hcju/proofwikiqa huggingface dataset:
Let $x \in X$. By Supremum of Singleton: :$\left\{ {x}\right\}$ admits a supremum and $\sup \left\{ {x}\right\} = x$ By definitions of subset and singleton: :$\left\{ {x}\right\} \subseteq X$ By Singleton is Finite: :$\left\{ {x}\right\}$ is a finite set. Thus by definition of finite suprema set: :$x \in \operatorname{finsups}\left({X}\right)$ {{qed}}
Let $X \subseteq Y$. Let $x \in X^\preceq$. By definition of lower closure of subset: :$\exists y \in X: x \preceq y$ By definition of subset: :$y \in Y$ Thus by definition of lower closure of subset: :$x \in Y^\preceq$ {{qed}}
By Set is Subset of Finite Suprema Set: :$X \subseteq \operatorname{finsups}\left({X}\right)$ By Lower Closure of Subset is Subset of Lower Closure: :$X^\preceq \subseteq \operatorname{finsups}\left({X}\right)^\preceq$ By Set is Subset of Lower Closure: :$X \subseteq X^\preceq$ Thus by Subset Relation is Transitive: :$X \subseteq \operatorname{finsups}\left({X}\right)^\preceq$ Let $I$ be a ideal in $L$ such that :$X \subseteq I$ Let $x \in {\operatorname{finsups}\left({X}\right)}^\preceq$ By definition of lower closure of subset: :$\exists y \in \operatorname{finsups}\left({X}\right): x \preceq y$ By definition of finite suprema set: :$\exists A \in \mathit{Fin}\left({X}\right): y = \sup A \land A$ admits a supremum where $\mathit{Fin}\left({X}\right)$ denotes the set of all finite subsets of $S$. By Subset Relation is Transitive: :$A \subseteq I$ By Directed in Join Semilattice with Finite Suprema: :$A \ne \varnothing \implies \sup A \in I$ By Supremum of Empty Set is Smallest Element and Bottom in Ideal: :$A = \varnothing \implies \sup A = \bot \in I$ where $\bot$ denotes the smallest element of $S$. So :$y \in I$ Thus by definition of lower set: :$x \in I$ {{qed}}
Let $x \in X$. By definition of reflexivity: :$x \preceq x$ Thus by definition of lower closure: :$x \in X^\preceq$ {{qed}}
From Sum of Sequence of Odd Cubes: :$1^3 + 3^3 + 5^3 + \cdots + \paren {2 m − 1}^3 = m^2 \paren {2 m^2 − 1}$ By the Theorem of Even Perfect Numbers: :$n = 2^{r - 1} \paren {2^r - 1}$ for some $r$. Setting $m = 2^{r - 2}$: :$m^2 = 2^{r - 1}$ and so: :$2 m^2 = 2^r$ and it follows that: :$\displaystyle n = \sum_{k \mathop = 1}^{2^{r - 2} } \paren {2 k - 1}^3$ and hence the result. When $n = 6$ we have: :$6 = 2^1 \paren {2^2 - 1}$ leading to $r = 2$ and thence $m^2 = 2$, at which point the formula fails to work. {{qed}}
Write $t^r = t^{p/q}$, and set $t > 1$. {{begin-eqn}} {{eqn | l  = t^{p/q}       | o = <       | r = K e^{a t}       | c = an Ansatz }} {{eqn | ll = \impliedby       | l = t^p       | o = <       | r = \paren {K e^{a t} }^q       | c = Rational Power is Strictly Increasing }} {{eqn | r = K^q e^{q a t}       | c = Exponential of Product }} {{eqn | r =  K' e^{a' t}       | c = $K^q = K', q a = a'$ }} {{end-eqn}} Recall from Polynomial is of Exponential Order Epsilon, $t^p < K'e^{a't}$ for ''any'' $a' > 0$, arbitrarily small in magnitude. Therefore the inequality $t^{p/q} < K e^{a t}$ has solutions of the same nature. {{qed}} Category:Exponential Order e4d45d6d8d4cdizv1nso1gtcjr75utz
