hcju/proofwikiqa huggingface dataset:
Let $\map \rem {n, m} = r$. We see that as $n$ increases by $1$, then so does $r$, except when $n = m-1$ in which case increasing $n$ by $1$ makes $r$ go to $0$. We also take into account the case where $m = 0$: So we can define $\rem$ by cases: :$\map \rem {n, m} = \begin{cases} 0 & : \map \rem {n - 1, m} = m - 1 \text { or } m = 0 \text { or } n = 0\\ \map \rem {n - 1, m} + 1 & :  \text {otherwise} \\ \end{cases}$ We remind ourselves of the following primitive recursive functions: * Signum function $\sgn$ * Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$ * Cut-off subtraction $m \, \dot - \, n$. We have: * $\map \sgn m = 1 \iff m \ne 0$ * $\map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} } = 1 \iff \map \rem {n, m} \ne m \, \dot - \, 1$. So: :$\map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} }  = 1 \iff m > 0 \land \map \rem {n, m} \ne m \, \dot - \, 1$. So we see that: :$\map \rem {n + 1, m} = \paren {\map \rem {n, m} + 1} \, \map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} }$ So $\rem$ is obtained by primitive recursion (over the first variable, which is allowed by Permutation of Variables of Primitive Recursive Function) from the primitive recursive functions: * Signum function $\sgn$ * Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$ * Cut-off subtraction $m \, \dot - \, n$ * Addition * Multiplication. Thus we can use Definition by Cases is Primitive Recursive and it follows that $\rem$ is primitive recursive. {{qed}} Category:Primitive Recursive Functions 9zoo0k2s3rxz1h0j34qmq5ur3u5iph3
We note that if $m \ne 0$ and $n = m q + r$, we have: :$\dfrac n m = q + \frac r m$ Also note that $\dfrac n m$ and $\dfrac r m$ are rational numbers and not necessarily natural numbers. Indeed, we have: :$0 \le \dfrac r m < 1$ So if $m > 0$ then $\map {\operatorname {quot} } {n, m}$ is the floor $\floor {\dfrac n m}$ of $\dfrac n m$. So we have: :$\map {\operatorname {quot} } {n, m} = \begin{cases} 0 & : m = 0 \\ \floor {\dfrac n m} & : m \ne 0 \end{cases}$ Then we see that for $m \ne 0$: :$\floor {\dfrac {n + 1} m} = \begin {cases} \floor {\dfrac n m} + 1 & : m \divides \paren {n + 1} \\ \floor {\dfrac n m} & : \text {otherwise} \end {cases}$ So for $m \ne 0$: :$\map {\operatorname {quot} } {n + 1, m} = \begin{cases} \map {\operatorname {quot} } {n, m} + 1 & : \map \rem {n + 1, m} = 0 \\ \map {\operatorname {quot} } {n, m} & : \map \rem {n + 1, m} \ne 0 \end{cases}$ Now note that: :$\map {\overline {\sgn} } {\map \rem {n + 1, m} } = \begin{cases} 1 & : \map \rem {n + 1, m} = 0 \\ 0 & : \map \rem {n + 1, m} \ne 0 \end{cases}$ So the $\operatorname {quot}$ is defined as: :$\map {\operatorname {quot} } {0, m} = 0$ :$\map {\operatorname {quot} } {n + 1, m} = \map \sgn m \map {\operatorname {quot} } {n, m} + \map {\overline {\sgn} } {\map \rem {n + 1, m} }$ (note that the factor $\map \sgn m$ is needed to cover the case where $m = 0$). Thus $\operatorname {quot}$ is obtained  by primitive recursion (over the first variable, which is allowed by Permutation of Variables of Primitive Recursive Function) from the primitive recursive functions: :Signum function $\sgn$ :Remainder $\rem$ :Addition :Multiplication. So it follows that $\operatorname {quot}$ is primitive recursive. {{qed}} Category:Primitive Recursive Functions mtvzwief8480pk2jxmsosxrnawnk465
Immediate from Tarski's Geometry is Complete and the corollary to Gödel's First Incompleteness Theorem. {{qed}} Category:Tarski's Geometry ew232a0m24zm6l7im6b8stu8ykwizrx
{{begin-eqn}} {{eqn|l = \mathcal R^\leftrightarrow       |r = \mathcal R \cup \mathcal R^{-1}      |c =  Definition of Symmetric Closure }} {{eqn|r = \mathcal R \cup \mathcal R      |c = Inverse of Symmetric Relation is Symmetric }} {{eqn|r = \mathcal R      |c = Union is Idempotent }} {{end-eqn}} {{qed}} Category:Relation Theory ntp7xd7skcscvnph1k9bn2m9iccb53e
Consider the game of Matching Pennies. Recall its payoff table: {{:Definition:Matching Pennies/Payoff Table}} Trivially, by inspection, this has no entry which is the smallest entry in its row and the largest entry in its column. {{qed}}
Recall the payoff table: {{:Definition:Eluding Game/Payoff Table}} Trivially, by inspection, this has no entry which is the smallest entry in its row and the largest entry in its column.
