avewright/proofwiki-math huggingface dataset:
Theorem: Principle of Mathematical Induction/One-Based

Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$.

Proof:
Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the Well-Ordering Principle $S$ contains a minimal element $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$.
As $k + 1$ is the minimal element of $S$ it follows that all $n$ such that $n < k + 1 = s$ are not in $S$
But this means that $\map P n$ is true for all $n < s$
But by $(2)$ it follows that $\map P s$ is true.
That is, $s \notin S$.
This contradicts our assertion that $s \in S$.
Hence our assumption that $S \ne \O$ is false.
It follows by Proof by Contradiction that $S = \O$.
So $\map P n$ holds for all $n \in \N_{>0}$.
{{Qed}}
Theorem: Principle of Finite Induction/Zero-Based

Let $S \subseteq \N$ be a subset of the natural numbers.
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N : n \in S \implies n + 1 \in S$
Then:
:$S = \N$

Proof:
Consider $\N$ defined as a naturally ordered semigroup.
The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup.
{{qed}}
Theorem: Principle of Finite Induction/One-Based

Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$

Proof:
Consider $\N$ defined as a naturally ordered semigroup.
The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result.
{{qed}}
Theorem: Principle of Finite Induction/One-Based

Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
Then:
:$S = \N_{>0}$

Proof:
Let $T$ be the set of all $1$-based natural numbers not in $S$:
:$T = \N_{>0} \setminus S$
{{AimForCont}} $T$ is non-empty.
From the Well-Ordering Principle, $T$ has a smallest element.
Let this smallest element be denoted $a$.
We have been given that $1 \in S$.
So:
:$a > 1$
and so:
:$0 < a - 1 < a$
As $a$ is the smallest element of $T$, it follows that:
:$a - 1 \notin T$
That means $a - 1 \in S$.
But then by hypothesis:
:$\paren {a - 1} + 1 \in S$
But:
:$\paren {a - 1} + 1 = a$
and so $a \notin T$.
This contradicts our assumption that $a \in T$.
It follows by Proof by Contradiction that $T$ has no such smallest element.
Hence it follows that $T$ can have no elements at all.
That is:
:$\N_{>0} \setminus S = \O$
That is:
:$S = \N_{>0}$
Theorem: Second Principle of Finite Induction/Zero-Based

Let $S \subseteq \N$ be a subset of the natural numbers.
Suppose that:
:$(1): \quad 0 \in S$
:$(2): \quad \forall n \in \N: \paren {\forall k: 0 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N$

Proof:
Define $T$ as:
:$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 0: n \in T$
Firstly, we have that $0 \in T$ {{iff}} the following condition holds:
:$\forall k: 0 \le k \le 0 \implies k \in S$
Since $0 \in S$, it thus follows that $0 \in T$.
Now suppose that $n \in T$; that is:
:$\forall k: 0 \le k \le n \implies k \in S$
By $(2)$, this implies:
:$n + 1 \in S$
Thus, we have:
:$\forall k: 0 \le k \le n + 1 \implies k \in S$
{{MissingLinks|Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor for $\N$}}
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction:
:$\forall n \ge 0: n \in T$
That is:
:$T = \N$
and as $S \subseteq \N$ it follows that:
:$S = N$
{{Qed}}
Category:Second Principle of Finite Induction
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Theorem: Second Principle of Finite Induction/One-Based

Let $S \subseteq \N_{>0}$ be a subset of the $1$-based natural numbers.
Suppose that:
:$(1): \quad 1 \in S$
:$(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$
Then:
:$S = \N_{>0}$