Proof by induction: For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$ === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^1 \paren {2 j - 1}^3       | r = 1^3       | c =  }} {{eqn | r = 1^2 \paren {2 \times 1^2 - 1}       | c =  }} {{end-eqn}} and $\map P 1$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 1}^k \paren {2 j - 1}^3 = k^2 \paren {2 k^2 − 1}$ Then we need to show: :$\displaystyle \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}^3 = \paren {k + 1}^2 \paren {2 \paren {k + 1}^2 − 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}^3       | r = \sum_{j \mathop = 1}^k \paren {2 j - 1}^3 + \paren {2 k + 1}^3       | c =  }} {{eqn | r = k^2 \paren {2 k^2 − 1} + \paren {2 k + 1}^3       | c = Induction Hypothesis }} {{eqn | r = 2 k^4 + 8 k^3 + 11 k^2 + 6 k + 1       | c = multiplying out }} {{eqn | r = \paren {k + 1}^2 \paren {2 k^2 + 4 k + 1}       | c = extracting $\paren {k + 1}^2$ as a factor }} {{eqn | r = \paren {k + 1}^2 \paren {2 \paren {k + 1}^2 - 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 1}: \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = n^2 \paren {2 n^2 − 1}$ {{qed}}
By definition of ideal in ordered set: :$I$ is non-empty and lower. By definition of non-empty set: :$\exists x: x \in I$ By definition of smallest element: :$\bot \preceq x$ Thus by definition of lower set: :$\bot \in I$ {{qed}}
For $t > 0$, $t^r$ is continuous. At $t = 0$, defining $0^r = 0$, the function is continuous from the right: {{begin-eqn}} {{eqn | l = \lim_{t \mathop \to 0^+} t^r       | r = \lim_{t \mathop \to 0^+} \map \exp {r \ln t} }} {{eqn | r = \map \exp {\lim_{t \mathop \to 0^+} r \ln t}       | c = Exponential Function is Continuous }} {{eqn | r = 0       | c = Logarithm Tends to Negative Infinity, Exponential Tends to Zero and Infinity }} {{end-eqn}} Fix $t > 1$. By the Archimedean Principle, there is a natural number $n$ such that $n > r$. By Real Power Function on Base Greater than One is Strictly Increasing, $t^r < t^n$. Recall from Polynomial is of Exponential Order Epsilon, $t^n < K e^{at}$ for ''any'' $a > 0$, arbitrarily small in magnitude. Therefore the inequality $t^r < K e^{at}$ has solutions of the same nature. {{qed}} Category:Exponential Order mz9jnhn3zq5p851haa8k3mk9h6q0o6z
By Finite Suprema Set and Lower Closure is Smallest Ideal: :$X \subseteq {\operatorname{finsups}\left({X}\right)}^\preceq$ By definition of non-empty set: :${\operatorname{finsups}\left({X}\right)}^\preceq$ is a non-empty set. We will prove that :$\operatorname{finsups}\left({X}\right)$ is directed. Let $x, y \in \operatorname{fininfs}\left({X}\right)$ By definition of finite suprema set: :$\exists A \in \mathit{Fin}\left({X}\right): x = \sup A \land A$ admits a supremum and :$\exists B \in \mathit{Fin}\left({X}\right): y = \sup B \land B$ admits an supremum where $\mathit{Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$. Define $C = A \cup B$. By Union of Subsets is Subset: :$C \subseteq X$ By Union of Finite Sets is Finite: :$C$ is finite. Then :$C \in \mathit{Fin}\left({X}\right)$ By Existence of Non-Empty Finite Suprema in Join Semilattice: :$C \ne \varnothing \implies C$ admits a supremum. By Union is Empty iff Sets are Empty: :$C = \varnothing \implies A = \varnothing$ So :$C$ admits a supremum. By definition of finite suprema set: :$\sup C \in \operatorname{finsups}\left({X}\right)$ By Set is Subset of Union: :$A \subseteq C$ and $B \subseteq C$ Thus by Supremum of Subset: :$x \preceq \sup C$ and $y \preceq \sup C$ Hence $\operatorname{finsups}\left({X}\right)$ is directed. {{qed|lemma}} By Directed iff Lower Closure Directed: :${\operatorname{finsups}\left({X}\right)}^\preceq$ is directed. By Lower Closure is Lower Set: :${\operatorname{finsups}\left({X}\right)}^\preceq$ is lower. Hence ${\operatorname{finsups}\left({X}\right)}^\preceq$ is ideal in $P$. {{qed}}
