avewright/proofwiki-math huggingface dataset:
Theorem: Multiplicative Identity for Quaternions

In the set of quaternions $\mathbb H$, the element:
:$\mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$
serves as the identity element for quaternion multiplication.
This element is written $\mathbf 1$.

Proof:
Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$.
From the definition of quaternion multiplication:
{{begin-eqn}}
{{eqn | l = \mathbf x \mathbf 1
      | r = \paren {a \cdot 1 - b \cdot 0 - c \cdot 0 - d \cdot 0} \mathbf 1
      | c = 
}}
{{eqn | o = +
      | r = \paren {a \cdot 0 + b \cdot 1 + c \cdot 0 - d  \cdot 0} \mathbf i
      | c = 
}}
{{eqn | o = +
      | r = \paren {a \cdot 0 - b \cdot 0 + c \cdot 1 + d \cdot 0} \mathbf j
      | c = 
}}
{{eqn | o = +
      | r = \paren {a \cdot 0 + b \cdot 0 - c \cdot 0 + d \cdot 1} \mathbf k
      | c = 
}}
{{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
      | c = 
}}
{{eqn | r = \mathbf x
      | c = 
}}
{{end-eqn}}
Similarly for $\mathbf 1 \mathbf x$.
{{qed}}
Category:Quaternions
bj6kz4l5i0oylvv53hoxsmzktikilvu
Theorem: Quaternions form Skew Field

The set $\H$ of quaternions forms a '''skew field''' under the operations of addition and multiplication.

Proof:
From Ring of Quaternions is Ring we have that $\H$ forms a ring.
From Multiplicative Identity for Quaternions we have that $\mathbf 1$ is the identity for quaternion multiplication.
From Multiplicative Inverse of Quaternion we have that every element of $\H$ except $\mathbf 0$ has an inverse under quaternion multiplication.
So $\H \setminus \set 0 = \H^*$ is a group.
Hence $\H$ forms a division ring.
But quaternion multiplication is specifically not commutative, for example:
:$\mathbf i \mathbf j = \mathbf k, \ \mathbf j \mathbf i = -\mathbf k$
So $\H$ forms a skew field under addition and multiplication.
{{qed}}
Theorem: Cancellation Law for Ring Product of Integral Domain

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $D^*$ denote $D \setminus \set {0_D}$, that is, $D$ without its zero.
Let $a \in D^*$.
Then:
:$\forall x, y \in D: a \circ x = a \circ y \implies x = y$
That is, all elements of $D^*$ are cancellable for the ring product.

Proof:
From the definition of integral domain, no elements of $D^*$ are zero divisors.
From Ring Element is Zero Divisor iff not Cancellable, it follows that all elements of $D^*$ are cancellable for the ring product $\circ$.
{{qed}}
Theorem: Gaussian Integers form Integral Domain

The ring of Gaussian integers $\struct {\Z \sqbrk i, +, \times}$ is an integral domain.

Proof:
The set of complex numbers $\C$ forms a field, which is by definition a division ring.
We have that $\Z \sqbrk i \subset \C$.
So from Cancellable Element is Cancellable in Subset, all non-zero elements of $\Z \sqbrk i$ are cancellable for complex multiplication.
The identity element for complex multiplication is $1 + 0 i$ which is in $\Z \sqbrk i$.
We also have that Complex Multiplication is Commutative.
From Identity of Cancellable Monoid is Identity of Submonoid, the identity element of $\struct {\Z \sqbrk i^*, \times}$ is the same as for $\struct {\C^*, \times}$.
So we have that:
:$\struct {\Z \sqbrk i, +, \times}$ is a commutative ring with unity
:All non-zero elements of $\struct {\Z \sqbrk i, +, \times}$ are cancellable.
Hence the result from definition of integral domain.
{{qed}}
Theorem: Set Intersection Not Cancellable

Let $S$ be a set and let $\powerset S$ be the power set of $S$.
Let $S_1, S_2, T \in \powerset S$.
Suppose that $S_1 \cap T = S_2 \cap T$.
Then it is not necessarily the case that $S_1 = S_2$.

