hcju/proofwikips huggingface dataset:
Handshake Lemma/Corollary
== Corollary to the Handshake Lemma == Let $G$ be a $\tuple {p, q}$-graph, which may be a multigraph or a loop-graph, or both.  The number of odd vertices in $G$ is even.  == Proof == Let $G$ be a $\tuple {p, q}$-graph. Consider the sum of the degrees of its vertices: :$\displaystyle K = \sum_{v \mathop \in V} \map {\deg_G} v$ From the Handshake Lemma: :$K = 2 \card E$ which is an even integer. Subtracting from $K$ the degrees of all even vertices, we are left with the sum of all degrees of odd vertices in $V$. That is: :$\displaystyle \paren {\sum_{v \mathop \in V} \map {\deg_G} v} - \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k} \map {\deg_G} v} = \paren {\sum_{v \mathop \in V : \map {\deg_G} v \mathop = 2 k + 1} \map {\deg_G} v}$ This must still be an even number, as it is equal to the difference of two even numbers. Because this is a sum of exclusively odd terms, there must be an even number of such terms for the sum on the  to be even. Hence the number of odd vertices in $G$ must be even.  == Sources == * {{BookReference|Introductory Graph Theory|1977|Gary Chartrand|prev = Definition:Odd Vertex (Graph Theory)|next = Definition:Regular Graph}}: Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Theorem $2.2$ * {{BookReference|Calculus Gems|1992|George F. Simmons|prev = Definition:Eulerian Circuit|next = Characteristics of Eulerian Graph}}: Chapter $\text {A}.21$: Euler ({{DateRange|1707|1783}}) * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Definition:Odd Vertex (Graph Theory)|next = Orthogonal Latin Squares of Order 6 do not Exist/Historical Note}}: The Bridges of Königsberg: $134$
Image of Composite Mapping/Corollary
== Corollary to Image of Composite Mapping ==  Let $f: S \to T$ and $g: R \to S$ be mappings. Then: :$\Img {f \circ g} \subseteq \Img f$  where: :$f \circ g$ denotes composition of $g$ and $f$ :$\Img f$ denotes image of $f$. == Proof == From Image of Composite Mapping, it holds that: :$\Img {f \circ g} = f \sqbrk {\Img g}$ where $f \sqbrk {\, \cdot \,}$ denotes image of subset. By definition of composite mapping: :$\Img g \subseteq \Dom f$ where $\Dom f$ denotes the domain of $f$. Now Image of Subset under Relation is Subset of Image: Corollary 2 yields: :$\Img {f \circ g} \subseteq \Img f$  == Sources == * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Domain of Composite Mapping|next = Composition of Mappings is Associative}}: $\S 1.14$: Composition of Functions
Relation between Two Ordinals/Corollary
== Corollary to Relation between Two Ordinals ==  Let $S$ and $T$ be ordinals. If $S \ne T$, then either $S$ is an initial segment of $T$, or vice versa.  == Proof 1 == {{:Relation between Two Ordinals/Corollary/Proof 1}} == Proof 2 == {{:Relation between Two Ordinals/Corollary/Proof 2}} == Proof 3 == {{:Relation between Two Ordinals/Corollary/Proof 3}}
Complement Union with Superset is Universe/Corollary
== Corollary to Complement Union with Superset is Universe ==  :$S \cup T = \mathbb U \iff \map \complement S \subseteq T$  where: :$S \subseteq T$ denotes that $S$ is a subset of $T$ :$S \cup T$ denotes the union of $S$ and $T$ :$\complement$ denotes set complement :$\mathbb U$ denotes the universal set. == Proof == Let $X = \map \complement S$. Then: {{begin-eqn}} {{eqn | l = X \subseteq T       | o = \iff       | r = \map \complement X \cup T = \mathbb U       | c = Complement Union with Superset is Universe }} {{eqn | ll= \leadsto       | l = \map \complement S \subseteq T       | o = \iff       | r = \map \complement {\map \complement S} \cup T = \mathbb U       | c = substituting $X = \map \complement S$ }} {{eqn | ll= \leadsto       | l = \map \complement S \subseteq T       | o = \iff       | r = S \cup T = \mathbb U       | c = Complement of Complement }} {{end-eqn}}  == Also see == * Empty Intersection iff Subset of Complement == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Cardinality of Set Union/Examples/Student Subjects/Mathematics and Chemistry|next = Empty Intersection iff Subset of Complement}}: $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(a)}$ * {{BookReference|Introduction to Topology|1975|Bert Mendelson|ed = 3rd|edpage = Third Edition|prev = Empty Intersection iff Subset of Complement|next = Set Complement inverts Subsets}}: Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(d)}$
Image of Subset under Relation is Subset of Image/Corollary 1
== Corollary to Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation from $S$ to $T$.  Let $C, D \subseteq T$. Then: :$C \subseteq D \implies \RR^{-1} \sqbrk C \subseteq \RR^{-1} \sqbrk D$ where $\RR^{-1} \sqbrk C$ is the preimage of $C$ under $\RR$.  == Proof == We have that $\RR^{-1}$ is itself a relation, by definition of inverse relation. The result follows directly from Image of Subset under Relation is Subset of Image.