Consider the game defined by the following payoff table: {{PayoffTable|table = $\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} } & B_1 & B_2 & B_3 \\ \hline A_1 & 1 & 2 & 4 \\ \hline A_2 & 4 & 2 & 1 \\ \hline \end{array}$}} This has two solutions: :$(1): \quad A: \left({1/3, 2/3}\right), B: \left({0, 1, 0}\right)$ :$(2): \quad A: \left({2/3, 1/3}\right), B: \left({0, 1, 0}\right)$ {{explain|The reason why the above are solutions is not indicated.}} It follows that the strategies: :$\left({\dfrac t 3, \dfrac {2 \left({1 - t}\right)} 3}\right), B: \left({0, 1, 0}\right)$ for all $0 \le t \le 1$ are also solutions. {{explain|Again, the explanation of this needs to wait till more learning has been done.}}
From Solution to Card Game with Bluffing, the optimum strategy for the [[Definition:Card Game with Bluffing|card game with bluffing, is: :For $A$, to use both pure strategies but favour $A_1$ by $2/3$ to $1/3$. :For $B$, to use both pure strategies but favour $B_1$ by $2/3$ to $1/3$. Hence the result by definition of completely mixed game.
The optimum strategy for matching pennies, for both players, is to randomise both their pure strategies. Hence the result by definition of completely mixed game.
Consider the game defined by the following payoff table: {{PayoffTable|table = $\begin{array} {r {{|}} c {{|}} } & B_1 & B_2 \\ \hline A_1 & 1 & 2 \\ \hline A_2 & 1 & 3  \\ \hline \end{array}$}} This has two solutions: :$(1): \quad A: \left({1, 0}\right), B: \left({1, 0}\right)$ :$(2): \quad A: \left({0, 1}\right), B: \left({1, 0}\right)$ {{explain|The reason why the above are solutions is not indicated.}} Thus both pure strategies for $A$ are optimal, but $A_1$ is dominated by $A_1$. {{qed}}
=== Sufficient Condition === Assume that $p$ is a prime element. By definition of the greatest element: :$p \preceq \top$ By definition of $\prec$: :$p \prec \top$ By definition of antisymmetry: :$\top \npreceq p$ By definition of lower closure of element: :$\top \notin p^\preceq$ By definition of relative complement: :$\top \in \relcomp S {p^\preceq}$ By definition :$X := \relcomp S {p^\preceq}$ is a non-empty set We will prove that :$X$ is filtered. Let $x, y \in X$. By definition of relative complement: :$x \notin p^\preceq$ and $y \notin p^\preceq$ By definition of lower closure of element: :$x \npreceq p$ and $y \npreceq p$ Then by definition of prime element: :$x \wedge y \npreceq p$ By definition of lower closure of element: :$x \wedge y \notin p^\preceq$ Thus by definition of relative complement: :$x \wedge y \in X$ Thus by Meet Precedes Operands: :$x \wedge y \preceq x$ and $x \wedge y \preceq y$ Thus :$\exists z \in X: z \preceq x \land z \preceq y$ {{qed|lemma}} We will prove that $X$ is an upper set. Let $x \in X, y \in S$ such that: :$x \preceq y$ By definition of relative complement: :$x \notin p^\preceq$ By definition of lower closure of element: :$x \npreceq p$ By definition of transitivity: :$y \npreceq p$ By definition of lower closure of element: :$y \notin p^\preceq$ Thud by definition of relative complement: :$y \in X$ {{qed|lemma}} Hence, by definition, $X$ is filter in $L$. {{qed|lemma}} === Necessary Condition === Assume that $\relcomp S {p^\preceq}$ is filter in $L$. Let $x, y \in S$ such that: :$x \wedge y \preceq p$ {{AimForCont}} that: :$x \npreceq p$ and $y \npreceq p$ By definition of lower closure of element: $x \notin p^\preceq$ and $y \notin p^\preceq$ By definition of relative complement: :$x \in \relcomp S {p^\preceq}$ and $y \in \relcomp S {p^\preceq}$ By definition of filtered: :$\exists z \in \relcomp S {p^\preceq}: z \preceq x \text { and } z \preceq y$ Then: $z \preceq x \wedge y$ By definition of transitivity: :$z \preceq p$ By definition of upper set: :$p \in \relcomp S {p^\preceq}$ By definition of reflexivity: :$p \preceq p$ By definition of lower closure of element: :$p \in p^\preceq$ This contradicts $p \in \relcomp S {p^\preceq}$. Hence it is not the case that $x \npreceq p$ and $y \npreceq p$. That is: :$x \preceq p$ or $y \preceq p$ and so, by definition, $p$ is a prime element of $\struct {S, \vee, \wedge, \preceq}$. {{qed}}