Proof:
Define $T$ as:
:$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
Since $n \le n$, it follows that $T \subseteq S$.
Therefore, it will suffice to show that:
:$\forall n \ge 1: n \in T$
Firstly, we have that $1 \in T$ {{iff}} the following condition holds:
:$\forall k: 1 \le k \le 1 \implies k \in S$
Since $1 \in S$, it thus follows that $1 \in T$.
Now suppose that $n \in T$; that is:
:$\forall k: 1 \le k \le n \implies k \in S$
By $(2)$, this implies:
:$n + 1 \in S$
Thus, we have:
:$\forall k: 1 \le k \le n + 1 \implies k \in S$
{{MissingLinks|Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor for $\N$}}
Therefore, $n + 1 \in T$.
Hence, by the Principle of Finite Induction:
:$\forall n \ge 1: n \in T$
That is:
:$T = \N_{>0}$
and as $S \subseteq \N_{>0}$ it follows that:
:$S = N_{>0}$
{{Qed}}
Theorem: Second Principle of Mathematical Induction/Zero-Based

Let $\map P n$ be a propositional function depending on $n \in \N$.
Suppose that:
:$(1): \quad \map P 0$ is true
:$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N$.

Proof:
For each $n \in \N$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 0 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N$.
It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\map P {n + 1}$ holds as well.
Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
Thus by the Principle of Mathematical Induction:
:$\map {P'} n$ holds for all $n \in \N$
as desired.
{{Qed}}
Theorem: Second Principle of Mathematical Induction/One-Based

Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$.

Proof:
For each $n \in \N_{>0}$, let $\map {P'} n$ be defined as:
:$\map {P'} n := \map P 1 \land \dots \land \map P n$
It suffices to show that $\map {P'} n$ is true for all $n \in \N_{>0}$.
It is immediate from the assumption $\map P 1$ that $\map {P'} 1$ is true.
Now suppose that $\map {P'} n$ holds.
By $(2)$, this implies that $\map P {n + 1}$ holds as well.
Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
Thus by the Principle of Mathematical Induction:
:$\map {P'} n$ holds for all $n \in \N_{>0}$
as desired.
{{Qed}}
Theorem: Second Principle of Mathematical Induction/One-Based

Let $\map P n$ be a propositional function depending on $n \in \N_{>0}$.
Suppose that:
:$(1): \quad \map P 1$ is true
:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
Then:
:$\map P n$ is true for all $n \in \N_{>0}$.

Proof:
Let $S \subseteq \N_{>0}$ containing those $n \in \N_{>0}$ for which $\map P n$ does not hold.
{{AimForCont}} $S \ne \O$.
Then by the Well-Ordering Principle $S$ contains a minimal element $s$.
We have that $s \ne 1$ because $\map P 1$ is true from $(1)$.
Thus there must exist some $k \in \N_{>0}$ such that $s = k + 1$.
As $k + 1$ is the minimal element of $S$ it follows that all $n$ such that $n < k + 1 = s$ are not in $S$
But this means that $\map P n$ is true for all $n < s$
But by $(2)$ it follows that $\map P s$ is true.
That is, $s \notin S$.
This contradicts our assertion that $s \in S$.
Hence our assumption that $S \ne \O$ is false.
It follows by Proof by Contradiction that $S = \O$.
So $\map P n$ holds for all $n \in \N_{>0}$.
{{Qed}}
Theorem: Null Sequences form Maximal Left and Right Ideal/Corollary 1

Then $\NN$ is a maximal ring ideal of $\CC$.

Proof:
By Null Sequences form Maximal Left and Right Ideal then $\NN$ is a maximal left ideal of $\CC$. 
A field is by definition a commutative ring.
In a commutative ring, a maximal left ideal is by definition a maximal ideal.
{{qed}}
Theorem: Non-Zero Integer has Finite Number of Divisors

Let $n \in \Z_{\ne 0}$ be a non-zero integer.
Then $n$ has a finite number of divisors.

Proof:
Let $S$ be the set of all divisors of $n$.
Then from Absolute Value of Integer is not less than Divisors:
:$\forall m \in S: -n \le m \le n$
Thus $S$ is finite.
{{qed}}
Category:Number Theory
Category:Divisors
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Theorem: Coprimality Relation is Non-Reflexive

:$\perp$ is non-reflexive.

