hcju/proofwikiqa huggingface dataset:
Suppose: :$\displaystyle x \in S \setminus \bigcup_{i \mathop \in I} T_i$ Note that by Set Difference is Subset we have that $x \in S$ (we need this later). Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \setminus \bigcup_{i \mathop \in I} T_i       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \notin       | r = \bigcup_{i \mathop \in I} T_i       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = \neg \leftparen {\exists i \in I: \leftparen x}       | o = \in       | r = \rightparen {\rightparen {T_i} }       | c = {{Defof|Union of Family}} }} {{eqn | ll= \leadstoandfrom       | l = \forall i \in I: \neg \leftparen x       | o = \in       | r = \rightparen {T_i}       | c = De Morgan's Laws (Predicate Logic) }} {{eqn | ll= \leadstoandfrom       | l = \forall i \in I: x       | o = \in       | r = S \setminus T_i       | c = {{Defof|Set Difference}}: note $x \in S$ from above }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}       | c = {{Defof|Intersection of Family}} }} {{end-eqn}} Therefore: :$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \complement \paren {\bigcap \mathbb T}       | r = \mathbb U \setminus \paren {\bigcap \mathbb T}       | c = {{Defof|Set Complement}} }} {{eqn | r = \bigcup_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \bigcup_{H \mathop \in \mathbb T} \complement \paren H       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | l = \complement \paren {\bigcup \mathbb T}       | r = \mathbb U \setminus \paren {\bigcup \mathbb T}       | c = {{Defof|Set Complement}} }} {{eqn | r = \bigcap_{H \mathop \in \mathbb T} \paren {\mathbb U \setminus H}       | c = De Morgan's Laws for Set Difference: Difference with Union }} {{eqn | r = \bigcap_{H \mathop \in \mathbb T} \complement \paren H       | c = {{Defof|Set Complement}} }} {{end-eqn}} {{qed}} {{Namedfor|Augustus De Morgan}}
Let the cardinality $\card I$ of the indexing set $I$ be $n$. Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition: :$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ as: :$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$ The proof of this is more amenable to proof by Principle of Mathematical Induction. For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$ $\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$. === Base Case === $\map P 2$ is the case: :$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$ which has been proved.  This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle S \setminus \bigcap_{i \mathop = 1}^k T_i = \bigcup_{i \mathop = 1}^k \paren {S \setminus T_i}$ === Induction Step === Now we need to show: :$\displaystyle S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i = \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$ This is our induction step: {{begin-eqn}} {{eqn | l = S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i       | r = S \setminus \paren {\bigcap_{i \mathop = 1}^k T_i \cap T_{k + 1} }       | c = Intersection is Associative }} {{eqn | r = \paren {S \setminus \bigcap_{i \mathop = 1}^k T_i} \cup \paren {S \setminus T_{k + 1} }       | c = Base case }} {{eqn | r = \paren {\bigcup_{i \mathop = 1}^k \paren {S \setminus T_i} } \cup \paren {S \setminus T_{k + 1} }       | c = Induction hypothesis }} {{eqn | r = \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}       | c = Union is Associative }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$ that is: :$\displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$ {{qed}}
Let the cardinality $\size I$ of the indexing set $I$ be $n$. Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition: :$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ as: :$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$ The proof of this is more amenable to proof by Principle of Mathematical Induction. For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$. $\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$. === Base Case === $\map P 2$ is the case: :$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$ which has been proved. