id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-11600 | Summation of i from 1 to n of Summation of j from 1 to i | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$ | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$
can be expressed as:
:$\ds \sum_{\map R i} \sum_{\map S {i, j} } a_{i j}$
where:
:$\map R i$ is the propositional function $1 \le i \le n$
:$\map S {i, j}$ is the propositional function $1 \le j \le i$
We wish to find a propositional function $\map {S'} j$ which is t... | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$ | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$
can be expressed as:
:$\ds \sum_{\map R i} \sum_{\map S {i, j} } a_{i j}$
where:
:$\map R i$ is the [[Definition:Propositional Function|propositional function]] $1 \le i \le n$
:$\map S {i, j}$ is the [[Definition:Propositional Function|propositional function]] $1 \l... | Summation of i from 1 to n of Summation of j from 1 to i/Proof 1 | https://proofwiki.org/wiki/Summation_of_i_from_1_to_n_of_Summation_of_j_from_1_to_i | https://proofwiki.org/wiki/Summation_of_i_from_1_to_n_of_Summation_of_j_from_1_to_i/Proof_1 | [
"Summations",
"Summation of i from 1 to n of Summation of j from 1 to i"
] | [] | [
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Exchange of Order of Summation with Dependency on Both Indices"
] |
proofwiki-11601 | Summation of i from 1 to n of Summation of j from 1 to i | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j}
| r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le i \le n}\right] \left[{1 \le j \le i}\right]
| c =
}}
{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le i \le n}\right]
| c =
}}
{{eqn | r = \sum_{i, ... | :$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$ | {{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j}
| r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le i \le n}\right] \left[{1 \le j \le i}\right]
| c =
}}
{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le i \le n}\right]
| c =
}}
{{eqn | r = \sum_{i, ... | Summation of i from 1 to n of Summation of j from 1 to i/Proof 2 | https://proofwiki.org/wiki/Summation_of_i_from_1_to_n_of_Summation_of_j_from_1_to_i | https://proofwiki.org/wiki/Summation_of_i_from_1_to_n_of_Summation_of_j_from_1_to_i/Proof_2 | [
"Summations",
"Summation of i from 1 to n of Summation of j from 1 to i"
] | [] | [] |
proofwiki-11602 | Permutation of Indices of Summation/Infinite Series | Let the fiber of truth of $R$ be infinite.
Let $\ds \sum_{\map R i} a_i$ be absolutely convergent.
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$ | This is a restatemtent of Manipulation of Absolutely Convergent Series: Permutation in the context of summations.
{{qed}} | Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Infinite Set|infinite]].
Let $\ds \sum_{\map R i} a_i$ be [[Definition:Absolutely Convergent Series|absolutely convergent]].
Then:
:$\ds \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$ | This is a restatemtent of [[Manipulation of Absolutely Convergent Series/Permutation|Manipulation of Absolutely Convergent Series: Permutation]] in the context of [[Definition:Summation by Propositional Function|summations]].
{{qed}} | Permutation of Indices of Summation/Infinite Series | https://proofwiki.org/wiki/Permutation_of_Indices_of_Summation/Infinite_Series | https://proofwiki.org/wiki/Permutation_of_Indices_of_Summation/Infinite_Series | [
"Summations"
] | [
"Definition:Fiber of Truth",
"Definition:Infinite Set",
"Definition:Absolutely Convergent Series"
] | [
"Manipulation of Absolutely Convergent Series/Permutation",
"Definition:Summation/Propositional Function"
] |
proofwiki-11603 | Sum of Summations equals Summation of Sum | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$ | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
| r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
| c = Definition of $b_i$ and $c_i$
}}
{{eqn | r = \sum_{\map R i} \paren {\sum_{j \mathop = 1}^2 a_{i j} }
| c = {{Defof|Summation}}
}}
{{eqn... | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let $\ds \sum_{\map R i} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [... | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
| r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
| c = Definition of $b_i$ and $c_i$
}}
{{eqn | r = \sum_{\map R i} \paren {\sum_{j \mathop = 1}^2 a_{i j} }
| c = {{Defof|Summation}}
}}
{{e... | Sum of Summations equals Summation of Sum | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum | [
"Summations"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Summation/Propositional Function",
"Definition:Fiber of Truth",
"Definition:Finite Set"
] | [
"Exchange of Order of Summation"
] |
proofwiki-11604 | Sum of Summations equals Summation of Sum | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$ | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
| r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
| c = by definition
}}
{{eqn | r = \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }
| c = {{Defof|Summation}}
}}
{{eqn |... | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let $\ds \sum_{\map R i} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [... | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
| r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
| c = by definition
}}
{{eqn | r = \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }
| c = {{Defof|Summation}}
}}
{{eqn... | Sum of Summations equals Summation of Sum/Infinite Sequence/Proof 1 | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum/Infinite_Sequence/Proof_1 | [
"Summations"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Summation/Propositional Function",
"Definition:Fiber of Truth",
"Definition:Finite Set"
] | [
"Exchange of Order of Summation/Finite and Infinite Series"
] |
proofwiki-11605 | Sum of Summations equals Summation of Sum | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \sum_{\map R i} x_i$ denote a summation over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$ | By definition, $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ are sequences in $\R$.
Hence the result as an instance of Sum Rule for Real Sequences.
{{qed}} | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let $\ds \sum_{\map R i} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [... | By definition, $\ds \sum_{\map R i} b_i$ and $\ds \sum_{\map R i} c_i$ are [[Definition:Real Sequence|sequences]] in $\R$.
Hence the result as an instance of [[Sum Rule for Real Sequences]].
{{qed}} | Sum of Summations equals Summation of Sum/Infinite Sequence/Proof 2 | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum | https://proofwiki.org/wiki/Sum_of_Summations_equals_Summation_of_Sum/Infinite_Sequence/Proof_2 | [
"Summations"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Summation/Propositional Function",
"Definition:Fiber of Truth",
"Definition:Finite Set"
] | [
"Definition:Real Sequence",
"Combination Theorem for Sequences/Real/Sum Rule"
] |
proofwiki-11606 | Sum of Summations over Overlapping Domains | :$\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j = \sum_{\map R j \mathop \lor \map S j} a_j + \sum_{\map R j \mathop \land \map S j} a_j$
where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively. | {{begin-eqn}}
{{eqn | l = \sum_{\map R j} a_j + \sum_{\map S j} a_j
| r = \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j} + \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j}
| c = {{Defof|Summation by Iverson's Convention}}
}}
{{eqn | r = \sum_{j \mathop \in \Z} a_j \paren {\sqbrk {\map R j} + \sqbrk {\map R j} }... | :$\ds \sum_{\map R j} a_j + \sum_{\map S j} a_j = \sum_{\map R j \mathop \lor \map S j} a_j + \sum_{\map R j \mathop \land \map S j} a_j$
where $\lor$ and $\land$ signify [[Definition:Disjunction|logical disjunction]] and [[Definition:Conjunction|logical conjunction]] respectively. | {{begin-eqn}}
{{eqn | l = \sum_{\map R j} a_j + \sum_{\map S j} a_j
| r = \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j} + \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j}
| c = {{Defof|Summation by Iverson's Convention}}
}}
{{eqn | r = \sum_{j \mathop \in \Z} a_j \paren {\sqbrk {\map R j} + \sqbrk {\map R j} }... | Sum of Summations over Overlapping Domains/Proof | https://proofwiki.org/wiki/Sum_of_Summations_over_Overlapping_Domains | https://proofwiki.org/wiki/Sum_of_Summations_over_Overlapping_Domains/Proof | [
"Summations",
"Sum of Summations over Overlapping Domains"
] | [
"Definition:Disjunction",
"Definition:Conjunction"
] | [
"Cardinality of Set Union"
] |
proofwiki-11607 | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2 | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | Let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j
| c =
}}
{{eqn | r = \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j
| c = Summation of i from 1 to n of Summation of j from 1 to i
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop... | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | Let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j
| c =
}}
{{eqn | r = \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j
| c = [[Summation of i from 1 to n of Summation of j from 1 to i]]
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \ma... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2/Proof_1 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Summation of i from 1 to n of Summation of j from 1 to i",
"Change of Index Variable of Summation",
"Sum of Summations over Overlapping Domains/Example",
"Sum of Summations equals Summation of Sum",
"General Distributivity Theorem",
"Change of Index Variable of Summation"
] |
proofwiki-11608 | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2 | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x... | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$
From [[Summation of Products of n Numbers taken m at a time with Repetitions]]:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} ... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2/Proof_2 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] |
proofwiki-11609 | Uniqueness of Continuously Differentiable Solution to Initial Value Problem | Let $D \subseteq \R^2$ be a region containing $\tuple {a, b}$.
Let $f: D \to \R$ be real-valued mapping such that $f$ and $\dfrac {\partial f} {\partial x}$ are continuous on $D$.
Consider the initial value problem:
:$\dfrac {\d x} {\d t} = \map f {t, x}$
:$\map x a = b$
Suppose the initial value problem above has a so... | {{ProofWanted|Is this the same as Picard's Little Theorem?}} | Let $D \subseteq \R^2$ be a [[Definition:Region of Plane|region]] containing $\tuple {a, b}$.
Let $f: D \to \R$ be [[Definition:Real Function of Two Variables|real-valued mapping]] such that $f$ and $\dfrac {\partial f} {\partial x}$ are [[Definition:Continuous Real-Valued Vector Function|continuous]] on $D$.
Conside... | {{ProofWanted|Is this the same as Picard's Little Theorem?}} | Uniqueness of Continuously Differentiable Solution to Initial Value Problem | https://proofwiki.org/wiki/Uniqueness_of_Continuously_Differentiable_Solution_to_Initial_Value_Problem | https://proofwiki.org/wiki/Uniqueness_of_Continuously_Differentiable_Solution_to_Initial_Value_Problem | [
"First Order ODEs"
] | [
"Definition:Region/Plane",
"Definition:Real Function/Two Variables",
"Definition:Continuous Real-Valued Vector Function",
"Definition:Initial Value Problem",
"Definition:Initial Value Problem",
"Definition:Differential Equation/Solution",
"Definition:Real Interval",
"Definition:Unique "
] | [] |
proofwiki-11610 | Exhausting Sequence of Sets on the Strictly Positive Real Numbers | For each $k \in \N$, let $S_k = \openint {\dfrac 1 k} k$.
Then $\sequence {S_k}_k$ is an exhausting sequence of sets on $\R_{>0}$. | To prove that $\sequence {S_k}_k$ is exhausting $\R_{>0}$, it is sufficient to show:
:$(1): \quad \forall k \in \N: S_k \subseteq S_{k + 1}$
:$(2): \quad \ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ | For each $k \in \N$, let $S_k = \openint {\dfrac 1 k} k$.
Then $\sequence {S_k}_k$ is an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]] on $\R_{>0}$. | To prove that $\sequence {S_k}_k$ is [[Definition:Exhausting Sequence of Sets|exhausting]] $\R_{>0}$, it is sufficient to show:
:$(1): \quad \forall k \in \N: S_k \subseteq S_{k + 1}$
:$(2): \quad \ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ | Exhausting Sequence of Sets on the Strictly Positive Real Numbers | https://proofwiki.org/wiki/Exhausting_Sequence_of_Sets_on_the_Strictly_Positive_Real_Numbers | https://proofwiki.org/wiki/Exhausting_Sequence_of_Sets_on_the_Strictly_Positive_Real_Numbers | [
"Real Numbers"
] | [
"Definition:Exhausting Sequence of Sets"
] | [
"Definition:Exhausting Sequence of Sets",
"Definition:Exhausting Sequence of Sets"
] |
proofwiki-11611 | Way Below in Ordered Set of Topology | Let $\struct {S, \tau}$ be a topological space.
Let $\struct {\tau, \preceq}$ be an ordered set
where $\preceq \mathop = \subseteq \restriction_{\tau \times \tau}$
Let $x, y \in \tau$.
Then:
:$x \ll y$ in $\struct {\tau, \preceq}$
{{iff}}
:for every set $F$ of open subsets of $S$: if $y \subseteq \bigcup F$, then:
::th... | === Sufficient Condition ===
Let
:$x \ll y$
Let $F$ be a set of open subsets of $S$ such that
:$y \subseteq \bigcup F$
By definition of subset:
:$F \subseteq \tau$
By proof of Topology forms Complete Lattice:
:$\sup F = \bigcup F$
By assumption:
:$y \preceq \sup F$
By Topology forms Complete Lattice:
:$\struct {\tau, \... | Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\struct {\tau, \preceq}$ be an [[Definition:Ordered Set|ordered set]]
where $\preceq \mathop = \subseteq \restriction_{\tau \times \tau}$
Let $x, y \in \tau$.
Then:
:$x \ll y$ in $\struct {\tau, \preceq}$
{{iff}}
:for every [[Defi... | === Sufficient Condition ===
Let
:$x \ll y$
Let $F$ be a [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$ such that
:$y \subseteq \bigcup F$
By definition of [[Definition:Subset|subset]]:
:$F \subseteq \tau$
By proof of [[Topology forms Complete Lattice]... | Way Below in Ordered Set of Topology | https://proofwiki.org/wiki/Way_Below_in_Ordered_Set_of_Topology | https://proofwiki.org/wiki/Way_Below_in_Ordered_Set_of_Topology | [
"Way Below Relation",
"Topology"
] | [
"Definition:Topological Space",
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Element is Way Below"
] | [
"Definition:Set of Sets",
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Subset",
"Topology forms Complete Lattice",
"Topology forms Complete Lattice",
"Definition:Complete Lattice",
"Way Below in Complete Lattice",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition... |
proofwiki-11612 | Continuous Inverse Theorem | Let $f$ be a real function defined on an interval $I$.
Let $f$ be strictly monotone and continuous on $I$.
Let $g$ be the inverse mapping to $f$.
Let $J := f \left[{I}\right]$ be the image of $I$ under $f$.
Then $g$ is strictly monotone and continuous on $J$. | From Strictly Monotone Real Function is Bijective, $f$ is a bijection.
From Inverse of Strictly Monotone Function, $f^{-1} : J \to I$ exists and is strictly monotone.
From Surjective Monotone Function is Continuous, $f^{-1}$ is continuous.
Hence the result.
{{qed}} | Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
Let $f$ be [[Definition:Strictly Monotone Real Function|strictly monotone]] and [[Definition:Continuous on Interval|continuous]] on $I$.
Let $g$ be the [[Definition:Inverse of Mapping|inverse mapping]] to ... | From [[Strictly Monotone Real Function is Bijective]], $f$ is a [[Definition:Bijection|bijection]].
From [[Inverse of Strictly Monotone Function]], $f^{-1} : J \to I$ exists and is [[Definition:Strictly Monotone Real Function|strictly monotone]].
From [[Surjective Monotone Function is Continuous]], $f^{-1}$ is [[Defi... | Continuous Inverse Theorem | https://proofwiki.org/wiki/Continuous_Inverse_Theorem | https://proofwiki.org/wiki/Continuous_Inverse_Theorem | [
"Real Analysis",
"Continuity",
"Named Theorems"
] | [
"Definition:Real Function",
"Definition:Real Interval",
"Definition:Strictly Monotone/Real Function",
"Definition:Continuous Real Function/Interval",
"Definition:Inverse of Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Strictly Monotone/Real Function",
"Definition:Continuous R... | [
"Strictly Monotone Real Function is Bijective",
"Definition:Bijection",
"Inverse of Strictly Monotone Function",
"Definition:Strictly Monotone/Real Function",
"Surjective Monotone Function is Continuous",
"Definition:Continuous Real Function/Interval"
] |
proofwiki-11613 | Logarithm is Strictly Concave | :$\ln x: x > 0$ strictly concave. | From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing on $x > 0$.
From Second Derivative of Natural Logarithm Function:
:$D^2 \ln x = -\dfrac 1 {x^2}$
Thus $D^2 \ln x$ is strictly negative on $x > 0$ (in fact is strictly negative for all $x \ne 0$).
Thus from Derivative of Monotone Function, $D \dfrac 1... | :$\ln x: x > 0$ [[Definition:Strictly Concave Real Function|strictly concave]]. | From [[Logarithm is Strictly Increasing]], $\ln x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $x > 0$.
From [[Second Derivative of Natural Logarithm Function]]:
:$D^2 \ln x = -\dfrac 1 {x^2}$
Thus $D^2 \ln x$ is [[Definition:Strictly Negative|strictly negative]] on $x > 0$ (in fact is... | Logarithm is Strictly Concave | https://proofwiki.org/wiki/Logarithm_is_Strictly_Concave | https://proofwiki.org/wiki/Logarithm_is_Strictly_Concave | [
"Logarithms"
] | [
"Definition:Strictly Concave Real Function"
] | [
"Logarithm is Strictly Increasing",
"Definition:Strictly Increasing/Real Function",
"Second Derivative of Natural Logarithm Function",
"Definition:Strictly Negative",
"Definition:Strictly Negative",
"Derivative of Monotone Function",
"Definition:Strictly Decreasing/Real Function",
"Real Function is St... |
proofwiki-11614 | Exponential is Strictly Convex | :The function $f \left({x}\right) = \exp x$ is strictly convex. | By definition, the exponential function is the inverse of the natural logarithm function.
