hcju/proofwikips huggingface dataset:
Continuous Image of Separable Space is Separable
== Definition == Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces. Let $f: T_1 \to T_2$ be a continuous mapping. If $T_1$ is separable, then so is the image $f \sqbrk {T_1}$. That is, separability is a continuous invariant. == Proof == By definition, $T_1 = \struct {S_1, \tau_1}$ is separable  there exists a countable subset $D \subset S_1$ which is (everywhere) dense in $T_1$. We need to show that if there exists a continuous mapping $f: T_1 \to T_2$, then $f \sqbrk {T_1}$ is also separable. That is, that there exists a countable subset of $f \sqbrk {S_1}$ which is dense in $T_2$. Let $x_2$ be any point in the image $f \sqbrk {S_1}$ of $S_1$ under $f$. Let $U \in \tau_2$ be an arbitrary open set such that $x_2 \in U$. By definition of image set, there exists some $x_1 \in S_1$ with $\map f {x_1} = x_2$. Since $f$ is continuous, $f^{-1} \sqbrk U$ is open in $T_1$. By definition of preimage, $x_1$ is in this set. We are given that $D$ is a countable subset of $S_1$ which is dense in $T_1$. By definition of dense, $D \cap f^{-1} \sqbrk U \ne \O$ Thus there exists some $d \in D$ such that $d \in f^{-1} \sqbrk U$. Therefore $\map f d \in U$. Since $U$ was arbitrary, it follows that $f \sqbrk D$ is dense in $T_2$. By Image of Countable Set under Mapping is Countable, $f \sqbrk D$ is countable. Hence $T_2$ is separable.  == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Closed Invariant|next = Continuous Image of Compact Space is Compact}}: Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Lindelöf Property is Preserved under Continuous Surjection|next = Compactness Properties Preserved under Projection Mapping}}: Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Urysohn's Lemma Converse
== Lemma == Let $T = \struct {S, \tau}$ be a topological space. Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \varnothing$. Then $T = \struct {S, \tau}$ is a $T_4$ space. == Proof == Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$. Let $C = \hointr 0 {\dfrac 1 4}$ and $D = \hointl {\dfrac 3 4} 1$. Then $C$ and $D$ are open in $\closedint 0 1$. {{explain|There's a link available somewhere for the above fact.}} Hence, $\map {f^{-1} } C$ and $\map {f^{-1} } D$ are open in $T$. Furthermore, by definition of Urysohn function, $A \subset \map {f^{-1} } C$ and $B \subset \map {f^{-1} } D$. Also, from Preimage of Intersection under Mapping: :$C \cap D = \O \implies \map {f^{-1} } C \cap \map {f^{-1} } D = \O$ Therefore, $T$ is a $T_4$ space.  == Also see == * Urysohn's Lemma {{Namedfor|Pavel Samuilovich Urysohn|cat = Urysohn}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Urysohn's Lemma|next = Existence of Urysohn Function does not guarantee Normal Space}}: Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Localization of Ring Exists/Lemma 1
== Lemma for Localization of Ring Exists == Let $A$ be a commutative ring with unity. Let $S \subseteq A$ be a multiplicatively closed subset with $0 \notin S$. Let $\sim$ be the relation defined on the Cartesian product $A \times S$ by: :$\left({a, s}\right) \sim \left({b, t}\right) \iff \exists u \in S: a t u = b s u$  The relation $\sim$ is an equivalence relation.  == Proof ==  Since by definition $1 \in S$ and $a s = a s$ for all $a \in A, s \in S$, it follows that $\sim$ is reflexive. Also $\sim$ is clearly symmetric because if $a t u = b s u$ then $b s u = a t u$ by symmetry of $=$. Lastly suppose that: : $\left({a, s}\right) \sim \left({b, t}\right)$ and: : $\left({b, t}\right) \sim \left({c, u}\right)$ Then there are $v, w \in S$ such that: :$a t v = b s v$ :$b u w = c t w$ Therefore: :$a u t v w = b u w s v = c t w s v$ and so: : $a u \left({t v w}\right) = c s \left({t v w}\right)$ Since $S$ is multiplicatively closed: : $\left({a, s}\right) \sim \left({c, u}\right)$ Thus $\sim$ is transitive.
