hcju/proofwikiqa huggingface dataset:
=== 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}} === Proof that Inverse Image is a Subring === Let $U \in \mathbb S$, that is, let $U$ be a subring of $\Img \phi$. From Preimage of Subring under Ring Homomorphism is Subring, we have that $\map {\paren {\phi^\to}^{-1} } U$ is a subring of $R$ such that $K \subseteq R$ and so: :$\map {\paren {\phi^\to}^{-1} } U \in \mathbb K$ {{qed|lemma}} === Proof that Image is a Subring === Let $V \in \mathbb K$, that is, a subring of $R$ containing $K$. From Ring Homomorphism Preserves Subrings, we have that $\map {\phi^\to} V$ is a subring of $S$ and so: :$\map {\phi^\to} V \in \mathbb S$ {{qed|lemma}} === Proof that $f$ is a Bijection === By Subset of Codomain is Superset of Image of Preimage, we have that: :$U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$ {{questionable|It doesn't, actually, it shows it's the superset. IIRC it needs to be a surjection for that. Check this to see whether $\phi^\to$ is indeed a surjection.}} We also have from Subset of Domain is Subset of Preimage of Image that: :$V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ Now let $r \in \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$. Thus $\map \phi r \in \map {\phi^\to} V$ and so: :$\exists v \in V: \map \phi r = \map \phi v$ So $\map \phi {r + \paren {-v} } = 0_S$ and so $r - v \in K$ by definition of kernel. So: :$\exists k \in K: r + k = v$ But by assumption, $K \subseteq V$ as $V \in \mathbb K$. So $k \in V$ and so it follows that $r \in V$ as well, by the fact that $V$ is subring and so closed for $+$. So: :$\map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V} \subseteq V$ Putting that together with $V \subseteq \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ and we see: :$V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ So we have: :$U = \map {\phi^\to} {\map {\paren {\phi^\to}^{-1} } U}$ :$V = \map {\paren {\phi^\to}^{-1} } {\map {\phi^\to} V}$ That is: :$\phi^\to \circ \paren {\phi^\to}^{-1} = I_\mathbb S$ :$\paren {\phi^\to}^{-1} \circ \phi^\to = I_\mathbb K$ where $I_\mathbb S$ and $I_\mathbb K$ are the identity mappings of $\mathbb S$ and $\mathbb K$ respectively. By Composite of Bijection with Inverse is Identity Mapping, it follows that $\phi^\to: \mathbb K \to \mathbb S$ is a bijection. {{qed}} === Proof of Preservation of Ideals === Let $V \in \mathbb K$ be an ideal of $R$. Let $U = \map {\phi^\to} V$. Then from Ring Epimorphism Preserves Ideals, $U$ is an ideal of $S$. Similarly, let $U = \map {\phi^\to} V$ be an ideal of $S$. Then by Preimage of Ideal under Ring Homomorphism is Ideal, $V = \map {\paren {\phi^\to}^{-1} } U$ is an ideal of $R$ such that $K \subseteq V$. Hence ideals are preserved in both directions. {{qed}}
Let $\displaystyle S = \bigcup_{i \mathop \in \N} S_i$. 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 $\displaystyle l = \max \set {i, j}$ 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}}
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 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}}
Follows directly from: :Subrings of Integers are Sets of Integer Multiples and: :Subgroup of Integers is Ideal. {{qed}}
Let $P \sqbrk R$ be the polynomial ring over $R$. Consider the injection $\phi: R \to P \sqbrk R$ defined as: :$\forall r \in R: \map \phi r = \sequence {r, 0, 0, \ldots}$ It is easily checked that $\phi$ is a ring monomorphism. So the set $\set {\sequence {r, 0, 0, \ldots}: r \in R}$ is a subring of $P \sqbrk R$ which is isomorphic to $R$. So we ''identify'' $r \in R$ with the sequence $\sequence {r, 0, 0, \ldots}$. Next we note that $P \sqbrk R$ contains the element $\sequence {0, 1, 0, \ldots}$ which we can call $x$. From the definition of ring product on the polynomial ring over $R$, we have that: :$x^2 = \sequence {0, 1, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 1, 0, 0, \ldots}$ :$x^3 = \sequence {0, 0, 1, 0, 0, \ldots} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, 0, 0, 1, 0, \ldots}$ and in general: :$x^n = \sequence {0, 1, 0, \ldots}^{n - 1} \odot \sequence {0, 1, 0, \ldots} = \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}$ for all $n \ge 1$. Hence we see that: {{begin-eqn}} {{eqn | l = \sequence {r_0, r_1, \ldots, r_n, 0, \ldots \ldots}       | r = \sequence {r_0, 0, 0, \ldots \ldots} \odot \sequence {1, 0, 0, \ldots}       | c =  }} {{eqn | ro= \oplus       | o =        | r = \sequence {r_1, 0, 0, \ldots \ldots} \odot \sequence {0, 1, 0, \ldots}       | c =  }} {{eqn | ro= \oplus       | o =        | r = \cdots       | c =  }} {{eqn | ro= \oplus       | o =        | r = \sequence {r_n, 0, 0, \ldots \ldots} \odot \sequence {0, \ldots \paren n \ldots, 0, 1, 0, \ldots}       | c =  }} {{eqn | r = r_0 \oplus r_1 \circ x \oplus r_2 \circ x^2 \oplus \ldots \oplus r_n \circ x^n       | c =  }} {{end-eqn}} So by construction, $R \sqbrk X$ is seen to be equivalent to $P \sqbrk R$. {{qed}} {{handwaving|The entirety of the above argument needs to be made properly rigorous.}} It can also be shown that this proof works for the general ring whether it be a ring with unity or not.