From Logarithm is Strictly Increasing, $\ln x$ is strictly increasing.
From Logarithm is Strictly Concave, $\ln x$ is strictly concave.
The result follows from Inverse of Strictly Increasing Strictly Concave Real Function is Stric... | :The [[Definition:Real Function|function]] $f \left({x}\right) = \exp x$ is [[Definition:Strictly Convex Real Function|strictly convex]]. | By definition, the [[Definition:Real Exponential Function|exponential function]] is the [[Definition:Inverse Mapping|inverse]] of the [[Definition:Natural Logarithm|natural logarithm function]].
From [[Logarithm is Strictly Increasing]], $\ln x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
... | Exponential is Strictly Convex | https://proofwiki.org/wiki/Exponential_is_Strictly_Convex | https://proofwiki.org/wiki/Exponential_is_Strictly_Convex | [
"Exponential Function"
] | [
"Definition:Real Function",
"Definition:Strictly Convex Real Function"
] | [
"Definition:Exponential Function/Real",
"Definition:Inverse Mapping",
"Definition:Natural Logarithm",
"Logarithm is Strictly Increasing",
"Definition:Strictly Increasing/Real Function",
"Logarithm is Strictly Concave",
"Definition:Strictly Concave Real Function",
"Inverse of Strictly Increasing Strict... |
proofwiki-11615 | Way Below in Complete Lattice | Let $\struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, y \in S$.
Then
:$x \ll y$
{{iff}}
:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in \map {\operatorname {Fin} } X: x \preceq \sup A$
where
:$\ll$ denotes the way below relation,
:$\map {\operatorname {Fin} } X$ denotes the set of all... | === Sufficient Condition ===
Let
:$x \ll y$
Let $X \subseteq S$ such that
:$y \preceq \sup X$
By Set of Finite Suprema is Directed:
:$\set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$ is directed.
By definition of union:
:$\bigcup \map {\operatorname {Fin} } X \setminus \set \O = X$
By Supremum of Supr... | Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $x, y \in S$.
Then
:$x \ll y$
{{iff}}
:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in \map {\operatorname {Fin} } X: x \preceq \sup A$
where
:$\ll$ denotes the [[Definition:Element is Way Below|way be... | === Sufficient Condition ===
Let
:$x \ll y$
Let $X \subseteq S$ such that
:$y \preceq \sup X$
By [[Set of Finite Suprema is Directed]]:
:$\set {\sup A: A \in \map {\operatorname {Fin} } X \land A \ne \O}$ is [[Definition:Directed Subset|directed]].
By definition of [[Definition:Set Union|union]]:
:$\bigcup \map {\o... | Way Below in Complete Lattice | https://proofwiki.org/wiki/Way_Below_in_Complete_Lattice | https://proofwiki.org/wiki/Way_Below_in_Complete_Lattice | [
"Way Below Relation",
"Complete Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Element is Way Below",
"Definition:Set of Sets",
"Definition:Finite Set",
"Definition:Subset"
] | [
"Set of Finite Suprema is Directed",
"Definition:Directed Subset",
"Definition:Set Union",
"Supremum of Suprema",
"Definition:Element is Way Below",
"Definition:Directed Subset",
"Definition:Element is Way Below"
] |
proofwiki-11616 | Continuous Midpoint-Convex Function is Convex | Let $f$ be a real function which is defined on a real interval $I$.
Let $f$ be midpoint-convex and continuous on $I$.
Then $f$ is convex. | Let $x, y \in I$ and let $t \in \closedint 0 1$.
Let $\sequence {t_n}_{n \mathop \in \N}$ be a sequence of rational numbers in $\closedint 0 1$ converging to $t$.
It follows that:
{{begin-eqn}}
{{eqn | l = \lim \limits_{n \mathop \to\infty} \paren {t_n x + \paren {1 - t_n} y}
| r = t x + \paren {1 - t} y
}}
{{end... | Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
Let $f$ be [[Definition:Midpoint-Convex|midpoint-convex]] and [[Definition:Continuous on Interval|continuous]] on $I$.
Then $f$ is [[Definition:Convex Real Function|convex]]. | Let $x, y \in I$ and let $t \in \closedint 0 1$.
Let $\sequence {t_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Rational Number|rational numbers]] in $\closedint 0 1$ [[Definition:Convergent Rational Sequence|converging]] to $t$.
It follows that:
{{begin-eqn}}
{{eqn | l = \lim \limit... | Continuous Midpoint-Convex Function is Convex | https://proofwiki.org/wiki/Continuous_Midpoint-Convex_Function_is_Convex | https://proofwiki.org/wiki/Continuous_Midpoint-Convex_Function_is_Convex | [
"Real Analysis"
] | [
"Definition:Real Function",
"Definition:Real Interval",
"Definition:Midpoint-Convex",
"Definition:Continuous Real Function/Interval",
"Definition:Convex Real Function"
] | [
"Definition:Sequence",
"Definition:Rational Number",
"Definition:Convergent Sequence/Rational Numbers",
"Midpoint-Convex Function is Rational Convex",
"Definition:Rational Convex",
"Continuous Mapping is Sequentially Continuous",
"Definition:Sequential Continuity/Domain",
"Definition:Rational Convex",... |
proofwiki-11617 | Way Below Compact is Topological Compact | Let $T = \struct {S, \tau}$ be a topological space.
Let $L = \struct {\tau, \preceq}$ be an ordered set where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
Let $x \in \tau$.
Then
:$x$ is compact in $L$ (it means: $x \ll x$)
{{iff}}
:$T_x$ is compact (topologically)
where $T_x = \struct {x, \tau_x}$ denot... | === Sufficient Condition ===
Let
:$x \ll x$
Let $F \subseteq \tau_x$ be a open cover of $x$.
By definition of cover:
:$x \subseteq \bigcup F$
By definition of topological space:
:$\forall y \in \tau: x \cap y \in \tau$
By definition of subset:
:$\tau_x \subseteq \tau$
By Subset Relation is Transitive:
:$F \subseteq \t... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $L = \struct {\tau, \preceq}$ be an [[Definition:Ordered Set|ordered set]] where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
Let $x \in \tau$.
Then
:$x$ is [[Definition:Compact Element|compact]] in $L$ (it mean... | === Sufficient Condition ===
Let
:$x \ll x$
Let $F \subseteq \tau_x$ be a [[Definition:Open Cover|open cover]] of $x$.
By definition of [[Definition:Cover of Set|cover]]:
:$x \subseteq \bigcup F$
By definition of [[Definition:Topological Space|topological space]]:
:$\forall y \in \tau: x \cap y \in \tau$
By defin... | Way Below Compact is Topological Compact | https://proofwiki.org/wiki/Way_Below_Compact_is_Topological_Compact | https://proofwiki.org/wiki/Way_Below_Compact_is_Topological_Compact | [
"Way Below Relation",
"Topology"
] | [
"Definition:Topological Space",
"Definition:Ordered Set",
"Definition:Compact Element",
"Definition:Compact Topological Space/Subspace",
"Definition:Topological Subspace"
] | [
"Definition:Open Cover",
"Definition:Cover of Set",
"Definition:Topological Space",
"Definition:Subset",
"Subset Relation is Transitive",
"Way Below in Ordered Set of Topology",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Cover of Set",
"Definition:Subcover/Finite",
"Definition:Com... |
proofwiki-11618 | Continuous Midpoint-Concave Function is Concave | Let $f$ be midpoint-concave and continuous on $I$.
Then $f$ is concave. | As $f$ is midpoint-concave, then $-f$ is midpoint-convex.
{{LinkWanted|Prove the above}}
From Continuous Midpoint-Convex Function is Convex, $-f$ is convex.
From Real Function is Concave iff its Negative is Convex, $f$ is concave.
{{qed}} | Let $f$ be [[Definition:Midpoint-Concave|midpoint-concave]] and [[Definition:Continuous on Interval|continuous]] on $I$.
Then $f$ is [[Definition:Concave Real Function|concave]]. | As $f$ is [[Definition:Midpoint-Concave|midpoint-concave]], then $-f$ is [[Definition:Midpoint-Convex|midpoint-convex]].
{{LinkWanted|Prove the above}}
From [[Continuous Midpoint-Convex Function is Convex]], $-f$ is [[Definition:Convex Real Function|convex]].
From [[Real Function is Concave iff its Negative is Co... | Continuous Midpoint-Concave Function is Concave | https://proofwiki.org/wiki/Continuous_Midpoint-Concave_Function_is_Concave | https://proofwiki.org/wiki/Continuous_Midpoint-Concave_Function_is_Concave | [
" Real Analysis"
] | [
"Definition:Midpoint-Concave",
"Definition:Continuous Real Function/Interval",
"Definition:Concave Real Function"
] | [
"Definition:Midpoint-Concave",
"Definition:Midpoint-Convex",
"Continuous Midpoint-Convex Function is Convex",
"Definition:Convex Real Function",
"Real Function is Concave iff its Negative is Convex",
"Definition:Concave Real Function"
] |
proofwiki-11619 | Power Function on Strictly Positive Base is Convex | Let $a \in \R_{>0}$ be a strictly positive real number.
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = a^x$
where $a^x$ denotes $a$ to the power of $x$.
Then $f$ is convex. | Let $x, y \in \R$.
Note that, from Power of Positive Real Number is Positive: Real Number:
:$\forall t \in \R: a^t > 0$.
So:
{{begin-eqn}}
{{eqn | l = a^{\paren {x + y} / 2}
| r = \sqrt {a^{x + y} }
| c = Exponent Combination Laws: Power of Power: Proof 2
}}
{{eqn | r = \sqrt {a^x a^y}
| c = Exponent ... | Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\map f x = a^x$
where $a^x$ denotes [[Definition:Power to Real Number|$a$ to the power of $x$]].
Then $f$ is [[Definition:Convex Real... | Let $x, y \in \R$.
Note that, from [[Power of Positive Real Number is Positive/Real Number|Power of Positive Real Number is Positive: Real Number]]:
:$\forall t \in \R: a^t > 0$.
So:
{{begin-eqn}}
{{eqn | l = a^{\paren {x + y} / 2}
| r = \sqrt {a^{x + y} }
| c = [[Exponent Combination Laws/Power of Power... | Power Function on Strictly Positive Base is Convex | https://proofwiki.org/wiki/Power_Function_on_Strictly_Positive_Base_is_Convex | https://proofwiki.org/wiki/Power_Function_on_Strictly_Positive_Base_is_Convex | [
"Real Analysis",
"Powers"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Real Function",
"Definition:Power (Algebra)/Real Number",
"Definition:Convex Real Function"
] | [
"Power of Positive Real Number is Positive/Real Number",
"Exponent Combination Laws/Power of Power/Proof 2",
"Exponent Combination Laws/Product of Powers/Proof 2",
"Cauchy's Mean Theorem",
"Definition:Midpoint-Convex",
"Power Function on Strictly Positive Base is Continuous/Real Power",
"Definition:Cont... |
proofwiki-11620 | Natural Logarithm as Derivative of Exponential at Zero | Let $\ln: \R_{>0}$ denote the real natural logarithm.
Then:
:$\ds \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$ | Fix $x \in \R_{>0}$.
Let $x > 1$.
From Power Function on Strictly Positive Base is Convex, $x^h$ is convex.
Thus for $0 < h < s$:
{{begin-eqn}}
{{eqn | l = \frac {x^h - a^0} {h - 0}
| o = \le
| m = \frac {x^s - a^0} {s - 0}
| c = {{Defof|Convex Real Function}}
}}
{{eqn | ll= \leadsto
| l = \frac... | Let $\ln: \R_{>0}$ denote the [[Definition:Real Natural Logarithm|real natural logarithm]].
Then:
:$\ds \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$ | Fix $x \in \R_{>0}$.
Let $x > 1$.
From [[Power Function on Strictly Positive Base is Convex]], $x^h$ is [[Definition:Convex Real Function|convex]].
Thus for $0 < h < s$:
{{begin-eqn}}
{{eqn | l = \frac {x^h - a^0} {h - 0}
| o = \le
| m = \frac {x^s - a^0} {s - 0}
| c = {{Defof|Convex Real Functio... | Natural Logarithm as Derivative of Exponential at Zero | https://proofwiki.org/wiki/Natural_Logarithm_as_Derivative_of_Exponential_at_Zero | https://proofwiki.org/wiki/Natural_Logarithm_as_Derivative_of_Exponential_at_Zero | [
"Natural Logarithms"
] | [
"Definition:Natural Logarithm/Positive Real"
] | [
"Power Function on Strictly Positive Base is Convex",
"Definition:Convex Real Function",
"Power Function on Strictly Positive Base is Convex",
"Exponent Combination Laws/Negative Power",
"Order of Real Numbers is Dual of Order of their Negatives",
"Definition:Increasing/Real Function",
"Power Function o... |
proofwiki-11621 | Derivative of Exponential Function/Proof 5/Lemma | :$\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is increasing. | First:
{{begin-eqn}}
{{eqn | l = n
| o = >
| r = \ceiling {\size x}
}}
{{eqn | ll= \leadsto
| l = n
| o = >
| r = -x
| c = Negative of Absolute Value and Real Number is between Ceiling Functions
}}
{{eqn | ll= \leadsto
| l = \frac n {n + x}
| o = >
| r = 0
}}
{{end... | :$\forall x \in \R : n \ge \ceiling {\size x} \implies \sequence {\dfrac n {n + x} \paren {1 + \dfrac x n}^n}$ is [[Definition:Increasing Real Sequence|increasing]]. | First:
{{begin-eqn}}
{{eqn | l = n
| o = >
| r = \ceiling {\size x}
}}
{{eqn | ll= \leadsto
| l = n
| o = >
| r = -x
| c = [[Negative of Absolute Value]] and [[Real Number is between Ceiling Functions]]
}}
{{eqn | ll= \leadsto
| l = \frac n {n + x}
| o = >
| r = 0
... | Derivative of Exponential Function/Proof 5/Lemma | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_5/Lemma | https://proofwiki.org/wiki/Derivative_of_Exponential_Function/Proof_5/Lemma | [
"Derivative of Exponential Function"
] | [
"Definition:Increasing/Sequence/Real Sequence"
] | [
"Negative of Absolute Value",
"Real Number is between Ceiling Functions",
"Real Number is between Ceiling Functions",
"Negative of Absolute Value/Corollary 1",
"Power of Strictly Positive Real Number is Strictly Positive/Positive Integer",
"Definition:Positive/Real Number",
"Real Number Ordering is Comp... |
proofwiki-11622 | Permutation of Indices | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let the fiber of truth of $R$ be finite.
Let $\pi$ is a permutation on the fiber of truth of $R$.
Then: | {{begin-eqn}}
{{eqn | l = \sum_{\map R {\map \pi j} } a_{\map \pi j}
| r = \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
| c = {{Defof|Summation by Iverson's Convention}}
}}
{{eqn | r = \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}
... | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Finite Set|finite]].
Let $\pi$ is a [[Definition:Permutation|permutation]] on the [[Definition:Fib... | {{begin-eqn}}
{{eqn | l = \sum_{\map R {\map \pi j} } a_{\map \pi j}
| r = \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
| c = {{Defof|Summation by Iverson's Convention}}
}}
{{eqn | r = \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}
... | Permutation of Indices of Summation/Proof | https://proofwiki.org/wiki/Permutation_of_Indices | https://proofwiki.org/wiki/Permutation_of_Indices_of_Summation/Proof | [
"Summations",
"Continued Products"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Fiber of Truth",
"Definition:Finite Set",
"Definition:Permutation",
"Definition:Fiber of Truth"
] | [
"Change of Index Variable of Summation"
] |
proofwiki-11623 | Change of Index Variable of Product | :$\ds \prod_{\map R i} a_i = \prod_{\map R j} a_j$ | Let $S = \set {i \in \Z: \map R i}$.
Let $T = \set {j \in \Z: \map R j}$.
Let $i \in S$.
Then $\map R i$.
Let $j = i$.
By Leibniz's Law, $\map R j$.
Thus $i \in T$.
By definition of subset:
:$S \subseteq T$
Similarly, let $j \in T$.
Then $\map R j$.
Let $i = j$.
By Leibniz's Law, $\map R i$.
Thus $j \in S$.
By definiti... | :$\ds \prod_{\map R i} a_i = \prod_{\map R j} a_j$ | Let $S = \set {i \in \Z: \map R i}$.
Let $T = \set {j \in \Z: \map R j}$.
Let $i \in S$.
