}$
Hence the result by definition of normal subgroup.
{{qed}}
Theorem: Abelianization of Group is Abelian

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $G^{\mathrm {ab} }$ denote the abelianization of $G$.
Then $G^{\mathrm {ab} }$ is an abelian group.

Proof:
{{Recall|Abelianization of Group}}
{{:Definition:Abelianization of Group}}
From Derived Subgroup is Normal, $\sqbrk {G, G}$ is a normal subgroup of $G$.
Hence the above definition is valid.
By definition of derived subgroup:
:$\forall x, y \in G: \sqbrk {x, y} \in \sqbrk {G, G}$
where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.
Hence from Quotient Group is Abelian iff All Commutators in Divisor it follows that $G^{\mathrm {ab} }$ is an abelian group.
{{qed}}
Theorem: Matrix Multiplication on Diagonal Matrices is Commutative

Let $\mathbf A$ and $\mathbf B$ be diagonal matrices.
Then:
:$\mathbf A \mathbf B = \mathbf B \mathbf A$
where $\mathbf A \mathbf B$ denotes (conventional) matrix product.

Proof:
{{begin-eqn}}
{{eqn | l = \mathbf A
      | o = :=
      | r = \sqbrk {a_{ij} }_n
      | c = 
}}
{{eqn | l = \mathbf B
      | o = :=
      | r = \sqbrk {b_{ij} }_n
      | c = 
}}
{{end-eqn}}
Note that the orders of $\mathbf A$ and $\mathbf B$ must be equal in order for matrix product to be defined.
Then we have:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf B
      | r = \sqbrk {c_{ij} }_n
      | c = where $c_{ij}$ denotes the $\tuple {i, j}$th element of $\mathbf A \mathbf B$
}}
{{eqn | r = \begin {cases} a_{i i} b_{i i} & i = j \\ 0 & i \ne j \end {cases}
      | c = Product of Diagonal Matrices is Diagonal
}}
{{eqn | r = \begin {cases} b_{i i} a_{i i} & i = j \\ 0 & i \ne j \end {cases}
      | c = Commutative Law of Multiplication
}}
{{eqn | r = \mathbf B \mathbf A
      | c = Product of Diagonal Matrices is Diagonal
}}
{{end-eqn}}
{{qed}}
Theorem: Group Commutators are Commuting Elements

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\sqbrk {g, h}$ commutes with $\sqbrk {h, g}$, in the sense that:
:$\sqbrk {g, h} \circ \sqbrk {h, g} = \sqbrk {h, g} \circ \sqbrk {g, h}$

Proof:
From Inverse of Group Commutator:
:$\forall g, h \in G: \sqbrk {g, h} = \sqbrk {h, g}^{-1}$
The result follows from Group Element Commutes with Inverse.
{{qed}}
Theorem: Inverse of Group Commutator

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then $\sqbrk {g, h}$ is the inverse of $\sqbrk {h, g}$:
:$\sqbrk {g, h} = \sqbrk {h, g}^{-1}$

Proof:
{{begin-eqn}}
{{eqn | l = \sqbrk {g, h} \circ \sqbrk {h, g}
      | r = \paren {g^{-1} \circ h^{-1} \circ g \circ h} \circ \paren {h^{-1} \circ g^{-1} \circ h \circ g}
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = \paren {g^{-1} \circ h^{-1} \circ g \circ h} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}^{-1}
      | c = Inverse of Group Product: General Result
}}
{{eqn | r = e
      | c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {g^{-1} \circ h^{-1} \circ g \circ h}^{-1} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}
      | c = {{Group-axiom|3}}
}}
{{eqn | r = \paren {h^{-1} \circ g^{-1} \circ h \circ g} \circ \paren {g^{-1} \circ h^{-1} \circ g \circ h}
      | c = Inverse of Group Product: General Result
}}
{{eqn | r = \sqbrk {h, g} \circ \sqbrk {g, h}
      | c = {{Defof|Commutator of Group Elements}}
}}
{{end-eqn}}
{{qed}}
Category:Group Commutators
Category:Inverse Elements
9vvaf7mzbaeikmv7cqe9wd6txc0lnqg
Theorem: Commutator of Group Element with Identity is Identity

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\sqbrk {g, h}$ denote the commutator of $g$ and $h$.
Then:
:$\sqbrk {g, e} = e = \sqbrk {e, g}$