Proof:
Proof by Counterexample:
Let $S = \set {1, 2, 3}$.
Let $T = \set 3$.
Let $S_1 = \set {1, 3}, S_2 = \set {2, 3}$
Then $S_1 \cap T = S_2 \cap T = \set 3$ but $S_1 \ne S_2$.
{{qed}}
Category:Set Intersection
brmjm7pmkl21sk8q4nujb4mkne5r398
Theorem: Sum of Triangular Matrices

Let $\mathbf A = \left[{a}\right]_{n}, \mathbf B = \left[{b}\right]_{n}$ be square matrices of order $n$.
Let $\mathbf C = \mathbf A + \mathbf B$ be the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, then so is $\mathbf C$.
If $\mathbf A$ and $\mathbf B$ are lower triangular matrices, then so is $\mathbf C$.

Proof:
From the definition of matrix addition, we have:
:$\forall i, j \in \left[{1 .. n}\right]: c_{ij} = a_{ij} + b_{ij}$
If $\mathbf A$ and $\mathbf B$ are upper triangular matrices, we have:
:$\forall i > j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i > j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself upper triangular.
Similarly, if $\mathbf A$ and $\mathbf B$ are lower triangular matrices, we have:
:$\forall i < j: a_{ij} = b_{ij} = 0$
Hence:
:$\forall i < j: c_{ij} = a_{ij} + b_{ij} = 0 + 0 = 0$
and so $\mathbf C$ is itself lower triangular.
{{Qed}}
Category:Triangular Matrices
Category:Matrix Entrywise Addition
h6dwze7055xyqeo5fwlnt4ak9rikwy0
Theorem: Negative of Triangular Matrix

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
If $\mathbf A$ is an upper triangular matrix, then so is $-\mathbf A$.
If $\mathbf A$ is a lower triangular matrix, then so is $-\mathbf A$.

Proof:
From the definition of negative matrix, we have:
:$\forall i, j \in \closedint 1 n: \sqbrk {-a}_{i j} = -a_{i j}$
If $\mathbf A$ is an upper triangular matrix, we have:
:$\forall i > j: a_{i j} = 0$
Hence:
:$\forall i > j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
and so $-\mathbf A$ is itself upper triangular.
Similarly, if $\mathbf A$ is a lower triangular matrix, we have:
:$\forall i < j: a_{i j} = 0$
Hence:
:$\forall i < j: \sqbrk {-a}_{i j} = -a_{i j} = 0$
and so $-\mathbf A$ is itself lower triangular.
{{Qed}}
Category:Negative Matrices
Category:Triangular Matrices
ivy08olfqqzixab53xua25egndy4xer
Theorem: Triangular Matrices forms Subring of Square Matrices

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} n$ be the order $n$ square matrix space over a ring $R$.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
Let $\map {U_R} n$ be the set of upper triangular matrices of order $n$ over $R$.
Then $\map {U_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.
Similarly, let $\map {L_R} n$ be the set of lower triangular matrices of order $n$ over $R$.
Then $\map {L_R} n$ forms a subring of $\struct {\map {\MM_R} n, +, \times}$.

Proof:
From Negative of Triangular Matrix, if $\mathbf B \in \map {U_R} n$ then $-\mathbf B \in \map {U_R} n$.
Then from Sum of Triangular Matrices, if $\mathbf A, -\mathbf B \in \map {U_R} n$ then $\mathbf A + \paren {-\mathbf B} \in \map {U_R} n$.
From Product of Triangular Matrices, if $\mathbf A, \mathbf B \in \map {U_R} n$ then $\mathbf A \mathbf B \in \map {U_R} n$.
The result follows from the Subring Test.
The same argument can be applied to matrices in $\map {L_R} n$.
{{qed}}
Theorem: Natural Numbers form Subsemiring of Integers

The semiring of natural numbers $\struct {\N, +, \times}$ forms a subsemiring of the ring of integers $\struct {\Z, +, \times}$.