Then $\map R i$.
Let $j = i$.
By [[Definition:Leibniz's Law|Leibniz's Law]], $\map R j$.
Thus $i \in T$.
By definition of [[Definition:Subset|subset]]:
:$S \subseteq T$
Similarly, let $j \in T$.
Then $\map R j$.
Let $i =... | Change of Index Variable of Product | https://proofwiki.org/wiki/Change_of_Index_Variable_of_Product | https://proofwiki.org/wiki/Change_of_Index_Variable_of_Product | [
"Continued Products"
] | [] | [
"Axiom:Leibniz's Law",
"Definition:Subset",
"Axiom:Leibniz's Law",
"Definition:Subset",
"Definition:Set Equality"
] |
proofwiki-11624 | Translation of Index Variable of Summation/Corollary | :$\ds \sum_{j \mathop = m}^n a_j = \sum_{j \mathop = m + c}^{n + c} a_{j - c}$
where $c$ is an integer constant which is not dependent upon $j$. | {{proof wanted}}
Category:Summations
aracbiz9djucfdwdm6iryfcxhnaoxhg | :$\ds \sum_{j \mathop = m}^n a_j = \sum_{j \mathop = m + c}^{n + c} a_{j - c}$
where $c$ is an [[Definition:Integer|integer]] [[Definition:Constant|constant]] which is not dependent upon $j$. | {{proof wanted}}
[[Category:Summations]]
aracbiz9djucfdwdm6iryfcxhnaoxhg | Translation of Index Variable of Summation/Corollary | https://proofwiki.org/wiki/Translation_of_Index_Variable_of_Summation/Corollary | https://proofwiki.org/wiki/Translation_of_Index_Variable_of_Summation/Corollary | [
"Summations"
] | [
"Definition:Integer",
"Definition:Constant"
] | [
"Category:Summations"
] |
proofwiki-11625 | Permutation of Indices of Product | :$\ds \prod_{\map R j} a_j = \prod_{\map R {\map \pi j} } a_{\map \pi j}$ | {{begin-eqn}}
{{eqn | l = \prod_{\map R {\map \pi j} } a_{\map \pi j}
| r = \prod_{j \mathop \in \Z} {a_{\map \pi j} }^{\sqbrk {\map R {\map \pi j} } }
| c = {{Defof|Continued Product by Iverson's Convention}}
}}
{{eqn | r = \prod_{j \mathop \in \Z} \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} \sqbrk {... | :$\ds \prod_{\map R j} a_j = \prod_{\map R {\map \pi j} } a_{\map \pi j}$ | {{begin-eqn}}
{{eqn | l = \prod_{\map R {\map \pi j} } a_{\map \pi j}
| r = \prod_{j \mathop \in \Z} {a_{\map \pi j} }^{\sqbrk {\map R {\map \pi j} } }
| c = {{Defof|Continued Product by Iverson's Convention}}
}}
{{eqn | r = \prod_{j \mathop \in \Z} \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} \sqbrk {... | Permutation of Indices of Product | https://proofwiki.org/wiki/Permutation_of_Indices_of_Product | https://proofwiki.org/wiki/Permutation_of_Indices_of_Product | [
"Continued Products"
] | [] | [
"Definition:Power (Algebra)/Integer/Knuth Notation",
"Change of Index Variable of Product"
] |
proofwiki-11626 | Product of Products over Overlapping Domains | :$\ds \prod_{\map R j} a_j \prod_{\map S j} a_j = \paren {\prod_{\map R j \mathop \lor \map S j} a_j} \paren {\prod_{\map R j \mathop \land \map S j} a_j}$
where $\lor$ and $\land$ signify logical disjunction and logical conjunction respectively. | {{begin-eqn}}
{{eqn | l = \prod_{\map R j} a_j \times \prod_{\map S j} a_j
| r = \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map R j} } \times \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map S j} }
| c = {{Defof|Continued Product by Iverson's Convention}}
}}
{{eqn | r = \prod_{j \mathop \in \Z} a_j^{\paren {\sqbrk {\... | :$\ds \prod_{\map R j} a_j \prod_{\map S j} a_j = \paren {\prod_{\map R j \mathop \lor \map S j} a_j} \paren {\prod_{\map R j \mathop \land \map S j} a_j}$
where $\lor$ and $\land$ signify [[Definition:Disjunction|logical disjunction]] and [[Definition:Conjunction|logical conjunction]] respectively. | {{begin-eqn}}
{{eqn | l = \prod_{\map R j} a_j \times \prod_{\map S j} a_j
| r = \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map R j} } \times \prod_{j \mathop \in \Z} a_j^{\sqbrk {\map S j} }
| c = {{Defof|Continued Product by Iverson's Convention}}
}}
{{eqn | r = \prod_{j \mathop \in \Z} a_j^{\paren {\sqbrk {\... | Product of Products over Overlapping Domains | https://proofwiki.org/wiki/Product_of_Products_over_Overlapping_Domains | https://proofwiki.org/wiki/Product_of_Products_over_Overlapping_Domains | [
"Continued Products"
] | [
"Definition:Disjunction",
"Definition:Conjunction"
] | [
"Exponent Combination Laws/Product of Powers",
"Cardinality of Set Union"
] |
proofwiki-11627 | Topology is Locally Compact iff Ordered Set of Topology is Continuous | Let $T = \struct {S, \tau}$ be a topological space.
Let $L = \struct {\tau, \preceq}$ be the ordered set where $\preceq$ is the subset relation.
Then
:$(1): \quad T$ is locally compact implies $L$ is continuous
:$(2): \quad T$ is $T_3$ space and $L$ is continuous implies $T$ is locally compact | === Condition $(1)$ ===
Let $T$ be locally compact.
By Topology forms Complete Lattice:
:$L$ is complete lattice.
Thus by Way Below Closure is Directed in Bounded Below Join Semilattice:
:$\forall x \in \tau: x^\ll$ is directed
Thus by definition:
:$L$ is up-complete.
Let $x \in \tau$.
We will prove that
:$x \subseteq ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $L = \struct {\tau, \preceq}$ be the [[Definition:Ordered Set|ordered set]] where $\preceq$ is the [[Definition:Subset Relation|subset relation]].
Then
:$(1): \quad T$ is [[Definition:Locally Compact Space|locally compact]] impl... | === Condition $(1)$ ===
Let $T$ be [[Definition:Locally Compact Space|locally compact]].
By [[Topology forms Complete Lattice]]:
:$L$ is [[Definition:Complete Lattice|complete lattice]].
Thus by [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
:$\forall x \in \tau: x^\ll$ is [[Definition:Directed... | Topology is Locally Compact iff Ordered Set of Topology is Continuous | https://proofwiki.org/wiki/Topology_is_Locally_Compact_iff_Ordered_Set_of_Topology_is_Continuous | https://proofwiki.org/wiki/Topology_is_Locally_Compact_iff_Ordered_Set_of_Topology_is_Continuous | [
"Locally Compact Spaces",
"Continuous Lattices",
"Topology",
"Locally Compact Spaces"
] | [
"Definition:Topological Space",
"Definition:Ordered Set",
"Definition:Subset Relation",
"Definition:Locally Compact Space",
"Definition:Continuous Ordered Set",
"Definition:T3 Space",
"Definition:Continuous Ordered Set",
"Definition:Locally Compact Space"
] | [
"Definition:Locally Compact Space",
"Topology forms Complete Lattice",
"Definition:Complete Lattice",
"Way Below Closure is Directed in Bounded Below Join Semilattice",
"Definition:Directed Subset",
"Definition:Up-Complete",
"Definition:Open Set/Topology",
"Definition:Locally Compact Space",
"Defini... |
proofwiki-11628 | Sum of Sequence of Power by Index | :$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
for $x \ne 1$. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j x^j
| r = x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}
| c =
}}
{{eqn | r = x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j
| c =
}}
{{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mat... | :$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
for $x \ne 1$. | {{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j x^j
| r = x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}
| c =
}}
{{eqn | r = x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j
| c =
}}
{{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mat... | Sum of Sequence of Power by Index/Proof 1 | https://proofwiki.org/wiki/Sum_of_Sequence_of_Power_by_Index | https://proofwiki.org/wiki/Sum_of_Sequence_of_Power_by_Index/Proof_1 | [
"Arithmetic-Geometric Sequences",
"Sum of Sequence of Power by Index"
] | [] | [
"Sum of Geometric Sequence"
] |
proofwiki-11629 | Sum of Sequence of Power by Index | :$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
for $x \ne 1$. | From Sum of Arithmetic-Geometric Sequence:
:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} x^j = \frac {a \paren {1 - x^{n + 1} } } {1 - x} + \frac {x d \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j x^j
| r = \frac {0 \paren {1 - x^{n ... | :$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
for $x \ne 1$. | From [[Sum of Arithmetic-Geometric Sequence]]:
:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} x^j = \frac {a \paren {1 - x^{n + 1} } } {1 - x} + \frac {x d \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop = 0}^n j x^j
| r = \frac {0 \paren {1... | Sum of Sequence of Power by Index/Proof 2 | https://proofwiki.org/wiki/Sum_of_Sequence_of_Power_by_Index | https://proofwiki.org/wiki/Sum_of_Sequence_of_Power_by_Index/Proof_2 | [
"Arithmetic-Geometric Sequences",
"Sum of Sequence of Power by Index"
] | [] | [
"Sum of Arithmetic-Geometric Sequence"
] |
proofwiki-11630 | Reciprocal of Real Exponential | :$\dfrac 1 {\map \exp x} = \map \exp {-x}$ | {{begin-eqn}}
{{eqn | l = \map \exp 0
| r = 1
| c = Exponential of Zero
}}
{{eqn | ll= \leadsto
| l = \map \exp {x - x}
| r = 1
| c = as $\forall x \in \R : x - x = 0$
}}
{{eqn | ll= \leadsto
| l = \map \exp {-x} \, \map \exp x
| r = 1
| c = Exponential of Sum of Real Num... | :$\dfrac 1 {\map \exp x} = \map \exp {-x}$ | {{begin-eqn}}
{{eqn | l = \map \exp 0
| r = 1
| c = [[Exponential of Zero]]
}}
{{eqn | ll= \leadsto
| l = \map \exp {x - x}
| r = 1
| c = as $\forall x \in \R : x - x = 0$
}}
{{eqn | ll= \leadsto
| l = \map \exp {-x} \, \map \exp x
| r = 1
| c = [[Exponential of Sum of Re... | Reciprocal of Real Exponential | https://proofwiki.org/wiki/Reciprocal_of_Real_Exponential | https://proofwiki.org/wiki/Reciprocal_of_Real_Exponential | [
"Exponential Function",
"Reciprocals"
] | [] | [
"Exponential of Zero",
"Exponential of Sum/Real Numbers",
"Exponential of Real Number is Strictly Positive"
] |
proofwiki-11631 | Telescoping Series/Example 2 | Let $\sequence {b_n}$ be a sequence in $\R$.
Let $\sequence {a_n}$ be a sequence whose terms are defined as:
:$a_k = b_k - b_{k - 1}$
Then:
:$\ds \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = m}^n a_k
| r = \sum_{k \mathop = m}^n \paren {b_k - b_{k - 1} }
| c =
}}
{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m}^n b_{k - 1}
| c =
}}
{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m - 1}^{n - 1} b_k
| c = Translatio... | Let $\sequence {b_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
Let $\sequence {a_n}$ be a [[Definition:Real Sequence|sequence]] whose [[Definition:Term of Sequence|terms]] are defined as:
:$a_k = b_k - b_{k - 1}$
Then:
:$\ds \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = m}^n a_k
| r = \sum_{k \mathop = m}^n \paren {b_k - b_{k - 1} }
| c =
}}
{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m}^n b_{k - 1}
| c =
}}
{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m - 1}^{n - 1} b_k
| c = [[Translat... | Telescoping Series/Example 2 | https://proofwiki.org/wiki/Telescoping_Series/Example_2 | https://proofwiki.org/wiki/Telescoping_Series/Example_2 | [
"Telescoping Series"
] | [
"Definition:Real Sequence",
"Definition:Real Sequence",
"Definition:Term of Sequence"
] | [
"Translation of Index Variable of Summation"
] |
proofwiki-11632 | Repunit Integer as Product of Base - 1 by Increasing Digit Integer | {{begin-eqn}}
{{eqn | l = 9 \times 0 + 1
| r = 1
}}
{{eqn | l = 9 \times 1 + 2
| r = 11
}}
{{eqn | l = 9 \times 12 + 3
| r = 111
}}
{{eqn | l = 9 \times 123 + 4
| r = 1111
}}
{{eqn | l = 9 \times 1234 + 5
| r = 11111
}}
{{eqn | l = 9 \times 12345 + 6
| r = 111111
}}
{{eqn | o = \ldot... | A specific instance of the general result:
{{:Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result}}
where $b = 10$.
{{qed}} | {{begin-eqn}}
{{eqn | l = 9 \times 0 + 1
| r = 1
}}
{{eqn | l = 9 \times 1 + 2
| r = 11
}}
{{eqn | l = 9 \times 12 + 3
| r = 111
}}
{{eqn | l = 9 \times 123 + 4
| r = 1111
}}
{{eqn | l = 9 \times 1234 + 5
| r = 11111
}}
{{eqn | l = 9 \times 12345 + 6
| r = 111111
}}
{{eqn | o = \ldot... | A specific instance of the [[Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result|general result]]:
{{:Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result}}
where $b = 10$.
{{qed}} | Repunit Integer as Product of Base - 1 by Increasing Digit Integer | https://proofwiki.org/wiki/Repunit_Integer_as_Product_of_Base_-_1_by_Increasing_Digit_Integer | https://proofwiki.org/wiki/Repunit_Integer_as_Product_of_Base_-_1_by_Increasing_Digit_Integer | [
"Repunits"
] | [] | [
"Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result"
] |
proofwiki-11633 | Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result | :$\ds \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1 = \sum_{j \mathop = 0}^n b^j$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1
| c =
}}
{{eqn | r = n \paren {b - 1} \sum_{j \mathop = 0}^n b^j - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1
| c =
}}
{{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1... | :$\ds \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1 = \sum_{j \mathop = 0}^n b^j$ | {{begin-eqn}}
{{eqn | o =
| r = \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1
| c =
}}
{{eqn | r = n \paren {b - 1} \sum_{j \mathop = 0}^n b^j - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1
| c =
}}
{{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1... | Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result | https://proofwiki.org/wiki/Repunit_Integer_as_Product_of_Base_-_1_by_Increasing_Digit_Integer/General_Result | https://proofwiki.org/wiki/Repunit_Integer_as_Product_of_Base_-_1_by_Increasing_Digit_Integer/General_Result | [
"Recreational Mathematics"
] | [] | [
"Sum of Geometric Sequence",
"Sum of Sequence of Power by Index",
"Sum of Geometric Sequence"
] |
proofwiki-11634 | Exponential of Real Number is Strictly Positive/Proof 5/Lemma | :$\forall x \in \R: \exp x \ne 0$ | This proof assumes the definition of $\exp$ as the solution to an initial value problem.
That is, suppose $\exp$ satisfies:
:$(1): \quad D_x \exp x = \exp x$
:$(2): \quad \map \exp 0 = 1$
on $\R$.
{{AimForCont}} $\exists \alpha \in \R: \exp \alpha = 0$.
Suppose that $\alpha > 0$.
Let $J = \closedint 0 \alpha$.
From Exp... | :$\forall x \in \R: \exp x \ne 0$ | This proof assumes the [[Definition:Exponential Function/Real/Differential Equation|definition of $\exp$ as the solution to an initial value problem]].
That is, suppose $\exp$ satisfies:
:$(1): \quad D_x \exp x = \exp x$
:$(2): \quad \map \exp 0 = 1$
on $\R$.
{{AimForCont}} $\exists \alpha \in \R: \exp \alpha = 0$.
... | Exponential of Real Number is Strictly Positive/Proof 5/Lemma | https://proofwiki.org/wiki/Exponential_of_Real_Number_is_Strictly_Positive/Proof_5/Lemma | https://proofwiki.org/wiki/Exponential_of_Real_Number_is_Strictly_Positive/Proof_5/Lemma | [
"Exponential of Real Number is Strictly Positive"
] | [] | [
"Definition:Exponential Function/Real/Differential Equation",
"Exponential Function is Continuous/Real Numbers/Proof 3",
"Definition:Continuous Real Function/Subset",
"Max and Min of Function on Closed Real Interval",
"Taylor's Theorem/One Variable",
"Power over Factorial",
"Definition:Contradiction",
... |
proofwiki-11635 | Product of Products | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{R \paren i} x_i$ denote a continued product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \prod_{R \paren i} \paren {b_i c_i} = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}$ | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \prod_{R \paren i} \paren {b_i c_i}
| r = \prod_{R \paren i} \paren {a_{i 1} a_{i 2} }
| c = by definition
}}
{{eqn | r = \prod_{R \paren i} \paren {\prod_{1 \mathop \le j \mathop \le 2} a_{i j} }
| c = {{Defof|Continued Produc... | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let $\ds \prod_{R \paren i} x_i$ denote a [[Definition:Continued Product by Propositional Function|continued product]] over $R$.
