hcju/proofwikiqa huggingface dataset:
From Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is a cyclic group From Cyclic Group is Abelian, $\struct {n \Z, +}$ is abelian. From Integer Multiples Closed under Multiplication and Integer Multiplication is Associative, we have that $\struct {n \Z, \times}$ is a semigroup. From Integer Multiplication Distributes over Addition it follows that $\struct {n \Z, +, \times}$ is a ring. From Integer Multiplication is Commutative we have that $\struct {n \Z, +, \times}$ is a commutative ring. So $\struct {n \Z, +, \times}$ is a commutative ring. {{qed|lemma}} The unity of $\struct {\Z, +, \times}$ is $1$. But unless $n = 1$, we have that $1 \notin n \Z$. As $\struct {\Z, +, \times}$ is an integral domain, all its elements are cancellable, by the Cancellation Law of Ring Product of Integral Domain. We have that Subrings of Integers are Sets of Integer Multiples. From Identity of Cancellable Monoid is Identity of Submonoid it follows that $\struct {n \Z, +, \times}$ has no unity, because if it did, that unity would be $1$. {{qed}}
From Matrix Form of Quaternion we have that: :$\mathbf x_1 \mathbf x_2 = \begin{bmatrix} a_1 + b_1 i & c_1 + d_1 i \\ -c_1 + d_1 i & a_1 - b_1 i \end{bmatrix} \begin{bmatrix} a_2 + b_2 i & c_2 + d_2 i \\ -c_2 + d_2 i & a_2 - b_2 i \end{bmatrix}$ Let $\mathbf x_1 \mathbf x_2 = \begin{bmatrix} p_{11} & p_{12} \\ p_{21} & p_{22} \end{bmatrix} = \begin{bmatrix} a + b i & c + d i \\ -c + d i & a - b i \end{bmatrix}$. Throughout we use the definition of conventional matrix product. So: {{begin-eqn}} {{eqn | l=p_{11}       | r=\left({a_1 + b_1 i}\right) \left({a_2 + b_2 i}\right) + \left({c_1 + d_1 i}\right) \left({-c_2 + d_2 i}\right)       | c= }} {{eqn | r=a_1 a_2 + b_1 a_2 i + a_1 b_2 i + b_1 b_2 i^2 - c_1 c_2 - d_1 c_2 i + c_1 d_2 i + d_1 d_2 i^2       | c= }} {{eqn | r=\left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) + \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i       | c= }} {{eqn | r=a + b i       | c= }} {{end-eqn}} {{begin-eqn}} {{eqn | l=p_{12}       | r=\left({a_1 + b_1 i}\right) \left({c_2 + d_2 i}\right) + \left({c_1 + d_1 i}\right) \left({a_2 - b_2 i}\right)       | c= }} {{eqn | r=a_1 c_2 + b_1 c_2 i + a_1 d_2 i + b_1 d_2 i^2 + c_1 a_2 + d_1 a_2 i - c_1 b_2 i - d_1 b_2 i^2       | c= }} {{eqn | r=\left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) + \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) i       | c= }} {{eqn | r=c + d i       | c= }} {{end-eqn}} {{begin-eqn}} {{eqn | l=p_{21}       | r=\left({-c_1 + d_1 i}\right) \left({a_2 + b_2 i}\right) + \left({a_1 - b_1 i}\right) \left({-c_2 + d_2 i}\right)       | c= }} {{eqn | r=-c_1 a_2 + d_1 a_2 i - c_1 b_2 i + d_1 b_2 i^2 - a_1 c_2 + b_1 c_2 i + a_1 d_2 i - b_1 d_2 i^2       | c= }} {{eqn | r=-\left({a_1 c_2 - b_1 d_2 + c_1 a_2 + d_1 b_2}\right) + \left({a_1 d_2 + b_1 c_2 - c_1 b_2 + d_1 a_2}\right) i       | c= }} {{eqn | r=-c + d i       | c= }} {{end-eqn}} {{begin-eqn}} {{eqn | l=p_{22}       | r=\left({-c_1 + d_1 i}\right) \left({c_2 + d_2 i}\right) + \left({a_1 - b_1 i}\right) \left({a_2 - b_2 i}\right)       | c= }} {{eqn | r=-c_1 c_2 + d_1 c_2 i - c_1 d_2 i + d_1 d_2 i^2 + a_1 a_2 - b_1 a_2 i - a_1 b_2 i + b_1 b_2 i^2       | c= }} {{eqn | r=\left({a_1 a_2 - b_1 b_2 - c_1 c_2 - d_1 d_2}\right) - \left({a_1 b_2 + b_1 a_2 + c_1 d_2 - d_1 c_2}\right) i       | c= }} {{eqn | r=a - b i       | c= }} {{end-eqn}} Hence the result from Matrix Form of Quaternion. {{qed}} Category:Quaternions 5ppc4f8h3le29g9xdsoky7nc4e6u7im