We note that if $\set {a, b, c}$ is such a set, then $\set {k a, k b, k c}$ satisfy the same properties trivially. Hence the specification that $\set {a, b, c}$ is a coprime set. We have that: :$5 \times 1 = 2 + 3$ so $1 \divides 2 + 3$ :$2 \times 2 = 1 + 3$ so $2 \divides 1 + 3$ :$1 \times 3 = 1 + 2$ so $3 \divides 1 + 2$ It remains to be shown that this is the only such set. We are to find all the sets $\set {a, b, c}$ such that: :$a \divides b + c$ :$b \divides a + c$ :$c \divides a + b$ where $a \ne b$, $a \ne c$ and $b \ne c$. {{WLOG}}, suppose $a < b < c$. Since $2 c > a + b > 0$ and $c \divides a + b$, we must have $a + b = c$. Then it follows that: :$a \divides \paren {b + a + b}$ :$b \divides \paren {a + a + b}$ which reduces to: :$a \divides 2 b$ :$b \divides 2 a$ Suppose $b$ is odd. Then by Euclid's Lemma, we would have $b \divides a$. By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction. Thus $b$ is even. Suppose $a$ is even. Then $a, b, c$ are all even. So $\gcd \set {a, b, c} \ne 1$, which is a contradiction. Therefore it must be the case that $a$ is odd. Then by Euclid's Lemma, we have: :$a \divides \dfrac b 2$ and: :$\dfrac b 2 \divides a$ By Absolute Value of Integer is not less than Divisors, this gives: :$\dfrac b 2 = a$ Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$. Hence the set $\set {1, 2, 3}$ is obtained. {{qed}}
Let $n$ be such a positive integer with corresponding divisors $a, b, c$ such that: :$a + b + c = n$ We note that the set $\set {k a, k b, k c}$ satisfy the same properties trivially as divisors of $k n$. Hence the specification that $\set {a, b, c}$ is a coprime set. {{WLOG}}, suppose $a < b < c$. Since $a, b, c$ are strictly positive, $n \ne c$. Suppose $\dfrac n c \ge 3$. Then: {{begin-eqn}} {{eqn | l = n       | r = a + b + c }} {{eqn | o = <       | r = c + c + c }} {{eqn | r = n }} {{end-eqn}} which is a contradiction. Hence: {{begin-eqn}} {{eqn | l = \frac n c       | r = 2 }} {{eqn | ll = \leadsto       | l = a + b + c       | r = 2 c }} {{eqn | ll = \leadsto       | l = a + b       | r = c }} {{eqn | ll = \leadsto       | l = n       | r = 2 a + 2 b }} {{end-eqn}} Since $a, b$ are divisors of $n$: :$a \divides \paren {2 a + 2 b}$ :$b \divides \paren {2 a + 2 b}$ which reduces to: :$a \divides 2 b$ :$b \divides 2 a$ Suppose $b$ is odd. Then by Euclid's Lemma, we would have $b \divides a$. By Absolute Value of Integer is not less than Divisors, this gives $b \le a$, which is a contradiction. Thus $b$ is even. Suppose $a$ is even. Then $a, b, c$ are all even. So $\gcd \set {a, b, c} \ne 1$, which is a contradiction. Therefore it must be the case that $a$ is odd. Then by Euclid's Lemma, we have: :$a \divides \dfrac b 2$ and: :$\dfrac b 2 \divides a$ By Absolute Value of Integer is not less than Divisors, this gives: :$\dfrac b 2 = a$ Because $\gcd \set {a, b, c} = 1$, we must have $a = 1$. Hence the set $\set {1, 2, 3}$ is obtained. {{qed}}