Localization of Ring Exists/Lemma 2
== Lemma ==  The operations $+$ and $\cdot$ are well defined on $A_S$.  {{explain|Put this into the context in which it is relevant so as to make the page standalone.}} == Proof == Let: :$a / s = c / u$ :$b / t = d / v$ be two sets of representatives for two distinct equivalence classes in $A_S$. We have $w, z \in S$ such that: :$\left({a u - c s}\right) w = 0$ and :$\left({b v - d t}\right) z = 0$ Therefore: :$z w \left[{\left({a u - c s}\right) w - \left({b v - d t}\right) z}\right] = 0$ and: :$z w \left[{\left({a t + b s}\right) u v - \left({c v + d u}\right)s t}\right]$ So: :$\left({a t + b s, s t}\right) \sim \left({c v + d u, u v}\right)$ That is: :$\dfrac {a t + b s} {s t} = \dfrac {c v + d u} {u v}$ For multiplication, with $z, w$ as above we have: :$\left({a b u v - c d s t}\right)z w = 0$ So: :$\left({a b, s t}\right) \sim \left({d c, u v}\right)$ and: :$\dfrac {a b} {s t} = \dfrac {d c} {u v}$
Localization of Ring Exists/Lemma 3
{{refactor|The definitions of the entities need to be included here as they are on the main page.}} == Lemma for Localization of Ring Exists ==  The map $h \left({a / s}\right) = g \left({a}\right) g \left({s}\right)^{-1}$ is a well defined ring homomorphism $A_S \to B$.  == Proof == Let $a / s = b / t \in A_S$. Then there is $u \in S$ such that $a t u = b s u$. Therefore, by the homomorphism property, and the fact that $g \left[{S}\right] \subseteq B^\times$: :$g \left({a}\right) g \left({t}\right) = g \left({b}\right) g \left({s}\right) \to g \left({a}\right) g \left({s}\right)^{-1} = g \left({b}\right) g \left({t}\right)^{-1}$ We have that $g$ is a ring homomorphism. As $B$ is a commutative ring it follows that $h$ is also a ring homomorphism.
Equivalence Relation on Cauchy Sequences
== Lemma == Let $\struct {X, d}$ be a metric space. Let $\CC \sqbrk X$ denote the set of all Cauchy sequences of elements from $X$. Let a relation $\sim$ be defined on $\CC \sqbrk X$ by: :$\displaystyle \sequence {x_n} \sim \sequence {y_n}  \iff \lim_{n \mathop \to \infty} \map d {x_n, y_n} = 0$ Then $\sim$ is an equivalence relation on $\CC \sqbrk X$. == Proof == We must show that $\sim$ is :reflexive, :symmetric and :transitive on $\CC \sqbrk X$. To this end, let $\sequence {x_n}, \sequence {y_n}, \sequence {z_n} \in \CC \sqbrk X$ be arbitrary. For each $n \in \N$ we have that $\map d {x_n, x_n} = 0$ by metric space axiom '''M1'''. Therefore $\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, x_n} = 0$. This shows that $\sequence {x_n} \sim \sequence {x_n}$. Thus $\sim$ is reflexive. {{qed|lemma}} By metric space axiom '''M3''', $\map d {x_n, y_n} = \map d {y_n, x_n}$ for each $n \in \N$. Therefore: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \lim_{n \mathop \to \infty} \map d {y_n, x_n}$ So $\sequence {x_n} \sim \sequence {y_n}$ implies that $\sequence {y_n} \sim \sequence {x_n}$. Thus $\sim$ is symmetric. {{qed|lemma}} Finally, by metric space axiom '''M2''', $\map d {x_n, z_n} \le \map d {x_n, y_n} + \map d {y_n, z_n}$ for each $n \in \N$. Therefore, by the sum rule for limits of sequences: :$\displaystyle \lim_{n \mathop \to \infty} \map d {x_n, z_n} \le \lim_{n \mathop \to \infty} \map d {x_n, y_n} + \lim_{n \mathop \to \infty} \map d {y_n, z_n}$ Thus $\sequence {x_n} \sim \sequence {y_n}$ and $\sequence {y_n} \sim \sequence {z_n}$ together imply that $\sequence {x_n} \sim \sequence {z_n}$. Thus $\sim$ is transitive. {{qed|lemma}} So $\sim$ is shown to be reflexive, symmetric and transitive, and therefore an equivalence relation on $\CC \sqbrk X$.