Taking the abelian group axioms in turn: === G0: Closure === Let: : $\mathbf x_1 = a_1 \mathbf 1 + b_1 \mathbf i + c_1 \mathbf j + d_1 \mathbf k$ : $\mathbf x_2 = a_2 \mathbf 1 + b_2 \mathbf i + c_2 \mathbf j + d_2 \mathbf k$ be quaternions. By definition of quaternion addition: :$\mathbf x_1 + \mathbf x_2 = \left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$ So as $a_1, a_2, b_1, b_2$ etc. are all elements of $\R$, then so are $a_1 + a_2, b_1 + b_2$ etc. So $\left({a_1 + a_2}\right) \mathbf 1 + \left({b_1 + b_2}\right) \mathbf i + \left({c_1 + c_2}\right) \mathbf j + \left({d_1 + d_2}\right) \mathbf k$ is a quaternion. Hence $\left({\mathbb H, +}\right)$ is closed. {{qed|lemma}} === G1: Associativity === From Matrix Form of Quaternion, we can express a quaternion $\mathbf x$ in the form of a matrix: :$\mathbf x = \begin{bmatrix} a + bi & c + di \\ -c + di & a - bi \end{bmatrix}$ We have that Matrix Entrywise Addition is Associative. It follows that quaternion addition is also associative {{qed|lemma}} === G2: Identity === The identity element of $\left({\mathbb H, +}\right)$ is: :$\mathbf 0 = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k$ as can be seen: {{begin-eqn}} {{eqn | l = \mathbf 0 + \mathbf x       | r = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k       | c =  }} {{eqn | r = \left({0 + a}\right) \mathbf 1 + \left({0 + b}\right) \mathbf i + \left({0 + c}\right) \mathbf j + \left({0 + d}\right) \mathbf k       | c =  }} {{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k       | c =  }} {{eqn | r = \mathbf x       | c =  }} {{eqn | r = \left({a + 0}\right) \mathbf 1 + \left({b+ 0}\right) \mathbf i + \left({c + 0}\right) \mathbf j + \left({d + 0}\right) \mathbf k       | c =  }} {{eqn | r = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k + 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k       | c =  }} {{eqn | r = \mathbf x + \mathbf 0       | c =  }} {{end-eqn}} {{qed|lemma}} === G3: Inverses === Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion. The inverse of $\left({\mathbb H, +}\right)$ is: :$- \mathbf x = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k$ as can be seen: {{begin-eqn}} {{eqn | l = \mathbf -x + \mathbf x       | r = -a \mathbf 1 + -b \mathbf i + -c \mathbf j + -d \mathbf k + a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k       | c =  }} {{eqn | r = \left({-a + a}\right) \mathbf 1 + \left({-b + b}\right) \mathbf i + \left({-c + c}\right) \mathbf j + \left({-d + d}\right) \mathbf k       | c =  }} {{eqn | r = 0 \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k       | c =  }} {{eqn | r = \mathbf 0       | c =  }} {{end-eqn}} and similarly for $\mathbf x + \mathbf -x$. {{qed|lemma}} === Commutativity === Commutativity follows from Real Addition is Commutative. {{qed|lemma}} Thus all the abelian group axioms are seen to be fulfilled. {{qed}} Category:Quaternions Category:Abelian Groups f9okedoso5y46sqvj3orbt8k7gacxw9