=== Sufficient Condition === Suppose that :$p$ is prime element Let $x \in S$ such that :$p \prec x$ By definition of Boolean lattice: :$L$ is complemented distributive lattice. By definition of complemented: :$\exists y \in S: y$ is complement of $x$. By definition of complement :$x \wedge y = \bot$ By definition of smallest element: :$x \wedge y \preceq p$ By definition of prime element: :$x \preceq p$ or $y \preceq p$ Then by definition of antisymmetry: :$y \prec x$ By definition of $\prec$: :$y \preceq x$ By definition of complement: :$x \vee y = \top$ Thus by Preceding iff Join equals Larger Operand: :$x = \top$ {{qed|lemma}} === Necessary Condition === Suppose that :$\forall x \in S: \left({ p \prec x \implies x = \top }\right)$ Let $x, y \in S$ such that :$x \wedge y \preceq p$ {{begin-eqn}} {{eqn | l = p | r = p \vee \left({x \wedge y}\right) | c = Preceding iff Join equals Larger Operand }} {{eqn | r = \left({p \vee x}\right) \wedge \left({p \vee y}\right) | c = definition of distributive lattice }} {{end-eqn}} Aiming for a contradiction suppose that :$x \npreceq p$ and $y \npreceq p$ By Preceding iff Join equals Larger Operand: :$p \ne p \vee y$ and $p \ne p \vee x$ By Join Succeeds Operands: :$p \preceq p \vee y$ and $p \preceq p \vee x$ By definition of $\prec$: :$p \prec p \vee y$ and $p \prec p \vee x$ By assumption: :$p \vee y = \top$ and $p \vee x = \top$ By Meet is Idempotent: :$p = \top$ This contradicts $p \ne \top$ {{qed}}
=== Sufficient Condition === Let $p$ be a prime element. By definition of continuous: :$\forall x \in S: x^\ll$ is directed We will prove that :$\forall x \in S: \left({ x \in \complement_S\left({p^\preceq}\right) \implies \exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x }\right)$ Let $x \in S$ such that :$x \in \complement_S\left({p^\preceq}\right)$ By definition of relative complement: :$x \notin p^\preceq$ By definition of lower closure of element: :$x \npreceq p$ By Axiom of Approximation in Up-Complete Semilattice: :$\exists y \in S: y \ll x \land y \npreceq p$ By definition of lower closure of element: :$y \notin p^\preceq$ By definition of relative complement: :$y \in \complement_S\left({p^\preceq}\right)$ Thus :$\exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x$ {{qed|lemma}} By Prime Element iff Complement of Lower Closure is Filter: :$F := \complement_S\left({p^\preceq}\right)$ is way below open filter in $L$. By definitions of antisymmetry and lower closure of element: :$\lnot \exists y \in S: y \in F \land y \prec p$ By Relative Complement of Relative Complement:: :$\complement_S\left({F}\right) = p^\preceq$ By definitions of reflexivity and lower closure of element: :$p \in \complement_S\left({F}\right)$ Thus by definition of greatest element: :$p = \max \left({\complement_S\left({F}\right)}\right)$ {{qed|lemma}} === Necessary Condition === Let :there exists a way below open filter $F$ in $L$: $p = \max \left({ \complement_S\left({F}\right) }\right)$ By Maximal Element of Complement of Filter is Meet Irreducible: :$p$ is meet irreducible. Thus by Prime Element iff Meet Irreducible in Distributive Lattice: :$p$ is a prime element. {{qed}}