Field Adjoined Algebraic Elements is Algebraic
== Definition == Let $L / K$ be a field extension and $S \subseteq L$ a subset. If each $x \in S$ is algebraic over $K$ then $K \left({S}\right)$ is algebraic over $K$. == Proof == Let $S \subseteq L$ be arbitrary, and $x \in K \left({S}\right)$. By Field Adjoined Set $x \in K \left({S}\right)$  $x \in K \left({\alpha_1, \ldots, \alpha_n}\right)$ for some $\alpha_1, \ldots, \alpha_n \in S$. We have that $K \left({\alpha_1, \ldots, \alpha_n}\right) / K$ is finite by Finitely Generated Algebraic Extension is Finite. Moreover a Finite Field Extension is Algebraic. Therefore $x$ is algebraic over $K$.
Vinogradov's Theorem/Minor Arcs/Lemma 1
== Lemma ==  For $\beta \in \R$, define: :$\norm \beta := \min \set {\size {n - \beta}: n \in \Z}$ Then: :$\displaystyle \forall \alpha \in \R: \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \min \set {N_2 - N_1, \frac 1 {2 \norm \alpha} }$  == Proof == The bound $N_2 - N_1$ is trivial: Since $\size {\map e {\alpha k} } = 1$ for all $k$, by the Triangle Inequality: :$\displaystyle \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \sum_{k \mathop = N_1}^{N_2} 1 = N_2 - N_1$ To show the second bound we evaluate the sum as a geometric series. We have: :$\map e {\alpha k} = \map e \alpha^k$ So by Sum of Geometric Sequence: {{begin-eqn}} {{eqn | l = \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} }       | r = \size {\frac {\map e {\alpha \paren {N_2 + 1} } - \map e {\alpha N_1} } {\map e \alpha - 1} } }} {{eqn | r = \frac 2 {\size {\map e \alpha - 1} }       | o = \le }} {{end-eqn}} By the polar form of a complex number: :$\map e \alpha - 1 = \map e {\alpha / 2} \paren {\map e {\alpha / 2} - \map e {-\alpha / 2} }$ and: :$\map e {\alpha / 2} - \map e {-\alpha / 2} = \map \exp {\pi i \alpha} - \map \exp {-\pi i \alpha} = 2 i \map \sin {\pi \alpha}$ Therefore: :$\displaystyle \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \frac 1 {\size {\map \sin {\pi \alpha} } }$ We know that the sine function is concave on $\closedint 0 {\pi / 2}$. So: :$\map \sin {\pi \alpha} \ge 2 \alpha$ for $\alpha \in \closedint 0 {1/2}$. By definition there exists $n \in \Z$ such that: :$\alpha = n + \norm \alpha$ and: :$\norm \alpha \in \closedint 0 {\pi/2}$. So: {{begin-eqn}} {{eqn | l = \size {\map \sin {\pi \alpha} }       | r = \size {\sin {\pi \norm \alpha} }       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = \map \sin {\pi \norm \alpha}       | c = as $\sin x > 0$ for $x \in \closedint 0 {\pi/2}$ }} {{eqn | o = \ge       | r = 2 \norm \alpha       | c =  }} {{end-eqn}} This completes the proof.