Proof:
Proof by Counterexample:
We have from GCD of Integer and Divisor:
:$\gcd \set {n, n} = n$
and so, for example:
:$\gcd \set {2, 2} = 2$
and so:
:$2 \not \perp 2$
Hence $\perp$ is not reflexive.
But we also note that:
:$\gcd \set {1, 1} = 1$
and so:
:$1 \perp 1$
demonstrating that $\perp$ is not antireflexive either.
The result follows by definition of non-reflexive relation.
{{qed}}
Theorem: Coprimality Relation is Symmetric

:$\perp$ is symmetric.

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \perp
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \gcd \set {x, y}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \gcd \set {y, x}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | o = \perp
      | r = x
      | c = 
}}
{{end-eqn}}
Hence the result by definition of symmetric relation.
{{qed}}
Theorem: Coprimality Relation is not Antisymmetric

:$\perp$ is not antisymmetric.

Proof:
Proof by Counterexample:
We have:
:$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$
and so:
:$3 \perp 5$ and $5 \perp 3$
However, it is not the case that $3 = 5$.
The result follows by definition of antisymmetric relation.
{{qed}}
Category:Coprime Integers
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Theorem: Coprimality Relation is Non-Transitive

:$\perp$ is non-transitive.

Proof:
Proof by Counterexample:
We have:
{{begin-eqn}}
{{eqn | l = \gcd \set {2, 3}
      | r = 1
      | c = 
}}
{{eqn | l = \gcd \set {3, 4}
      | r = 1
      | c = 
}}
{{eqn | l = \gcd \set {2, 4}
      | r = 2
      | c = 
}}
{{end-eqn}}
Hence we have:
:$2 \perp 3$ and $3 \perp 4$
However, it is not the case that $2 \perp 4$.
Thus $\perp$ is not transitive.
Then we have:
{{begin-eqn}}
{{eqn | l = \gcd \set {2, 3}
      | r = 1
      | c = 
}}
{{eqn | l = \gcd \set {3, 5}
      | r = 1
      | c = 
}}
{{eqn | l = \gcd \set {2, 5}
      | r = 1
      | c = 
}}
{{end-eqn}}
:$2 \perp 3$ and $3 \perp 5$
and also:
:$2 \perp 5$
Thus $\perp$ is not antitransitive either.
The result follows by definition of non-transitive relation.
{{qed}}
Theorem: Ring of Polynomials over Reals is not Field

Let $\R \sqbrk X$ be the ring of polynomials in an indeterminate $X$ over $\R$.
Then $\R \sqbrk X$ is not a field.

Proof:
Consider the polynomial $x + 1$ in $\R \sqbrk X$.
There exists no polynomial $\map f x$ such that:
:$\paren {x + 1} \map f x = 1$
This is because the {{LHS}} has degree $1$, and the {{RHS}} has degree $0$.
{{qed}}
Theorem: Even Integers form Commutative Ring

Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a commutative ring.
However, $\struct {2 \Z, +, \times}$ is not an integral domain.

Proof:
From Integer Multiples form Commutative Ring, $\struct {2 \Z, +, \times}$ is a commutative ring.
As $2 \ne 1$, we also have from Integer Multiples form Commutative Ring that $\struct {2 \Z, +, \times}$ has no unity.
Hence by definition it is not an integral domain.
{{qed}}
Theorem: Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers

Let $2 \Z$ be the set of even integers.
Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$.

Proof:
From Subrings of Integers are Sets of Integer Multiples, a ring of the form $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
$\struct {2 \Z, +, \times}$ is such an example.
{{qed}}
Theorem: Gaussian Integers does not form Subfield of Complex Numbers

The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
is not a subfield of $\C$.