Let $n + 2$ be expressed as $3 r$ for some positive integer $r$. Then: {{begin-eqn}} {{eqn | l = m       | r = \paren {3 r - 2} + \paren {3 r - 1} + 3 r       | c =  }} {{eqn | r = 9 r - 3       | c =  }} {{eqn | r = 9 \paren {r - 1} + 6       | c =  }} {{end-eqn}} The result follows from Digital Root of Number equals its Excess over Multiple of 9. {{qed}}
{{begin-eqn}} {{eqn | l = \size {t^\phi}       | r = t^{\map \Re \phi}       | c = Modulus of Positive Real Number to Complex Power is Positive Real Number to Power of Real Part }} {{end-eqn}} The result follows from Real Power is of Exponential Order Epsilon. {{qed}} Category:Exponential Order fmibd8v9dkf2rrq46kvp0cbd1rhtziv
This follows by mutatis mutandis of the proof of Finite Subset Bounds Element of Finite Infima Set and Upper Closure. {{qed}}
When $p = 3$: :$\map {q_3} 2 = \dfrac {2^{3 - 1} - 1} 3 = 1$ which is square. When $p = 7$: :$\map {q_7} 2 = \dfrac {2^{7 - 1} - 1} 7 = \dfrac {63} 7 = 9$ which is square. To show that these are the only ones, we observe that since $p$ is an odd prime, write: :$p = 2 n + 1$ for $n \ge 1$. Let $\map {q_p} 2$ be a square. Then $2^{p - 1} - 1 = p x^2$ for some integer $x$. Note that: :$2^{p - 1} - 1 = 2^{2 n} - 1 = \paren {2^n - 1} \paren {2^n + 1}$ and we have: :$\gcd \set {2^n - 1, 2^n + 1} = \gcd \set {2^n - 1, 2} = 1$ so $2^n - 1$ and $2^n + 1$ are coprime. Hence there are $2$ cases: === Case $1$: $p \divides 2^n - 1$ === By Divisor of One of Coprime Numbers is Coprime to Other: :$\gcd \set {\dfrac {2^n - 1} p, 2^n + 1} = 1$ Hence both the numbers are squares. In particular we have: :$\exists k \in \Z: 2^n + 1 = k^2$ by 1 plus Power of 2 is not Perfect Power except 9, the only solution to the equation above is: :$n = k = 3$ This gives $p = 2 n + 1 = 7$. {{qed|lemma}} === Case $2$: $p \divides 2^n + 1$ === By Divisor of One of Coprime Numbers is Coprime to Other: :$\gcd \set {\dfrac {2^n + 1} p, 2^n - 1} = 1$ Hence both the numbers are squares. In particular we have: :$\exists k \in \Z: 2^n - 1 = k^2$ For $n > 1$: :$2^n - 1 \equiv 3 \pmod 4$ which by Square Modulo 4 is never a square. Hence we must have $n = 1$. This gives $p = 2 n + 1 = 3$. {{qed|lemma}} We have already shown that $p = 3, 7$ gives square Fermat quotients. Hence the result. {{qed}}
{{AimForCont}} $x$ and $y$ are both odd such that: :$\exists z \in \Z: x^2 + y^2 = z^2$ Then: :$x^2 + y^2 \equiv 1 + 1 \equiv 2 \pmod 4$ But from Square Modulo 4: :$z^2 \equiv 0 \pmod 4$ or $z^2 \equiv 1 \pmod 4$ Thus $x^2 + y^2$ can not be square. It follows by Proof by Contradiction that  $x$ and $y$ cannot both be odd. {{qed}} Category:Pythagorean Triples o1vtbu7f6w8n47oqvi0eo9kkmzt1yyy
Let $x, y \in \bigcup A$. By definition of union: :$\exists X \in A: x \in X$ and :$\exists Y \in A: y \in Y$ By assumption: :$\exists Z \in A: X \cup Y \subseteq Z$ By definition of union: :$x, y \in X \cup Y$ By definition of subset: :$x, y \in Z$ By assumption: :$Z$ is directed. By definition of directed subset: :$\exists z \in Z: x \preceq z \land y \preceq z$ Thus by definition of union: :$z \in \bigcup A$ Thus :$x \preceq z \land y \preceq z$ Hence $\bigcup A$ is directed. {{qed}}