Proof:
We have that Natural Numbers form Commutative Semiring.
From Natural Numbers are Non-Negative Integers we have that $\N$ is a subset of $\Z$.
Hence the result, from the definition of subsemiring.
{{qed}}
Category:Natural Numbers
Category:Integers
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Theorem: Integers form Subdomain of Rationals

The integral domain of integers $\left({\Z, +, \times}\right)$ forms a subdomain of the field of rational numbers.

Proof:
The rational numbers are defined as the quotient field of the integers.
From its method of construction, it follows that the integers $\Z$ are a subset of the rational numbers $\Q$.
Hence the result, from the definition of subdomain.
{{qed}}
Theorem: Real Numbers form Subfield of Complex Numbers

The field of real numbers $\struct {\R, +, \times}$ forms a subfield of the field of complex numbers $\struct {\C, +, \times}$.

Proof:
From Additive Group of Reals is Normal Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
From Multiplicative Group of Reals is Normal Subgroup of Complex, $\struct {\R, \times}$ is a subgroup of $\struct {\C, \times}$.
The result follows from the Subfield Test via the One-Step Subgroup Test and Two-Step Subgroup Test.
{{qed}}
Theorem: Sum of All Ring Products is Additive Subgroup

Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S + T$ be defined as subset product.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then both $S + T$ and $S T$ are additive subgroups of $\struct {R, +, \circ}$.

Proof:
As $\struct {R, +}$ is abelian (from the definition of a ring), we have:
:$S + T = T + S$
from Subset Product of Commutative is Commutative.
So from Subset Product of Subgroups it follows that $S + T$ is an additive subgroup of $\struct {R, +, \circ}$.
Let $x, y \in S T$.
We have that $\struct {S T, +}$ is closed.
So $x + y \in S T$.
So, if $\ds y = \sum s_i \circ t_i \in S T$, it follows that:
:$\ds -y = \sum \paren {-s_i} \circ t_i \in S T$
By the Two-Step Subgroup Test, we have that $S T$ is an additive subgroup of $\struct {R, +, \circ}$.
{{qed}}
Theorem: Sum of All Ring Products is Closed under Addition

Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}$ and $\struct {T, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $\struct {S T, +}$ is a closed subset of $\struct {R, +}$.

Proof:
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$
for some $s_i, t_i, j, k$, etc.
By renaming the indices, we can express $x_2$ as:
:$\ds x_2 = \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i$
and hence:
:$\ds x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j + 1}^{j + k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$
So $x_1 + x_2 \in S T$ and $\struct {S T, +}$ is shown to be closed.
{{qed}}
Theorem: Sum of All Ring Products is Associative

Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {S, +}, \struct {T, +}, \struct {U, +}$ be additive subgroups of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then:
:$\paren {S T} U = S \paren {T U}$

Proof:
We have by definition that $S T$ is made up of all finite sums of elements of the form $s \circ t$ where $s \in S, t \in T$.
From Sum of All Ring Products is Closed under Addition, this set is closed under ring addition.
Therefore, so are $\paren {S T} U$ and $S \paren {T U}$.
Let $z \in \paren {S T} U$.
Then $z$ is a finite sum of elements in the form $x \circ u$ where $x \in ST$ and $u \in U$.
So $x$ is a finite sum of elements in the form $s \circ t$ where $s \in S, t \in T$.
Therefore $z$ is a finite sum of elements in the form $\paren {s \circ t} \circ u$ where $s \in S, t \in T, u \in U$.
As $\struct {R, +, \circ}$ is a ring, $\circ$ is associative.
So $z$ is a finite sum of elements in the form $s \circ \paren {t \circ u}$ where $s \in S, t \in T, u \in U$.
So these elements all belong to $S \paren {T U}$.
Since $S \paren {T U}$ is closed under addition, $z \in S \paren {T U}$.
So:
:$\paren {S T} U \subseteq S \paren {T U}$
By a similar argument in the other direction:
:$S \paren {T U} \subseteq \paren {S T} U $
and so by definition of set equality:
:$\paren {S T} U = S \paren {T U}$
{{qed}}
Theorem: Sum of Ring Products is Subring of Commutative Ring

Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {S, +, \circ}$ and $\struct {T, +, \circ}$ be subrings of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $S T$ is a subring of $\struct {R, +, \circ}$.