Recursively define a sequence $\langle C_n \rangle$ in the power set of $S$ as follows: : $C_0 = S \setminus T$, the difference between $S$ and $T$. : $C_{n+1} = f \left[{C_n}\right]$, the image of $C_n$ under $f$. Let $\displaystyle C := \bigcup_{n \mathop \in \N} C_n$. Define a mapping $h: S \to T$ as follows: : $\displaystyle h \left({x}\right) = \begin{cases} f \left({x}\right) & : x \in C \\ x & : x \notin C \end{cases}$ === $h$ is a mapping from $S$ to $T$ === By Law of Excluded Middle, for each $x \in S$: : $x \in C$ or: : $x \notin C$ Thus the construction produces a mapping on S. It remains to be shown that: : $\forall x \in S: h \left({x}\right) \in T$ Let $x \in C$. Then: : $h \left({x}\right) = f \left({x}\right) \in T$ Let $x \notin C$. Then by Set is Subset of Union: : $x \notin C_0$ As $x \in S$: : $x \in S \setminus C_0$ by the definition of set difference. By Relative Complement of Relative Complement: : $S \setminus C_0 = T$ Thus $x \in T$ by the definition of subset. As $h \left({x}\right) = x$ in this case: :$h \left({x}\right) \in T$ {{qed|lemma}} === $h$ is injective === Let $x, y \in S$. Suppose that $h \left({x}\right) = h \left({y}\right)$. By Law of Excluded Middle for Two Variables: : $\left({x \in C}\right) \land \left({y \in C}\right)$ or: : $\left({x \notin C}\right) \land \left({y \notin C}\right)$ or: : $\left({x \in C}\right) \land \left({y \notin C}\right)$ or: : $\left({x \notin C}\right) \land \left({y \in C}\right)$ Let $x, y \in C$. Then $x = y$ because $f$ is injective. Let $x, y \notin C$. Then $x = y$ by Identity Mapping is Injection. Let $x \in C$ and $y \notin C$. Then $x \in C_n$ for some $n$ by the definition of union. By Set is Subset of Union: :$h \left({x}\right) = f \left({x}\right) \in C_{n+1} \subseteq C$  Thus $h \left({x}\right) \in C$. :$h \left({y}\right) = y$ by the definition of $h$. Since $y \notin C$, this contradicts the assumption that $h \left({x}\right) = h \left({y}\right)$. The argument for the case of $x \notin C$ and $y \in C$ is identical to the preceding. Thus $h \left({x}\right) = h \left({y}\right) \implies x = y$ for all $x, y \in S$, so $h$ is injective. {{qed|lemma}} === $h$ is surjective === Let $y \in T$. By Law of Excluded Middle: : $y \in C$ or: : $y \notin C$ Let $y \notin C$. Then: : $h \left({y}\right) = y$ Let $y \in C$. Then as $y \notin C_0 = S \setminus T$: : $\exists n \in \N_{>0}: y \in C_{n+1}$ Thus: : $\exists x \in C_n: y = f \left({x}\right)$ By the definition of union: : $x \in C$ Thus: : $h \left({x}\right) = f \left({x}\right) = y$ Thus: :$\forall y \in T: \exists x \in S: h \left({x}\right) = y$ Thus by definition, $h$ is surjective. {{qed}}
Define a mapping $E: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ as: : $E \left({K}\right) = S \setminus \left({T \setminus f \left[{K}\right]}\right)$ where $f \left[{K}\right]$ is the image of $K$ under $f$. Then: : $E \left({K}\right) = \left({S \setminus T}\right) \cup f \left[{K}\right]$ By Image of Subset under Relation is Subset of Image and the corollary to Set Union Preserves Subsets, $E$ is increasing. Thus by Knaster-Tarski Lemma: Power Set, $E$ has a fixed point $X$. Then by the definition of fixed point: : $E \left({X}\right) = X$ That is: : $\left({S \setminus \left({T \setminus f \left[{X}\right]}\right)}\right) = X$ Taking the set difference from $S$: : $T \setminus f \left[{X}\right] = S \setminus X$ Let $f'$ be the restriction of $f$ to $X \times f \left[{X}\right]$. By Injection to Image is Bijection, $f'$ is a bijection. By Identity Mapping is Bijection, the identity mapping $I_{S \mathop \setminus X}$ from $S \setminus X$ to $T \setminus f \left[{X}\right]$ is a bijection. Thus by Union of Bijections with Disjoint Domains and Codomains is Bijection, $g = f' \cup i$ is the bijection we seek. {{qed}}