Let $x \in \bigcup A, y \in S$ such that :$y \preceq x$ By definition of union: :$\exists X \in A: x \in X$ By assumption: :$X$ is a lower set. By definition of lower set: :$y \in X$ Thus by definition of union: :$y \in \bigcup A$ {{qed}}
By Upper Closure of Element is Filter: :$x^\succeq$ is a filter. Wr will prove that :$I \cap x^\succeq = \varnothing$ Aiming for a contradiction suppose that :$\exists y: y \in I \cap x^\succeq$ By definition of intersection: :$y \in I$ and $y \in x^\succeq$ By definition of upper closure of element: :$x \preceq y$ By definition of lower set: :$x \in I$ This contradicts $x \notin I$ {{qed|lemma}} Thus by If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter: :there exists prime ideal $P$ in $L$: $I \subseteq P$ and $P \cap x^\succeq = \varnothing$ By definition of reflexivity: :$x \preceq x$ By definition of upper closure of element: :$x \in x^\succeq$ Thus by definitions of intersection and empty set: :$x \notin P$ {{qed}}
{{begin-eqn}} {{eqn | l = \int_{0^+}^{ \mathop \to +\infty} \ln t \,  e^{-t} \, \mathrm d t       | r = \int_{0^+}^{ \mathop \to +\infty} t^{1-1} \ln t \,  e^{-t} \, \mathrm d t }} {{eqn | r = \Gamma' \left({1}\right)       | c = Derivative of Gamma Function }} {{eqn | r = -\gamma       | c = Derivative of Gamma Function at $1$ }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Suppose: :$F$ is proper. By definition of proper subset: :$F \subseteq S$ and $F \ne S$ By definitions of set equality and subset: :$\exists x: x \in S \land x \notin F$ By definition of smallest element: :$\bot \preceq x$ Thus by definition of upper set: :$\bot \notin F$ {{qed|lemma}} === Necessary Condition === Suppose $\bot \notin F$ By definition of set equality: :$F \ne S$ Hence $F$ is proper filter. {{qed}}
By Dual Distributive Lattice is Distributive: :$L^{-1}$ is a distributive lattice where $L^{-1} = \left({S, \succeq}\right)$ denotes the dual of $L$. By Filter is Ideal in Dual Ordered Set: :$I' := F$ as an ideal in $L^{-1}$. By Ideal is Filter in Dual Ordered Set: :$F' := I$ as a filter on $L^{-1}$. By assumption: :$I' \cap F' = \varnothing$ By If Ideal and Filter are Disjoint then There Exists Prime Ideal Including Ideal and Disjoint from Filter: :there exists prime ideal $P'$ in $L^{-1}$: $I' \subseteq P'$ and $P' \cap F' = \varnothing$ By Prime Filter is Prime Ideal in Dual Lattice: :$P := P'$ as a prime filter on $L$. Thus there exists prime filter $P$ on $L$: $F \subseteq P$ and $P \cap I = \varnothing$ {{qed}}
=== Sufficient Condition === Let $L$ be a distributive lattice. By Dual of Lattice Ordering is Lattice Ordering: :$L^{-1}$ is lattice. Let $x, y, z \in S$. $\vee'$ and $\wedge'$ denotes join and meet in $L^{-1}$. Thus {{begin-eqn}} {{eqn | l = x \wedge' \left({y \vee' z}\right) | r = x \wedge' \left({y \wedge z}\right) | c = Join is Dual to Meet }} {{eqn | r = x \vee \left({y \wedge z}\right) | c = Join is Dual to Meet }} {{eqn | r = \left({x \vee y}\right) \wedge \left({x \vee z}\right) | c = definition of distributive lattice }} {{eqn | r = \left({x \wedge' y}\right) \wedge \left({x \wedge' z}\right) | c = Join is Dual to Meet }} {{eqn | r = \left({x \wedge' y}\right) \vee' \left({x \wedge' z}\right) | c = Join is Dual to Meet }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === This follows by mutatis mutandis. {{qed}}