Proof:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = \sqbrk {g, e}
      | r = g^{-1} \circ e^{-1} \circ g \circ e
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = g^{-1} \circ e \circ g \circ e
      | c = Identity is Self-Inverse
}}
{{eqn | r = g^{-1} \circ g
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = e
      | c = {{Defof|Inverse Element}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = \sqbrk {e, g}
      | r = e^{-1} \circ g^{-1} \circ e \circ g
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = e^ \circ g^{-1} \circ e \circ g
      | c = Identity is Self-Inverse
}}
{{eqn | r = g \circ g^{-1}
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = e
      | c = {{Defof|Inverse Element}}
}}
{{end-eqn}}
{{qed}}
Category:Group Commutators
Category:Inverse Elements
16770ysr99sxmmi1liln5jvqlje9bg8
Theorem: Derived Subgroup is Subgroup

Let $G$ be a group whose identity is $e$.
Let $\sqbrk {G, G}$ denote the derived subgroup of $G$.
Then $\sqbrk {G, G}$ is indeed a subgroup of $G$.

Proof:
Recall the definition of $\sqbrk {G, G}$:
:$\sqbrk {G, G}$ is the subgroup of $G$ generated by all its commutators.
Recall also the definition of the commutator of $g, h \in G$:
:$\sqbrk {g, h} = g^{-1} h^{-1} g h$
We note that from Commutator of Group Element with Identity is Identity:
:$\sqbrk {e, e} = e$
and so $e \in \sqbrk {G, G}$
Thus:
:$\sqbrk {G, G} \ne \O$
Let $g, h \in G$ be arbitrary.
We have by definition of inverse element that:
:$\sqbrk {g, h} = \sqbrk {g, h} e$
and so $\sqbrk {g, h} \in \sqbrk {G, G}$.
Let $g_1, h_1, g_2, h_2 \in G$ be arbitrary.
By definition of $\sqbrk {G, G}$:
:$\sqbrk {g_1, h_1} \in \sqbrk {G, G}$
and also:
:$\sqbrk {g_2, h_2} \in \sqbrk {G, G}$
Hence by definition of generated subgroup:
:$\sqbrk {g_1, h_1} \sqbrk {g_2, h_2} \in \sqbrk {G, G}$
Hence by the Two-Step Subgroup Test, $\sqbrk {G, G}$ is a subgroup of $G$.
{{qed}}
Category:Derived Subgroups
9ocogc8i8gbzx2pm5fw44e7y3tgux4z
Theorem: Prime Sum of Squares of 3 Prime Numbers

Let $a$, $b$ and $c$ be prime numbers with the property that:
:$a^2 + b^2 + c^2 = p$
where $p$ is a prime number.
Then either $1$ or $2$ of $a$, $b$ and $c$ is equal to $3$.