From Quaternion Addition forms Abelian Group, $\mathbb H$ forms an abelian group under quaternion addition. From the definition it is clear that quaternion multiplication is closed. We have from Matrix Form of Quaternion that quaternions can be expressed in matrix form. From Quaternion Multiplication we have that quaternion multiplication can be expressed in terms of matrix multiplication, which is associative. So $\mathbb H$ forms a semigroup under quaternion multiplication. Finally, we have that Matrix Multiplication Distributes over Matrix Addition. So all the conditions are fulfilled for $\mathbb H$ to be a ring. {{qed}}
From Matrix Form of Quaternion it is clear that the quaternions $\mathbb H$ can be expressed in matrix form, as elements of $\mathcal M_\C \left({2}\right)$. Thus $\mathbb H \subseteq \mathcal M_\C \left({2}\right)$. As the quaternions form a ring, the result follows by definition of subring. {{qed}}
From the definition of quaternion multiplication: {{begin-eqn}} {{eqn | l = \mathbf x \overline {\mathbf x}       | r = \paren {a^2 - b \paren {-b} - c \paren {-c} - d \paren {-d} } \mathbf 1       | c =  }} {{eqn | o =        | ro= +       | r = \paren {a \paren {-b} + b a + c \paren {-d} - d \paren {-c} } \mathbf i       | c =  }} {{eqn | o =        | ro= +       | r = \paren {a \paren {-c} - b \paren {-d} + c a + d \paren {-b} } \mathbf j       | c =  }} {{eqn | o =        | ro= +       | r = \paren {a \paren {-d} + b \paren {-c} - c \paren {-b} + d a} \mathbf k       | c =  }} {{eqn | r = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-a b + b a - c d + d c} \mathbf i       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-a c + b d + c a - d b} \mathbf j       | c =  }} {{eqn | o =        | ro= +       | r = \paren {-a d - b c + c b + d a} \mathbf k       | c =  }} {{end-eqn}} We have that $a, b, c, d \in \R$. So from Real Multiplication is Commutative, their products commute. So the terms in $\mathbf i, \mathbf j, \mathbf k$ vanish. The proof that $\overline {\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$ is similar. Hence the result. {{qed}}
From Multiplicative Identity for Quaternions, we need to show that $\lambda \overline{\mathbf x} \mathbf x = \mathbf 1 = \mathbf x \lambda \overline{\mathbf x}$. From Product of Quaternion with Conjugate we have that: :$\overline{\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$ Using the definition of $\lambda$ from above: {{begin-eqn}} {{eqn | l = \lambda \overline{\mathbf x} \mathbf x       | r = \paren {\frac 1 {a^2 + b^2 + c^2 + d^2} } \mathbf 1 \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1       | c =  }} {{eqn | r = \paren {\frac {a^2 + b^2 + c^2 + d^2} {a^2 + b^2 + c^2 + d^2} } \mathbf 1       | c = {{Defof|Quaternion Multiplication}} }} {{eqn | r = \mathbf 1       | c =  }} {{end-eqn}} Unless all of $a, b, c, d = 0$ we have that $a^2 + b^2 + c^2 + d^2 \ne 0$. Therefore $\lambda$ is defined for all quaternions not equal to $\mathbf 0$. It follows from Left Inverse for All is Right Inverse that $\mathbf x \lambda \overline{\mathbf x} = \mathbf 1$. Hence the result. {{qed}}
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
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}}
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}}
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}}
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
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
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}}
