hcju/proofwikiqa huggingface dataset:
Let: :$z_1 = x_1 + i y_1$ :$z_2 = x_2 + i y_2$ :$z_3 = x_3 + i y_3$ Then: {{begin-eqn}} {{eqn | l = z_1 \times \paren {z_2 + z_3}       | r = \paren {x_1 + i y_1} \times \paren {\paren {x_2 + i y_2} + \paren {x_3 + i y_3} }       | c =  }} {{eqn | r = \paren {x_1 + i y_1} \times \paren {\paren {x_2 + x_3} + i \paren {y_2 + y_3} }       | c = {{Defof|Complex Addition}} }} {{eqn | r = x_1 \left({y_2 + y_3}\right) - y_1 \left({x_2 + x_3}\right)       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = x_1 y_2 + x_1 y_3 - y_1 x_2 - y_1 x_3       | c = Real Multiplication Distributes over Addition }} {{eqn | r = x_1 y_2 - y_1 x_2 + x_1 y_3 - y_1 x_3       | c = Real Addition is Commutative }} {{eqn | r = z_1 \times z_2 + z_1 \times z_3       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =       | c = $S$ is closed }} {{eqn | o = \leadsto       | c = $\forall z \in \C: z$ is a limit point of $S \implies z \in S$       | cc= {{Defof|Closed Set (Complex Analysis)|Closed Set}}  }} {{eqn | o = \leadsto       | c = $\forall z \in \C: z \notin S \implies z$ is not a limit point of $S$       | cc= Rule of Transposition }} {{eqn | o = \leadsto       | c = $\forall z \in \C \setminus S \implies z$ is not a limit point of $S$       | cc= {{Defof|Relative Complement}} }} {{eqn | o = \leadsto       | c = $\forall z \in \C \setminus S:$ there exists a deleted $\epsilon$-neighborhood $\map {N_\epsilon} z \setminus \set z$ of $z$ entirely in $S$       | cc= {{Defof|Limit Point (Complex Analysis)|Limit Point}} }} {{eqn | o = \leadsto       | c = $\C \setminus S$ is open in $\C$       | cc= {{Defof|Open Set (Complex Analysis)|Open Set}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =       | c = $S$ is open in $\C$       | cc= }} {{eqn | o = \leadsto       | c = $\forall z \in S$: there exists a deleted $\epsilon$-neighborhood $\map {N_\epsilon} z \setminus \set z$ of $z$ entirely in $S$       | cc= {{Defof|Open Set (Complex Analysis)|Open Set}} }} {{eqn | o = \leadsto       | c = $\forall z \in \C: z \in S \implies z$ is not a limit point of $S$       | cc= {{Defof|Limit Point (Complex Analysis)|Limit Point}} }} {{eqn | o = \leadsto       | c = $\forall z \in \C: z$ is a limit point of $S \implies z \notin S$       | cc= Rule of Transposition }} {{eqn | o = \leadsto       | c = $\forall z \in \C: z$ is a limit point of $S \implies z \in \C \setminus S$       | cc= {{Defof|Relative Complement}} }} {{eqn | o = \leadsto       | c = $\C \setminus S$ is closed in $\C$       | cc= {{Defof|Closed Set (Complex Analysis)|Closed Set}} }} {{end-eqn}} {{qed}}
Let $\displaystyle z \in \bigcup_{k \mathop = 1}^n S_k$. Then by definition of finite union: :$\exists k \in \set {1, 2, \ldots, n}: z \in S_k$ By definition of open set: :$\exists \epsilon \in \R_{>0}: \map {N_\epsilon} z \subseteq S_k$ where $\map {N_\epsilon} z$ is the $\epsilon$-neighborhood of $z$ for $\epsilon$. By Set is Subset of Union: :$S_k \subseteq S$ Thus by Subset Relation is Transitive: :$\map {N_\epsilon} z \subseteq S$ As $z$ is any arbitrary element of $S$, it follows that $S$ is an open set of $\C$ by definition. {{qed}}