Proof:
First we note that sets of $3$ prime numbers with this property are plentiful.
For example, these are all such sets for $a, b, c < 20$:
{{begin-eqn}}
{{eqn | l = 2^2 + 2^2 + 3^2
      | r = 17
      | c = 
}}
{{eqn | l = 3^2 + 3^2 + 5^2
      | r = 43
      | c = 
}}
{{eqn | l = 3^2 + 3^2 + 7^2
      | r = 67
      | c = 
}}
{{eqn | l = 3^2 + 3^2 + 11^2
      | r = 139
      | c = 
}}
{{eqn | l = 3^2 + 3^2 + 17^2
      | r = 307
      | c = 
}}
{{eqn | l = 3^2 + 3^2 + 19^2
      | r = 379
      | c = 
}}
{{eqn | l = 3^2 + 5^2 + 5^2
      | r = 59
      | c = 
}}
{{eqn | l = 3^2 + 5^2 + 7^2
      | r = 93
      | c = 
}}
{{eqn | l = 3^2 + 7^2 + 7^2
      | r = 107
      | c = 
}}
{{eqn | l = 3^2 + 7^2 + 11^2
      | r = 179
      | c = 
}}
{{eqn | l = 3^2 + 7^2 + 13^2
      | r = 227
      | c = 
}}
{{eqn | l = 3^2 + 7^2 + 17^2
      | r = 347
      | c = 
}}
{{eqn | l = 3^2 + 7^2 + 19^2
      | r = 419
      | c = 
}}
{{eqn | l = 3^2 + 11^2 + 11^2
      | r = 251
      | c = 
}}
{{eqn | l = 3^2 + 11^2 + 17^2
      | r = 467
      | c = 
}}
{{eqn | l = 3^2 + 11^2 + 19^2
      | r = 491
      | c = 
}}
{{eqn | l = 3^2 + 13^2 + 13^2
      | r = 347
      | c = 
}}
{{eqn | l = 3^2 + 13^2 + 17^2
      | r = 467
      | c = 
}}
{{eqn | l = 3^2 + 17^2 + 17^2
      | r = 587
      | c = 
}}
{{eqn | l = 3^2 + 17^2 + 19^2
      | r = 659
      | c = 
}}
{{end-eqn}}
And indeed, the number of instances of $3$ in all the above is either $1$ or $2$.
First we note that if $a = b = c = 3$ then $a^2 + b^2 + c^2 = 27$ which is not prime.
It remains to demonstrate that if none of $a$, $b$ and $c$ are equal to $3$, then $a^2 + b^2 + c^2$ is not prime.
Let $q$ be a prime number such that $q \ne 3$.
Then by definition of prime number:
:$3 \nmid q$
where $\nmid$ denotes non-divisibility.
Then from Square Modulo 3:
:$q^2 \pmod 3 = 1$
Suppose none of $a$, $b$ and $c$ is equal to $3$.
Then:
{{begin-eqn}}
{{eqn | l = a^2
      | o = \equiv
      | r = 1
      | rr= \pmod 3
      | c = 
}}
{{eqn | l = b^2
      | o = \equiv
      | r = 1
      | rr= \pmod 3
      | c = 
}}
{{eqn | l = c^2
      | o = \equiv
      | r = 1
      | rr= \pmod 3
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^2 + b^2 + c^2
      | o = \equiv
      | r = 1 + 1 + 1
      | rr= \pmod 3
      | c = Modulo Addition is Well-Defined
}}
{{eqn | ll= \leadsto
      | l = a^2 + b^2 + c^2
      | o = \equiv
      | r = 0
      | rr= \pmod 3
      | c = {{Defof|Congruence Modulo Integer}}
}}
{{eqn | ll= \leadsto
      | l = a^2 + b^2 + c^2
      | o = \equiv
      | r = 0
      | rr= \pmod 3
      | c = {{Defof|Congruence Modulo Integer}}
}}
{{end-eqn}}
That is:
:$3 \divides a^2 + b^2 + c^2$
where $\divides$ denotes divisibility.
Hence if none of $a$, $b$ and $c$ is equal to $3$, $a^2 + b^2 + c^2$ is not prime number.
The result follows.
{{qed}}
Category:Prime Numbers
Category:3
ddsg8tfgmg9k1cx0rfbcehr4agjc61b
Theorem: Topological Space is Connected iff any Proper Non-Empty Subset has Non-Empty Boundary

Let $\struct {X, \tau}$ be a topological space.
Then $\struct {X, \tau}$ is connected {{iff}} for each proper non-empty subset $S \subseteq X$, we have $\partial S \ne \O$.

Proof:
From Connected iff no Proper Clopen Sets, we have that:
:$\struct {X, \tau}$ is connected {{iff}} there exists no proper non-empty clopen set $S \subseteq X$.
From Set is Clopen iff Boundary is Empty, we have that: 
:$S \subseteq X$ is clopen {{iff}} $\partial S = \O$.
Hence we have:
:$\struct {X, \tau}$ is connected {{iff}} there exists no proper non-empty set $S \subseteq X$ such that $\partial S = \O$.
That is:
:$\struct {X, \tau}$ is connected {{iff}} for each proper non-empty set $S \subseteq X$, we have $\partial S \ne \O$.
{{qed}}
Category:Connected Topological Spaces
Category:Set Boundaries
ckgov1jhpfc4ipykencsno8a8qqck1v
Theorem: Finite Direct Sum of Noetherian Module is Noetherian

Let $A$ be a commutative ring with unity.
Let $n \in \N_{>0}$.
Let $M_1, \ldots, M_n$ be $A$-Noetherian modules.
Then the direct sum:
:$\ds \bigoplus_{i \mathop = 1}^n M_i$
is an $A$-Noetherian module.