User:Yohji kinomoto
User:Yohji kinomoto 2 10774 56649 2011-06-08T20:58:56Z Yohji kinomoto 404 Created page with "'''<big><big>日本への栄光</big>></big>'''" wikitext text/x-wiki '''<big><big>日本への栄光</big>></big>''' jekumnbz28ww4r0l5dep1423smbmjhk
Vinogradov's Theorem/Major Arcs/Lemma 1
== Lemma ==  Let $\phi$ be the Euler $\phi$ function. Let $\mu$ be the Möbius function. Let $c_q$ be the Ramanujan sum modulo $q$. Let $P, N \ge 1$. {{Explain|what are $P$ and $N$: real, integer, ...?}} Let: :$\displaystyle \map {\SS_P} N := \sum_{q \mathop \le P} \frac {\map \mu q \map {c_q} N} {\map \phi q^3}$ :$\displaystyle \map \SS N := \lim_{P \mathop \to \infty} \map {\SS_P} N$ Then: :$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon -1} }$ and $\SS$ has the Euler product: :$\displaystyle \map \SS N = \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} } \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} }$ where: :$p \nmid N$ denotes that $p$ is not a divisor of $N$ :$p \divides N$ denotes that $p$ is a divisor of $N$.  == Proof == We have: {{begin-eqn}} {{eqn | l = \size {\map {c_q} N}       | o = \le        | r = \sum_{\substack {1 \mathop \le a \mathop \le q \\ \map \gcd {a, q} \mathop = 1} } 1       | c = Triangle Inequality }} {{eqn | r = \map \phi q       | c = Definition of Euler $\phi$ Function }} {{end-eqn}} Trivially we have $\size {\map \mu q} \le 1$. Therefore: {{begin-eqn}} {{eqn | l = \size {\map \SS N - \map {\SS_P} N}       | r = \sum_{q \mathop > P} \frac {\size {\map \mu q \map {c_q} N} } {\map \phi q^3}       | c =  }} {{eqn | o = \le       | r = \sum_{q \mathop > P} \frac 1 {\map \phi q^2}       | c =  }} {{eqn | o = \le       | r = \sum_{q \mathop > P} q^{\epsilon - 2}       | c = for large $P$ and $\epsilon \in \R_{>0}$, by Asymptotic Growth of Euler Phi Function }} {{eqn | o = \le       | r = P^{\delta - 1} \sum_{q \mathop > P} q^{\epsilon - \delta - 1}       | c = for any $\delta > \epsilon$ }} {{end-eqn}} Therefore by Convergence of Powers of Reciprocals: :$\map \SS N = \map {\SS_P} N + \map \OO {P^{\epsilon - 1} }$ as claimed. By Euler Product, and because $\map \mu {p^k} = 0$ for $k > 1$: :$\displaystyle \map \SS N = \prod_p \paren {1 + \frac {\map \mu q \map {c_q} N} {\map \phi q^3} }$ Now for a prime $p$ we have: :$\map \mu p = -1$ :$\map \phi p = p - 1$ We also have Kluyver's Formula for Ramanujan's Sum: :$\displaystyle \map {c_p} n = \sum_{d \mathop \divides \map \gcd {p, n} } \rd \map \mu {\frac p d}$ Let $p \divides N$. Then: $\map \gcd {p, N} = p$ which gives: :$\map {c_p} N = p \map \mu 1 + \map \mu p = p - 1$ Let $p \nmid N$. Then: :$\map \gcd {p, N} = 1$ and so: :$\map {c_p} N = -1$ Therefore: :$\displaystyle \map \SS N = \prod_{p \mathop \divides N} \paren {1 - \frac 1 {\paren {p - 1}^2} } \prod_{p \mathop \nmid N} \paren {1 + \frac 1 {\paren {p - 1}^3} }$ This completes the proof.