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$. Suppose $z$ is not a boundary point of $S$. Then there exists an $\epsilon$-neighborhood $\map {N_\epsilon} z$ of $z$ such that either: :$(1): \quad$ All elements of $\map {N_\epsilon} z$ are in $S$ or :$(2): \quad$ All elements of $\map {N_\epsilon} z$ are not in $S$. If $(1)$, then $z \in S$ which is a contradicts $z \notin S$. If $(2)$, then $z$ is not a limit point of $S$, which also contradicts the conditions on $z$. {{qed}}
The two directions of this theorem are respectively addressed on: :Soundness Theorem for Semantic Tableaus :Completeness Theorem for Semantic Tableaus {{qed}}
Let $\mathbf A$ be a WFF of propositional logic. The Semantic Tableau Algorithm applied to $\neg \mathbf A$ yields a completed tableau for $\neg \mathbf A$. By Corollary 2 to Soundness and Completeness of Semantic Tableaus, this completed tableau decides if $\mathbf A$ is a tautology. {{qed}}
We will prove inductively the following claim for every node $t$ of $T$: :If all leaves that are descendants of $t$ are marked closed, then $U \left({t}\right)$ is unsatisfiable. By the Semantic Tableau Algorithm, we know this statement to hold for the leaf nodes themselves. For, a leaf $t$ is marked closed iff $U \left({t}\right)$ contains a complementary pair. The assertion follows from Set of Literals Satisfiable iff No Complementary Pairs. Inductively, suppose that all children of a node $t$ satisfy the mentioned condition. If all descendant leaf nodes of $t$ are marked closed, this evidently holds for the children $t',t''$ of $t$ as well. Hence by hypothesis, $U \left({t'}\right)$ and $U \left({t''}\right)$ are unsatisfiable. Let $\mathbf B$ be the WFF used by the Semantic Tableau Algorithm at $t$. Let $\mathbf B_1, \mathbf B_2$ be the formulas added to $t'$ and $t''$. First, the case that $\mathbf B$ is an $\alpha$-formula. Then $t' = t''$, and $\mathbf B$ is semantically equivalent to $\mathbf B_1 \land \mathbf B_2$. It follows that if: :$v \models_{\mathrm{BI}} U \left({t}\right)$ for some boolean interpretation $v$, then also: :$v \models_{\mathrm{BI}} U \left({t'}\right)$ which contradicts our hypothesis. Thus, $U \left({t}\right)$ is unsatisfiable. Next, the case that $\mathbf B$ is a $\beta$-formula. Then $\mathbf B$ is semantically equivalent to $\mathbf B_1 \lor \mathbf B_2$. It follows that if: :$v \models_{\mathrm{BI}} U \left({t}\right)$ for some boolean interpretation $v$, then also one of the following must hold: :$v \models_{\mathrm{BI}} U \left({t'}\right)$ :$v \models_{\mathrm{BI}} U \left({t''}\right)$ which contradicts our hypothesis. Thus, $U \left({t}\right)$ is unsatisfiable. {{handwaving|"It follows", is obvious, and is tedious to write down}} This proves our claim: :If all leaves that are descendants of $t$ are marked closed, then $U \left({t}\right)$ is unsatisfiable. Applying this claim to the root node of $T$, we obtain the desired result. {{qed}}
The proof proceeds by contradiction. That is, we assume that some open tableau $T$ for $\mathbf A$ exists, and prove that $\mathbf A$ must be satisfiable. This claim is proved by inductively establishing the following claim for all nodes $t$ of $T$: :If a descendant leaf node of $t$ is marked open, then $U \left({t}\right)$ is satisfiable. By the Semantic Tableau Algorithm, this statement is immediate for leaf nodes. For, sought satisfiability is the condition for a leaf to be marked open. Inductively suppose that all children of $t$ satisfy the mentioned condition. Now suppose that $t$ has a