Image of Subset under Mapping is Subset of Image
== Corollary to Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$.  Let $A, B \subseteq S$ such that $A \subseteq B$. Then the image of $A$ is a subset of the image of $B$: :$A \subseteq B \implies f \sqbrk A \subseteq f \sqbrk B$ This can be expressed in the language and notation of direct image mappings as: :$\forall A, B \in \powerset S: A \subseteq B \implies \map {f^\to} A \subseteq \map {f^\to} B$  == Proof == As $f: S \to T$ is a mapping, it is also a relation, and thus: :$f \subseteq S \times T$ The result follows directly from Image of Subset under Relation is Subset of Image.  == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Image of Preimage under Mapping/Corollary|next = Image of Subset under Relation is Subset of Image/Corollary 3}}: $\S 2.2$: Homomorphisms: $\text{(ii)}$ * {{BookReference|Undergraduate Topology|1971|Robert H. Kasriel|prev = Mapping Images are Disjoint only if Domains are Disjoint|next = Image of Union under Mapping}}: $\S 1.10$: Functions: Exercise $5 \ \text{(c)}$ * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Direct Image Mapping of Mapping is Empty iff Argument is Empty|next = Image of Subset under Relation is Subset of Image/Corollary 3}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(i)}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Definition:Surjection/Also known as|next = Image of Intersection under Mapping/Proof 1}}: $\S 21.3$: The image of a subset of the domain; surjections * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Preimage of Set Difference under Mapping|next = Image of Union under Mapping}}: $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(e)}$
Image of Subset under Relation is Subset of Image/Corollary 3
== Corollary of Image of Subset under Relation is Subset of Image == Let $S$ and $T$ be sets. Let $f: S \to T$ be a mapping from $S$ to $T$.  Let $C, D \subseteq T$. Then: :$C \subseteq D \implies f^{-1} \sqbrk C \subseteq f^{-1} \sqbrk D$ This can be expressed in the language and notation of inverse image mappings as: :$\forall C, D \in \powerset T: C \subseteq D \implies \map {f^\gets} C \subseteq \map {f^\gets} D$  == Proof == As $f: S \to T$ is a mapping, it is also a relation, and thus so is its inverse: :$f^{-1} \subseteq T \times S$ The result follows directly from Image of Subset under Relation is Subset of Image.  == Sources == * {{BookReference|Rings, Modules and Linear Algebra|1970|B. Hartley|author2 = T.O. Hawkes|prev = Image of Subset under Relation is Subset of Image/Corollary 2|next = Fourth Isomorphism Theorem}}: $\S 2.2$: Homomorphisms: $\text{(ii)}$ * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Image of Subset under Relation is Subset of Image/Corollary 2|next = Image of Union under Mapping/Proof 1}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1: \ \text{(j)}$ * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Image of Preimage of Subset under Surjection equals Subset|next = Preimage of Union under Mapping}}: $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(a)}$
Subset of Preimage under Relation is Preimage of Subset/Corollary
== Corollary to Subset of Preimage under Relation is Preimage of Subset == Let $f: S \to T$ be a mapping. Let $X \subseteq S, Y \subseteq T$. Then:  :$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$ This can be expressed in the language and notation of direct image mappings and inverse image mappings as: :$\forall X \in \powerset S, Y \in \powerset T: X \subseteq \map {f^\gets} Y \iff \map {f^\to} X \subseteq Y$  == Proof == Let $f: S \to T$ be a mapping. As a mapping is also a relation, it follows that $f$ is a relation and so: :$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$ holds on the strength of Subset of Preimage under Relation is Preimage of Subset.  == Sources == * {{BookReference|Set Theory and Abstract Algebra|1975|T.S. Blyth|prev = Complement of Preimage equals Preimage of Complement|next = Definition:Composition of Mappings/Definition 1}}: $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1 \ \text{(v)}$
Monotone Convergence Theorem
 * Monotone Convergence Theorem (Real Analysis) * Monotone Convergence Theorem (Measure Theory) mbc419yt8xmiqu5qcn58fczbv7yeyti
Cardinality of Cartesian Product/Corollary