In the below: :$r_i$ denotes the initial state of row $i$ :$r_j$ denotes the initial state of row $j$ :$r_i'$ denotes the state of row $i$ after having had the latest elementary row operation applied :$r_j'$ denotes the state of row $j$ after having had the latest elementary row operation applied. $(1)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$: :$r_j \to r_j + r_i$ After this operation: {{begin-eqn}} {{eqn | l = r_i'       | r = r_i }} {{eqn | l = r_j'       | r = r_i + r_j }} {{end-eqn}} {{qed|lemma}} $(2)$: Apply $\text {ERO} 2$ to row $i$ for $\lambda = -1$: :$r_i \to r_i + \paren {-r_j}$ After this operation: {{begin-eqn}} {{eqn | l = r_i'       | r = r_i - \paren {r_i + r_j} }} {{eqn | r = -r_j }} {{eqn | l = r_j'       | r = r_i + r_j }} {{end-eqn}} {{qed|lemma}} $(3)$: Apply $\text {ERO} 2$ to row $j$ for $\lambda = 1$: :$r_j \to r_j + r_i$ After this operation: {{begin-eqn}} {{eqn | l = r_i'       | r = -r_j }} {{eqn | l = r_j'       | r = r_i + r_j - r_j }} {{eqn | r = r_i }} {{end-eqn}} {{qed|lemma}} $(4)$: Apply $\text {ERO} 1$ to row $i$ for $\lambda = -1$: :$r_i \to -r_i$ After this operation: {{begin-eqn}} {{eqn | l = r_i'       | r = -\paren {-r_j} }} {{eqn | r = r_j }} {{eqn | l = r_j'       | r = r_i }} {{end-eqn}} {{qed|lemma}} Thus, after all the $4$ elementary row operations have been applied, we have: {{begin-eqn}} {{eqn | l = r_i'       | r = r_j }} {{eqn | l = r_j'       | r = r_i }} {{end-eqn}} Hence the result. {{qed}}
By Singleton of Bottom is Ideal: :$I := \left\{ {\bot}\right\}$ is an ideal in $L$. where $\bot$ denotes the bottom of $L$. We will prove that :$F \cap I = \varnothing$ Let $x \in I$. By definition of singleton: :$x = \bot$ Thus by Bottom not in Proper Filter: :$x \notin F$ {{qed|lemma}} By If Ideal and Filter are Disjoint then There Exists Prime Filter Including Filter and Disjoint from Ideal: :there exists a prime filter $G$ on $L$: $F \subseteq G$ and $G \cap I = \varnothing$ By definition of singleton: :$\bot \in I$ By definitions of empty set and intersection: :$\bot \notin G$ By Bottom not in Proper Filter: :$G$ is proper. Thus by Proper and Prime iff Ultrafilter in Boolean Lattice: :$G$ is ultrafilter. Thus $F \subseteq G$ {{qed}}
=== Sufficient Condition === The result follows by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal. {{qed|lemma}} === Necessary Condition === Suppose :for every prime ideal $P$ in $L$: :$y \preceq \sup P \implies x \in P$ We will prove that: :for every ideal $I$ in $L$: $y \preceq \sup I \implies x \in I$ Let $I$ be an ideal in $L$ such that :$y \preceq \sup I$ Aiming for a contradiction suppose that :$x \notin I$ By If Element Does Not Belong to Ideal then There Exists Prime Ideal Including Ideal and Excluding Element: :there exists a prime ideal $P$ in $L$: $I \subseteq P$ and $x \notin P$ By Supremum of Subset: :$\sup I \preceq \sup P$ By definition of transitivity: :$y \preceq \sup P$ By assumption: :$x \in P$ Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$x \ll y$ {{qed}}
Let $x, y \in S$ such that :$x \wedge y \in p^\preceq$ By definition of lower closure of element: :$x \wedge y \preceq p$ By Characterization of Prime Ideal: :$x \preceq p$ or $y \preceq p$ Thus by definition of lower closure of element: :$x \in p^\preceq$ or $y \in p^\preceq$ {{qed}}