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle S \setminus \bigcup_{i \mathop = 1}^k T_i = \bigcap_{i \mathop = 1}^k \paren {S \setminus T_i}$ === Induction Step === Now we need to show: :$\displaystyle S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$ This is our induction step: {{begin-eqn}} {{eqn | l = S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i       | r = S \setminus \paren {\bigcup_{i \mathop = 1}^k T_i \cup T_{k + 1} }       | c = Union is Associative }} {{eqn | r = \paren {S \setminus \bigcup_{i \mathop = 1}^k T_i} \cap \paren {S \setminus T_{k + 1} }       | c = Base case }} {{eqn | r = \paren {\bigcap_{i \mathop = 1}^k \paren {S \setminus T_i} } \cap \paren {S \setminus T_{k + 1} }       | c = Induction hypothesis }} {{eqn | r = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}       | c = Intersection is Associative }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$ that is: :$\displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$ {{qed}}
Suppose: :$\displaystyle x \in S \setminus \bigcup \mathbb T$ Note that by Set Difference is Subset we have that $x \in S$ (we need this later). Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \setminus \bigcup \mathbb T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \notin       | r = \bigcup \mathbb T       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = \neg (\exists T' \in  \mathbb T: (x       | o = \in       | r = T'))       | c = {{Defof|Union of Set of Sets}} }} {{eqn | ll= \leadstoandfrom       | l = \forall T' \in  \mathbb T: \neg (x       | o = \in       | r = T')       | c = De Morgan's Laws (Predicate Logic) }} {{eqn | ll= \leadstoandfrom       | l = \forall T' \in  \mathbb T: x       | o = \in       | r = S \setminus T'       | c = {{Defof|Set Difference}}: note $x \in S$ from above }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}       | c = {{Defof|Intersection of Set of Sets}} }} {{end-eqn}} Therefore: : $\displaystyle S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in S \setminus \paren {T_1 \cap T_2} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cap T_2} }       | c = {{Defof|Set Difference}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 \land x \in T_2} }       | c = {{Defof|Set Intersection}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {x \notin T_1 \lor x \notin T_2}       | c = De Morgan's Laws: Disjunction of Negations }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S \land x \notin T_1}) \lor \paren {x \in S \land x \notin T_2}       | c = Rule of Distribution }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {S \setminus T_1} \cup \paren {S \setminus T_2}       | c = {{Defof|Set Union}} and {{Defof|Set Difference}} }} {{end-eqn}} By definition of set equality: : $S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$ {{qed}} {{Namedfor|Augustus De Morgan}}
{{begin-eqn}} {{eqn | o =        | r = x \in S \setminus \paren {T_1 \cup T_2} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {x \notin \paren {T_1 \cup T_2} }       | c = {{Defof|Set Difference}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {\neg \paren {x \in T_1 \lor x \in T_2} }       | c = {{Defof|Set Union}} }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S} \land \paren {x \notin T_1 \land x \notin T_2}       | c = De Morgan's Laws: Conjunction of Negations }} {{eqn | o = \leadstoandfrom       | r = \paren {x \in S \land x \notin T_1} \land \paren {x \in S \land x \notin T_2}       | c = Rules of Idempotence, Commutation and Association }} {{eqn | o = \leadstoandfrom       | r = x \in \paren {S \setminus T_1} \cap \paren {S \setminus T_2}       | c = {{Defof|Set Intersection}} and {{Defof|Set Difference}} }} {{end-eqn}} By definition of set equality: :$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$ {{qed}} {{Namedfor|Augustus De Morgan}}
To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$. Let $f$ be the mapping from the parity ring $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the ring of integers modulo $2$ $\struct {\Z_2, +_2, \times_2}$: :$f: \struct {\set {\text{even}, \text{odd} }, +, \times} \to \struct {\Z_2, +_2, \times_2}$: ::$\forall x \in R: \map f x = \begin{cases} 0 & : x = \text{even} \\ 1 & : x = \text{odd}  \end{cases}$ The bijective nature of $f$ is apparent: :$f^{-1}: \struct {\Z_2, +_2, \times_2} \to \struct {\set {\text{even}, \text{odd} }, +, \times}$: ::$\forall x \in \Z_2: \map {f^{-1} } x = \begin{cases} \text{even} & : x = 0 \\ \text{odd}  & : x = 1 \end{cases}$ Thus the following equations can be checked: {{begin-eqn}} {{eqn | ll= 0 +_2 0 =       | l = \map f {\text{even} } +_2 \map f {\text{even} }       | r = \map f {\text{even} + \text{even} }       | rr= = 0 }} {{eqn | ll= 0 +_2 1 =       | l = \map f {\text{even} } +_2 \map f {\text{odd} }       | r = \map f {\text{even} + \text{odd} }       | rr= = 1 }} {{eqn | ll= 1 +_2 0 =       | l = \map f {\text{odd} } +_2 \map f {\text{even} }       | r = \map f {\text{odd} + \text{even} }       | rr= = 1 }} {{eqn | ll= 1 +_2 1 =       | l = \map f {\text{odd} } +_2 \map f {\text{odd} }       | r = \map f {\text{odd} + \text{odd} }       | rr= = 0 }} {{end-eqn}} and: {{begin-eqn}} {{eqn | ll= 0 \times_2 0 =       | l = \map f {\text{even} } \times_2 \map f {\text{even} }       | r = \map f {\text{even} \times \text{even} }       | rr= = 0 }} {{eqn | ll= 0 \times_2 1 =       | l = \map f {\text{even} } \times_2 \map f {\text{odd} }       | r = \map f {\text{even} \times \text{odd} }       | rr= = 0 }} {{eqn | ll= 1 \times_2 0 =       | l = \map f {\text{odd} } \times_2 \map f {\text{even} }       | r = \map f {\text{odd} \times \text{even} }       | rr= = 0 }} {{eqn | ll= 1 \times_2 1 =       | l = \map f {\text{odd} } \times_2 \map f {\text{odd} }       | r = \map f {\text{odd} \times \text{odd} }       | rr= = 1 }} {{end-eqn}} {{qed}} These results can be determined from their Cayley tables: === Cayley Tables for Parity Ring === {{:Parity Ring/Cayley Tables}} === Cayley Tables for $\Z_2$ === {{:Ring of Integers Modulo 2/Cayley Tables}}
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring: :$\struct {\set {\text {even}, \text {odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$ the ring of integers modulo $2$. The result follows from: :Modulo Addition is Associative and: :Isomorphism Preserves Associativity. {{qed}}
Let $a, b, c \in R$. That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$. By definition of odd: :$\text{odd} = 2 m + 1$ for some $m \in \Z$. By definition of even: :$\text{even} = 2 n + 0$ for some $n \in \Z$. Thus we can define the mapping $f: R \to \Z$ as: :$\forall x \in R: f \left({x}\right) := \begin{cases} 0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$ Thus an element of $R$ can be expressed as an arbitrary integer of the form: :$x = 2 k + f \left({x}\right)$ where: :$k \in \Z$ is an integer :$f \left({x}\right)$ is either $0$ or $1$ according to whether $x$ is even or odd. Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities. Then: {{begin-eqn}} {{eqn | l = \left({a +_2 b}\right) +_2 c       | r = \left({2 r + f \left({a}\right) + 2 s + f \left({b}\right)}\right) + 2 t + f \left({c}\right)       | c = where $r, s, t \in \Z$ }} {{eqn | r = 2 r + f \left({a}\right) + \left({2 s + f \left({b}\right) + 2 t + f \left({c}\right)}\right)       | c = Integer Addition is Associative }} {{eqn | r = a +_2 \left({b +_2 c}\right)       | c = Definition of odd and even }} {{end-eqn}} The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$: :$\left({a + b}\right) + c = a + \left({b + c}\right)$ {{qed}}
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring: :$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$ the ring of integers modulo $2$. The result follows from: :Modulo Addition is Commutative and: :Isomorphism Preserves Commutativity. {{qed}}
Let $a, b \in R$. That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$. By definition of odd: :$\text{odd} = 2 m + 1$ for some $m \in \Z$. By definition of even: :$\text{even} = 2 n + 0$ for some $n \in \Z$. Thus we can define the mapping $f: R \to \Z$ as: :$\forall x \in R: \map f x := \begin{cases} 0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$ Thus an element of $R$ can be expressed as an arbitrary integer of the form: :$x = 2 k + \map f x$ where: :$k \in \Z$ is an integer :$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd. Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities. Then: {{begin-eqn}} {{eqn | l = a +_2 b       | r = 2 r + \map f a + 2 s + \map f b       | c = where $r, s \in \Z$ }} {{eqn | r = 2 s + \map f b + 2 r + \map f a       | c = Integer Addition is Commutative }} {{eqn | r = b +_2 a       | c = Definition of odd and even }} {{end-eqn}} The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$: :$a + b = b + a$ {{qed}}