Proof:
From Sum of All Ring Products is Additive Subgroup we have that $\struct {S T, +}$ is an additive subgroup of $R$.
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$
for some $s_i, t_i, s_j, t_j, m, n$, etc.
Then:
{{begin-eqn}}
{{eqn | l = x_1 \circ x_2
      | r = \paren {\sum_{i \mathop = 1}^m s_i \circ t_i} \circ \paren {\sum_{j \mathop = 1}^n s'_j \circ t'_j}
      | c = 
}}
{{eqn | r = \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {s_i \circ s'_j} \circ \paren {t_i \circ t'_j}
      | c = as $\circ$ is commutative
}}
{{end-eqn}}
So $x_1 \circ x_2 \in S T$ and the result follows from the Subring Test.
{{qed}}
Theorem: Complex Numbers form Subfield of Quaternions

The field of complex numbers $\struct {\C, +, \times}$ is isomorphic to the subfields of the quaternions $\struct {\mathbb H, +, \times}$ whose underlying subsets are:
:$(1): \quad \mathbb H_\mathbf i = \set {a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: c = d = 0}$
:$(2): \quad \mathbb H_\mathbf j = \set {a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = d = 0}$
:$(3): \quad \mathbb H_\mathbf k = \set {a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k \in \mathbb H: b = c = 0}$
That is:
:$(1): \quad \mathbb H_\mathbf i = \set {a \mathbf 1 + b \mathbf i \in \mathbb H}$
:$(2): \quad \mathbb H_\mathbf j = \set {a \mathbf 1 + c \mathbf j \in \mathbb H}$
:$(3): \quad \mathbb H_\mathbf k = \set {a \mathbf 1 + d \mathbf k \in \mathbb H}$

Proof:
Let $\phi_i: \mathbb H_\mathbf i \to \C$  be defined as:
:$\forall \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \map {\phi_i} {\mathbf x} = a + b i$
where in this context $i$ is the imaginary unit.
Similarly we can define $\phi_j$ and $\phi_k$:
:$\forall \mathbf x = \mathbf a \mathbf 1 + c \mathbf j \in \mathbb H_\mathbf j: \map {\phi_j} {\mathbf x} = a + c i$
:$\forall \mathbf x = \mathbf a \mathbf 1 + d \mathbf k \in \mathbb H_\mathbf k: \map {\phi_k} {\mathbf x} = a + d i$
Theorem: Ring Homomorphism whose Kernel contains Ideal

Let $R$ be a ring.
Let $J$ be an ideal of $R$.
Let $\nu: R \to R / J$ be the quotient epimorphism.
Let $\phi: R \to S$ be a ring homomorphism such that:
:$J \subseteq \map \ker \phi$
where $\map \ker \phi$ is the kernel of $\phi$.
Then there exists a unique ring homomorphism $\psi: R / J \to S$ such that:
:$\phi = \psi \circ \nu$
where $\circ$ denotes composition of mappings.
:$<nowiki>\begin {xy} \xymatrix@L + 2mu@ + 1em {
 R \ar[rr]^*{\nu}
   \ar[rdrd]_*{\phi}
& & R / J \ar[dd]_*{\psi} \\
\\
& & S 
} \end {xy}</nowiki>$
Also:
:$\map \ker \psi = \map \ker \phi / J$