Let the [[Definition:Fiber of Truth|fiber of ... | Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
Then:
{{begin-eqn}}
{{eqn | l = \prod_{R \paren i} \paren {b_i c_i}
| r = \prod_{R \paren i} \paren {a_{i 1} a_{i 2} }
| c = by definition
}}
{{eqn | r = \prod_{R \paren i} \paren {\prod_{1 \mathop \le j \mathop \le 2} a_{i j} }
| c = {{Defof|Continued Prod... | Product of Products | https://proofwiki.org/wiki/Product_of_Products | https://proofwiki.org/wiki/Product_of_Products | [
"Continued Products"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Continued Product/Propositional Function",
"Definition:Fiber of Truth",
"Definition:Finite Set"
] | [
"Exchange of Order of Product"
] |
proofwiki-11636 | Summation of General Logarithms | Let $R: \Z \to \set {\T, \F}$ be a propositional function on the set of integers.
Let $\ds \prod_{\map R i} a_i$ denote a product over $R$.
Let the fiber of truth of $R$ be finite.
Then:
:$\ds \map {\log_b} {\prod_{\map R i} a_i} = \sum_{\map R i} \log_b a_i$ | The proof proceeds by induction.
First let:
:$S := \set {a_i: \map R i}$
We have that $S$ is finite.
Hence the contents of $S$ can be well-ordered, by Finite Totally Ordered Set is Well-Ordered.
Let $S$ have $m$ elements, identified as:
:$S = \set {s_1, s_2, \ldots, s_m}$
For all $n \in \Z_{\ge 0}$ such that $n \le m$,... | Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
Let $\ds \prod_{\map R i} a_i$ denote a [[Definition:Continued Product by Propositional Function|product]] over $R$.
Let the [[Definition:Fiber of Truth|fiber of truth]] of $... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
First let:
:$S := \set {a_i: \map R i}$
We have that $S$ is [[Definition:Finite Set|finite]].
Hence the contents of $S$ can be [[Definition:Well-Ordered Set|well-ordered]], by [[Finite Totally Ordered Set is Well-Ordered]].
Let $S$ have $m$ [[... | Summation of General Logarithms | https://proofwiki.org/wiki/Summation_of_General_Logarithms | https://proofwiki.org/wiki/Summation_of_General_Logarithms | [
"Summations",
"Continued Products",
"Logarithms"
] | [
"Definition:Propositional Function",
"Definition:Integer",
"Definition:Continued Product/Propositional Function",
"Definition:Fiber of Truth",
"Definition:Finite Set"
] | [
"Principle of Mathematical Induction",
"Definition:Finite Set",
"Definition:Well-Ordered Set",
"Finite Totally Ordered Set is Well-Ordered",
"Definition:Element",
"Definition:Proposition",
"Natural Logarithm of 1 is 0",
"Principle of Mathematical Induction"
] |
proofwiki-11637 | Way Below Closure is Directed in Bounded Below Join Semilattice | Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $x \in S$.
Then
:$x^\ll$ is directed. | By Bottom is Way Below Any Element:
:$\bot \ll x$
By definition of way below closure:
:$\bot \in x^\ll$
Thus by definition:
:$x^\ll$ is a non-empty set.
Let $y, z \in x^\ll$
By definition of way below closure:
:$y \ll x$ and $z \ll x$
By Join is Way Below if Operands are Way Below
:$y \vee z \ll x$
By definition of way... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $x \in S$.
Then
:$x^\ll$ is [[Definition:Directed Subset|directed]]. | By [[Bottom is Way Below Any Element]]:
:$\bot \ll x$
By definition of [[Definition:Way Below Closure|way below closure]]:
:$\bot \in x^\ll$
Thus by definition:
:$x^\ll$ is a [[Definition:Non-Empty Set|non-empty set]].
Let $y, z \in x^\ll$
By definition of [[Definition:Way Below Closure|way below closure]]:
:$y \ll... | Way Below Closure is Directed in Bounded Below Join Semilattice | https://proofwiki.org/wiki/Way_Below_Closure_is_Directed_in_Bounded_Below_Join_Semilattice | https://proofwiki.org/wiki/Way_Below_Closure_is_Directed_in_Bounded_Below_Join_Semilattice | [
"Join and Meet Semilattices",
"Way Below Relation"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Directed Subset"
] | [
"Bottom is Way Below Any Element",
"Definition:Way Below Closure",
"Definition:Non-Empty Set",
"Definition:Way Below Closure",
"Join is Way Below if Operands are Way Below",
"Definition:Way Below Closure",
"Join Succeeds Operands",
"Definition:Directed Subset"
] |
proofwiki-11638 | Bottom is Way Below Any Element | Let $\left({S, \preceq}\right)$ be a bounded below ordered set.
Let $x \in S$.
Then
:$\bot \ll x$
where $\bot$ denotes the smallest element in $S$. | Let $D$ be directed subset of $S$ such that
:$D$ admits a supremum and $x \preceq \sup D$
Because $D$ is non-empty, therefore
:$\exists a: a \in D$
By definition of smallest element:
:$\bot \preceq a$
Thus by definition of way below relation;
:$\bot \ll x$
{{qed}} | Let $\left({S, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Ordered Set|ordered set]].
Let $x \in S$.
Then
:$\bot \ll x$
where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $S$. | Let $D$ be [[Definition:Directed Subset|directed subset]] of $S$ such that
:$D$ admits a [[Definition:Supremum of Set|supremum]] and $x \preceq \sup D$
Because $D$ is [[Definition:Non-Empty Set|non-empty]], therefore
:$\exists a: a \in D$
By definition of [[Definition:Smallest Element|smallest element]]:
:$\bot \prec... | Bottom is Way Below Any Element | https://proofwiki.org/wiki/Bottom_is_Way_Below_Any_Element | https://proofwiki.org/wiki/Bottom_is_Way_Below_Any_Element | [
"Way Below Relation"
] | [
"Definition:Bounded Below",
"Definition:Ordered Set",
"Definition:Smallest Element"
] | [
"Definition:Directed Subset",
"Definition:Supremum of Set",
"Definition:Non-Empty Set",
"Definition:Smallest Element",
"Definition:Element is Way Below"
] |
proofwiki-11639 | Equivalence of Definitions of Real Exponential Function/Proof 2 | The following definitions of the exponential function are equivalent. | From Derivative of Exponential Function and Exponential of Zero, each definition of $\exp$ satisfies the following:
:$(1): \quad D_x \exp = \exp$
:$(2): \quad \map \exp 0 = 1$
on $\R$.
From Exponential Function is Well-Defined, such a solution is unique.
Thus they all are all equivalent.
{{qed}} | The following definitions of the [[Definition:Real Exponential Function|exponential function]] are [[Definition:Logical Equivalence|equivalent]]. | From [[Derivative of Exponential Function]] and [[Exponential of Zero]], each definition of $\exp$ satisfies the following:
:$(1): \quad D_x \exp = \exp$
:$(2): \quad \map \exp 0 = 1$
on $\R$.
From [[Exponential Function is Well-Defined/Real/Proof 5|Exponential Function is Well-Defined]], such a solution is [[Definiti... | Equivalence of Definitions of Real Exponential Function/Proof 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Real_Exponential_Function/Proof_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Real_Exponential_Function/Proof_2 | [
"Equivalence of Definitions of Exponential Function"
] | [
"Definition:Exponential Function/Real",
"Definition:Logical Equivalence"
] | [
"Derivative of Exponential Function",
"Exponential of Zero",
"Exponential Function is Well-Defined/Real/Proof 5",
"Definition:Unique",
"Definition:Logical Equivalence"
] |
proofwiki-11640 | Product to n of Product to Index | :$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j = \prod_{i \mathop = 0}^n {a_i}^{n + 2}$ | Let:
{{begin-eqn}}
{{eqn | l = P_1
| o = :=
| r = \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j
}}
{{eqn | l = P_2
| o = :=
| r = \prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = P_1 P_2
| r = \paren {\prod_{i \mathop = 0}^n \... | :$\ds \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j = \prod_{i \mathop = 0}^n {a_i}^{n + 2}$ | Let:
{{begin-eqn}}
{{eqn | l = P_1
| o = :=
| r = \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j
}}
{{eqn | l = P_2
| o = :=
| r = \prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = P_1 P_2
| r = \paren {\prod_{i \mathop = 0}^n... | Product to n of Product to Index | https://proofwiki.org/wiki/Product_to_n_of_Product_to_Index | https://proofwiki.org/wiki/Product_to_n_of_Product_to_Index | [
"Continued Products"
] | [] | [
"Product of Products over Overlapping Domains"
] |
proofwiki-11641 | Bernoulli's Inequality/Corollary/General Result | For all $n \in \Z_{\ge 0}$:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
where $0 < a_j < 1$ for all $j$. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 1}^0 \paren {1 - a_j}
| r = 1
| c = {{Defof|Vacuous Product}... | For all $n \in \Z_{\ge 0}$:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
where $0 < a_j < 1$ for all $j$. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \prod_{j \mathop = 1}^n \paren {1 - a_j} \ge 1 - \sum_{j \mathop = 1}^n a^j$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 1... | Bernoulli's Inequality/Corollary/General Result | https://proofwiki.org/wiki/Bernoulli's_Inequality/Corollary/General_Result | https://proofwiki.org/wiki/Bernoulli's_Inequality/Corollary/General_Result | [
"Bernoulli's Inequality",
"Summations",
"Continued Products"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11642 | Product of Sequence of 1 minus Reciprocal of Squares | For all $n \in \Z_{\ge 1}$:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
It is first noted that $n = 0$ is excluded because in that case $\dfrac {n + 1} {2 n}$ is undefined.
$\map P 1$ is the other edge case:
{{beg... | For all $n \in \Z_{\ge 1}$:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
It is first noted that $n = 0$ is excluded because in that case $\dfrac... | Product of Sequence of 1 minus Reciprocal of Squares/Proof 1 | https://proofwiki.org/wiki/Product_of_Sequence_of_1_minus_Reciprocal_of_Squares | https://proofwiki.org/wiki/Product_of_Sequence_of_1_minus_Reciprocal_of_Squares/Proof_1 | [
"Product of Sequence of 1 minus Reciprocal of Squares",
"Continued Products"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Product of Sequence of 1 minus Reciprocal of Squares",
"Principle of Mathematical Induction"
] |
proofwiki-11643 | Product of Sequence of 1 minus Reciprocal of Squares | For all $n \in \Z_{\ge 1}$:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$ | We have:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} }
| r = \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} }
| c = Difference of Two Squares
}}
{{eqn | r = \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + 1} } ... | For all $n \in \Z_{\ge 1}$:
:$\ds \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$ | We have:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} }
| r = \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} }
| c = [[Difference of Two Squares]]
}}
{{eqn | r = \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + ... | Product of Sequence of 1 minus Reciprocal of Squares/Proof 2 | https://proofwiki.org/wiki/Product_of_Sequence_of_1_minus_Reciprocal_of_Squares | https://proofwiki.org/wiki/Product_of_Sequence_of_1_minus_Reciprocal_of_Squares/Proof_2 | [
"Product of Sequence of 1 minus Reciprocal of Squares",
"Continued Products"
] | [] | [
"Difference of Two Squares",
"Product of Products",
"Gamma Difference Equation"
] |
proofwiki-11644 | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3 | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | Let:
{{begin-eqn}}
{{eqn | n = a
| l = A
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k
| c =
}}
{{eqn | r = \sum_{a \mathop \le i \mathop \le j \mathop \le k \mathop \le b} x_i x_j x_k
| c =
}}
{{eqn | n = b
| r = \sum_{i \mathop = ... | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | Let:
{{begin-eqn}}
{{eqn | n = a
| l = A
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k
| c =
}}
{{eqn | r = \sum_{a \mathop \le i \mathop \le j \mathop \le k \mathop \le b} x_i x_j x_k
| c =
}}
{{eqn | n = b
| r = \sum_{i \mathop =... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3/Proof 1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3/Proof_1 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [] |
proofwiki-11645 | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3 | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cd... | :$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$
From [[Summation of Products of n Numbers taken m at a time with Repetitions]]:
:$\ds \sum_{a \mathop \le j_1 \mathop... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3/Proof 2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3/Proof_2 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] |
proofwiki-11646 | Summation of Product of Differences | :$\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_j - u_k} \paren {v_j - v_k} = n \sum_{j \mathop = 1}^n u_j v_j - \sum_{j \mathop = 1}^n u_j \sum_{j \mathop = 1}^n v_j$ | Take the Binet-Cauchy Identity:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
Make the follo... | :$\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {u_j - u_k} \paren {v_j - v_k} = n \sum_{j \mathop = 1}^n u_j v_j - \sum_{j \mathop = 1}^n u_j \sum_{j \mathop = 1}^n v_j$ | Take the [[Binet-Cauchy Identity]]:
:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
Make th... | Summation of Product of Differences | https://proofwiki.org/wiki/Summation_of_Product_of_Differences | https://proofwiki.org/wiki/Summation_of_Product_of_Differences | [
"Summations"
] | [] | [
"Binet-Cauchy Identity"
] |
proofwiki-11647 | Product of Summations | :$\ds \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j} = \sum_{1 \mathop \le i_1 \mathop , \mathop \ldots \mathop , i_n \mathop \le m} a_{i_1 1} \cdots a_{i_n n}$ | {{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j}
| r = \sum_{i \mathop = 1}^m a_{i 1} \sum_{i \mathop = 1}^m a_{i 2} \cdots \sum_{i \mathop = 1}^m a_{i n}
| c = {{Defof|Continued Product}}
}}
{{eqn | r = \sum_{i_1 \mathop = 1}^m a_{i_1 1} \sum_{i_2 \mathop = 1}^m a_{i_2 2} \... | :$\ds \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j} = \sum_{1 \mathop \le i_1 \mathop , \mathop \ldots \mathop , i_n \mathop \le m} a_{i_1 1} \cdots a_{i_n n}$ | {{begin-eqn}}
{{eqn | l = \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j}
| r = \sum_{i \mathop = 1}^m a_{i 1} \sum_{i \mathop = 1}^m a_{i 2} \cdots \sum_{i \mathop = 1}^m a_{i n}
| c = {{Defof|Continued Product}}
}}
{{eqn | r = \sum_{i_1 \mathop = 1}^m a_{i_1 1} \sum_{i_2 \mathop = 1}^m a_{i_2 2} \... | Product of Summations | https://proofwiki.org/wiki/Product_of_Summations | https://proofwiki.org/wiki/Product_of_Summations | [
"Summations",
"Continued Products"
] | [] | [
"General Distributivity Theorem"
] |
proofwiki-11648 | Summation of Powers over Product of Differences | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds S_n := \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds ... | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds S_n := \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} ... | Summation of Powers over Product of Differences/Proof 1 | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences/Proof_1 | [
"Summation of Powers over Product of Differences",
"Summations",
"Continued Products"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Summation/Vacuous Summation",
"Definition:Continued Product",
"Definition:Continued Product/Vacuous Product",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Summation ... |
proofwiki-11649 | Summation of Powers over Product of Differences | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | By Cauchy's Residue Theorem:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\size z \mathop = R} \dfrac {z^r \rd z} {\paren {z - z_1} \cdots \paren {z - z... | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | By [[Cauchy's Residue Theorem]]:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\size z \mathop = R} \dfrac {z^r \rd z} {\paren {z - z_1} \cdots \paren {... | Summation of Powers over Product of Differences/Proof 2 | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences/Proof_2 | [
"Summation of Powers over Product of Differences",
"Summations",
"Continued Products"
] | [] | [
"Cauchy's Residue Theorem",
"Definition:Laurent Series",
"Definition:Integration/Integrand",
"Definition:Uniform Convergence",
"Definition:Primitive (Calculus)/Integration"
] |
proofwiki-11650 | Summation of Powers over Product of Differences | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | === Definition ===
{{begin-eqn}}
{{eqn | l = \map p x
| r = \prod_{k \mathop = 1}^n \paren {x - x_k}