Axiom:Axiom of Limitation of Size
== Axiom ==  For any class $\CC$, a set $x$ such that $x = \CC$ exists  there is no bijection between $\CC$ and the universe.  {{explain|Check whether "universal class" is meant here}} Limitation of Size 7zgunwpct6381rvcoas04h8zhnakky0
Axiom:Class Comprehension Schema
== Axiom ==  For any formula $\phi$ containing no quantifiers over classes, there is a class $A$ such that: :$\forall x: \paren {x \in A \iff \map \phi x}$  Class Comprehension Schema bkn58cjsj43c7fdu63cibngfneffbfv
Bijection between Specific Elements
== Lemma == Let $A$ and $B$ be sets. Let $f: A \leftrightarrow B$ be a bijection. Then given $a \in A$ and $b \in B$, there exists another bijection $g: A \leftrightarrow B$ such that $\map g a = b$. == Proof == Let $\map f a = b'$. Since $f$ is surjective, there is $a' \in A$ such that $\map f {a'} = b$. We define the mapping $g: A \to B$ as: :$\map g x = \begin {cases} b & : x = a \\ b' & : x = a' \\ \map f x & : x \in A \setminus \set {a, a'} \end {cases}$ === $g$ is Surjective === Let $y \in B \setminus \set {b, b'}$. Then: :$\exists x \in A: \map f x = y$ and so by definition of $g$: :$\exists x \in A: \map g x = y$ Otherwise: :$y = b \implies b = \map g a$ :$y = b' \implies b = \map g {a'}$ and so $y$ is seen to be surjective. === $g$ is Injective === Suppose $\map g x = \map g {x'}$. If $\map g x, \map g {x'} \in B \setminus \set {b, b'}$ then $\map f x = \map f {x'}$. As $f$ is bijective and so injective it follows that $x = x'$. If $\map g x, \map g {x'} \in \set {b, b'}$ then either: :$\map g x = \map g {x'} = b$ and so $x = x' = a$ or :$\map g x = \map g {x'} = b'$ and so $x = x' = a'$ For the other two cases: :$\map g x \in \set {b, b'}, \map g {x'} \notin \set {b, b'}$ :$\map g x \notin \set {b, b'}, \map g {x'} \in \set {b, b'}$ it is clear that $\map g x \ne \map g {x'}$. So $g$ has been shown to be injective. === $g$ is Bijective === So $g$ is both injective and surjective, and so by definition a bijection.  == Comment == What this means is that given any two equivalent sets, it is always possible to find a bijection that connects any given pair of elements, one in one set and one in the other.
Tietze Extension Theorem/Lemma
== Lemma == Let $T = \left({S, \tau}\right)$ be a topological space which is normal. Let $A \subseteq S$ be closed in $T$.  Let $f: A \to \R$ be a continuous mapping such that $\left|{f \left({x}\right)}\right| \le 1$. Then there exists a continuous mapping $g: S \to \R$ such that: : $\forall x \in S: \left|{g \left({x}\right)}\right| \le \dfrac 1 3$ : $\forall x \in A: \left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$  == Proof == The sets $f^{−1} \left({\left({−\infty \,.\,.\, −\dfrac 1 3}\right]}\right)$ and $f^{−1} \left({\left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)$ are disjoint and closed in $A$. Since $A$ is closed, they are also closed in $S$. We have that $S$ is normal. Then by: : Urysohn's Lemma and: : the fact that $\left[{0 \,.\,.\, 1}\right]$ is homeomorphic to $\left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$ there exists a continuous mapping $g: S \to \left[{−\dfrac 1 3 \,.\,.\, \dfrac 1 3}\right]$ such that: :$g \left({f^{-1} \left({ \left({-\infty \,.\,.\, -\dfrac 1 3}\right]}\right)}\right) = -\dfrac 1 3$ and :$g \left({f^{-1} \left({ \left[{\dfrac 1 3 \,.\,.\, \infty}\right)}\right)}\right) = \dfrac 1 3$ Thus: : $\forall x \in S: \left|{g \left({x}\right)}\right| \le \dfrac 1 3$ Now if $−1 \le f \left({x}\right) \le −\dfrac 1 3$, then: :$g \left({x}\right) = −\dfrac 1 3$ and thus: :$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Similarly if $\dfrac 1 3 \le f \left({x}\right) \le 1$, then: :$g \left({x}\right) = \dfrac 1 3$ and thus: :$\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Finally, for $\left|{f \left({x}\right)}\right| \le \dfrac 1 3$ we have that: :$\left|{g \left({x}\right)}\right| \le \dfrac 1 3$ and so: : $\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$ Hence, for all $x \in S$: : $\left|{f \left({x}\right) − g \left({x}\right)}\right| \le \dfrac 2 3$  {{MissingLinks|Some of the statements above need to be justified.}} {{Namedfor|Heinrich Franz Friedrich Tietze}} == Sources == {{Planetmath|id = 5566|title = Tietze extension theorem}}