From Integers under Multiplication form Semigroup, $\struct {\Z, \times}$ is a countably infinite semigroup. Then we have that the Integers are Countably Infinite. The criteria for $\struct {\Z, \times}$ to be a countably infinite semigroup are seen to be satisfied.  {{Qed}} Category:Integer Multiplication Category:Examples of Semigroups ofhk6z2f8lxht1xp6vezhh6swr7te1j
Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$. $\boxminus$ is the congruence relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested. Integer addition is defined as: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$ We have that: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$ Also: :$\forall a, b, c, d \in \N: \eqclass {a, b} {} + \eqclass {c, d} {} = \eqclass {a + c, b + d} {}$ But: :$a + c \in \N, b + d \in \N$ So: :$\forall a, b, c, d \in \N: \eqclass {a + c, b + d} {} \in \Z$ Therefore integer addition is closed. {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$. $\boxtimes$ is the congruence relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested. From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the isomorphic copy of $0 \in \N$. So, we need to show that: :$\forall a, b, c \in \N: \eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ Thus: {{begin-eqn}} {{eqn | l = \eqclass {a, b} {} + \eqclass {c, c} {}       | r = \eqclass {a + c, b + c} {}       | c =  }} {{eqn | r = \eqclass {a, b} {}       | c = Construction of Inverse Completion: Members of Equivalence Classes }} {{end-eqn}} So: :$\eqclass {a, b} {} + \eqclass {c, c} {} = \eqclass {a, b} {}$ The identity $\eqclass {a, b} {} = \eqclass {c, c} {} + \eqclass {a, b} {}$ is demonstrated similarly. {{qed}}
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers. That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$. $\boxtimes$ is the congruence relation defined on $\N \times \N$ by: :$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$ In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested. From the method of construction, the element $\eqclass {a, a + x} {}$ has an inverse $\eqclass {a + x, a} {}$ where $a$ and $x$ are elements of the natural numbers $\N$. Thus: {{begin-eqn}} {{eqn | l = \eqclass {a, a + x} {} + \eqclass {a + x, a} {}       | r = \eqclass {a + a + x, a + x + a} {}       | c =  }} {{eqn | r = \eqclass {a, a} {}       | c = Construction of Inverse Completion: Members of Equivalence Classes }} {{eqn | r = \eqclass {a + x + a , a + a + x} {}       | c =  }} {{eqn | r = \eqclass {a + x, a} {} + \eqclass {a, a + x} {}       | c =  }} {{end-eqn}} So $\eqclass {a, a + x} {}$ has the inverse $\eqclass {a + x, a} {}$. {{explain|Needs more detail explaining the connection to the natural numbers. The Inverse Completion page is okay, but the link to what's going on here is hard to follow.}} {{qed}}
Recall that $\struct {\Q, +, \times}$ is a field. As a field is also by definition a division ring, which is an example of a ring, the result follows. {{qed}}
From Real Numbers under Addition form Abelian Group, $\struct {\R, +}$ is an abelian group. We also have that: :Real Multiplication is Closed: ::$\forall x, y \in \R: x \times y \in \R$ :Real Multiplication is Associative: ::$\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$ Thus $\struct {\R, +}$ is a semigroup. Finally we have that Real Multiplication Distributes over Addition: {{begin-eqn}} {{eqn | lo= \forall x, y, z \in \R:       | l = x \times \paren {y + z}       | r = x \times y + x \times z }} {{eqn | l = \paren {y + z} \times x       | r = y \times x + z \times x }} {{end-eqn}} Hence the result, by definition of ring. {{qed}}