Proof:
Suppose $\phi = \psi \circ \nu$.
Let $J + x$ be an arbitrary element of $R / J$.
Then:
:$(1) \quad \map \psi {J + x} = \psi \circ \map \nu x = \map \phi x$
So there is only one possible way to define $\psi$.
Now suppose $J + x = J + x'$.
Then $x + \paren {-x'} \in J$.
So $x + \paren {-x'} \in \map \ker \phi$ as $J \subseteq \map \ker \phi$.
That is, $\map \phi {x + \paren {-x'} } = 0_S$.
So $\map \phi x = \map \phi {x'}$ and so $\psi$ as defined in $(1)$ is well-defined.
Now suppose $J + x, J + y \in R / J$.
We have:
{{begin-eqn}}
{{eqn | l = \map \psi {\paren {J + x} + \paren {J + y} }
      | r = \map \psi {J + \paren {x + y} }
      | c = {{Defof|Quotient Ring}}
}}
{{eqn | r = \map \phi {x + y}
      | c = Definition of $\psi$ in $(1)$ above
}}
{{eqn | r = \map \phi x + \map \phi y
      | c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \psi {J + x} + \map \psi {J + y}
      | c = 
}}
{{end-eqn}}
So $\psi$ preserves ring addition.
Then:
{{begin-eqn}}
{{eqn | l = \map \psi {\paren {J + x} \paren {J + y} }
      | r = \map \psi {J + \paren {x y} }
      | c = {{Defof|Quotient Ring}}
}}
{{eqn | r = \map \phi {x y}
      | c = Definition of $\psi$ in $(1)$ above
}}
{{eqn | r = \map \phi x \map \phi y
      | c = {{Defof|Ring Homomorphism}}
}}
{{eqn | r = \map \psi {J + x} \map \psi {J + y}
      | c = 
}}
{{end-eqn}}
So $\psi$ preserves ring product, and so $\psi$ is a ring homomorphism.
Finally:
{{begin-eqn}}
{{eqn | l = \map \psi {J + x}
      | r = 0_S
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi x
      | r = 0_S
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \in
      | r = \map \ker \phi
      | c = 
}}
{{end-eqn}}
So:
:$\map \ker \psi = \map \ker \phi / J$
{{qed}}
Theorem: Decay Equation

The first order ordinary differential equation:
:$\dfrac {\d y} {\d x} = k \paren {y_a - y}$
where $k \in \R: k > 0$
has the general solution:
:$y = y_a + C e^{-k x}$
where $C$ is an arbitrary constant.
If $y = y_0$ at $x = 0$, then:
:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
This differential equation is known as the '''decay equation'''.

Proof:
{{begin-eqn}}
{{eqn | l = \frac {\d y} {\d x}
      | r = -k \paren {y - y_a}
      | c =
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d y} {y - y_a}
      | r = -\int k \rd x
      | c = Solution to Separable Differential Equation
}}
{{eqn | ll= \leadsto
      | l = \map \ln {y - y_a}
      | r = -k x + C_1
      | c = Primitive of Reciprocal and Derivatives of Function of $a x + b$
}}
{{eqn | ll= \leadsto
      | l = y - y_a
      | r = e^{-k x + C_1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = y_a + C e^{-k x}
      | c = where we put $C = e^{C_1}$
}}
{{end-eqn}}
This is our general solution.
{{qed|lemma}}
Suppose we have the initial condition:
:$y = y_0$ when $x = 0$
Then:
:$y_0 = y_a + C e^{-k \cdot 0} = y_a + C$
and so:
:$C = y_0 - y_a$
Hence the solution:
:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
{{qed}}
Theorem: Relation on Empty Set is Equivalence

Let $S = \O$, that is, the empty set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Then $\RR$ is the null relation and is an equivalence relation.