By Lower Closure is Prime Ideal for Prime Element: :$p^\preceq$ is prime ideal. By Supremum of Lower Closure of Element: :$ \sup \left({ p^\preceq }\right) = p$ Hence $p$ is pseudoprime. {{qed}}
The fact that it is impossible to construct a regular heptagon using a compass and straightedge construction is demonstrated in Compass and Straightedge Construction for Regular Heptagon does not exist. From Construction of Equilateral Triangle, an equilateral triangle can be constructed. From Inscribing Square in Circle, for example, a square can be constructed. From Inscribing Regular Pentagon in Circle, a regular pentagon can be constructed. From Inscribing Regular Hexagon in Circle, a regular hexagon can be constructed. Thus all regular polygons with $3$ to $6$ sides can be so constructed, but not one with $7$ sides. {{qed}}
From Maximum Period of Reciprocal of Prime, the maximum period of $\dfrac 1 p$ is $p - 1$. :$\dfrac 1 2 = 0 \cdotp 5$: not recurring. :$\dfrac 1 3 = 0 \cdotp \dot 3$: recurring with period $1$. :$\dfrac 1 5 = 0 \cdotp 2$: not recurring. :$\dfrac 1 7 = 0 \cdotp \dot 14285 \dot 7$: recurring with period $6$. {{qed}}
By definition of pseudoprime element: :there exists a prime ideal $P$ in $L$: $p = \sup P$ By definition of way below closure: :$\inf A \in p^\ll$ By definition of reflexivity: :$p \preceq \sup P$ By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum: :$p^\ll \subseteq P$ By definition of subset: :$\inf A \in P$ Thus by Characterization of Prime Ideal by Finite Infima :$\exists a \in A: a \in P$ By definition of up-complete: :$P$ admits a supremum. By definition of supremum: :$p$ is upper bound for $P$. Thus by definition of upper bound: :$a \preceq p$ {{qed}}
Let $e$ be the elementary row operation such that $e \paren {\mathbf X} = \mathbf Y$. Then this result follows immediately from Elementary Row Operations as Matrix Multiplications: :$e \paren {\mathbf X} = \mathbf {E X} = \mathbf Y$ where $\mathbf E = e \paren {\mathbf I}$. {{qed}}
From Product of Cyclotomic Polynomials: :$\map {\Phi_p} x \map {\Phi_1} x = x^p - 1$ Thus from Sum of Geometric Sequence: :$\map {\Phi_p} x = \dfrac {x^p - 1} {x - 1} = x^{p - 1} + x^{p - 2} + \cdots + x + 1$ {{qed}} Category:Examples of Cyclotomic Polynomials luxpvhzthn8nahhtjocbazudx6g9wgl
From the Polynomial Factor Theorem and Complex Roots of Unity in Exponential Form: :$\displaystyle x^n - 1 = \prod_\zeta \left({x - \zeta}\right)$ where the product runs over all complex $n$th roots of unity. In the {{LHS}}, each factor $x - \zeta$ appears exactly once, in the factorization of $\Phi_d \left({x}\right)$ where $d$ is the order of $\zeta$. {{explain|The above statement needs justification.}} Thus the polynomials are equal. {{qed}}
By definition: :$\displaystyle \map {\Phi_1} x = \prod_\zeta \paren {x - \zeta}$ where the product runs over all primitive complex first roots of unity. A root of unity has order $1$ {{iff}} it equals $1$. {{explain|The above statement needs justification, with reference to the definition of the order of a root of unity.}} Hence the only factor is $x - 1$. {{MissingLinks|Factor of Polynomial -- for which we appear not to have a page.}} Category:Examples of Cyclotomic Polynomials czxoo4jr9iujpwf7gtp0n8wu4kr6pct