Finitely Many Reduced Associated Quadratic Irrationals
== Lemma == Let $n \in \Z: n > 0$. There are finitely many reduced quadratic irrationals which are associated to $n$. == Proof == <!-- This was taken from Continued Fraction Expansion of Irrational Square Root and may require some tinkering. --> By definition, we can write an arbitrary reduced irrational as $\alpha = \dfrac{\sqrt n + P} Q$. By definition, $\alpha > 1$ and its conjugate $\tilde{\alpha} > -1$. So we know $\alpha + \tilde{\alpha} = \dfrac {2P} Q > 0$. Hence with the assumption $Q > 0$ we have $P > 0$. Since $\tilde{\alpha} < 0$ we also have $P < \sqrt n$. Also, since $\alpha > 1$ by assumption, we have $Q < P + \sqrt n < 2 \sqrt n$. Thus there are finitely many choices for both $P$ and $Q$, forcing finitely many reduced quadratic irrationals associated to a fixed $n$. This amount is strictly bounded above by $2n$. {{explain|Why the word "strictly" is needed here - a quantity is bounded above or it's not. Also, clarify "amount".}}
Expansion of Associated Reduced Quadratic Irrational
== Lemma == Let $\alpha$ be a reduced quadratic irrational which is associated to $n$. Let $\alpha$ be transformed into its integer part and fractional part via: :$\alpha = \left \lfloor {\alpha} \right \rfloor + \dfrac 1{\alpha'}$ Then the resulting quadratic irrational $\alpha'$ is also reduced and associated to $n$. == Proof == Let $\alpha = \dfrac{\sqrt n + P} Q$ and $X = \left \lfloor {\alpha} \right \rfloor$ Then we have: :$\dfrac 1 {\alpha'} = \dfrac {\sqrt n - \left({Q X - P}\right)} Q$ Because $\sqrt n$ is irrational, we must have: :$\dfrac 1 {\alpha'} > 0$ Because $\dfrac 1 {\alpha'}$ is the fractional part: :$0 < \dfrac 1 {\alpha'} < 1 \implies \alpha' > 1$ Transforming: {{begin-eqn}} {{eqn | l = \alpha'       | r = \frac Q {\sqrt n - \left({Q X - P}\right)} \cdot \frac {\sqrt n + \left({Q X - P}\right)} {\sqrt n + \left({Q X - P}\right)}       | c =  }} {{eqn | r = \frac {\sqrt n + \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}       | c =  }} {{end-eqn}} we have: :$P' = Q X - P$ and: :$Q' = \dfrac 1 Q \left({n - \left({Q X - P}\right)^2}\right)$ To show $Q' \in \Z$: We have that: :$n - \left({Q X - P}\right)^2 \equiv n - P^2 \pmod Q$ Because $\alpha$ is associated to $n$, $Q$ must divide this quantity. Hence $Q'$ is an integer as defined. Because $X = \left \lfloor{\dfrac{\sqrt n + P} Q}\right \rfloor$ is an integer and $\alpha$ is irrational: :$X < \dfrac {\sqrt n + P} Q$ Hence: :$P' = QX - P < \sqrt n $ forcing: :$\tilde{\alpha}' < 0$ Because $\alpha > 1$: :$X \ge 1 \iff 0 \le X - 1$ Thus: {{begin-eqn}} {{eqn | l = \tilde \alpha = \frac {P - \sqrt n} Q       | o = <       | r = 0 \le X - 1       | c =  }} {{eqn | ll= \leadsto       | l = Q       | o = <       | r = \sqrt n + \left({Q X - P}\right)       | c =  }} {{eqn | ll= \leadsto       | l = Q \left({\sqrt n - \left({Q X - P}\right)}\right)       | o = <       | r = n - \left({Q X - P}\right)^2       | c =  }} {{eqn | ll= \leadsto       | l = -\tilde \alpha' = \frac {\sqrt n - \left({Q X - P}\right)} {\frac 1 Q \left({n - \left({Q X - P}\right)^2}\right)}       | o = <       | r = 1       | c =  }} {{end-eqn}} Hence $\tilde \alpha' > -1$ and $\alpha'$ is reduced. Because $Q' = \dfrac 1 Q \left({n - \left({P'}\right)^2}\right)$: :$n - \left({P'}\right)^2 \equiv Q Q' \equiv 0 \pmod {Q'}$ Hence $\alpha'$ is associated to $n$. Thus $\alpha'$ is both reduced and associated to $n$.