descendant marked open leaf. Then obviously some child $t'$ of $t$ also has a descendant marked open leaf. By induction hypothesis, $U \left({t'}\right)$ is satisfiable. First, the case that the formula $\mathbf B$ used at $t$ by the Semantic Tableau Algorithm is an $\alpha$-formula. Then $U \left({t'}\right)$ is obtained from $U \left({t}\right)$ by replacing $\mathbf B$ by $\mathbf B_1$ and $\mathbf B_2$. Here, $\mathbf B_1$ and $\mathbf B_2$ are the formulas such that $\mathbf B$ is semantically equivalent to $\mathbf B_1 \land \mathbf B_2$. It follows that if: :$v \models_{\mathrm{BI}} U \left({t'}\right)$ for some boolean interpretation $v$, then also: :$v \models_{\mathrm{BI}} U \left({t}\right)$ by the boolean interpretation of $\land$. Consequently, $U \left({t}\right)$ is satisfiable. Next, the case that $\mathbf B$ is a $\beta$-formula. Then $\mathbf B$ is semantically equivalent to $\mathbf B_1 \lor \mathbf B_2$. Then, by the Rule of Addition: :$\mathbf B_1 \models_{\mathrm{BI}} \mathbf B$ :$\mathbf B_2 \models_{\mathrm{BI}} \mathbf B$ Consequently, by the definitions of $U \left({t'}\right)$ and $U \left({t''}\right)$: :$U \left({t'}\right) \models_{\mathrm{BI}} U \left({t}\right)$ :$U \left({t''}\right) \models_{\mathrm{BI}} U \left({t}\right)$ By hypothesis, one of $U \left({t'}\right)$ and $U \left({t''}\right)$ must be satisfiable. Hence, so must $U \left({t}\right)$ be. This establishes our claim. Applying the claim to the root node, i.e., to the whole tableau, we obtain the result. {{qed}}
Let $\map P z$ be expressed as: :$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ Then: {{begin-eqn}} {{eqn | l = \overline {\map P z}       | r = \overline {a_n z^n + a_{n-1} z^{n - 1} + \cdots + a_1 z + a_0}       | c =  }} {{eqn | r = \overline {a_n z^n} + \overline {a_{n - 1} z^{n - 1} } + \cdots + \overline {a_1 z} + \overline {a_0}       | c = Sum of Complex Conjugates }} {{eqn | r = \overline {a_n} \overline {z^n} + \overline {a_{n - 1} } \overline {z^{n - 1} } + \cdots + \overline {a_1} \overline z + \overline {a_0}       | c = Product of Complex Conjugates }} {{eqn | r = a_n \overline {z^n} + a_{n - 1} \overline {z^{n - 1} } + \cdots + a_1 \overline z + a_0       | c = Complex Number equals Conjugate iff Wholly Real }} {{eqn | r = a_n \overline z^n + a_{n - 1} \overline z^{n - 1} + \cdots + a_1 \overline z + a_0       | c = Product of Complex Conjugates/General Result }} {{eqn | r = \map P {\overline z}       | c =  }} {{end-eqn}} {{qed}}
Let $z = x + i y \in \C$ be an arbitrary complex number. {{AimForCont}} the contrary: {{begin-eqn}} {{eqn | l = \size x + \size y       | o = >       | r = \sqrt 2 \cmod z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\size x + \size y}^2       | o = >       | r = 2 \cmod z^2       | c = squaring both sides }} {{eqn | ll= \leadsto       | l = \size x^2 + 2 \size x \, \size y + \size y^2       | o = >       | r = 2 \cmod z^2       | c = multiplying out }} {{eqn | ll= \leadsto       | l = x^2 + 2 \size x \, \size y + y^2       | o = >       | r = 2 \cmod z^2       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = x^2 + 2 \size x \, \size y + y^2       | o = >       | r = 2 \paren {x^2 + y^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = 2 \size x \, \size y       | o = >       | r = x^2 + y^2       | c =  }} {{eqn | ll= \leadsto       | l = 2 \size x \, \size y       | o = >       | r = \size x^2 + \size y^2       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = \size x^2 - 2 \size x \, \size y + \size y^2       | o = <       | r = 0       | c = rearranging }} {{eqn | ll= \leadsto       | l = \paren {\size x - \size y}^2       | o = <       | r = 0       | c = factoring }} {{end-eqn}} But as $\size x$ and $\size y$ are both real this cannot happen. Thus our initial assumption $\size x + \size y > \sqrt 2 \cmod z$ is false. Hence the result. {{qed}}
{{begin-eqn}} {{eqn|l = \left({p \oplus q}\right) \oplus q      |o = \dashv \vdash      |r = p \oplus \left({q \oplus q}\right)      |c = Exclusive Or is Associative }} {{eqn|o = \dashv \vdash      |r = p \oplus \bot      |c = Exclusive Or with Itself }} {{eqn|o = \dashv \vdash      |r = p      |c = Exclusive Or with Contradiction }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \paren {p \iff q} \iff q       | o = \dashv \vdash       | r = p \iff \paren {q \iff q}       | c = Biconditional is Associative }} {{eqn | o = \dashv \vdash       | r = p \iff \top       | c = Biconditional with Itself }} {{eqn | o = \dashv \vdash       | r = p       | c = Biconditional with Tautology }} {{end-eqn}} {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the appropriate truth values match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \implies & q & p & \iff & (p & \land & q) \\ \hline F & T & F & F & T & F & F & F \\ F & T & T & F & T & F & F & T \\ T & F & F & T & F & T & F & F \\ T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the appropriate truth values match for all boolean interpretations. $\begin{array}{|ccc||ccccc|} \hline p & \implies & q & q & \iff & (p & \lor & q) \\ \hline F & T & F & F & T & F & F & F \\ F & T & T & T & T & F & T & T \\ T & F & F & F & F & T & T & F \\ T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
We apply the Method of Truth Tables. As can be seen by inspection, the appropriate truth values match for all boolean interpretations. $\begin{array}{|ccc||ccccccc|} \hline p & \land & q & (p & \iff & q) & \iff & (p & \lor & q) \\ \hline F & F & F & F & T & F & F & F & F & F \\ F & F & T & F & F & T & F & F & T & T \\ T & F & F & T & F & F & F & T & T & F \\ T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$ {{qed}}
Let $v_T$ be the boolean interpretation that assigns $T$ to each propositional symbol. Then it follows by the nature of the truth functions for $\land$ and $\lor$ that: :$\map {v_T} {\mathbf A} = T$ for each WFF $\mathbf A$ comprising only $\land$ and $\lor$. On the other hand: :$\map {v_T} {\neg p} = F$ Therefore, $\neg p$ cannot be expressed in terms of $\land$ and $\lor$. Hence, $\set {\land, \lor}$ is not functionally complete. {{qed}}
Let $\mathbf A$ be a WFF comprising only the logical connectives $\neg$ and $\iff$. The central claim to this proof is that: :The number of models for $\mathbf A$ is even. In this claim, we disregard the obvious exceptions of the form $\neg \cdots \neg p$. Firstly, note that including any propositional symbols not occurring in $\mathbf A$ in the truth table for $\mathbf A$ simply doubles the number of boolean interpretations modelling $\mathbf A$. Next, it is obvious that if the number of models of $\mathbf A$ is even, then this also holds for $\neg \mathbf A$. For, by Count of Truth Functions, the