}}
{{eqn | l = \map {p_m} x
| r = \prod_{\substack {k \mathop = 1 \\ k \mathop \ne m} }^n \paren {x - x_k}
| c = where $1 \le m \le n$
}}
{{eqn | l = A
| r = <nowiki>\begin{pmatrix} 1 & x_1 ... | :$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$ | === Definition ===
{{begin-eqn}}
{{eqn | l = \map p x
| r = \prod_{k \mathop = 1}^n \paren {x - x_k}
}}
{{eqn | l = \map {p_m} x
| r = \prod_{\substack {k \mathop = 1 \\ k \mathop \ne m} }^n \paren {x - x_k}
| c = where $1 \le m \le n$
}}
{{eqn | l = A
| r = <nowiki>\begin{pmatrix} 1 & x_1 ... | Summation of Powers over Product of Differences/Proof 3 | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences | https://proofwiki.org/wiki/Summation_of_Powers_over_Product_of_Differences/Proof_3 | [
"Summation of Powers over Product of Differences",
"Summations",
"Continued Products"
] | [] | [
"Definition:Distinct/Plural",
"Definition:Cofactor/Element",
"Effect of Elementary Row Operations on Determinant",
"Laplace Expansion Theorem for Determinants",
"Value of Vandermonde Determinant/Formulation 1",
"Laplace Expansion Theorem for Determinants",
"Square Matrix with Duplicate Rows has Zero Det... |
proofwiki-11651 | Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r | :$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } } = 1$
where $1 \le m \le n$ and $x$ is arbitrary. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } }
| r = \sum_{k \mathop = 1}^n \paren {\dfrac {k^{n - 1} + \ma... | :$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } } = 1$
where $1 \le m \le n$ and $x$ is arbitrary. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \paren {k - r} } }
| r = \sum_{k \mathop = 1}^n \paren {\dfrac {k^{n - 1} + \ma... | Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r | https://proofwiki.org/wiki/Summation_by_k_of_Product_by_r_of_x_plus_k_minus_r_over_Product_by_r_less_k_of_k_minus_r | https://proofwiki.org/wiki/Summation_by_k_of_Product_by_r_of_x_plus_k_minus_r_over_Product_by_r_less_k_of_k_minus_r | [
"Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r",
"Summations",
"Continued Products"
] | [] | [
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Summation of Powers over Product of Differences"
] |
proofwiki-11652 | Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example | :$\dfrac {x \paren {x - 2} \paren {x - 3} } {\paren {-1} \paren {-2} \paren {-3} } + \dfrac {\paren {x + 1} \paren {x - 1} \paren {x - 2} } {\paren 1 \paren {-1} \paren {-2} } + \dfrac {\paren {x + 2} x \paren {x - 1} } {\paren 2 \paren 1 \paren {-1} } + \dfrac {\paren {x + 3} \paren {x + 1} x} {\paren 3 \paren 2 \pare... | This is an example of Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r:
:$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \n... | :$\dfrac {x \paren {x - 2} \paren {x - 3} } {\paren {-1} \paren {-2} \paren {-3} } + \dfrac {\paren {x + 1} \paren {x - 1} \paren {x - 2} } {\paren 1 \paren {-1} \paren {-2} } + \dfrac {\paren {x + 2} x \paren {x - 1} } {\paren 2 \paren 1 \paren {-1} } + \dfrac {\paren {x + 3} \paren {x + 1} x} {\paren 3 \paren 2 \pare... | This is an example of [[Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r]]:
:$\ds \sum_{k \mathop = 1}^n \paren {\dfrac {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \paren {x + k - r} } {\ds \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \matho... | Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example | https://proofwiki.org/wiki/Summation_by_k_of_Product_by_r_of_x_plus_k_minus_r_over_Product_by_r_less_k_of_k_minus_r/Example | https://proofwiki.org/wiki/Summation_by_k_of_Product_by_r_of_x_plus_k_minus_r_over_Product_by_r_less_k_of_k_minus_r/Example | [
"Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r"
] | [] | [
"Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r"
] |
proofwiki-11653 | Inverse of Cauchy Matrix | Let $C_n$ be the square Cauchy matrix of order $n$:
:<nowiki>$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_... | '''Preliminaries''':
Vandermonde Matrix Identity for Cauchy Matrix supplies matrix equation
:$(1): \quad -C = PV_x^{-1} V_y Q^{-1}$
:Definitions of symbols:
::<nowiki>$V_x = \paren {\begin{smallmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & \vdots & \ddo... | Let $C_n$ be the [[Definition:Square Matrix|square]] [[Definition:Cauchy Matrix|Cauchy matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
:<nowiki>$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfra... | '''Preliminaries''':
[[Vandermonde Matrix Identity for Cauchy Matrix]] supplies matrix equation
:$(1): \quad -C = PV_x^{-1} V_y Q^{-1}$
:Definitions of symbols:
::<nowiki>$V_x = \paren {\begin{smallmatrix}
1 & 1 & \cdots & 1 \\
x_1 & x_2 & \cdots & x_n \\
\vdots & \vdots ... | Inverse of Cauchy Matrix | https://proofwiki.org/wiki/Inverse_of_Cauchy_Matrix | https://proofwiki.org/wiki/Inverse_of_Cauchy_Matrix | [
"Cauchy Matrices"
] | [
"Definition:Matrix/Square Matrix",
"Definition:Cauchy Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix"
] | [
"Vandermonde Matrix Identity for Cauchy Matrix",
"Definition:Polynomial/Complex Numbers",
"Inverse of Matrix Product",
"Transpose of Matrix Product",
"Matrix Product with Adjugate Matrix",
"Definition:Cofactor",
"Laplace Expansion Theorem for Determinants",
"Value of Vandermonde Determinant/Formulatio... |
proofwiki-11654 | Sum of Elements in Inverse of Combinatorial Matrix | Let $C_n$ be the combinatorial matrix of order $n$ given by:
:$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$
Let $C_n^{-1}$ be its inverse, from Inverse of Combinatorial Matrix:
:$b_{i j} = \dfrac {-y + \delta_{i j} \... | All $n^2$ elements of $C_n^{-1}$ have a term $\dfrac {-y} {x \paren {x + n y} }$.
Further to this, the $n$ elements on the main diagonal contribute an extra $\dfrac {x + n y} {x \paren {x + n y} }$ to the total.
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
| r = n^2 \dfrac {-... | Let $C_n$ be the [[Definition:Combinatorial Matrix|combinatorial matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
:$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$
Let $C_n^{-1}$ be its [[Definit... | All $n^2$ [[Definition:Element of Matrix|elements]] of $C_n^{-1}$ have a term $\dfrac {-y} {x \paren {x + n y} }$.
Further to this, the $n$ elements on the [[Definition:Main Diagonal|main diagonal]] contribute an extra $\dfrac {x + n y} {x \paren {x + n y} }$ to the total.
Hence:
{{begin-eqn}}
{{eqn | l = \sum_{1 \ma... | Sum of Elements in Inverse of Combinatorial Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Combinatorial_Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Combinatorial_Matrix | [
"Combinatorial Matrix"
] | [
"Definition:Combinatorial Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix",
"Inverse of Combinatorial Matrix",
"Definition:Kronecker Delta",
"Definition:Matrix/Element"
] | [
"Definition:Matrix/Element",
"Definition:Matrix/Diagonal/Main"
] |
proofwiki-11655 | Sum of Elements in Inverse of Vandermonde Matrix | Let $V_n$ be the Vandermonde matrix of order $n$ given by:
:<nowiki>$V_n = \begin{pmatrix}
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^n & x_2^n & \cdots & x_n^n
\end{pmatrix}$</nowiki>
Let $V_n^{-1}$ be its inverse, from Inverse of Vandermonde Matrix:
:<... | From Sum of Elements of Nonsingular Matrix, the sum of elements in $V_n^{-1}$ is:
:$1 - \map \det {V_n^{-1} } \map \det {V_n - J_n}$
The plan is to expand $\map \det {V_n - J_n}$ and simplify.
The method is efficiently communicated in the $3 \times 3$ case:
{{begin-eqn}}
{{eqn | l = \map \det {V_3 - J_3}
| r = <n... | Let $V_n$ be the [[Definition:Vandermonde Matrix/Formulation 2|Vandermonde matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
:<nowiki>$V_n = \begin{pmatrix}
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^n & x_2^n & \cdots & x_n^n
\end{p... | From [[Sum of Elements of Nonsingular Matrix]], the sum of elements in $V_n^{-1}$ is:
:$1 - \map \det {V_n^{-1} } \map \det {V_n - J_n}$
The plan is to expand $\map \det {V_n - J_n}$ and simplify.
The method is efficiently communicated in the $3 \times 3$ case:
{{begin-eqn}}
{{eqn | l = \map \det {V_3 - J_3}
... | Sum of Elements in Inverse of Vandermonde Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Vandermonde_Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Vandermonde_Matrix | [
"Vandermonde Matrices"
] | [
"Definition:Vandermonde Matrix/Formulation 2",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix",
"Inverse of Vandermonde Matrix",
"Definition:Matrix/Element"
] | [
"Sum of Elements of Nonsingular Matrix",
"Definition:Ones Matrix",
"Effect of Elementary Row Operations on Determinant",
"Effect of Elementary Row Operations on Determinant",
"Effect of Elementary Row Operations on Determinant"
] |
proofwiki-11656 | Sum of Elements in Inverse of Cauchy Matrix | Let $C_n$ be the Cauchy matrix of order $n$ given by:
:<nowiki>$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots & \vdots & \ddots & \vdots \\
\dfrac 1 {x_m + y_1} & \dfrac 1 {x_m ... | It suffices to prove the Theorem for Cauchy matrix:
{{begin-eqn}}
{{eqn | l = C
| r = <nowiki>\begin{pmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n}... | Let $C_n$ be the [[Definition:Cauchy Matrix|Cauchy matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
:<nowiki>$C_n = \begin{bmatrix}
\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\
\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\
\vdots... | It suffices to prove the Theorem for Cauchy matrix:
{{begin-eqn}}
{{eqn | l = C
| r = <nowiki>\begin{pmatrix}
\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\
\dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n... | Sum of Elements in Inverse of Cauchy Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Cauchy_Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Cauchy_Matrix | [
"Cauchy Matrices",
"Summations",
"Continued Products",
"Summation of Powers over Product of Differences"
] | [
"Definition:Cauchy Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix",
"Inverse of Cauchy Matrix",
"Definition:Matrix/Element"
] | [
"Vandermonde Matrix Identity for Cauchy Matrix",
"Inverse of Matrix Product",
"Definition:Inverse Matrix",
"Transpose of Matrix Product",
"Definition:Symmetric Function/Elementary",
"Viète's Formulas",
"Viète's Formulas",
"Summation of Powers over Product of Differences"
] |
proofwiki-11657 | Hilbert Matrix is Cauchy Matrix | A Hilbert matrix is a special case of a Cauchy matrix. | By definition of Hilbert matrix, the element $a_{i j}$ is:
:$a_{i j} = \dfrac 1 {i + j - 1}$
For all $i, j \in \Z$ such that $1 \le i \le n$ and $1 \le j \le n$, let:
:$x_i = i$
:$y_j = j - 1$
Then:
:$a_{i j} = \dfrac 1 {x_i + y_j}$
The result follows by definition of a Cauchy matrix.
{{qed}} | A [[Definition:Hilbert Matrix|Hilbert matrix]] is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]]. | By definition of [[Definition:Hilbert Matrix|Hilbert matrix]], the [[Definition:Element of Matrix|element]] $a_{i j}$ is:
:$a_{i j} = \dfrac 1 {i + j - 1}$
For all $i, j \in \Z$ such that $1 \le i \le n$ and $1 \le j \le n$, let:
:$x_i = i$
:$y_j = j - 1$
Then:
:$a_{i j} = \dfrac 1 {x_i + y_j}$
The result follows b... | Hilbert Matrix is Cauchy Matrix | https://proofwiki.org/wiki/Hilbert_Matrix_is_Cauchy_Matrix | https://proofwiki.org/wiki/Hilbert_Matrix_is_Cauchy_Matrix | [
"Hilbert Matrices",
"Cauchy Matrices"
] | [
"Definition:Hilbert Matrix",
"Definition:Cauchy Matrix"
] | [
"Definition:Hilbert Matrix",
"Definition:Matrix/Element",
"Definition:Cauchy Matrix"
] |
proofwiki-11658 | Inverse of Hilbert Matrix | Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Then its inverse $H_n^{-1} = \sqbrk b_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + ... | From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
From Inverse of Cauchy Matrix, the inverse of the square Cauchy matrix of order $n$ is:
:$\begin{bmatrix} b_{i ... | Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Then its [[Definition:Inverse Matrix|inverse]] $H_n^{-1} = \sqbrk b_n$ can be specified as:
:$\begin{bmatrix... | From [[Hilbert Matrix is Cauchy Matrix]], $H_n$ is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]]:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
From [[Inverse of Cauchy Matrix]], the [[Definition:Inverse Matrix|inverse... | Inverse of Hilbert Matrix | https://proofwiki.org/wiki/Inverse_of_Hilbert_Matrix | https://proofwiki.org/wiki/Inverse_of_Hilbert_Matrix | [
"Hilbert Matrices"
] | [
"Definition:Hilbert Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix"
] | [
"Hilbert Matrix is Cauchy Matrix",
"Definition:Cauchy Matrix",
"Inverse of Cauchy Matrix",
"Definition:Inverse Matrix",
"Definition:Matrix/Square Matrix",
"Definition:Cauchy Matrix",
"Definition:Matrix/Square Matrix/Order",
"Product of Products",
"Definition:Factor",
"Definition:Expression",
"Tr... |
proofwiki-11659 | Elements of Inverse of Hilbert Matrix are Integers | Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its inverse $H_n^{-1}$.
All the elements of $H_n^{-1}$ are integers. | From Inverse of Hilbert Matrix, $H_n^{-1} = \begin {bmatrix} b \end{bmatrix}_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \pare... | Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its [[Definition:Inverse Matrix|inverse]] $H_n^{-1}$.
All the [[Definition:Element of Matrix|eleme... | From [[Inverse of Hilbert Matrix]], $H_n^{-1} = \begin {bmatrix} b \end{bmatrix}_n$ can be specified as:
:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! ... | Elements of Inverse of Hilbert Matrix are Integers | https://proofwiki.org/wiki/Elements_of_Inverse_of_Hilbert_Matrix_are_Integers | https://proofwiki.org/wiki/Elements_of_Inverse_of_Hilbert_Matrix_are_Integers | [
"Hilbert Matrices"
] | [
"Definition:Hilbert Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix",
"Definition:Matrix/Element",
"Definition:Integer"
] | [
"Inverse of Hilbert Matrix",
"Definition:Integer",
"Binomial Coefficient is Integer"
] |
proofwiki-11660 | Sum of Elements in Inverse of Hilbert Matrix | Let $H_n$ be the Hilbert matrix of order $n$:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its inverse $H_n^{-1}$.
All the elements of $H_n^{-1}$ are integers.
The sum of all the elements $b_{i j}$ of $H_n^{-1}$ is:
:$\ds \sum_{1 \mathop \le i, \ j \mathop \le n... | From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
| r = \sum_{k \mathop =... | Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
Consider its [[Definition:Inverse Matrix|inverse]] $H_n^{-1}$.
All the [[Definition:Element of Matrix|eleme... | From [[Hilbert Matrix is Cauchy Matrix]], $H_n$ is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]]:
:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
:$x_i = i$
:$y_j = j - 1$
Then:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} ... | Sum of Elements in Inverse of Hilbert Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Hilbert_Matrix | https://proofwiki.org/wiki/Sum_of_Elements_in_Inverse_of_Hilbert_Matrix | [
"Hilbert Matrices"
] | [
"Definition:Hilbert Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Inverse Matrix",
"Definition:Matrix/Element",
"Definition:Integer",
"Definition:Matrix/Element"
] | [
"Hilbert Matrix is Cauchy Matrix",
"Definition:Cauchy Matrix",
"Sum of Elements in Inverse of Cauchy Matrix",
"Closed Form for Triangular Numbers"
] |
proofwiki-11661 | Suprema of two Real Sets | Let $S$ and $T$ be real sets.
Let $S$ and $T$ admit suprema.
Then:
: $\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$ | === Necessary Condition ===
Let $\sup S \le \sup T$.
The aim is to establish that:
:$\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
Let $\epsilon \in \R_{>0}$.
Let:
:$S' = \set {s \in S : \exists t \in T : s \le t}$
There are two cases for an $s \in S$:
:$s \in S'$, in other words: $\... | Let $S$ and $T$ be [[Definition:Real Number|real sets]].
Let $S$ and $T$ admit [[Definition:Supremum of Subset of Real Numbers|suprema]].
Then:
: $\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$ | === Necessary Condition ===
Let $\sup S \le \sup T$.
The aim is to establish that:
:$\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
Let $\epsilon \in \R_{>0}$.