=== Sufficient Condition === Assume that :$I$ is a prime ideal. Let $x, y \in S$ such that :$x \wedge y \in I$ By definition of relative complement: :$x \wedge y \notin \relcomp S I$ By definition of prime ideal: :$\relcomp S I$ is filter in $L$. By Filtered in Meet Semilattice: :$x \notin \relcomp S I$ or $y \notin \relcomp S I$ Thus by definition of relative complement: :$x \in I$ or $y \in I$ {{qed|lemma}} === Necessary Condition === Assume that :$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$ Define $F := \relcomp S I$. By definition of proper subset: :$\exists x_0 \in S: x_0 \notin I$ By definition of relative complement: :$x_0 \in F$ Thus by definition :$F$ is a non-empty set. We will prove that :$F$ is filtered. Let $x, y \in F$. By definition of relative complement: :$x \notin I$ and $y \notin I$ By assumption: :$x \wedge y \notin I$ By definition of relative complement: :$x \wedge y \in F$ By Meet Precedes Operands: :$x \wedge y \preceq x$ and $x \wedge y \preceq y$ Thus :$\exists z \in F: z \preceq x \land z \preceq y$ {{qed|lemma}} We will prove that: :$F$ is an upper set. Let $x \in F, y \in S$ such that :$x \preceq y$ By definition of lower set: :$y \in I \implies x \in I$ By definition of relative complement: :$x \notin I$ Then :$y \notin I$ Thus by definition of relative complement: :$y \in F$ {{qed|lemma}} Thus by definition: :$F$ is a filter in $L$. {{qed}}
By definition, a strictly competitive game is a game in which the interests of each player are diametrically opposed. Recall the payoff table of Matching Pennies: {{:Definition:Matching Pennies/Payoff Table}} It can be seen by inspection that exchanging $\text A$ and $\text B$ results in exactly the same entries in the payoff table. From the nature of a payoff table it follows that the fortunes of each player are opposite and equal. Hence the result. {{qed}}
As $e$ equalises $f$ and $g$, $f \circ e = g \circ e$. Since $e$ is epic, it follows that $f = g$. Then in the equaliser diagram:  :::$\begin{xy}\xymatrix{  E    \ar[r]^*{e} &  A   \ar[r]<2pt>^*{f}   \ar[r]<-2pt>_*{g}  &  B \\  A   \ar@{.>}[u]^*{k}   \ar[ur]_*{\operatorname{id}_A} }\end{xy}$ We have that: :$f \circ 1_A = g \circ 1_A$ so there is a unique $k$ with $e \circ k = \operatorname{id}_A$. Then: {{begin-eqn}} {{eqn | l = e \circ k \circ e       | r = \operatorname {id}_A \circ e       | c =  }} {{eqn | r = e       | c =  }} {{eqn | r = e       | c = e \circ \operatorname{id}_E }} {{end-eqn}} By Equalizer is Monomorphism, it follows that $k \circ e = \operatorname{id}_E$. This gives $k$ as an inverse to $e$. Thus $e$ is an isomorphism. {{qed}}
We have that: :$A$ is a a set of extended real numbers :$A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$. From this follows by Infimum of Real Subset: :$\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$ Let $\epsilon \in \R_{>0}$. Then: {{begin-eqn}} {{eqn | l = \map \inf {A \cap \R}       | o = \in       | r = \R }} {{eqn | ll= \leadsto       | l = \exists x       | o = \in       | r = \paren {A \cap \R}: x - b < \epsilon       | c = Infimum of Subset of Real Numbers is Arbitrarily Close }} {{end-eqn}} {{qed}} Category:Extended Real Numbers 35yrzle13fpml1vc5r1lhrs1egxujts