total number of boolean interpretations is even. Lastly, to consider $\mathbf A = \mathbf A_1 \iff \mathbf A_2$. Suppose that the claim were verified for $\mathbf A_1$ and $\mathbf A_2$. Then we have the following identities: :$\set {v \mid \map v {\mathbf A_1} = T} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T}$ :$\set {v \mid \map v {\mathbf A_1} = F} = \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = T}$ :$\set {v \mid \map v {\mathbf A_2} = T} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = T}$ :$\set {v \mid \map v {\mathbf A_2} = F} = \set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = F} \cup \set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F}$ The sets on the {{RHS}} are disjoint, while the sets on the {{LHS}} have an even number of elements. Therefore, the parity of the cardinalities of the sets on the {{RHS}} must be equal. Then combining the first and fourth equations, the sets: :$\set {v \mid \map v {\mathbf A_1} = T, \map v {\mathbf A_2} = T}$ :$\set {v \mid \map v {\mathbf A_1} = F, \map v {\mathbf A_2} = F}$  have equal parity. Together, these two constitute the set of models for $\mathbf A$, as can be seen from the boolean interpretation of $\iff$. Therefore, the number of models for $\mathbf A$ is even. This completes the proof of the claim. On the other hand, the truth table of $p \implies q$ has three entries $F$, and one $T$. Therefore, $p \implies q$ cannot be expressed in terms of $\neg$ and $\iff$. Hence, $\set {\neg, \iff}$ is not functionally complete. {{qed}}
Let $\phi$ be a $\mathcal L$-formula. Suppose that $T \left({\mathcal F}\right) \models_{\mathscr M} \phi$. By definition of $T \left({\mathcal F}\right)$: :$\mathcal F \models_{\mathscr M} \psi$ for every $\psi \in T \left({\mathcal F}\right)$. Hence by definition of semantic consequence: :$\mathcal F \models_{\mathscr M} T \left({\mathcal F}\right)$ By Semantic Consequence is Transitive, it follows that: :$\mathcal F \models_{\mathscr M} \phi$ Finally, by definition of $T \left({\mathcal F}\right)$: :$\phi \in T \left({\mathcal F}\right)$. Since $\phi$ was arbitrary, the result follows. {{qed}}
Let $\mathcal M$ be an $\mathscr M$-structure. By assumption, if $\mathcal M$ is a model of $\mathcal F$, it is one of $\mathcal G$ as well. But any model of $\mathcal G$ is also a model of $\mathcal H$. In conclusion, any model of $\mathcal F$ is also a model of $\mathcal H$. Hence the result, by definition of semantic consequence. {{qed}} Category:Formal Semantics 3n79d69nwfm43gnzhstbnkng6sx70f5
{{begin-eqn}} {{eqn | l = \cmod {z - a} \cmod {z - a}       | r = a^2       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {x - a}^2 + y^2} \paren {\paren {x + a}^2 + y^2}       | r = a^4       | c = {{Defof|Complex Modulus}} }} {{eqn | ll= \leadsto       | l = \paren {x^2 - 2 a x + a^2 + y^2} \paren {x^2 + 2 a x + a^2 + y^2}       | r = a^4       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\paren {x^2 + a^2 + y^2} - 2 a x} \paren {\paren {x^2 + a^2 + y^2} + 2 a x}       | r = a^4       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + a^2 + y^2}^2 - 4 a^2 x^2       | r = a^4       | c = Difference of Two Squares }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2 + a^4 + 2 \paren {x^2 + y^2} a^2 - 4 a^2 x^2       | r = a^4       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2 + 2 a^2 \paren {x^2 - 2 x^2 + y^2}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2 + 2 a^2 \paren {y^2 - x^2}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2       | r = 2 a^2 \paren {x^2 - y^2}       | c =  }} {{end-eqn}} which is the equation defining a lemniscate of Bernoulli with major semiaxis $a \sqrt 2$. {{qed}}