Proof:
Proof by Counterexample:
We have that:
:$2 + 0 i \in \Z \sqbrk i$
However there is no $z \in \Z \sqbrk i$ such that:
:$x \paren {2 + 0 i} = 1 + 0 i$
So, by definition, $\Z \sqbrk i$ is not a field.
Thus $\Z \sqbrk i$ is not a subfield of $\C$.
{{qed}}
Theorem: Ideals of Ring of Integers Modulo m

Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\Z_m, +, \times}$ denote the ring of integers modulo $m$.
The ideals of $\struct {\Z_m, +, \times}$ are of the form:
:$d \Z / m \Z$
where $d$ is a divisor of $m$.

Proof:
Let $J$ be an ideal of $\struct {\Z_m, +, \times}$.
Recall from Ring of Integers Modulo m is Ring that $\struct {\Z_m, +}$ is the additive group of integers modulo $m$.
From Integers Modulo $m$ under Addition form Cyclic Group, $\struct {\Z_m, +}$ is a cyclic group.
By definition of ideal, $\struct {J, +}$ is a subgroup of $\struct {\Z_m, +}$.
Let $\struct {G, +}$ be a subgroup of $\struct {\Z_m, +}$.
From Subgroup of Cyclic Group is Cyclic $\struct {G, +}$ is a cyclic group.
Hence by definition of cyclic group, $\struct {G, +}$ can be generated by $d$ and denoted $\gen d$, where $d \divides m$.
By Lagrange's Theorem, for a finite cyclic group of order $k$, the order of every subgroup is a divisor of $k$.
By Bijection from Divisors to Subgroups of Cyclic Group there is exactly one subgroup for each divisor.
It follows that all ideals of $\struct {\Z_m, +, \times}$ are of form $\gen d$, where $d$ is a positive divisor of $m$.
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1

Then the quotient ring $\CC \,\big / \NN$ is a field.

Proof:
By Quotient Ring of Cauchy Sequences is Division Ring, $\CC \,\big / \NN$ is a division ring.
By {{Corollary|Cauchy Sequences form Ring with Unity}}, $\CC$ is a commutative ring with unity.
By Quotient Ring of Commutative Ring is Commutative, $\CC \,\big / \NN$ is a commutative division ring, that is, a field.
{{qed}}
Theorem: Quotient Ring of Cauchy Sequences is Normed Division Ring/Corollary 1

Then $\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a valued field.

Proof:
By Quotient Ring of Cauchy Sequences is Normed Division Ring, $\CC \,\big / \NN$ is a normed division ring.
By {{Corollary|Quotient Ring of Cauchy Sequences is Division Ring|1}}, $\CC \,\big / \NN$ is a field.
The result follows.
{{qed}}
Theorem: Cauchy Sequence Converges Iff Equivalent to Constant Sequence

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $\CC$ be the ring of Cauchy sequences over $R$
Let $\NN$ be the set of null sequences.
Let $\\CC \,\big / \NN$ be the quotient ring of Cauchy sequences of $\CC$ by the maximal ideal $\NN$.
Let $\sequence {x_n} \in \CC$.
Then $\sequence {x_n}$ converges in $\struct {R, \norm {\,\cdot\,} }$ {{iff}}:
:$\exists a \in R: \sequence {x_n} \in \sequence {a, a, a, \dotsc} + \NN$
where $\sequence {a, a, a, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {a, a, a, \dotsc}$.

Proof:
By definition, $\sequence {x_n}$ converges to $a \in R$ {{iff}} $\ds \lim_{n \mathop \to \infty} \norm {x_n - a} = 0$
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n - a}
      | r = 0
}}
{{eqn | ll= \leadstoandfrom
      | l = \sequence {x_n - a}
      | o = \in
      | r = \NN
      | c = {{Defof|Null Sequence in Normed Division Ring}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sequence {x_n} - \sequence {a, a, a, \dotsc}
      | o = \in
      | r = \NN
      | c = {{Defof|Ring of Cauchy Sequences|Addition in Ring of Cauchy Sequences}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sequence {x_n}
      | o = \in
      | r = \sequence {a, a, a, \dotsc} + \NN
      | c = Element in Left Coset iff Product with Inverse in Subgroup
}}
{{end-eqn}}
{{qed}}
Category:Normed Division Rings
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Theorem: Homomorphism of Ring Subtraction