Let:
:$S' = \set {s \in S : \exists t \in T : s \le t}$
There are two cases for an $s \in S$:
:$s \in S'$, in other ... | Suprema of two Real Sets | https://proofwiki.org/wiki/Suprema_of_two_Real_Sets | https://proofwiki.org/wiki/Suprema_of_two_Real_Sets | [
"Real Analysis"
] | [
"Definition:Real Number",
"Definition:Supremum of Set/Real Numbers"
] | [
"Definition:Upper Bound of Set/Real Numbers",
"Definition:Upper Bound of Set/Real Numbers",
"Definition:Supremum of Set/Real Numbers",
"Definition:Upper Bound of Set/Real Numbers",
"Definition:Element",
"Supremum of Subset of Real Numbers is Arbitrarily Close",
"Supremum of Subset of Real Numbers is Arb... |
proofwiki-11662 | Sequential Continuity is Equivalent to Continuity in the Reals | Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f : A \to \R$ be a real function.
Then $f$ is continuous at $c$ {{iff}}:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$. | === Sufficient Condition ===
{{:Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition}}{{qed|lemma}} | Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
Let $c \in A$.
Let $f : A \to \R$ be a [[Definition:Real Function|real function]].
Then $f$ is [[Definition:Continuous Real Function|continuous]] at $c$ {{iff}}:
:for each [[Definition:Real Sequence|sequence]] $\s... | === [[Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition|Sufficient Condition]] ===
{{:Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition}}{{qed|lemma}} | Sequential Continuity is Equivalent to Continuity in the Reals | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals | [
"Real Analysis",
"Continuity",
"Sequential Continuity",
"Sequential Continuity is Equivalent to Continuity in the Reals"
] | [
"Definition:Subset",
"Definition:Real Number",
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers"
] | [
"Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition"
] |
proofwiki-11663 | Rising Factorial as Quotient of Factorials | :$x^{\overline n} = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!} = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}$ | {{begin-eqn}}
{{eqn | l = x^{\overline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + n - 1}
| c =
}}
{{eqn | r = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!}
| c = {{Defof|Factorial}}
}}... | :$x^{\overline n} = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!} = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}$ | {{begin-eqn}}
{{eqn | l = x^{\overline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + n - 1}
| c =
}}
{{eqn | r = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!}
| c = {{Defof|Factorial}}
}}... | Rising Factorial as Quotient of Factorials | https://proofwiki.org/wiki/Rising_Factorial_as_Quotient_of_Factorials | https://proofwiki.org/wiki/Rising_Factorial_as_Quotient_of_Factorials | [
"Rising Factorials",
"Factorials"
] | [] | [
"Gamma Function Extends Factorial"
] |
proofwiki-11664 | Equivalence of Definitions for Alternating Bilinear Mapping on Ring of Characteristic Not 2 | Let $R$ be a commutative ring.
Let $\struct {A_R, \oplus}$ be an algebra over $R$.
Let $R$ have a characteristic not equal to $2$.
Then the following definitions for alternating bilinear mappings are equivalent:
:$\oplus$ is an alternating bilinear mapping {{iff}} for all $a \in A_R$, $a \oplus a = 0$
:$\oplus$ is an a... | Let $\oplus$ be a bilinear mapping with the property that:
:$\forall a \in A_R: a \oplus a = 0$
Then for all $u, v \in A_R$:
{{begin-eqn}}
{{eqn | l = 0
| r = \paren {u + v} \oplus \paren {u + v}
}}
{{eqn | r = u \oplus \paren {u + v} + v \oplus \paren {u + v}
}}
{{eqn | r = u \oplus v + v \oplus u = 0
}}
{{end-... | Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
Let $\struct {A_R, \oplus}$ be an [[Definition:Algebra over Ring|algebra over $R$]].
Let $R$ have a [[Definition:Characteristic of Ring|characteristic]] not equal to $2$.
Then the following definitions for [[Definition:Alternating Bilinear Mapping/Chara... | Let $\oplus$ be a [[Definition:Bilinear Mapping|bilinear mapping]] with the property that:
:$\forall a \in A_R: a \oplus a = 0$
Then for all $u, v \in A_R$:
{{begin-eqn}}
{{eqn | l = 0
| r = \paren {u + v} \oplus \paren {u + v}
}}
{{eqn | r = u \oplus \paren {u + v} + v \oplus \paren {u + v}
}}
{{eqn | r = u \o... | Equivalence of Definitions for Alternating Bilinear Mapping on Ring of Characteristic Not 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_for_Alternating_Bilinear_Mapping_on_Ring_of_Characteristic_Not_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_for_Alternating_Bilinear_Mapping_on_Ring_of_Characteristic_Not_2 | [
"Alternating Bilinear Mappings"
] | [
"Definition:Commutative Ring",
"Definition:Algebra over Ring",
"Definition:Characteristic of Ring",
"Definition:Alternating Bilinear Mapping/Characteristic Not 2",
"Definition:Alternating Bilinear Mapping/Characteristic Not 2",
"Definition:Alternating Bilinear Mapping/Characteristic Not 2"
] | [
"Definition:Bilinear Mapping",
"Definition:Bilinear Mapping",
"Definition:Characteristic of Ring",
"Category:Alternating Bilinear Mappings"
] |
proofwiki-11665 | Way Below if Between is Compact Set in Ordered Set of Topology | Let $T = \struct {S, \tau}$ be a topological space.
Let $L = \struct {\tau, \preceq}$ be an ordered set
where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
Let $x, y \in \tau$ such that:
:$\exists H \subseteq S: x \subseteq H \subseteq y \land H$ is compact
Then:
:$x \ll y$ | Let $D$ be a directed subset of $\tau$ such that:
:$y \preceq \sup D$
By proof of Topology forms Complete Lattice:
:$\ds y \subseteq \bigcup D$
By Subset Relation is Transitive:
:$\ds H \subseteq \bigcup D$
By definition:
:$D$ is open cover of $H$
By definition of compact:
:$H$ has finite subcover $\GG$ of $D$
By Direc... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $L = \struct {\tau, \preceq}$ be an [[Definition:Ordered Set|ordered set]]
where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
Let $x, y \in \tau$ such that:
:$\exists H \subseteq S: x \subseteq H \subseteq y \land... | Let $D$ be a [[Definition:Directed Subset|directed subset]] of $\tau$ such that:
:$y \preceq \sup D$
By proof of [[Topology forms Complete Lattice]]:
:$\ds y \subseteq \bigcup D$
By [[Subset Relation is Transitive]]:
:$\ds H \subseteq \bigcup D$
By definition:
:$D$ is [[Definition:Open Cover|open cover]] of $H$
By ... | Way Below if Between is Compact Set in Ordered Set of Topology | https://proofwiki.org/wiki/Way_Below_if_Between_is_Compact_Set_in_Ordered_Set_of_Topology | https://proofwiki.org/wiki/Way_Below_if_Between_is_Compact_Set_in_Ordered_Set_of_Topology | [
"Way Below Relation",
"Topology"
] | [
"Definition:Topological Space",
"Definition:Ordered Set",
"Definition:Compact"
] | [
"Definition:Directed Subset",
"Topology forms Complete Lattice",
"Subset Relation is Transitive",
"Definition:Open Cover",
"Definition:Compact",
"Definition:Subcover/Finite",
"Directed iff Finite Subsets have Upper Bounds",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Union i... |
proofwiki-11666 | Locally Compact iff Open Neighborhood contains Compact Set | Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is locally compact
{{iff}}:
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is compact
where $K^\circ$ denotes the interior of $K$. | === Sufficient Condition ===
Let $T$ be locally compact.
Let $U \in \tau, x \in U$.
Since $T$ is locally compact there exists a neighborhood basis $\BB$ of $x$ consisting of compact sets.
From Set is Open iff Neighborhood of all its Points, $U$ is a neighborhood of $x$.
By definition of a neighborhood basis:
:$\exists... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Then:
:$T$ is [[Definition:Locally Compact Space|locally compact]]
{{iff}}:
:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
where... | === Sufficient Condition ===
Let $T$ be [[Definition:Locally Compact Space|locally compact]].
Let $U \in \tau, x \in U$.
Since $T$ is [[Definition:Locally Compact Space|locally compact]] there exists a [[Definition:Neighborhood Basis|neighborhood basis]] $\BB$ of $x$ consisting of [[Definition:Compact Topological Su... | Locally Compact iff Open Neighborhood contains Compact Set | https://proofwiki.org/wiki/Locally_Compact_iff_Open_Neighborhood_contains_Compact_Set | https://proofwiki.org/wiki/Locally_Compact_iff_Open_Neighborhood_contains_Compact_Set | [
"Locally Compact Spaces"
] | [
"Definition:Topological Space",
"Definition:Locally Compact Space",
"Definition:Compact Topological Space/Subspace",
"Definition:Interior (Topology)"
] | [
"Definition:Locally Compact Space",
"Definition:Locally Compact Space",
"Definition:Neighborhood Basis",
"Definition:Compact Topological Space/Subspace",
"Set is Open iff Neighborhood of all its Points",
"Definition:Neighborhood (Topology)",
"Definition:Neighborhood Basis",
"Definition:Neighborhood (T... |
proofwiki-11667 | Cauchy-Binet Formula | Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf B$ be an $n \times m$ matrix.
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
Let $\mathbf B_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ ma... | Let $\tuple {k_1, k_2, \ldots, k_m}$ be an ordered $m$-tuple of integers.
Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the sign of $\tuple {k_1, k_2, \ldots, k_m}$.
Let $\tuple {l_1, l_2, \ldots, l_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ except for $k_i$ and $k_j$ having been transposed.
Then from Tra... | Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
Let $\mathbf B$ be an [[Definition:Matrix|$n \times m$ matrix]].
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j... | Let $\tuple {k_1, k_2, \ldots, k_m}$ be an [[Definition:Ordered Tuple|ordered $m$-tuple]] of [[Definition:Integer|integers]].
Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the [[Definition:Sign of Ordered Tuple|sign]] of $\tuple {k_1, k_2, \ldots, k_m}$.
Let $\tuple {l_1, l_2, \ldots, l_m}$ be the same as $\tupl... | Cauchy-Binet Formula | https://proofwiki.org/wiki/Cauchy-Binet_Formula | https://proofwiki.org/wiki/Cauchy-Binet_Formula | [
"Matrix Theory",
"Determinants",
"Cauchy-Binet Formula"
] | [
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix",
"Definition:Matrix/Row",
"Definition:Determinant/Matrix"
] | [
"Definition:Ordered Tuple",
"Definition:Integer",
"Definition:Sign of Ordered Tuple",
"Transposition is of Odd Parity"
] |
proofwiki-11668 | Auxiliary Relation is Congruent | Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\RR$ be relation on $S$ satisfying conditions $(2)$ and $(3)$ of auxiliary relation.
Then:
:$\forall x, y, z, u \in S: \tuple {x, z} \in \RR \land \tuple {y, u} \in \RR \implies \tuple {x \vee y, z \vee u} \in \RR$ | Let $x, y, z, u \in S$ such that:
:$\tuple {x, z} \in \RR \land \tuple {y, u} \in \RR$
By definition of reflexivity:
:$x \preceq x$ and $y \preceq y$
By Join Succeeds Operands:
:$z \preceq z \vee u$ and $u \preceq z \vee u$
By condition $(2)$ of auxiliary relation:
:$\tuple {x, z \vee u} \in \RR$ and $\tuple {y, z \vee... | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\RR$ be [[Definition:Relation|relation]] on $S$ satisfying conditions $(2)$ and $(3)$ of [[Definition:Auxiliary Relation|auxiliary relation]].
Then:
:$\forall x, y, z, u \in S: \tup... | Let $x, y, z, u \in S$ such that:
:$\tuple {x, z} \in \RR \land \tuple {y, u} \in \RR$
By definition of [[Definition:Reflexivity|reflexivity]]:
:$x \preceq x$ and $y \preceq y$
By [[Join Succeeds Operands]]:
:$z \preceq z \vee u$ and $u \preceq z \vee u$
By condition $(2)$ of [[Definition:Auxiliary Relation|auxiliar... | Auxiliary Relation is Congruent | https://proofwiki.org/wiki/Auxiliary_Relation_is_Congruent | https://proofwiki.org/wiki/Auxiliary_Relation_is_Congruent | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Relation",
"Definition:Auxiliary Relation"
] | [
"Definition:Reflexivity",
"Join Succeeds Operands",
"Definition:Auxiliary Relation",
"Definition:Auxiliary Relation"
] |
proofwiki-11669 | Auxiliary Relation is Transitive | Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\RR$ be relation on $S$ satisfying conditions $(1)$ and $(2)$ of auxiliary relation.
Then
:$\RR$ is a transitive relation. | Let $x, y, z \in S$ such that
:$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR$
By definition of reflexivity:
:$z \preceq z$
By condition $(1)$ of auxiliary relation:
:$x \preceq y$
Thus by condition $(2)$ of auxiliary relation:
:$\tuple {x, z} \in \RR$
Thus by definition
:$\RR$ is a transitive relation.
{{qed}} | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\RR$ be [[Definition:Relation|relation]] on $S$ satisfying conditions $(1)$ and $(2)$ of [[Definition:Auxiliary Relation|auxiliary relation]].
Then
:$\RR$ is a [[Definition:Transiti... | Let $x, y, z \in S$ such that
:$\tuple {x, y} \in \RR \land \tuple {y, z} \in \RR$
By definition of [[Definition:Reflexivity|reflexivity]]:
:$z \preceq z$
By condition $(1)$ of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$x \preceq y$
Thus by condition $(2)$ of [[Definition:Auxiliary Relation|auxiliary r... | Auxiliary Relation is Transitive | https://proofwiki.org/wiki/Auxiliary_Relation_is_Transitive | https://proofwiki.org/wiki/Auxiliary_Relation_is_Transitive | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Relation",
"Definition:Auxiliary Relation",
"Definition:Transitive Relation"
] | [
"Definition:Reflexivity",
"Definition:Auxiliary Relation",
"Definition:Auxiliary Relation",
"Definition:Transitive Relation"
] |
proofwiki-11670 | Cauchy-Binet Formula/Example/m equals 1 | Let $\mathbf A = \sqbrk a_{1 n}$ be a row matrix with $n$ columns.
and $\mathbf B = \sqbrk b_{n 1}$ be a column matrix with $n$ rows.
Let $\mathbf A \mathbf B$ be the (conventional) matrix product of $\mathbf A$ and $\mathbf B$.
Then:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{j \mathop = 1}^n a_j b_j$
where:
:$a_j$... | The Cauchy-Binet Formula gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \mathop \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times ... | Let $\mathbf A = \sqbrk a_{1 n}$ be a [[Definition:Row Matrix|row matrix]] with $n$ [[Definition:Column of Matrix|columns]].
and $\mathbf B = \sqbrk b_{n 1}$ be a [[Definition:Column Matrix|column matrix]] with $n$ [[Definition:Row of Matrix|rows]].
Let $\mathbf A \mathbf B$ be the [[Definition:Matrix Product (Conven... | The [[Cauchy-Binet Formula]] gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \mathop \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
:$... | Cauchy-Binet Formula/Example/m equals 1 | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/m_equals_1 | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/m_equals_1 | [
"Determinants",
"Cauchy-Binet Formula"
] | [
"Definition:Row Matrix",
"Definition:Matrix/Column",
"Definition:Column Matrix",
"Definition:Matrix/Row",
"Definition:Matrix Product (Conventional)",
"Definition:Matrix/Element",
"Definition:Matrix/Element"
] | [
"Cauchy-Binet Formula",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix",
"Definition:Matrix/Row",
"Determinant/Examples/Order 1"
] |
proofwiki-11671 | Preceding is Auxiliary Relation | Let $\left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Then
:$\preceq$ is an auxiliary relation. | :$\forall x, y \in S: x \preceq y \implies x \preceq y$
Then condition $(1)$ of auxiliary relation is satisfied.
By definition of transitivity:
:$\forall x, y, z, u \in S: x \preceq y \preceq z \preceq u \implies x \preceq u$
Then the condition $(2)$ of auxiliary relation is satisfied.
By definition of supremum:
:$\fo... | Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Then
:$\preceq$ is an [[Definition:Auxiliary Relation|auxiliary relation]]. | :$\forall x, y \in S: x \preceq y \implies x \preceq y$
Then condition $(1)$ of [[Definition:Auxiliary Relation|auxiliary relation]] is satisfied.
By definition of [[Definition:Transitivity|transitivity]]:
:$\forall x, y, z, u \in S: x \preceq y \preceq z \preceq u \implies x \preceq u$
Then the condition $(2)$ of [[D... | Preceding is Auxiliary Relation | https://proofwiki.org/wiki/Preceding_is_Auxiliary_Relation | https://proofwiki.org/wiki/Preceding_is_Auxiliary_Relation | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Auxiliary Relation"
] | [
"Definition:Auxiliary Relation",
"Definition:Transitive",
"Definition:Auxiliary Relation",
"Definition:Supremum of Set",
"Definition:Auxiliary Relation",
"Definition:Smallest Element",
"Definition:Auxiliary Relation",
"Definition:Auxiliary Relation"
] |
proofwiki-11672 | Cauchy-Binet Formula/Example/Matrix by Transpose | Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf A^\intercal$ be the transpose $\mathbf A$.