== Corollary to Cardinality of Cartesian Product == Let $S \times T$ be the cartesian product of two sets $S$ and $T$ which are both finite. Then:  :$\card {S \times T} = \card {T \times S}$  where $\card {S \times T}$ denotes the cardinality of $S \times T$. == Proof 1 == {{:Cardinality of Cartesian Product/Corollary/Proof 1}} == Proof 2 == {{:Cardinality of Cartesian Product/Corollary/Proof 2}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Surjection from Finite Set to Itself is Permutation|next = Definition:Binary Operation}}: Chapter $3$. Mappings: Exercise $9 \ \text {(i)}$ * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition|prev = Definition:Countable-Dimensional Real Cartesian Space|next = Definition:Finite Set}}: $1$: Set Theory and Logic: $\S 5$: Cartesian Products: Exercise $1$
Modulo Subtraction is Well-Defined
== Corollary to Modulo Addition is Well-Defined == Let $m \in \Z$ be an integer. Let $\Z_m$ be the set of integers modulo $m$. The modulo subtraction operation on $\Z_m$, defined by the rule:  :$\eqclass a m -_m \eqclass b m = \eqclass {a - b} m$ is a well-defined operation.  That is: :If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a - x \equiv b - y \pmod m$. == Proof == We have: {{begin-eqn}} {{eqn | l = \eqclass a m -_m \eqclass b m       | r = \eqclass {a - b} m       | c =  }} {{eqn | r = \eqclass {a + \paren {-b} } m       | c =  }} {{eqn | r = \eqclass a m +_m \eqclass {-b} m       | c =  }} {{end-eqn}} The result follows from the fact that Modulo Addition is Well-Defined for all integers.  == Examples == {{:Modulo Subtraction is Well-Defined/Examples}} == Sources == * {{BookReference|Sets and Groups|1965|J.A. Green|prev = Modulo Addition is Well-Defined/Proof 2|next = Modulo Multiplication is Well-Defined}}: $\S 2.6$. Algebra of congruences: $\text{(ii)}$ * {{BookReference|Number Theory|1971|George E. Andrews|prev = Modulo Addition is Well-Defined/Proof 1|next = Modulo Multiplication is Well-Defined/Proof 1}}: $\text {4-1}$ Basic Properties of Congruences: Theorem $\text {4-2}$ * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined}}: $\S 14.3 \ \text {(ii)}$: Congruence modulo $m$ ($m \in \N$) * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined}}: $\S 1.2.4$: Integer Functions and Elementary Number Theory: Law $\text{A}$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Modulo Addition is Well-Defined|next = Modulo Multiplication is Well-Defined|entry = congruence}}
Single Instruction URM Programs/Identity Function
Single Instruction URM Programs/Identity Function 0 17866 99431 2012-07-29T21:49:10Z Prime.mover 59 Created page with "== Theorem ==  The identity function $I_\N: \N \to \N$ defined as: :$\forall n \in \N: I_\N \l..." wikitext text/x-wiki == Theorem ==  The identity function $I_\N: \N \to \N$ defined as: :$\forall n \in \N: I_\N \left({n}\right) = n$  is URM computable by a single-instruction URM program. == Proof == Any of the following URM programs compute the identity function: {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $Z \left({m}\right)$ |} ... where $m \ne 1$. This sets the value $0$ into $R_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $S \left({m}\right)$ |} ... where $m \ne 1$. The input $n$ is in $R_1$ when the program starts. The program adds $1$ to $r_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $C \left({j, m}\right)$ |} ... where $m \ne 1$. The input $n$ is in $R_1$ when the program starts. The program copies $r_j$ to $r_m$ and then stops. {| |- ! align="right" | Line !! ! align="left" | Command |- | align="right" | $1$ || | align="left" | $C \left({1, 1}\right)$ |} The input $n$ is in $R_1$ when the program starts. The program copies $r_1$ to $r_1$, i.e. to itself, and then stops. In none of these programs is $R_1$ affected, and so no change is effected to the input, which is returned as the output unchanged. Hence they all compute the identity function.  Note that the latter program, consisting entirely of $C \left({1, 1}\right)$, is a direct implementation of the projection function program $\operatorname{pr}^1_1: \N \to \N$. This latter function is the usual way of implementing the identity function as it is well-defined and obvious, and guaranteed to have no other side-effects when embedded in a larger program.