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 0io25lnk5ngnlb90vw3fp3kag33fsht
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}}
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$. From Multiplicative Group of Reals is 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}}
As $\left({R, +}\right)$ 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 $\left({R, +, \circ}\right)$. Let $x, y \in S T$. We have that $\left({S T, +}\right)$ is closed. So $x + y \in S T$. So, if $\displaystyle y = \sum s_i \circ t_i \in S T$, it follows that: : $\displaystyle -y = \sum \left({-s_i}\right) \circ t_i \in S T$ By the Two-Step Subgroup Test, we have that $S T$ is an additive subgroup of $\left({R, +, \circ}\right)$. {{qed}}
Let $x_1, x_2 \in S T$. Then: : $\displaystyle x_1 = \sum_{i \mathop = 1}^j s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^k s_i \circ t_i$ for some $s_i, t_i, j, k$, etc. By renaming the indices, we can express $x_2$ as: : $\displaystyle x_2 = \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i$ and hence: :$\displaystyle x_1 + x_2 = \sum_{i \mathop = 1}^j s_i \circ t_i + \sum_{i \mathop = j+1}^{j+k} s_i \circ t_i = \sum_{i \mathop = 1}^k s_i \circ t_i$ So $x_1 + x_2 \in S T$ and $\left({S T, +}\right)$ is shown to be closed. {{qed}}
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 $\left({S T}\right) U$ and $S \left({T U}\right)$. Let $z \in \left({S T}\right) U$. 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 $\left({s \circ t}\right) \circ u$ where $s \in S, t \in T, u \in U$. As $\left({R, +, \circ}\right)$ os a ring, $\circ$ is associative. So $z$ is a finite sum of elements in the form $s \circ \left({t \circ u}\right)$ where $s \in S, t \in T, u \in U$. So these elements all belong to $S \left({T U}\right)$. Since $S \left({T U}\right)$ is closed under addition, $z \in S \left({T U}\right)$. So: :$\left({S T}\right) U \subseteq S \left({T U}\right)$ By a similar argument in the other direction,  $S \left({T U}\right) \subseteq \left({S T}\right) U $ and so by definition of set equality: :$\left({S T}\right) U = S \left({T U}\right)$ {{qed}}
From Sum of All Ring Products is Additive Subgroup we have that $\left({S T, +}\right)$ is an additive subgroup of $R$. Let $x_1, x_2 \in S T$. Then: : $\displaystyle x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$ for some $s_i, t_i, s_j, t_j, m, n$, etc. Then: {{begin-eqn}} {{eqn | l = x_1 \circ x_2       | r = \left({\sum_{i \mathop = 1}^m s_i \circ t_i}\right) \circ \left({\sum_{j \mathop = 1}^n s'_j \circ t'_j}\right)       | c =  }} {{eqn | r = \sum_{ {1 \mathop  \le i \mathop  \le m} \atop {1 \mathop  \le j \mathop  \le n} } \left({s_i \circ s'_j}\right) \circ \left({t_i \circ t'_j}\right)       | 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}}