{{begin-eqn}} {{eqn | l = \tan \theta \cot \theta       | r = \frac {\sin \theta} {\cos \theta} \cot \theta       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \frac {\sin \theta} {\cos \theta} \frac {\cos \theta} {\sin \theta}       | c = Cotangent is Cosine divided by Sine }} {{eqn | r = \frac {\sin \theta} {\sin \theta} \frac {\cos \theta} {\cos \theta}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Tangent Function Category:Cotangent Function 46j184iwn1vblxlg7w7srw5q3g4jcc2
We have: {{begin-eqn}} {{eqn | l = \frac {k \pi} {2 m} + \frac {\paren {m - k} \pi} {2 m}       | r = \frac {\paren {k + m - k} \pi} {2 m}       | c =  }} {{eqn | r = \frac \pi 2       | c =  }} {{end-eqn}} That means: :$(1): \quad \dfrac \pi 2 - \dfrac {k \pi} {2 m} = \dfrac {\paren {m - k} \pi} {2 m}$ Let $m$ be odd. Then $m - 1$ is even, and so: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m}       | r = \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \prod_{k \mathop = \frac {m - 1} 2 + 1}^{m - 1} \cot \frac {k \pi} {2 m}       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \cot \frac {\paren {m - k} \pi} {2 m}       | c =  }} {{eqn | r = \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \map \cot {\frac \pi 2 - \frac {k \pi} {2 m} }       | c = from $(1)$ }} {{eqn | r = \prod_{k \mathop = 1}^{\frac {m - 1} 2} \cot \frac {k \pi} {2 m} \tan \frac {k \pi} {2 m}       | c = Cotangent of Complement equals Tangent }} {{eqn | r = 1       | c = Product of Tangent and Cotangent }} {{end-eqn}} Now suppose $m$ is even. Then $m - 1$ is odd, and so: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^{m - 1} \cot \frac {k \pi} {2 m}       | r = \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} } \frac m 2 \frac \pi {2 m} \paren {\prod_{k \mathop = \frac m 2 + 1}^{m - 1} \cot \frac {k \pi} {2 m} }       | c =  }} {{eqn | r = \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \cot \frac {\paren {m - k} \pi} {2 m} } \frac \pi 4       | c =  }} {{eqn | r = \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \map \cot {\frac \pi 2 - \frac {k \pi} {2 m} } } \cot \frac \pi 4       | c = from $(1)$ }} {{eqn | r = \paren {\prod_{k \mathop = 1}^{\frac m 2 - 1} \cot \frac {k \pi} {2 m} \tan \frac {k \pi} {2 m} } \cot \frac \pi 4       | c = Cotangent of Complement equals Tangent }} {{eqn | r = 1 \cot \frac \pi 4       | c = Product of Tangent and Cotangent }} {{eqn | r = 1       | c = Cotangent of $45 \degrees$ }} {{end-eqn}} {{qed}}
{{AimForCont}} $S \setminus T$ were countable. By definition of relative complement: :$S \setminus T = \relcomp S T$ Thus from Union with Relative Complement: :$\paren {S \setminus T} \cup T = S$ But from Finite Union of Countable Sets is Countable it follows that $S$ is countable. From this contradiction it follows that $S \setminus T$ is uncountable. {{qed}} {{LEM|Union with Relative Complement}} Category:Uncountable Sets Category:Countable Sets 6ihlmj25zzqiff0xm1y2mrt578y9wfj