Cancellability of Congruences/Corollary 1
== Corollary to Cancellability of Congruences ==  Let $c$ and $n$ be coprime integers, that is: :$c \perp n$ Then: :$c a \equiv c b \pmod n \implies a \equiv b \pmod n$  where $\equiv$ denotes congruence. == Proof 1 == {{:Cancellability of Congruences/Corollary 1/Proof 1}} == Proof 2 == {{:Cancellability of Congruences/Corollary 1/Proof 2}} == Proof 3 == {{:Cancellability of Congruences/Corollary 1/Proof 3}}
Euler Phi Function of Integer/Corollary
== Corollary to Euler Phi Function of Integer == Let $n \in \Z_{>0}$ be a (strictly) positive integer. Let $p$ be a prime number. Let $\map \phi n$ denote the Euler $\phi$ function of $n$.  Let $p$ be a divisor of $n$. Then $p - 1$ is a divisor of $\phi \left({n}\right)$.  == Proof == From Euler Phi Function of Integer: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ Let $n$ be expressed as its prime decomposition: :$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}, p_1 < p_2 < \ldots < p_r$ Then: {{begin-eqn}} {{eqn | l = \phi \left({n}\right)       | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} \paren {1 - \dfrac 1 {p_i} }       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} \paren {\dfrac {p_i - 1} {p_i} }       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}       | c =  }} {{eqn | r = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i - 1} \paren {p_i - 1}       | c =  }} {{end-eqn}} Hence the result.
Euler Phi Function of Prime Power/Corollary
== Corollary to Euler Phi Function of Prime Power == Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function. Then:  :$\map \phi {2^k} = 2^{k-1}$  == Proof == We have that: :$\displaystyle 1 - \frac 1 2 = \frac {2 - 1} 2 = \frac 1 2$ It follows from Euler Phi Function of Prime Power: :$\map \phi {2^k} = \paren {\dfrac 1 2} 2^k = 2^{k - 1}$
Abelian Group Factored by Prime/Corollary
== Corollary to Abelian Group Factored by Prime ==  Any finite abelian group $G$ can be factored as follows: Let $\order G = \displaystyle \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorisation of the order of $G$. Then we have $G = \displaystyle \prod_{i \mathop = 1}^k H_i$, where $H_i = \set {x \in G : x^{p_i^{n_i} } = e}$. This factorisation is unique up to ordering of the factors.  == Proof == Let $\displaystyle \prod_{i \mathop = 1}^k p_i^{n_i}$ be the prime factorisation of $\order G$. We proceed by induction on $k$. === Basis for the induction === For $n = 1$, the statement is trivial. === Induction Hypothesis === Now we assume the theorem is true for abelian groups whose order has $k - 1$ distinct prime factors. === Induction Step === Apply Abelian Group Factored by Prime to $G$ and $p_1$.  By definition, $H = H_1$. Also, the resulting $K$ has $\order K = \displaystyle \prod_{i \mathop = 2}^k p_i^{n_i}$. Therefore, it satisfies the induction hypothesis. It follows that $G = H_1 \times \displaystyle \prod_{i \mathop = 2}^k H_i$. From: :Subgroup of Abelian Group is Normal :the definition of Sylow $p$-subgroup all the $H_i$ are normal Sylow $p$-subgroups. From Sylow $p$-Subgroup is Unique iff Normal, the factorisation is unique up to ordering of the factors.