From Complex Numbers under Addition form Abelian Group, $\struct {\C, +}$ is an abelian group. We also have that: :Complex Multiplication is Closed: ::$\forall x, y \in \C: x \times y \in \C$ :Complex Multiplication is Associative: ::$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$ Thus $\struct{\C, +}$ is a semigroup. Finally we have that Complex Multiplication Distributes over Addition: {{begin-eqn}} {{eqn | lo= \forall x, y, z \in \C:       | l = x \times \paren {y + z}       | r = x \times y + x \times z }} {{eqn | l = \paren {y + z} \times x       | r = y \times x + z \times x }} {{end-eqn}} Hence the result, by definition of ring. {{qed}}
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. From Zero of Inverse Completion of Integral Domain, for any $k \in \Z^*$, the element $\dfrac {0_D} k$ of $\Q$ serves as the zero of $\struct {\Q, +, \times}$. Hence $\dfrac 0 k$ is the identity for $\left({\Q, +}\right)$: {{begin-eqn}} {{eqn | l = \frac a b + \frac 0 k       | r = \frac {a \times k + b \times 0} {b \times k}       | c =  }} {{eqn | r = \frac {a \times k} {b \times k}       | c =  }} {{eqn | r = \frac a b       | c = Equal Elements of Field of Quotients }} {{end-eqn}} Similarly for $\dfrac 0 k + \dfrac a b$. Next we note that it ''is'' a zero: {{begin-eqn}} {{eqn | l = \frac a b \times \frac 0 k       | r = \frac {a \times 0} {b \times k}       | c =  }} {{eqn | r = \frac 0 {b \times k}       | c =  }} {{eqn | r = \frac {0 \times b} {k \times b}       | c =  }} {{eqn | r = \frac 0 k       | c = Equal Elements of Field of Quotients }} {{end-eqn}} Hence we ''define'' the zero of $\struct {\Q, +, \times}$ as $0$ and identify it with the set of all elements of $\Q$ of the form $\dfrac 0 k$ where $k \in \Z^*$. {{qed}} Category:Rational Addition 7xbsf9voemuqibzxl3t8s0tzo1ramkn
From Complex Addition Identity is Zero, the identity element for $\struct {\C, +}$ is $0 + 0 i$. Then: {{begin-eqn}} {{eqn | o =         | r = \paren {x + i y} + \paren {-x - i y}       | c =  }} {{eqn | r = \paren {x - x} + i \paren {y - y}       | c =  }} {{eqn | r = 0 + 0 i       | c =  }} {{end-eqn}} Similarly for $\paren {-x - i y} + \paren {x + i y}$. {{qed}}
We have that the field of quotients $\struct {K, +, \circ}$ of an integral domain is its inverse completion. Thus we have: :$\forall x_1, x_2 \in D, y_1, y_2 \in D^*: \dfrac {x_1} {y_1} = \dfrac {x_2} {y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$ So: {{begin-eqn}} {{eqn | l = \paren {p \circ k} \circ q       | r = p \circ \paren {q \circ k}       | c = as $D$ is an integral domain: $\circ$ is commutative and associative }} {{eqn | ll= \leadsto       | l = \frac p q       | r = \frac {p \circ k} {q \circ k}       | c =  }} {{end-eqn}} Hence the result. Note that in order for $\dfrac {p \circ k} {q \circ k}$ to be defined, $q \circ k \ne 0_D$, that is, $k \ne 0_D$. {{qed}} Category:Integral Domains Category:Inverse Completions Category:Fields of Quotients 2vqbfdk1ql793ab95ow031s3pvw1stq
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $+$ is well-defined and closed on $\Q$. {{qed}}
Let $x = \dfrac a b$ where $b \ne 0$. We take the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. From Existence of Field of Quotients, we have that the inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$: {{begin-eqn}} {{eqn | l = \frac a b + \frac {-a} b       | r = \frac {a \times b + b \times \paren {-a} } {b \times b}       | c =  }} {{eqn | r = \frac {b \times \paren {a + \paren {-a} } } {b \times b}       | c =  }} {{eqn | r = \frac {b \times 0} {b \times b}       | c =  }} {{eqn | r = \frac 0 {b \times b}       | c =  }} {{eqn | r = 0       | c = Rational Addition Identity is Zero }} {{end-eqn}} From Negative of Division Product, we have that: :$\displaystyle -\frac a b = \frac {-a} b = \frac a {-b}$ So $\dfrac a b$ has a unique and unambiguous inverse for $+$. {{qed}} Category:Rational Addition Category:Inverse Elements beycyloc3bpkkhd2qgufhmeb4pfzacf