Let $\phi_i: \mathbb H_\mathbf i \to \C$  be defined as: :$\forall \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = a + b i$ where in this context $i$ is the imaginary unit. Similarly we can define $\phi_j$ and $\phi_k$: :$\forall \mathbf x = \mathbf a \mathbf 1 + c \mathbf j \in \mathbb H_\mathbf j: \phi_j \left({\mathbf x}\right) = a + c i$ :$\forall \mathbf x = \mathbf a \mathbf 1 + d \mathbf k \in \mathbb H_\mathbf k: \phi_k \left({\mathbf x}\right) = a + d i$ === Proof of Bijectivity === First note that each of $\phi_i, \phi_j, \phi_k$ are bijections, as follows: * Injections: Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$. Then: :$\phi_i \left({x_1}\right) = \phi_i \left({x_2}\right) \implies a_1 + b_1 i = a_2 + b_2 i \implies a_1 = a_2, b_1 = b_2 \implies \mathbf x_1 = \mathbf x_2$ and similarly with $\phi_j$ and $\phi_k$. * Surjections: Let $x = a + b i \in \C$. Then: : $\exists \mathbf x = \mathbf a \mathbf 1 + b \mathbf i \in \mathbb H_\mathbf i: \phi_i \left({\mathbf x}\right) = x$ and similarly with $\phi_j$ and $\phi_k$. Thus it is established that $\phi_i, \phi_j, \phi_k$ are bijections. {{qed|lemma}} === Proof of Morphism Property === Let $\mathbf x_1 = \mathbf a_1 \mathbf 1 + b_1 \mathbf i, \mathbf x_2 = \mathbf a_2 \mathbf 1 + b_2 \mathbf i$. First we show that $+$ has the morphism property under $\phi_i$: {{begin-eqn}} {{eqn | l=\phi_i \left({x_1}\right) + \phi_i \left({x_2}\right)       | r=\phi_i \left({a_1 \mathbf 1 + b_1 \mathbf i}\right) + \phi_i \left({a_2 \mathbf 1 + b_2 \mathbf i}\right)       | c= }} {{eqn | r=\left({a_1 + b_1 i}\right) + \left({a_2 + b_2 i}\right)       | c= }} {{eqn | r=\left({a_1 + a_2}\right) + \left({b_1 + b_2}\right) i       | c= }} {{eqn | r=\phi_i \left({x_1 + x_2}\right)       | c= }} {{end-eqn}} and similarly for $\phi_j$ and $\phi_k$. Next we show that $\times$ has the morphism property under $\phi_i$: {{begin-eqn}} {{eqn | l=\phi_i \left({x_1 \times x_2}\right)       | r=\phi_i \left({a_1 a_2 - b_1 b_2 \mathbf i}\right) + \phi_i \left({a_1 b_2 + b_1 a_2 \mathbf i}\right)       | c=by quaternion multiplication - most terms are zero }} {{eqn | r=\left({a_1 a_2 - b_1 b_2}\right) + \left({a_1 b_2 + b_1 a_2}\right) i       | c= }} {{eqn | r=\left({a_1 + b_1 i}\right) \times \left({a_2 + b_2 i}\right)       | c=by complex multiplication }} {{eqn | r=\phi_i \left({x_1}\right) \times \phi_i \left({x_2}\right)       | c= }} {{end-eqn}} Similarly for  $\phi_j$ and $\phi_k$. So we have shown that $\phi_i, \phi_j, \phi_k$ are ring homomorphisms. {{qed|lemma}} So all of these mappings are bijective homomorphisms, that is, isomorphisms. {{qed}}
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}}
{{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 = Separation of Variables }} {{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}}
As $S = \varnothing$, we have from Cartesian Product is Empty iff Factor is Empty that $S \times S = \varnothing$. Then it follows that $\mathcal R \subseteq S \times S = \varnothing$. === Reflexivity === From the definition: :$\mathcal R = \varnothing \implies \forall x \in S: \left({x, x}\right) \notin \mathcal R$ But as $\neg \, \exists x \in S$ it follows vacuously that $\mathcal R$ is reflexive. {{qed|lemma}} === Symmetry === It follows vacuously that: :$\left({x, y}\right) \in \mathcal R \implies \left({y, x}\right) \in \mathcal R$ and so $\mathcal R$ is symmetric. {{qed|lemma}} === Transitivity === It follows vacuously that: :$\left({x, y}\right), \left({y, z}\right) \in \mathcal R \implies \left({x, z}\right) \in \mathcal R$ and so $\mathcal R$ is transitive. {{qed|lemma}} It follows from the definition that $\mathcal R$ is an equivalence relation. {{qed}} Category:Relation Theory Category:Equivalence Relations Category:Empty Set ayuky9nnbh7tsmmmcvwaj4wgkcpliug