Proof:
As $S = \O$, we have from Cartesian Product is Empty iff Factor is Empty that $S \times S = \O$.
Then it follows that $\RR \subseteq S \times S = \O$.
Theorem: Fourth Isomorphism Theorem

Let $\phi: R \to S$ be a ring homomorphism.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
Let $\mathbb K$ be the set of all subrings of $R$ which contain $K$ as a subset.
Let $\mathbb S$ be the set of all subrings of $\Img \phi$.
Let $\phi^\to: \powerset R \to \powerset S$ be the direct image mapping of $\phi$.
Then its restriction $\phi^\to: \mathbb K \to \mathbb S$ is a bijection.
Also:
:$(1): \quad \phi^\to$ and its inverse both preserve subsets.
:$(2): \quad \phi^\to$ and its inverse both preserve ideals:
:::: If $J$ is an ideal of $R$, then $\map {\phi^\to} J$ is an ideal of $S$
:::: If $J'$ is an ideal of $S$, then $\map {\paren {\phi^\to}^{-1} } {J'}$ is an ideal of $R$

Proof:
=== Proof of Preservation of Subsets ===
From Subset Maps to Subset, we have:
:$(a) \quad \forall X, X' \in \mathbb K: X \subseteq X' \implies \map {\phi^\to} X \subseteq \map {\phi^\to} {X'}$
:$(b) \quad \forall Y, Y' \in \mathbb S: Y \subseteq Y' \implies \map {\paren {\phi^\to}^{-1} } Y \subseteq \map {\paren {\phi^\to}^{-1} } {Y'}$
So $\phi^\to$ and its inverse both preserve subsets, and $(1)$ has been demonstrated to hold.
{{qed|lemma}}
Theorem: Increasing Union of Subrings is Subring

Let $R$ be a ring.
Let $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots \subseteq S_i \subseteq \ldots$ be subrings of $R$.
Then the increasing union $S$:
:$\ds S = \bigcup_{i \mathop \in \N} S_i$
is a subring of $R$.

Proof:
Let $\ds S = \bigcup_{i \mathop \in \N} S_i$.
Clearly $0_R \in S$.
Let $a, b \in S$.
Then $\exists i, j \in \N: a \in S_i, b \in S_j$.
From the construction, we have that either of $S_i$ and $S_j$ contains the other.
Let $l = \max \set {i, j}$ so $a, b \in S_l$.
Then $a + \paren {-b} \in S_l$ and $a b \in S_l$, as $S_l$ is a subring.
Thus $a + \paren {-b} \in S$ and $a b \in S$.
By the Subring Test, $S$ is a subring of $R$.
{{qed}}
Theorem: Subrings of Integers are Sets of Integer Multiples

Let $\struct {\Z, +, \times}$ be the integral domain of integers.
The subrings of $\struct {\Z, +, \times}$ are the rings of integer multiples:
:$\struct {n \Z, +, \times}$
where $n \in \Z: n \ge 0$.
There are no other subrings of $\struct {\Z, +, \times}$ but these.

Proof:
From Integer Multiples form Commutative Ring, it is clear that $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.
We also note that when $n = 0$, we have:
:$\struct {n \Z, +, \times} = \struct {0, +, \times}$
which is the null ring.
When $n = 1$, we have:
:$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$
From Null Ring and Ring Itself are Subrings, these extreme cases are also subrings of $\struct {\Z, +, \times}$.
From Subgroups of Additive Group of Integers, the only additive subgroups of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.
So there can be no subrings of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their additive group.
Hence the result.
{{qed}}
Theorem: Subring of Integers is Ideal

Every subring of $\struct {\Z, +, \times}$ is an ideal of the ring $\struct {\Z, +, \times}$.

Proof:
Follows directly from:
:Subrings of Integers are Sets of Integer Multiples
and:
:Subgroup of Integers is Ideal.
{{qed}}
Theorem: Polynomial Ring of Sequences is Ring

Let $R$ be a ring.
Let $P \sqbrk R$ be the polynomial ring over sequences in $R$.
Then $P \sqbrk R$ is itself a ring.

Proof:
We have by definition of polynomial ring over sequences in $R$ that:
:$P \sqbrk R = \set {\sequence {r_0, r_1, r_2, \ldots} }$
where each $r_i \in R$, and all but a finite number of terms is zero.