Follows directly from the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. So $\struct {\Q, +, \times}$ is a field, and therefore a priori $\times$ is well-defined and closed on $\Q$. {{qed}}
From the definition of rational numbers, there exists four integers $p$, $q$, $r$, $s$, where: :$\displaystyle q \ne 0$ :$\displaystyle s \ne 0$ :$\displaystyle \frac p q = x$ :$\displaystyle \frac r s = y$ We have that: :$\displaystyle p \times r \in \Z$ :$\displaystyle q \times s \in \Z$ Since $q \ne 0$ and $s \ne 0$, we have that $q \times s \ne 0$. Therefore, by the definition of rational numbers: :$\displaystyle x \times y = \frac {p \times r} {q \times s} \in \Q$. Hence the result. {{qed}}
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. From the properties of the quotient structure, elements of $\Q$ of the form $\dfrac p p$ where $p \ne 0$ act as the identity for multiplication. From Equal Elements of Field of Quotients, we have that: :$\displaystyle \frac p p = \frac {1 \times p} {1 \times p} = \frac 1 1$ Hence $\dfrac p p$ is the identity for $\struct {\Q, \times}$: {{begin-eqn}} {{eqn | l = \frac a b \times \frac p p       | r = \frac {a \times p} {b \times p}       | c =  }} {{eqn | r = \frac a b       | c = Equal Elements of Field of Quotients }} {{end-eqn}} Similarly for $\dfrac p p \times \dfrac a b$. Hence we ''define'' the unity of $\struct {\Q, +, \times}$ as $1$ and identify it with the set of all elements of $\Q$ of the form $\dfrac p p$ where $ \in \Z^*$. {{qed}} Category:Rational Multiplication biysaxcjqkesm7cn7xak6e3e1dz6nqb
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers. From Rational Multiplication Identity is One, the identity for $\struct {\Q, \times}$ is $1 = \dfrac 1 1 = \dfrac p p$ where $p \in \Z$ and $p \ne 0$. From the definition of product inverses in field of quotients, we have that: :$\paren {\dfrac p q}^{-1} = \dfrac q p$ which can be demonstrated directly: {{begin-eqn}} {{eqn | l = \frac p q \times \frac q p       | r = \frac {p \times q} {q \times p}       | c =  }} {{eqn | r = \frac {p \times q} {p \times q}       | c =  }} {{eqn | r = \frac 1 1       | c = Equal Elements of Field of Quotients }} {{end-eqn}} Note that this is defined only when $p \ne 0$ and $q \ne 0$. Now let $x = \dfrac p q \in \Q$. We define $\dfrac 1 x$ as $1 \times \paren {\dfrac p q}^{-1}$. Hence: {{begin-eqn}} {{eqn | l = \frac 1 x       | r = \frac 1 1 \times \paren {\frac p q}^{-1}       | c =  }} {{eqn | r = \frac 1 1 \times \frac q p       | c =  }} {{eqn | r = \frac {1 \times q} {1 \times p}       | c =  }} {{eqn | r = \frac q p       | c =  }} {{end-eqn}} {{qed}} Category:Rational Multiplication Category:Inverse Elements 7hiyuzjr3bjoa27wg3xduwqvvmxanph
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real addition, $x + y$ is defined as $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$. Let $\sequence {0_n}$ be such that $\forall i: 0_n = 0$. Then we have: {{begin-eqn}} {{eqn | l = \eqclass {\sequence {0_n} } {} + \eqclass {\sequence {x_n} } {}       | r = \eqclass {\sequence {0_n + x_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {0 + x_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n} } {}       | c =  }} {{end-eqn}} Similarly for $\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {0_n} } {}$. Thus the identity element of $\struct {\R, +}$ is the real number $0$. {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as: :$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$ Let $\sequence {1_n}$ be such that $\forall i: 1_n = 1$. Then we have: {{begin-eqn}} {{eqn | l = \eqclass {\sequence {1_n} } {} \times \eqclass {\sequence {x_n} } {}       | r = \eqclass {\sequence {1_n \times x_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {1 \times x_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {x_n} } {}       | c =  }} {{end-eqn}} Similarly for $\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {1_n} } {}$. So the identity element of $\struct {\R_{\ne 0}, \times}$ is the real number $1$. {{qed}}