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
Let $\mathbf A^\intercal_{j_1 j_2 \ldots j_m}$ den... | The Cauchy-Binet Formula gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matri... | Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] $\mathbf A$.
Let $1 \le j_1, j_2, \ldots, j_m \le n$.
Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Col... | The [[Cauchy-Binet Formula]] gives:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
:$\mathbf ... | Cauchy-Binet Formula/Example/Matrix by Transpose | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/Matrix_by_Transpose | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/Matrix_by_Transpose | [
"Cauchy-Binet Formula"
] | [
"Definition:Matrix",
"Definition:Transpose of Matrix",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix",
"Definition:Matrix/Row",
"Definition:Determinant/Matrix"
] | [
"Cauchy-Binet Formula",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix",
"Definition:Matrix/Row",
"Definition:Transpose of Matrix",
"Definition:Matrix",
"Cauchy-Binet Formula",
"Definition:Matrix/Square Matrix",
"Definition:Transpo... |
proofwiki-11673 | Preceding is Top in Ordered Set of Auxiliary Relations | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map {\it Aux} L$ be the set of all auxiliary relations on $S$.
Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ordered set
where $\precsim \mathop = \subseteq \restriction_{\map {\it Aux} L \times \map {\it Aux} L}$
Then
:$\preceq \mat... | By Preceding is Auxiliary Relation:
:$\preceq \mathop \in \map {\it Aux} L$
By definition:
:$\preceq$ is lower bound for $\O$ in $P$
We will prove that:
:$\forall R \in \map {\it Aux} L: R$ is lower bound for $\O \implies R \mathop \precsim \preceq$
Let $R \in \map {\it Aux} L$
By condition $(i)$ of definition of auxil... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\it Aux} L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ... | By [[Preceding is Auxiliary Relation]]:
:$\preceq \mathop \in \map {\it Aux} L$
By definition:
:$\preceq$ is [[Definition:Lower Bound of Set|lower bound]] for $\O$ in $P$
We will prove that:
:$\forall R \in \map {\it Aux} L: R$ is [[Definition:Lower Bound of Set|lower bound]] for $\O \implies R \mathop \precsim \prec... | Preceding is Top in Ordered Set of Auxiliary Relations | https://proofwiki.org/wiki/Preceding_is_Top_in_Ordered_Set_of_Auxiliary_Relations | https://proofwiki.org/wiki/Preceding_is_Top_in_Ordered_Set_of_Auxiliary_Relations | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Ordered Set",
"Definition:Greatest Element"
] | [
"Preceding is Auxiliary Relation",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Auxiliary Relation",
"Definition:Infimum of Set",
"Infimum of Empty Set is Greatest Element"
] |
proofwiki-11674 | Bottom Relation is Auxiliary Relation | Let $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $B = \left\{ {\left({\bot, x}\right): x \in S}\right\}$
where $\bot$ denotes the smallest element in $L$.
Then
:$B$ is auxiliary relation. | By definition of smallest element:
:$\forall x \in S: \bot \preceq x$
Thus by definition of $B$:
:$\forall x, y \in S: \left({x, y}\right) \in B \implies x \preceq y$
We will prove that
:$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in B \land z \preceq u \implies \left({x, u}\right) \in B$
Let $x, ... | Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $B = \left\{ {\left({\bot, x}\right): x \in S}\right\}$
where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
Then
:$B$ is [[Definition:Auxiliar... | By definition of [[Definition:Smallest Element|smallest element]]:
:$\forall x \in S: \bot \preceq x$
Thus by definition of $B$:
:$\forall x, y \in S: \left({x, y}\right) \in B \implies x \preceq y$
We will prove that
:$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in B \land z \preceq u \implies \... | Bottom Relation is Auxiliary Relation | https://proofwiki.org/wiki/Bottom_Relation_is_Auxiliary_Relation | https://proofwiki.org/wiki/Bottom_Relation_is_Auxiliary_Relation | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Smallest Element",
"Definition:Auxiliary Relation"
] | [
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Antisymmetric Relation",
"Join is Idempotent",
"Definition:Auxiliary Relation"
] |
proofwiki-11675 | Bottom Relation is Bottom in Ordered Set of Auxiliary Relations | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map {\it Aux} L$ be the set of all auxiliary relations on $S$.
Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ordered set where:
:$\precsim \mathop = \subseteq \restriction_{ \map {\it Aux} L \times \map {\it Aux} L}$
Let $B = \set {\... | By Bottom Relation is Auxiliary Relation:
:$B \in \map {\it Aux} L$
By definition:
:$B$ is an upper bound for $\O$ in $P$
We will prove that:
:$\forall R \in \map {\it Aux} L: R$ is an upper bound for $\O \implies B \mathop \precsim R$
Let $R \in \map {\it Aux} L$.
By condition $(iv)$ of definition of auxiliary relatio... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\it Aux} L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
Let $P = \struct {\map {\it Aux} L, \precsim}$ be an ... | By [[Bottom Relation is Auxiliary Relation]]:
:$B \in \map {\it Aux} L$
By definition:
:$B$ is an [[Definition:Upper Bound of Set|upper bound]] for $\O$ in $P$
We will prove that:
:$\forall R \in \map {\it Aux} L: R$ is an [[Definition:Upper Bound of Set|upper bound]] for $\O \implies B \mathop \precsim R$
Let $R \i... | Bottom Relation is Bottom in Ordered Set of Auxiliary Relations | https://proofwiki.org/wiki/Bottom_Relation_is_Bottom_in_Ordered_Set_of_Auxiliary_Relations | https://proofwiki.org/wiki/Bottom_Relation_is_Bottom_in_Ordered_Set_of_Auxiliary_Relations | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Ordered Set",
"Definition:Smallest Element"
] | [
"Bottom Relation is Auxiliary Relation",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Auxiliary Relation",
"Definition:Supremum of Set",
"Supremum of Empty Set is Smallest Element"
] |
proofwiki-11676 | Inverse of Conditional is Converse of Contrapositive | Let $p \implies q$ be a conditional.
Then the inverse of $p \implies q$ is the converse of its contrapositive. | The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The contrapositive of $p \implies q$ is:
:$\neg q \implies \neg p$
The converse of $\neg q \implies \neg p$ is:
:$\neg p \implies \neg q$
The two are seen to be equal.
{{qed}}
Category:Conditional
87yc00e6gp5kuh83oqd4glbg0a8ece4 | Let $p \implies q$ be a [[Definition:Conditional|conditional]].
Then the [[Definition:Inverse Statement|inverse]] of $p \implies q$ is the [[Definition:Converse Statement|converse]] of its [[Definition:Contrapositive Statement|contrapositive]]. | The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
:$\neg p \implies \neg q$
The [[Definition:Contrapositive Statement|contrapositive]] of $p \implies q$ is:
:$\neg q \implies \neg p$
The [[Definition:Converse Statement|converse]] of $\neg q \implies \neg p$ is:
:$\neg p \implies \neg q$
The two a... | Inverse of Conditional is Converse of Contrapositive | https://proofwiki.org/wiki/Inverse_of_Conditional_is_Converse_of_Contrapositive | https://proofwiki.org/wiki/Inverse_of_Conditional_is_Converse_of_Contrapositive | [
"Conditional"
] | [
"Definition:Conditional",
"Definition:Inverse Statement",
"Definition:Converse Statement",
"Definition:Contrapositive Statement"
] | [
"Definition:Inverse Statement",
"Definition:Contrapositive Statement",
"Definition:Converse Statement",
"Category:Conditional"
] |
proofwiki-11677 | Inverse of Conditional is Contrapositive of Converse | Let $p \implies q$ be a conditional.
Then the inverse of $p \implies q$ is the contrapositive of its converse. | The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The converse of $p \implies q$ is:
:$q \implies p$
The contrapositive of $q \implies p$ is:
:$\neg p \implies \neg q$
The two are seen to be equal.
{{qed}}
Category:Conditional
d44nh6jup1dia42nknek8d9zurgb9wf | Let $p \implies q$ be a [[Definition:Conditional|conditional]].
Then the [[Definition:Inverse Statement|inverse]] of $p \implies q$ is the [[Definition:Contrapositive Statement|contrapositive]] of its [[Definition:Converse Statement|converse]]. | The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
:$\neg p \implies \neg q$
The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
:$q \implies p$
The [[Definition:Contrapositive Statement|contrapositive]] of $q \implies p$ is:
:$\neg p \implies \neg q$
The two are seen to be equal.... | Inverse of Conditional is Contrapositive of Converse | https://proofwiki.org/wiki/Inverse_of_Conditional_is_Contrapositive_of_Converse | https://proofwiki.org/wiki/Inverse_of_Conditional_is_Contrapositive_of_Converse | [
"Conditional"
] | [
"Definition:Conditional",
"Definition:Inverse Statement",
"Definition:Contrapositive Statement",
"Definition:Converse Statement"
] | [
"Definition:Inverse Statement",
"Definition:Converse Statement",
"Definition:Contrapositive Statement",
"Category:Conditional"
] |
proofwiki-11678 | Converse of Conditional is Contrapositive of Inverse | Let $p \implies q$ be a conditional.
Then the converse of $p \implies q$ is the contrapositive of its inverse. | The converse of $p \implies q$ is:
:$q \implies p$
The inverse of $p \implies q$ is:
:$\neg p \implies \neg q$
The contrapositive of $\neg p \implies \neg q$ is:
:$\neg \neg q \implies \neg \neg p$
By Double Negation, the two are seen to be equal.
{{qed}}
Category:Conditional
qxlweqotkrviamlc86s2ga5z7j3idbk | Let $p \implies q$ be a [[Definition:Conditional|conditional]].
Then the [[Definition:Converse Statement|converse]] of $p \implies q$ is the [[Definition:Contrapositive Statement|contrapositive]] of its [[Definition:Inverse Statement|inverse]]. | The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
:$q \implies p$
The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
:$\neg p \implies \neg q$
The [[Definition:Contrapositive Statement|contrapositive]] of $\neg p \implies \neg q$ is:
:$\neg \neg q \implies \neg \neg p$
By [[Doub... | Converse of Conditional is Contrapositive of Inverse | https://proofwiki.org/wiki/Converse_of_Conditional_is_Contrapositive_of_Inverse | https://proofwiki.org/wiki/Converse_of_Conditional_is_Contrapositive_of_Inverse | [
"Conditional"
] | [
"Definition:Conditional",
"Definition:Converse Statement",
"Definition:Contrapositive Statement",
"Definition:Inverse Statement"
] | [
"Definition:Converse Statement",
"Definition:Inverse Statement",
"Definition:Contrapositive Statement",
"Double Negation/Formulation 1",
"Category:Conditional"
] |
proofwiki-11679 | Converse of Conditional is Inverse of Contrapositive | Let $p \implies q$ be a conditional.
Then the converse of $p \implies q$ is the inverse of its contrapositive. | The converse of $p \implies q$ is:
:$q \implies p$
The contrapositive of $p \implies q$ is:
:$\neg q \implies \neg p$
The inverse of $\neg q \implies \neg p$ is:
:$\neg \neg q \implies \neg \neg p$
By Double Negation, the two are seen to be equal.
{{qed}}
Category:Conditional
4dtdb2u7zqedssj9r5e1eereynfa6au | Let $p \implies q$ be a [[Definition:Conditional|conditional]].
Then the [[Definition:Converse Statement|converse]] of $p \implies q$ is the [[Definition:Inverse Statement|inverse]] of its [[Definition:Contrapositive Statement|contrapositive]]. | The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
:$q \implies p$
The [[Definition:Contrapositive Statement|contrapositive]] of $p \implies q$ is:
:$\neg q \implies \neg p$
The [[Definition:Inverse Statement|inverse]] of $\neg q \implies \neg p$ is:
:$\neg \neg q \implies \neg \neg p$
By [[Doub... | Converse of Conditional is Inverse of Contrapositive | https://proofwiki.org/wiki/Converse_of_Conditional_is_Inverse_of_Contrapositive | https://proofwiki.org/wiki/Converse_of_Conditional_is_Inverse_of_Contrapositive | [
"Conditional"
] | [
"Definition:Conditional",
"Definition:Converse Statement",
"Definition:Inverse Statement",
"Definition:Contrapositive Statement"
] | [
"Definition:Converse Statement",
"Definition:Contrapositive Statement",
"Definition:Inverse Statement",
"Double Negation/Formulation 1",
"Category:Conditional"
] |
proofwiki-11680 | Intersection of Auxiliary Relations is Auxiliary Relation | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\FF$ be a non-empty set of auxiliary relations on $S$.
Then
:$\ds \bigcap \FF$ is auxiliary relation. | By definition of non-empty set:
:$\exists \RR: \RR \in \FF$
We will prove that:
:$\ds \forall x, y \in S: \tuple {x, y} \in \bigcap \FF \implies x \preceq y$
Let $x, y \in S$ such that:
:$\ds \tuple {x, y} \in \bigcap \FF$
By definition of intersection:
:$\tuple {x, y} \in \RR$
Thus by definition of auxiliary relation:... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\FF$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set]] of [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
Then
:$\ds \bigcap \FF$ is [[Def... | By definition of [[Definition:Non-Empty Set|non-empty set]]:
:$\exists \RR: \RR \in \FF$
We will prove that:
:$\ds \forall x, y \in S: \tuple {x, y} \in \bigcap \FF \implies x \preceq y$
Let $x, y \in S$ such that:
:$\ds \tuple {x, y} \in \bigcap \FF$
By definition of [[Definition:Set Intersection|intersection]]:
:$... | Intersection of Auxiliary Relations is Auxiliary Relation | https://proofwiki.org/wiki/Intersection_of_Auxiliary_Relations_is_Auxiliary_Relation | https://proofwiki.org/wiki/Intersection_of_Auxiliary_Relations_is_Auxiliary_Relation | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Non-Empty Set",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Auxiliary Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Set Intersection",
"Definition:Auxiliary Relation",
"Definition:Set Intersection",
"Definition:Auxiliary Relation",
"Definition:Set Intersection",
"Definition:Set Intersection",
"Definition:Auxiliary Relation",
"Definition:Set Intersection",
"Definition:Auxi... |
proofwiki-11681 | Ordered Set of Auxiliary Relations is Complete Lattice | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $\map {\operatorname {Aux} } L$ be the set of all auxiliary relations on $S$.
Let $P = \struct {\map {\operatorname {Aux} } L, \precsim}$ be an ordered set
where $\precsim \mathop = \subseteq \restriction_{\map {\operatorname {Aux} } L \times... | Let $X \subseteq \map {\operatorname {Aux} } L$
In the case when $X \ne \O$:
By Intersection of Auxiliary Relations is Auxiliary Relation:
:$\bigcap X \in \map {\operatorname {Aux} } L$
By Intersection is Largest Subset, $\bigcap X$ is the infimum of $X$.
In case when $X = \O$:
By proof of Preceding is Top in Ordered S... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\operatorname {Aux} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
Let $P = \struct {\map {\operatorname {A... | Let $X \subseteq \map {\operatorname {Aux} } L$
In the case when $X \ne \O$:
By [[Intersection of Auxiliary Relations is Auxiliary Relation]]:
:$\bigcap X \in \map {\operatorname {Aux} } L$
By [[Intersection is Largest Subset]], $\bigcap X$ is the [[Definition:Infimum of Set|infimum]] of $X$.
In case when $X = \O$:... | Ordered Set of Auxiliary Relations is Complete Lattice | https://proofwiki.org/wiki/Ordered_Set_of_Auxiliary_Relations_is_Complete_Lattice | https://proofwiki.org/wiki/Ordered_Set_of_Auxiliary_Relations_is_Complete_Lattice | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Ordered Set",
"Definition:Complete Lattice"
] | [
"Intersection of Auxiliary Relations is Auxiliary Relation",
"Intersection is Largest Subset",
"Definition:Infimum of Set",
"Preceding is Top in Ordered Set of Auxiliary Relations",
"Definition:Infimum of Set",
"Definition:Empty Set",
"Definition:Non-Empty Set",
"Definition:Infimum of Set",
"Lattice... |
proofwiki-11682 | Cauchy-Binet Formula/Example/m greater than n | Let $\mathbf A$ be an $m \times n$ matrix.
Let $\mathbf B$ be an $n \times m$ matrix.
Let $m > n$.
Then:
:$\map \det {\mathbf A \mathbf B} = 0$ | The Cauchy-Binet Formula gives:
:$(1): \quad \ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \tim... | Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
Let $\mathbf B$ be an [[Definition:Matrix|$n \times m$ matrix]].
Let $m > n$.