Let: :$I, J \in S_x$ :$J \subset I$ :$\map {\omega_f} I \in \R$ where: :$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ We need to prove: :$\map {\omega_f} J \in \R$ :$\map {\omega_f} J \le \map {\omega_f} I$ We intend to prove that $\map {\omega_f} J \in \R$. We start by proving that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above and non-empty. We have that $J$ is a subset of $I$. Therefore: :$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$. The statement $\map {\omega_f} I \in \R$ means that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum. Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above. Accordingly: :$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above as $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$. Therefore, $0 \in \set {\size {\map f y - \map f z}: y, z \in J \cap D}$ as $x \in J \cap D$. Accordingly: :$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty We have that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a real set as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$. We have shown that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty and bounded above. Therefore, $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum by the Continuum Property. In other words: {{begin-eqn}} {{eqn | l = \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}       | o = \in       | r = \R }} {{eqn | ll= \leadstoandfrom       | l = \map {\omega_f} J       | o = \in       | r = \R       | c = definition }} {{end-eqn}} We finished proving that $\map {\omega_f} J \in \R$. It remains to prove that $\map {\omega_f} J \le \map {\omega_f} I$. We have: :$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ :$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum :$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum Then: {{begin-eqn}} {{eqn | l = \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}       | o = \le       | r = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}       | c = Supremum of Subset }} {{eqn | ll= \leadstoandfrom       | l = \map {\omega_f} J       | o = \le       | r = \map {\omega_f} I       | c = definitions }} {{end-eqn}} {{qed}} Category:Real Analysis gnz0ox5scu4o4yjgcwbre4uui50klar
By long division: <pre>          0.0201030407111829      ---------------------- 9899 ) 199.0000000000000000        197 98        --- --          1 0200            9899          - ----             30100             29697             -----               40300               39596               -----                 70400                 69293                 -----                  11070                   9899                  -----                   11710                    9899                   -----                    18110                     9899                    -----                     82110                     79192                     -----                      29180                      19798                      -----                       93820                       89091                       -----                        4729 </pre> By Generating Function for Lucas Numbers: {{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 0} L_k 10^{-2 k - 2}       | r = \frac {2 - 10^{-2} } {1 - 10^{-2} - 10^{-4} } \times 10^{-2} }} {{eqn | r = \frac {2 \times 10^2 - 1} {10^4 - 10^2 - 1} }} {{eqn | r = \frac {199} {9899} }} {{end-eqn}} The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry. {{qed}}
By definition, a transcendental number (in this context) is a real number which is not an algebraic number. Recall that the Real Numbers are Uncountable. Also recall that the Algebraic Numbers are Countable. The result follows from Uncountable Set less Countable Set is Uncountable. {{qed}}
By definition, a transcendental number (in this context) is a real number which is not an algebraic number. Recall that the Real Numbers are Uncountable. Also recall that the Algebraic Numbers are Countable. Thus the subset of all real numbers which are not transcendental is countable. The result follows by the definition of '''almost all'''. {{qed}}