Proof:
By Direct Sum of Modules is Module, it is an $A$-module.
Thus we only need to show that it is Noetherian.
We prove it by induction.
For $n = 1$, there is nothing to prove.
Let $n \ge 2$.
Suppose that the claim is true for $n-1$, i.e.:
:$\ds \bigoplus_{i \mathop = 1}^{n-1} M_i$
is Noetherian.
Then, consider the short exact sequence:
:$\ds 0 \longrightarrow \bigoplus_{i \mathop = 1}^{n-1} M_i \stackrel {\alpha} {\longrightarrow} \bigoplus_{i \mathop = 1}^n M_i \stackrel {\beta} {\longrightarrow} M_n \longrightarrow 0$
where:
:$\map \alpha {\tuple {x_1, \ldots, x_{n-1} } } := \tuple {x_1, \ldots, x_{n-1}, 0}$
:$\map \beta { \tuple {x_1, \ldots, x_n} } := x_n$
Recall that $M_n$ is also Noetherian {{hypothesis}}.
Thus by Short Exact Sequence Condition of Noetherian Modules:
:$\ds \bigoplus_{i \mathop = 1}^n M_i$
is Noetherian.
{{qed}}
Category:Noetherian Modules
16o6f6aig9a9mrvp1mekgxc0badxmf6
Theorem: Complement of Closed Disk in Complex Plane is Path-Connected

Let $R > 0$. 
Let:
:$S_R = \set {z \in \C : \cmod z > R}$
Then $S_R$ is path-connected.

Proof:
{{ProofWanted|Someone else can take this, we should probably have a diagram}}
Category:Path-Connected Sets
942u636v6iw8dgrnuolq9kbar324so6
Theorem: Complement of Bounded Set in Complex Plane has at most One Unbounded Component

Let $S \subseteq \C$ be bounded.
Then $\C \setminus S$ has at most one unbounded component.

Proof:
If $\C \setminus S$ has no unbounded components, we are done.
Suppose that $\C \setminus S$ has at least one unbounded component.
We must show that it has at most one. 
Since $S$ is bounded, there exists $R > 0$ such that:
:$\cmod z \le R$ for all $z \in S$.
Let $C$ be an unbounded component of $\C \setminus S$.
Since $C$ is unbounded, there exists $z_0 \in C$ with $\cmod {z_0} > R$.
Hence:
:$\ds z_0 \in C \cap \set {z \in \C : \cmod z > R}$
From Complement of Closed Disk in Complex Plane is Path-Connected, we have that:
:$\set {z \in \C : \cmod z > R}$ is path-connected.
From Path-Connected Space is Connected:
:$\set {z \in \C : \cmod z > R}$ is connected.
Hence from Union of Connected Sets with Common Point is Connected, we have that:
:$C \cup \set {z \in \C : \cmod z > R}$ is connected.
From definition 3 of a component, we have that:
:$C$ is a maximal connected set.
Since:
:$C \subseteq C \cup \set {z \in \C : \cmod z > R}$
we therefore have:
:$C = C \cup \set {z \in \C : \cmod z > R}$
So we have:
:$\set {z \in \C : \cmod z > R} \subseteq C$
Let $C'$ be a component of $\C \setminus S$ that is distinct from $C$. 
Then $C'$ is disjoint from $C$.
In particular, $C'$ is disjoint from $\set {z \in \C : \cmod z > R}$.
So, we have:
:$C' \subseteq \C \setminus C \subseteq \set {z \in \C : \cmod z \le R}$
That is:
:$C'$ is bounded.
Hence:
:$C$ is the only unbounded component of $\C \setminus S$.
This was the demand.
{{qed}}
Category:Components (Topology)
lxtn7qjhai6eqocozbwz7go3rkchjvh
Theorem: Algebra over Field Embeds into Unitization as Ideal

Let $K$ be a field.
Let $A$ be an algebra over $K$. 
Let $A_+$ be the unitization of $A$.
Let:
:$A_0 = \set {\tuple {x, 0_K} : x \in A} \subseteq A_+$.
Then $A_0$ is an ideal in $A_+$.

Proof:
This follows from Ideal of Algebra over Field Embeds into Unitization as Ideal, since $A$ is an ideal of itself.
{{qed}}
Category:Unitizations of Algebras over Fields
rk50tujy4ssp9br568eau9doxpkyu32