From Real Numbers are Uncountable, $\R$ is an uncountable set. From Rational Numbers are Countably Infinite $\Q$ is countable. The result follows from Uncountable Set less Countable Set is Uncountable. {{qed}} {{AoC|Uncountable Set less Countable Set is Uncountable}}
First it is worth examining the degenerate case $n = 3$, where there are no such lines. In this case: {{begin-eqn}} {{eqn | l = \frac {n \csc^2 \frac \pi n} 4       | r = \frac 4 3 \csc^2 \frac \pi 3       | c =  }} {{eqn | r = \frac 3 4 \paren {\frac {2 \sqrt 3} 3}^2       | c = Cosecant of $60 \degrees$ }} {{eqn | r = 1       | c =  }} {{end-eqn}} The product $\displaystyle \prod_{k \mathop = 2}^{n - 2} A_0 A_k$ is vacuous here. By convention, such a vacuous product is defined as being equal to $1$. So the result holds for $n = 3$. :420px Let $n > 3$. The illustration above is a diagram of the case where $n = 7$. From Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle, the vertices of $P$ can be identified with the complex $n$th roots of unity. Thus: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 2}^{n - 2} A_1 A_k       | r = \prod_{k \mathop = 2}^{n - 2} \paren {1 - \alpha^k}       | c = where $\alpha = e^{2 i \pi / n}$ is the first complex $n$th root of unity }} {{eqn | r = \dfrac 1 {\paren {1 - \alpha} \paren {1 - \alpha^{n - 1} } } \prod_{k \mathop = 1}^{n - 1} \paren {1 - \alpha^k}       | c =  }} {{eqn | r = \dfrac n {\paren {1 - \alpha} \paren {1 - \alpha^{n - 1} } }       | c = Product of Differences between 1 and Complex Roots of Unity }} {{eqn | r = \dfrac n {\paren {1 - \paren {\cos \frac {2 \pi} n + i \sin \frac {2 \pi} n} } \paren {1 - \paren {\cos \frac {2 \paren {n - 1} \pi} n + i \sin \frac {2 \paren {n - 1} \pi} n} } }       | c = Definition of $\alpha$ }} {{eqn | r = \dfrac n {\paren {1 - \paren {\cos \frac {2 \pi} n + i \sin \frac {2 \pi} n} } \paren {1 - \paren {\cos \frac {2 \pi} n - i \sin \frac {2 \pi} n} } }       | c =  }} {{eqn | r = \dfrac n {\paren {1 - \cos \frac {2 \pi} n}^2 + \paren {\sin \frac {2 \pi} n}^2}       | c =  }} {{eqn | r = \dfrac n {1 - 2 \cos \frac {2 \pi} n + \paren {\cos \frac {2 \pi} n}^2 + \paren {\sin \frac {2 \pi} n}^2}       | c =  }} {{eqn | r = \dfrac n {2 - 2 \cos \frac {2 \pi} n}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \dfrac n {2 - 2 \paren {1 - 2 \sin^2 \frac \pi n} }       | c = Double Angle Formulas for Cosine: Corollary 2 }} {{eqn | r = \dfrac n {2 - 2 + 4 \sin^2 \frac \pi n}       | c =  }} {{eqn | r = \frac {n \csc^2 \frac \pi n} 4       | c = {{Defof|Cosecant}} }} {{end-eqn}} {{qed}}
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$. As $z_1 z_2$ is real: :$(1): \quad z_1 z_2 = x_1 x_2 - y_1 y_2$ and: :$(2): \quad x_1 y_2 + y_1 x_2 = 0$ So: {{begin-eqn}} {{eqn | l = \frac {\paren {z_1} } {\paren {\overline {z_2} } }       | r = \frac {x_1 + i y_1} {x_2 - i y_2}       | c = {{Defof|Complex Conjugate}} }} {{eqn | r = \frac {\paren {x_1 + i y_1} \paren {x_2 + i y_2} } {\paren {x_2 - i y_2} \paren {x_2 + i y_2} }       | c = multiplying top and bottom by $x_2 + i y_2$ }} {{eqn | r = \frac {\paren {x_1 + i y_1} \paren {x_2 + i y_2} } { {x_2}^2 + {y_2}^2}       | c = Difference of Two Squares and $i^2 = -1$ }} {{eqn | r = \frac {x_1 x_2 - y_1 y_2} { {x_2}^2 + {y_2}^2}       | c = from $(1)$ }} {{end-eqn}} So $z_1 / \overline {z_2} = p$ where $p = \dfrac {x_1 x_2 - y_1 y_2} { {x_2}^2 + {y_2}^2}$, which is real. {{qed}}