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring: :$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$ the ring of integers modulo $2$. The result follows from: :Modulo Multiplication is Associative and: :Isomorphism Preserves Associativity. {{qed}}
Let $a, b, c \in R$. That is, $a, b, c$ are all either $\text{even}$ or $\text{odd}$. By definition of odd: :$\text{odd} = 2 m + 1$ for some $m \in \Z$. By definition of even: :$\text{even} = 2 n + 0$ for some $n \in \Z$. Thus we can define the mapping $f: R \to \Z$ as: :$\forall x \in R: \map f x := \begin{cases} 0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$ Thus an element of $R$ can be expressed as an arbitrary integer of the form: :$x = 2 k + \map f x$ where: :$k \in \Z$ is an integer :$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd. Then: {{begin-eqn}} {{eqn | l = \paren {a \times b} \times c       | r = \paren {\paren {2 r + \map f a} \paren {2 s + \map f b} } \paren {2 t + \map f c}       | c = where $r, s, t \in \Z$ }} {{eqn | r = \paren {2 r + \map f a} \paren {\paren {2 s + \map f b} \paren {2 t + \map f c} }       | c = Integer Multiplication is Associative }} {{eqn | r = a \times \paren {b \times c}       | c = Definition of odd and even }} {{end-eqn}} {{qed}}
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring: :$\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$ the ring of integers modulo $2$. The result follows from: :Modulo Multiplication is Commutative and: :Isomorphism Preserves Associativity. {{qed}}
Let $a, b \in R$. That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$. By definition of odd: :$\text{odd} = 2 m + 1$ for some $m \in \Z$. By definition of even: :$\text{even} = 2 n + 0$ for some $n \in \Z$. Thus we can define the mapping $f: R \to \Z$ as: :$\forall x \in R: \map f x := \begin{cases} 0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$ Thus an element of $R$ can be expressed as an arbitrary integer of the form: :$x = 2 k + \map f x$ where: :$k \in \Z$ is an integer :$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd. Then: {{begin-eqn}} {{eqn | l = a \times b       | r = \paren {2 r + \map f a} \paren {2 s + \map f b}       | c = where $r, s \in \Z$ }} {{eqn | r = \paren {2 s + \map f b} \paren {2 r + \map f a}       | c = Integer Multiplication is Commutative }} {{eqn | r = b \times a       | c = Definition of odd and even }} {{end-eqn}} {{qed}}
The set of integers modulo $n$ is the set exemplified by the integers: :$\Z_n = \set {0, 1, \ldots, n - 1}$ The complex $n$th roots of unity is the set: :$R_n = \set {z \in \C: z^n = 1}$ From Complex Roots of Unity in Exponential Form: :$R_n = \set {1, e^{\theta / n}, e^{2 \theta / n}, \ldots, e^{\left({n - 1}\right) \theta / n} }$ where $\theta = 2 i \pi$. Let $z, w, \in R_n$. Then: : $\paren {z w}^n = z^n w^n = 1$ and so $z w \in R_n$. Thus $\struct {R_n, \times}$ is a closed algebraic structure. Consider the mapping $f: \Z_n \to R_n$ defined as: :$\forall r \in \Z_n: \map f r = e^{r \theta / n}$ which can be seen to be a bijection by inspection. Let $j, k \in \Z_n$. Then: {{begin-eqn}} {{eqn | l = \map f j \map f k       | r = e^{j \theta / n} e^{k \theta / n}       | c =  }} {{eqn | r = e^{j \theta / n + k \theta / n}       | c =  }} {{eqn | r = e^{\paren {j + k} \theta / n}       | c =  }} {{eqn | r = \map f {j +_n k}       | c =  }} {{end-eqn}} Thus $f$ is an isomorphism. {{qed}}
Let the mapping $f: \Z_4 \to U_\C$ be defined as: {{begin-eqn}} {{eqn | l = \map f 0       | r = 1 }} {{eqn | l = \map f 1       | r = i }} {{eqn | l = \map f 2       | r = -1 }} {{eqn | l = \map f 3       | r = -i }} {{end-eqn}} From Isomorphism by Cayley Table, the two Cayley tables can be compared by eye to ascertain that $f$ is an isomorphism: === Cayley Table of Integers Modulo $4$ === The Cayley table for $\struct {\Z_4, +_4}$ is as follows: {{:Modulo Addition/Cayley Table/Modulo 4}} === Cayley Table of Gaussian Integer Units === The Cayley table for $\struct {U_\C, \times}$ is as follows: {{:Group of Gaussian Integer Units/Cayley Table}}
From Gaussian Integer Units are 4th Roots of Unity: :$U_\C$ is the set consisting of the (complex) $4$th roots of $1$. The result follows from Isomorphism between Roots of Unity under Multiplication and Integers under Modulo Addition.