Measure is Subadditive/Corollary
Measure is Subadditive/Corollary 0 18086 100774 2012-08-06T07:40:12Z Lord Farin 560 Created page with "== Theorem ==  Let $\left({X, \Sigma, \mu}\right)$ be a measure space.  Let $E_1, \ldots, E_n \in \Sigma$.   Then:  :$\displaystyle \..." wikitext text/x-wiki == Theorem == Let $\left({X, \Sigma, \mu}\right)$ be a measure space.  Let $E_1, \ldots, E_n \in \Sigma$. Then: :$\displaystyle \mu \left({\bigcup_{k \mathop = 1}^n E_k}\right) \le \sum_{k \mathop = 1}^n \mu \left({E_k}\right)$.  == Proof == We have Measure is Subadditive. The result follows by an application of Finite Union of Sets in Subadditive Function.
Composite Functor is Functor
== Definition == Let $\mathbf C, \mathbf D$ and $\mathbf E$ be metacategories. Let $F: \mathbf C \to \mathbf D$ and $G: \mathbf D \to \mathbf E$ be covariant functors. Let $GF: \mathbf C \to \mathbf E$ be the composition of $G$ with $F$. Then $G F$ is also a covariant functor. == Proof == Let $f, g$ be morphisms of $\mathbf C$ such that $g \circ f$ is defined. Then: {{begin-eqn}} {{eqn | l = \map {G F} {g \circ f}       | r = \map G {\map F {g \circ f} }       | c = {{Defof|Composition of Functors}} }} {{eqn | r = \map G {F g \circ F f}       | c = $F$ is a Covariant Functor }} {{eqn | r = \map G {F g} \circ \map G {F f}       | c = $G$ is a functor }} {{eqn | r = G F g \circ G F f       | c = {{Defof|Composition of Functors}} }} {{end-eqn}} Also, for any object $C$ of $\mathbf C$: {{begin-eqn}} {{eqn | l = G F I_C       | r = \map G {F I_C}       | c = {{Defof|Composition of Functors}} }} {{eqn | r = G I_{F C}       | c = $F$ is a Covariant Functor }} {{eqn | r = I_{\map G {F C} }       | c = $G$ is a Covariant Functor }} {{eqn | r = I_{G F C}       | c = {{Defof|Composition of Functors}} }} {{end-eqn}} Hence $G F$ is shown to be a covariant functor.
Transitive Set is Proper Subset of Ordinal iff Element of Ordinal/Corollary
== Corollary to Transitive Set is Proper Subset of Ordinal iff Element of Ordinal ==  Let $A$ and $B$ be ordinals. Then: :$A \subsetneq B \iff A \in B$  == Proof == We have that an ordinal is transitive. The result follows directly from Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.  == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2=Wilson M. Zaring}}: $\S 7.8$
Absolute Value of Integer is not less than Divisors/Corollary
== Corollary to Absolute Value of Integer is not less than Divisors ==  Let $a, b \in \Z_{>0}$ be (strictly) positive integers. Let $a \divides b$. Then: :$a \le b$  == Proof == Follows directly from Absolute Value of Integer is not less than Divisors.  == Sources == * {{BookReference|Elements of Abstract Algebra|1966|Richard A. Dean|prev = Definition:Integer|next = Definition:Divisor of Integer}}: $\S 0.1$. Arithmetic: Theorem $1$ * {{BookReference|Introduction to Abstract Algebra|1969|C.R.J. Clapham|prev = Divisor Relation is Antisymmetric/Corollary/Proof 2|next = Division Theorem/Positive Divisor/Existence/Proof 3}}: Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(iii)}$
Euclid's Lemma for Prime Divisors/General Result
== Lemma ==  Let $p$ be a prime number. Let $\displaystyle n = \prod_{i \mathop = 1}^r a_i$. Then if $p$ divides $n$, it follows that $p$ divides $a_i$ for some $i$ such that $1 \le i \le r$. That is: :$p \divides a_1 a_2 \ldots a_n \implies p \divides a_1 \lor p \divides a_2 \lor \cdots \lor p \divides a_n$  == Proof 1 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 1}} == Proof 2 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 2}} == Proof 3 == {{:Euclid's Lemma for Prime Divisors/General Result/Proof 3}} {{Namedfor|Euclid}} == Sources == * {{BookReference |An Introduction to the Theory of Numbers|1979|G.H. Hardy|author2 = E.M. Wright|ed = 5th|edpage = Fifth Edition|prev = Euclid's Lemma for Prime Divisors|next = Expression for Integer as Product of Primes is Unique/Proof 1}}: $\text I$: The Series of Primes: $1.3$ Statement of the fundamental theorem of arithmetic