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism.
Then:
:$\forall a, b \in R_1: \map \phi {a -_1 b} = \map \phi a -_2 \map \phi b$
where $a -_1 b$ denotes subtraction of $b$ from $a$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \phi {a -_1 b}
      | r = \map \phi {a +_1 \paren {-b} }
      | c = {{Defof|Ring Subtraction}}
}}
{{eqn | r = \map \phi a +_2 \map \phi {-b}
      | c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \phi a +_2 \paren {-\map \phi b}
      | c = Ring Homomorphism Preserves Negatives
}}
{{eqn | r = \map \phi a -_2 \map \phi b
      | c = {{Defof|Ring Subtraction}}
}}
{{end-eqn}}
{{qed}}
Theorem: Limit of Modulo Operation/Limit 1

Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to 0} x \bmod y = 0$.

Proof:
By Range of Modulo Operation for Positive Modulus and Range of Modulo Operation for Negative Modulus we have:
:$-\size y < x \bmod y < \size y$
The result follows from the Squeeze Theorem for Functions.
{{qed}}
Theorem: Limit of Modulo Operation/Limit 2

Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$.

Proof:
As $y \to \infty$:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | m = x
      | mo= <
      | r = y
}}
{{eqn | ll= \leadsto
      | l = 0
      | o = \le
      | m = \frac x y
      | mo= <
      | r = 1
}}
{{eqn | ll= \leadsto
      | o =
      | m = \floor {\frac x y}
      | mo= =
      | r = 0
}}
{{end-eqn}}
Therefore by the definition of modulo operation:
{{begin-eqn}}
{{eqn | l = \lim_{y \mathop \to \infty} x \bmod y
      | r = \lim_{y \mathop \to \infty} x - y \floor {\dfrac x y}
}}
{{eqn | r = \lim_{y \mathop \to \infty} x - y \cdot 0
}}
{{eqn | r = x
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Limit of Modulo Operation

Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then holding $x$ fixed gives:
:$\ds \lim_{y \mathop \to 0} x \bmod y = 0$
:$\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$

Proof:
=== Limit 1 ===
{{:Limit of Modulo Operation/Limit 1}}
Theorem: Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary

Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0_R$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n} = \norm {x_m}$
=== Corollary ===
{{:Non-Null Cauchy Sequence in P-adic Norm is Eventually Stationary}}

Proof:
By Cauchy Sequence Is Eventually Bounded Away From Non-Limit then:
:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$
Let $N = \max \set {N_1, N_2}$.
Let $n, m \ge N$.
Then:
:$\norm {x_n - x_m} < C < \norm {x_n}$
By {{Corollary|Three Points in Ultrametric Space have Two Equal Distances|4}} then:
:$\norm {x_n} = \norm {x_m}$
The result follows.
{{qed}}
Theorem: Equivalence of Definitions of Unit of Ring

Let $\struct {R, +, \circ}$ be a ring with unity whose unity is $1_R$.
{{TFAE|def = Unit of Ring}}

Proof:
Let $\struct {R, +, \circ}$ be a ring with unity.
Theorem: Euclidean Valuation of Non-Unit is less than that of Product

Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $b, c \in D_{\ne 0}$.
Then:
:If $c$ is not a unit of $D$ then $\map \nu b < \map \nu {b c}$

Proof:
By principal ideal domain, $D$ is a principal ideal domain
Consider the principal ideal $U = \ideal b$ of $D$ generated by $b$.
As $\nu$ is a valuation function, we have:
:$\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$
As $U$ is a principal ideal:
:$\forall a \in \ideal b: \exists x \in D: a = x \times b$
and so:
:$\forall a \in U: \map \nu b \le \map \nu a$
Let $\map \nu b = \map \nu {b c}$.
Then from above:
:$\forall a \in U: \map \nu {b c} \le \map \nu a$
From the argument in Euclidean Domain is Principal Ideal Domain, we have that $U$ is also the ideal generated by $b c$.
{{explain|The above statement is not very clear and needs to be expanded.}}
Thus $a b$ is a divisor of $b$ itself, that is:
:$\exists y \in D: b = b c y$
Thus:
:$c y = 1$
and so $c$ is a unit of $D$.
So if $c$ is not a unit, then it must be the case that $\map \nu b < \map \nu {b c}$.
{{qed}}
Theorem: Element is Unit iff its Euclidean Valuation equals that of 1

Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $a \in D$.
Then:
:$a$ is a unit of $D$ {{iff}} $\map \nu a = \map \nu 1$

Proof:
For $a \in D$ we have that:
:$\map \nu 1 \le \map \nu {1 a} = \map \nu a$
by definition of Euclidean valuation.
Let $a$ be a unit of $D$.
Then:
:$\exists b \in D: a b = 1$
Then:
:$\map \nu a \le \map \nu {a b} = \map \nu 1$
and so:
:$\map \nu a = \map \nu 1$
{{qed|lemma}}
Let $\map \nu a = \map \nu 1$.
We can write this as:
:$\map \nu {1 a} = \map \nu 1$
and it follows from Euclidean Valuation of Non-Unit is less than that of Product that $a$ is a unit of $D$.
{{qed}}
Theorem: Gauss's Lemma on Primitive Polynomials over Ring

Let $R$ be a commutative ring with unity.
Let $f, g \in R \sqbrk X$ be primitive polynomials.
{{explain|Definition:Primitive Polynomial over Ring}}
Then $f g$ is primitive.

Proof:
{{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
Category:Polynomial Theory
Category:Gauss's Lemma (Polynomial Theory)
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Theorem: Rational Polynomial is Content Times Primitive Polynomial/Existence

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$.
Then:
:$\map f X = \cont f \, \map {f^*} X$
where:
:$\cont f$ is the content of $\map f X$
:$\map {f^*} X$ is a primitive polynomial.

Proof:
Consider the coefficients of $f$ expressed as fractions.
Let $k$ be any positive integer that is divisible by the denominators of all the coefficients of $f$.
Such a number is bound to exist: just multiply all those denominators together, for example.
Then $\map f X$ is a polynomial equal to $\dfrac 1 k$ multiplied by a polynomial with integral coefficients.
Let $d$ be the GCD of all these integral coefficients.
Then $\map f X$ is equal to $\dfrac h k$ multiplied by a primitive polynomial.
Theorem: Rational Polynomial is Content Times Primitive Polynomial/Uniqueness

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.
Let $\map f X \in \Q \sqbrk X$ be given.
Then there exist unique content $\cont f$ of $\map f X$ and primitive polynomial $\map {f^*} X$ such that:
:$\map f X = \cont f \, \map {f^*} X$

Proof:
Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are primitive.
Then:
:$\map g X = \dfrac a b \map f X$
where $\dfrac a b$ is some rational number which can be expressed as $\dfrac m n$ where $m$ and $n$ are coprime.
Then:
:$\map g X = \dfrac m n \map f X$
that is:
:$m \cdot \map f X = n \cdot \map g X$
Suppose $m > 1$.
Then from Euclid's Lemma $m$ has a divisor $p$ which does not divide $n$ (as $m \perp n$).
So $m$ must divide every coefficient of $g$.
But this can not be so, as $g$ is primitive, so $m = 1$.
In a similar way, $n = 1$.
So $f = g$ and $a = b$, so demonstrating uniqueness.
{{qed}}
Theorem: Content of Polynomial in Dedekind Domain is Multiplicative

Let $R$ be a Dedekind domain.
Let $f, g \in R \sqbrk X$ be polynomials.
Let $\cont f$ denote the content of $f$.
Then $\cont {f g} = \cont f \cont g$ is the product of $\cont f$ and $\cont g$.

Proof:
{{ProofWanted}}
Category:Content of Polynomial
Category:Gauss's Lemma (Polynomial Theory)
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