Axiom:Area Axioms
== Axioms of Area ==  When discussing areas of surfaces, we usually take the following statements to be axiomatic: :$(1) \quad$ The area of a square with a side of length one unit is defined to be one square unit. :$(2) \quad$ Let a surface be divided into a finite number of smaller non-overlapping surfaces. ::Let the smaller surfaces cover the entire larger surface. ::Then the sum of the areas of the smaller surfaces equals the area of the larger surface. :$(3) \quad$ Equal surfaces have equal areas.
Gelfond-Schneider Theorem/Lemma 1
Gelfond-Schneider Theorem/Lemma 1 0 11991 326204 175307 2017-11-11T10:19:20Z Barto 3079 "work to do" template wikitext text/x-wiki {{wtd|Links to Polynomial related results need to be resolved.}} == Lemma ==  Let $a_1 \left({t}\right), \ldots, a_n \left({t}\right)$ be non-zero polynomials in $\R \left[{t}\right]$ of degrees $d_1, \ldots, d_n$ respectively. Let $w_1, \ldots, w_n$ be pairwise distinct real numbers. Then: :$\displaystyle F \left({t}\right) = \sum_{j \mathop = 1}^n a_j \left({t}\right) e^{w_j t}$ has at most $d_1 + \cdots + d_n + n − 1$ real roots (counting multiplicities).  == Proof == By multiplying through by $e^{−w_n t}$ if necessary, we may suppose that $w_n = 0$ and that otherwise $w_j \ne 0$. Let $k = d_1 + \cdots + d_n + n$. We use strong induction on $k$. If $k = 1$, then $n = 1$ and $d_1 = 0$, and the lemma easily follows. Let $l \ge 2$ be such that the lemma holds whenever $k < l$. Suppose $k = l$. Let $N$ be the number of real roots of $F \left({t}\right)$. By Rolle's Theorem, the number of real roots of $F' \left({t}\right)$ is at least $N − 1$. On the other hand: :$\displaystyle F' \left({t}\right) = \sum_{j=1}^n b_j \left({t}\right) e^{w_j t}$ where: :$b_j \left({t}\right) = a'_j \left({t}\right) + w_j a_j \left({t}\right)$ Note that for $1 \le j \le n − 1$, we have that $b_j \left({t}\right)$ has degree exactly $d_j$. Also, since $w_n = 0$, either: : there are only $n−1$ non-zero polynomials $b_j \left({t}\right)$ in the expression for $F' \left({t}\right)$ above or: : there are $n$ such polynomials and the degree of $b_n \left({t}\right)$ is one less than the degree of $a_n \left({t}\right)$. We get from the induction hypothesis that $F' \left({t}\right)$ has at most $d_1 + \cdots + d_n + n − 2$ real roots. Hence, $N − 1 \le d_1 + \cdots + d_n + n − 2$, and the result follows. {{qed|lemma}}