By the definition of real number: :$\forall \epsilon > 0 : \exists t \in \N : \forall i > t: \size {x_i - x} < \epsilon$ {{explain|In order for this line to make sense, it needs the definition (or at least the relevant part) extracted and posted here, so the context of the convergence condition is clear.}} Let $\epsilon = \size x$. This is possible because $x \ne 0$, since it is required that $\epsilon > 0$. {{begin-eqn}} {{eqn | ll= \exists t \in \N : \forall i > t :       | l = \size {x_i - x}       | o = <       | r = \size x       | c =  }} {{eqn | lll=\leadsto       | ll= \exists t \in \N : \forall i > t :       | l = \size {x_i - x}       | o = <       | r = \size {x_i - x} + \size {x_i}       | c = Triangle Inequality for Real Numbers }} {{eqn | lll=\leadsto       | ll= \exists t \in \N : \forall i > t :       | l = \size {x_i}       | o = >       | r = 0       | c =  }} {{end-eqn}} Construct a sequence $\sequence {y_n}$ as follows: :$y_n = \begin {cases} \dfrac 1 {x_n} & n > t \\ 0 & n \le t \end {cases}$ Let $u_n$ denote the Heaviside step function with parameter $t$. We have: {{begin-eqn}} {{eqn | l = \eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {}       | r = \eqclass {\sequence {x_n y_n} } {}       | c =  }} {{eqn | r = \eqclass {\sequence {u_n} } {}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} Similarly for $\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$. So the inverse of $x \in \struct {\R_{\ne 0}, \times}$ is $x^{-1} = \dfrac 1 x = y$. {{qed}} Category:Real Multiplication Category:Inverse Elements 5yxnk8q1yhk8t3wadof1vjkb4mirq68
{{begin-eqn}} {{eqn | l = \paren {x + i y} \frac {x - i y} {x^2 + y^2}       | r = \frac {\paren {x \cdot x - y \cdot \paren {-y} } + i \paren {x \cdot \paren {-y} + x \cdot y} } {x^2 + y^2}       | c =  }} {{eqn | r = \frac {\paren {x^2 + y^2} + 0 i} {x^2 + y^2}       | c =  }} {{eqn | r = 1 + 0 i       | c =  }} {{end-eqn}} Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$. So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$. As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$. {{qed|lemma}} From the definition, the complex conjugate $\overline z$ of $z = x + i y$ is $x - i y$. From the definition of the modulus of a complex number, we have: :$\cmod z = \sqrt {a^2 + b^2}$ From Modulus in Terms of Conjugate, we have that: :$\cmod z^2 = z \overline z$ Hence the result: :$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$ {{qed}} {{expand|Add the polar and exponential forms of this}}
From Additive Group of Integers is Normal Subgroup of Reals, $\struct {\Z, +} \lhd \struct {\R, +}$. From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$. Thus $\struct {\Z, +} \le \struct {\C, +}$. From Complex Numbers under Addition form Abelian Group, $\struct {\C, +}$ is abelian. From Subgroup of Abelian Group is Normal, it follows that $\struct {\Z, +} \lhd \struct {\C, +}$. {{qed}} Category:Additive Group of Integers Category:Additive Group of Complex Numbers Category:Examples of Normal Subgroups ikeecuhel3tpbsr0yhe0vfstw909fqw
From Additive Group of Rationals is Subgroup of Reals, $\struct {\Q, +} \lhd \struct {\R, +}$. From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$. Thus $\struct {\Q, +} \le \struct {\C, +}$. From Complex Numbers under Addition form Abelian Group, $\struct {\C, +}$ is abelian. From Subgroup of Abelian Group is Normal, it follows that $\struct {\Q, +} \lhd \struct {\C, +}$. {{qed}} Category:Additive Group of Rational Numbers Category:Additive Group of Complex Numbers Category:Normal Subgroups 4boyza8yljy6ipg9wn24xyz3fvfxgft