We have that $i = \sqrt {-1}$ is the imaginary unit. Thus: {{begin-eqn}} {{eqn | l = 1^4       | o =        | rr= = 1 }} {{eqn | l = i^4       | r = \paren {-1}^2       | rr= = 1 }} {{eqn | l = \paren {-1}^4       | r = 1^2       | rr= = 1 }} {{eqn | l = \paren {-i}^4       | r = \paren {-1}^2 \cdot \paren {-1}^2       | rr= = 1 }} {{end-eqn}} So $\set {1, i, -1, -i}$ constitutes the set of the $4$th roots of unity. {{qed}} Category:Gaussian Integers Category:Complex Roots of Unity pwj7xje8bvnpqyul3894062zu76x0j9
=== Necessary Condition === Let $S$ and $T$ be isomorphic. Then by definition there exists an isomorphism $f: S \to T$. An isomorphism is a bijection by definition. Thus the existence of the posited bijection has been demonstrated. By the definition of set equivalence, $S$ and $T$ have the same cardinality. Let $\card S = \card T = n$. Let: :$S = \set {s_1, s_2, \ldots, s_n}$ and: :$T = \set {t_1, t_2, \ldots, t_n}$ be such that: :$\forall i \in \set {1, 2, \ldots, n}: \map f {s_i} = t_i$ Let $S \sqbrk {j, k}$ denote the entry of $S$ in row $j$ and column $k$. Thus: :$S \sqbrk {j, k} = s_j \circ s_k$ for some $s_j, s_k \in S$. We have that $f$ is an isomorphism. Thus: :$\map f {S \sqbrk {j, k} } = \map f {s_j} * \map f {s_k}$ But: :$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$ :$\map f {s_j} = t_j$ :$\map f {s_k} = t_k$ Thus the Cayley table of $\struct {T, *}$ has been generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$. {{qed|lemma}} === Sufficient Condition === Let there exist a bijection $f: S \to T$ such that the Cayley table of $\struct {T, *}$ can be generated from the Cayley table of $\struct {S, \circ}$ by replacing each entry of $S$ with its image under $f$. By the definition of set equivalence, $S$ and $T$ have the same cardinality. Let $\card S = \card T = n$. Let the rows and columns of the Cayley table of $\left({S, \circ}\right)$ both be numbered from $1$ to $n$. Let the entries heading the rows and columns be denoted $s_1, s_2, \ldots, s_n$. Let $S \sqbrk {j, k}$ denote the entry of $S$ in row $j$ and column $k$. By the method of construction: :$S \sqbrk {j, k} = s_j \circ s_k$ for some $s_j, s_k \in S$. Let $f$ be used to generate the Cayley table of $\struct {T, *}$ from the Cayley table of $\struct {S, \circ}$. Then for all $i, j \in \set {1, 2, \ldots, n}$: :$\map f {S \sqbrk {j, k} } = T \sqbrk {j, k}$ :$\map f {s_j} = t_j$ :$\map f {s_k} = t_k$ But by definition of the Cayley table of $\struct {T, *}$: :$T \sqbrk {j, k} = t_j * t_k$ Thus: :$\forall j, k \in \set {1, 2, \ldots, n}: \map f {s_j \circ s_k} = \map f {s_j} * \map f {s_k}$ demonstrating that $f$ is an isomorphism. {{qed}} Category:Cayley Tables Category:Isomorphisms pr241m4nkfod3fojbk28b02aad0395f
Let $f: \Z \to 2 \Z$ be the mapping defined as: :$\forall n \in \Z: \map f n = 2 n$ Let $m, n \in \Z$ such that $\map f m = \map f n$. {{begin-eqn}} {{eqn | l = \map f m       | r = \map f n }} {{eqn | ll= \leadsto       | l = 2 m       | r = 2 n       | c = Definition of $f$ }} {{eqn | ll= \leadsto       | l = m       | r = n       | c = dividing both sides by $2$ }} {{end-eqn}} Thus $f$ is an injection. Let $n \in 2 Z$. Then by definition: :$n = 2 r$ for some $r \in \Z$ That is: :$n = \map f r$ and so $f$ is a surjection. So $f$ is both an injection and a surjection. Hence by definition, $f$ is a bijection. {{qed}} Category:Examples of Bijections Category:Integers Category:Even Integers 1guf8fayzhy4b82sb3b08kplc5wrbl6
Let $f: \Z \to 2 \Z$ be the mapping: :$\forall n \in \Z: \map f n = 2 n$ From Bijection between Integers and Even Integers, $f$ is a bijection. Let $m, n \in \Z$. Then: {{begin-eqn}} {{eqn | l = \map f {m + n}       | r = 2 \paren {m + n}       | c = Definition of $f$ }} {{eqn | r = 2 m + 2 n       | c = Integer Multiplication Distributes over Addition }} {{eqn | r = \map f m + \map f n       | c = Integer Multiplication Distributes over Addition }} {{end-eqn}} Thus $f$ is an isomorphism by definition. {{qed}}
From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ has an identity element $0$. However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$. Thus the general element of $\struct {\N, +}$ has no inverse. Hence the result by definition of group. {{qed}}