Then:
:$\map \det {\mathbf A \mathbf B} = 0$ | The [[Cauchy-Binet Formula]] gives:
:$(1): \quad \ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]... | Cauchy-Binet Formula/Example/m greater than n | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/m_greater_than_n | https://proofwiki.org/wiki/Cauchy-Binet_Formula/Example/m_greater_than_n | [
"Cauchy-Binet Formula"
] | [
"Definition:Matrix",
"Definition:Matrix"
] | [
"Cauchy-Binet Formula",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Matrix",
"Definition:Matrix/Row",
"Definition:Set",
"Definition:Empty Set",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-11683 | Krattenthaler's Identity | :$\begin{vmatrix}
\paren {x + q_2} \paren {x + q_3} & \paren {x + p_1} \paren {x + q_3} & \paren {x + p_1} \paren {x + p_2} \\
\paren {y + q_2} \paren {y + q_3} & \paren {y + p_1} \paren {y + q_3} & \paren {y + p_1} \paren {y + p_2} \\
\paren {z + q_2} \paren {z + q_3} & \paren {z + p_1} \paren {z + q_3} & \paren {z + ... | {{begin-eqn}}
{{eqn | o =
| r = \begin{vmatrix}
\paren {x + q_2} \paren {x + q_3} & \paren {x + p_1} \paren {x + q_3} & \paren {x + p_1} \paren {x + p_2} \\
\paren {y + q_2} \paren {y + q_3} & \paren {y + p_1} \paren {y + q_3} & \paren {y + p_1} \paren {y + p_2} \\
\paren {z + q_2} \paren {z + q_3} & \paren {z +... | :$\begin{vmatrix}
\paren {x + q_2} \paren {x + q_3} & \paren {x + p_1} \paren {x + q_3} & \paren {x + p_1} \paren {x + p_2} \\
\paren {y + q_2} \paren {y + q_3} & \paren {y + p_1} \paren {y + q_3} & \paren {y + p_1} \paren {y + p_2} \\
\paren {z + q_2} \paren {z + q_3} & \paren {z + p_1} \paren {z + q_3} & \paren {z + ... | {{begin-eqn}}
{{eqn | o =
| r = \begin{vmatrix}
\paren {x + q_2} \paren {x + q_3} & \paren {x + p_1} \paren {x + q_3} & \paren {x + p_1} \paren {x + p_2} \\
\paren {y + q_2} \paren {y + q_3} & \paren {y + p_1} \paren {y + q_3} & \paren {y + p_1} \paren {y + p_2} \\
\paren {z + q_2} \paren {z + q_3} & \paren {z +... | Krattenthaler's Identity | https://proofwiki.org/wiki/Krattenthaler's_Identity | https://proofwiki.org/wiki/Krattenthaler's_Identity | [
"Determinants"
] | [
"Definition:Determinant/Matrix"
] | [
"Multiple of Row Added to Row of Determinant",
"Determinant with Row Multiplied by Constant",
"Multiple of Row Added to Row of Determinant",
"Determinant with Row Multiplied by Constant",
"Multiple of Row Added to Row of Determinant",
"Value of Vandermonde Determinant/Formulation 1"
] |
proofwiki-11684 | Floor Function/Examples/Floor of Root 2 | :$\floor {\sqrt 2} = 1$ | The decimal expansion of $\sqrt 2$ is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 \le \sqrt 2 < 2$
Hence $1$ is the floor of $\sqrt 2$ by definition.
{{qed}} | :$\floor {\sqrt 2} = 1$ | The [[Square Root of 2|decimal expansion of $\sqrt 2$]] is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 \le \sqrt 2 < 2$
Hence $1$ is the [[Definition:Floor Function|floor]] of $\sqrt 2$ by definition.
{{qed}} | Floor Function/Examples/Floor of Root 2 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Root_2 | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Root_2 | [
"Examples of Floor Function"
] | [] | [
"Square Root/Examples/2",
"Definition:Floor Function"
] |
proofwiki-11685 | Ceiling Function/Examples/Ceiling of Root 2 | :$\ceiling {\sqrt 2} = 2$ | The decimal expansion of $\sqrt 2$ is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 < \sqrt 2 \le 2$
Hence $2$ is the ceiling of $\sqrt 2$ by definition.
{{qed}} | :$\ceiling {\sqrt 2} = 2$ | The [[Square Root of 2|decimal expansion of $\sqrt 2$]] is:
:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
Thus:
:$1 < \sqrt 2 \le 2$
Hence $2$ is the [[Definition:Ceiling Function|ceiling]] of $\sqrt 2$ by definition.
{{qed}} | Ceiling Function/Examples/Ceiling of Root 2 | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_Root_2 | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_Root_2 | [
"Examples of Ceiling Function"
] | [] | [
"Square Root/Examples/2",
"Definition:Ceiling Function"
] |
proofwiki-11686 | Floor Function/Examples/Floor of One Half | :$\floor {\dfrac 1 2} = 0$ | We have that:
:$0 \le \dfrac 1 2$
and:
:$1 = \dfrac 2 2 > \dfrac 1 2$
Hence $0$ is the floor of $\dfrac 1 2$ by definition.
{{qed}} | :$\floor {\dfrac 1 2} = 0$ | We have that:
:$0 \le \dfrac 1 2$
and:
:$1 = \dfrac 2 2 > \dfrac 1 2$
Hence $0$ is the [[Definition:Floor Function|floor]] of $\dfrac 1 2$ by definition.
{{qed}} | Floor Function/Examples/Floor of One Half | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_One_Half | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_One_Half | [
"Examples of Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-11687 | Ceiling Function/Examples/Ceiling of Minus One Half | :$\ceiling {-\dfrac 1 2} = 0$ | We have:
:$0 \ge -\dfrac 1 2$
and:
:$-1 = -\dfrac 2 2 < -\dfrac 1 2$
Hence $0$ is the ceiling of $-\dfrac 1 2$ by definition.
{{qed}} | :$\ceiling {-\dfrac 1 2} = 0$ | We have:
:$0 \ge -\dfrac 1 2$
and:
:$-1 = -\dfrac 2 2 < -\dfrac 1 2$
Hence $0$ is the [[Definition:Ceiling Function|ceiling]] of $-\dfrac 1 2$ by definition.
{{qed}} | Ceiling Function/Examples/Ceiling of Minus One Half | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_Minus_One_Half | https://proofwiki.org/wiki/Ceiling_Function/Examples/Ceiling_of_Minus_One_Half | [
"Examples of Ceiling Function"
] | [] | [
"Definition:Ceiling Function"
] |
proofwiki-11688 | Floor Function/Examples/Floor of Minus One Half | :$\floor {-\dfrac 1 2} = -1$ | We have that:
:$-1 = -\dfrac 2 2 \le -\dfrac 1 2$
and:
:$0 > -\dfrac 1 2$
Hence $-1$ is the floor of $-\dfrac 1 2$ by definition.
{{qed}} | :$\floor {-\dfrac 1 2} = -1$ | We have that:
:$-1 = -\dfrac 2 2 \le -\dfrac 1 2$
and:
:$0 > -\dfrac 1 2$
Hence $-1$ is the [[Definition:Floor Function|floor]] of $-\dfrac 1 2$ by definition.
{{qed}} | Floor Function/Examples/Floor of Minus One Half | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Minus_One_Half | https://proofwiki.org/wiki/Floor_Function/Examples/Floor_of_Minus_One_Half | [
"Examples of Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-11689 | Infima of two Real Sets | Let $S$ and $T$ be sets of real numbers.
Let $S$ and $T$ admit infima.
Then:
: $\inf S \ge \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$ | Let:
: $-S = \set {-s: s \in S}$
: $-T = \set {-t: t \in T}$
Observe that:
: $s \in S \iff -s \in -S$
: $t \in T \iff -t \in -T$
We know that $\inf S$ and $\inf T$ exist.
The expression $\inf S \ge \inf T$ exists as $\inf S$ and $\inf T$ exist.
In other words, for fixed sets $S$ and $T$, $\inf S \ge \inf T$ is either t... | Let $S$ and $T$ be [[Definition:Set|sets]] of [[Definition:Real Number|real numbers]].
Let $S$ and $T$ admit [[Definition:Infimum of Subset of Real Numbers|infima]].
Then:
: $\inf S \ge \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$ | Let:
: $-S = \set {-s: s \in S}$
: $-T = \set {-t: t \in T}$
Observe that:
: $s \in S \iff -s \in -S$
: $t \in T \iff -t \in -T$
We know that $\inf S$ and $\inf T$ exist.
The [[Definition:Expression|expression]] $\inf S \ge \inf T$ exists as $\inf S$ and $\inf T$ exist.
In other words, for fixed [[Definition:S... | Infima of two Real Sets | https://proofwiki.org/wiki/Infima_of_two_Real_Sets | https://proofwiki.org/wiki/Infima_of_two_Real_Sets | [
"Real Analysis"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Infimum of Set/Real Numbers"
] | [
"Definition:Expression",
"Definition:Set",
"Definition:True",
"Definition:False",
"Suprema of two Real Sets",
"Definition:Set"
] |
proofwiki-11690 | Floor Function is Integer | Let $x$ be a real number.
Then the floor function of $x$ is an integer:
:$\floor x \in \Z$ | This is by definition of the floor function.
{{qed}} | Let $x$ be a [[Definition:Real Number|real number]].
Then the [[Definition:Floor Function|floor function]] of $x$ is an [[Definition:Integer|integer]]:
:$\floor x \in \Z$ | This is by definition of the [[Definition:Floor Function|floor function]].
{{qed}} | Floor Function is Integer | https://proofwiki.org/wiki/Floor_Function_is_Integer | https://proofwiki.org/wiki/Floor_Function_is_Integer | [
"Floor Function"
] | [
"Definition:Real Number",
"Definition:Floor Function",
"Definition:Integer"
] | [
"Definition:Floor Function"
] |
proofwiki-11691 | Floor of Number plus Integer | :$\forall n \in \Z: \floor x + n = \floor {x + n}$ | {{begin-eqn}}
{{eqn | l = \floor {x + n}
| o = \le
| m = x + n
| mo= <
| r = \floor {x + n} + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = \floor {x + n} - n
| o = \le
| m = x
| mo= <
| r = \floor {x + n} - n + 1
| c =
}}
{{eqn | ll= ... | :$\forall n \in \Z: \floor x + n = \floor {x + n}$ | {{begin-eqn}}
{{eqn | l = \floor {x + n}
| o = \le
| m = x + n
| mo= <
| r = \floor {x + n} + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = \floor {x + n} - n
| o = \le
| m = x
| mo= <
| r = \floor {x + n} - n + 1
| c =
}}
{{eqn | ll= ... | Floor of Number plus Integer | https://proofwiki.org/wiki/Floor_of_Number_plus_Integer | https://proofwiki.org/wiki/Floor_of_Number_plus_Integer | [
"Floor Function"
] | [] | [] |
proofwiki-11692 | Floor is between Number and One Less | :$x - 1 < \floor x \le x$
where $\floor x$ denotes the floor of $x$. | By definition of floor function:
:$\floor x \le x < \floor x + 1$
Thus by subtracting $1$:
:$x - 1 < \paren {\floor x + 1} - 1 = \floor x$
So:
:$\floor x \le x$
and:
:$x - 1 < \floor x$
as required.
{{qed}} | :$x - 1 < \floor x \le x$
where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$. | By definition of [[Definition:Floor Function|floor function]]:
:$\floor x \le x < \floor x + 1$
Thus by subtracting $1$:
:$x - 1 < \paren {\floor x + 1} - 1 = \floor x$
So:
:$\floor x \le x$
and:
:$x - 1 < \floor x$
as required.
{{qed}} | Floor is between Number and One Less | https://proofwiki.org/wiki/Floor_is_between_Number_and_One_Less | https://proofwiki.org/wiki/Floor_is_between_Number_and_One_Less | [
"Floor Function"
] | [
"Definition:Floor Function"
] | [
"Definition:Floor Function"
] |
proofwiki-11693 | Equivalence of Definitions of Floor Function | Let $x$ be a real number.
{{TFAE|def = Floor Function}} | === Definition 1 equals Definition 2 ===
Follows from Supremum of Set of Integers equals Greatest Element.
{{qed|lemma}} | Let $x$ be a [[Definition:Real Number|real number]].
{{TFAE|def = Floor Function}} | === Definition 1 equals Definition 2 ===
Follows from [[Supremum of Set of Integers equals Greatest Element]].
{{qed|lemma}} | Equivalence of Definitions of Floor Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Floor_Function | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Floor_Function | [
"Floor Function"
] | [
"Definition:Real Number"
] | [
"Supremum of Set of Integers equals Greatest Element",
"Supremum of Set of Integers equals Greatest Element"
] |
proofwiki-11694 | Ceiling Function is Integer | Let $x$ be a real number.
Then the ceiling function of $x$ is an integer:
:$\ceiling x \in \Z$ | This is by definition of the ceiling function.
{{qed}} | Let $x$ be a [[Definition:Real Number|real number]].
Then the [[Definition:Ceiling Function|ceiling function]] of $x$ is an [[Definition:Integer|integer]]:
:$\ceiling x \in \Z$ | This is by definition of the [[Definition:Ceiling Function|ceiling function]].
{{qed}} | Ceiling Function is Integer | https://proofwiki.org/wiki/Ceiling_Function_is_Integer | https://proofwiki.org/wiki/Ceiling_Function_is_Integer | [
"Ceiling Function"
] | [
"Definition:Real Number",
"Definition:Ceiling Function",
"Definition:Integer"
] | [
"Definition:Ceiling Function"
] |
proofwiki-11695 | Real Number is Integer iff equals Ceiling | :$x = \ceiling x \iff x \in \Z$ | Let $x = \ceiling x$.
As $\ceiling x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\ceiling x = \inf \set {m \in \Z: m \ge x}$
As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
:$x = \ceiling x$
{{qed}}
Catego... | :$x = \ceiling x \iff x \in \Z$ | Let $x = \ceiling x$.
As $\ceiling x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\ceiling x = \inf \set {m \in \Z: m \ge x}$
As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
:$x = \ceiling x$
{{qed}}
... | Real Number is Integer iff equals Ceiling | https://proofwiki.org/wiki/Real_Number_is_Integer_iff_equals_Ceiling | https://proofwiki.org/wiki/Real_Number_is_Integer_iff_equals_Ceiling | [
"Ceiling Function"
] | [] | [
"Category:Ceiling Function"
] |
proofwiki-11696 | Ceiling Function is Idempotent | :$\ceiling {\ceiling x} = \ceiling x$ | Let $y = \ceiling x$.
By Ceiling Function is Integer, $y$ is an integer.
Then from Real Number is Integer iff equals Ceiling:
:$\ceiling y = y$
So:
:$\ceiling {\ceiling x} = \ceiling x$
{{qed}} | :$\ceiling {\ceiling x} = \ceiling x$ | Let $y = \ceiling x$.
By [[Ceiling Function is Integer]], $y$ is an [[Definition:Integer|integer]].
Then from [[Real Number is Integer iff equals Ceiling]]:
:$\ceiling y = y$
So:
:$\ceiling {\ceiling x} = \ceiling x$
{{qed}} | Ceiling Function is Idempotent | https://proofwiki.org/wiki/Ceiling_Function_is_Idempotent | https://proofwiki.org/wiki/Ceiling_Function_is_Idempotent | [
"Ceiling Function"
] | [] | [
"Ceiling Function is Integer",
"Definition:Integer",
"Real Number is Integer iff equals Ceiling"
] |
proofwiki-11697 | Way Below Relation is Auxiliary Relation | Lrt $L = \left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Then
:$\ll$ is auxiliary relation
where $\ll$ denotes the way below relation. | By Way Below implies Preceding:
:$\forall x, y \in S: x \ll y \implies x \preceq y$
By Preceding and Way Below implies Way Below:
:$\forall x, y, z, u \in S: x \preceq y \ll z \preceq u \implies x \ll u$
By Join is Way Below if Operands are Way Below:
:$\forall x, y, z \in S: x \ll z \land y \ll z \implies x \vee y \ll... | Lrt $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Then
:$\ll$ is [[Definition:Auxiliary Relation|auxiliary relation]]
where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]]. | By [[Way Below implies Preceding]]:
:$\forall x, y \in S: x \ll y \implies x \preceq y$
By [[Preceding and Way Below implies Way Below]]:
:$\forall x, y, z, u \in S: x \preceq y \ll z \preceq u \implies x \ll u$
By [[Join is Way Below if Operands are Way Below]]:
:$\forall x, y, z \in S: x \ll z \land y \ll z \implie... | Way Below Relation is Auxiliary Relation | https://proofwiki.org/wiki/Way_Below_Relation_is_Auxiliary_Relation | https://proofwiki.org/wiki/Way_Below_Relation_is_Auxiliary_Relation | [
"Auxiliary Relations",
"Way Below Relation"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Auxiliary Relation",
"Definition:Element is Way Below"
] | [
"Way Below implies Preceding",
"Preceding and Way Below implies Way Below",
"Join is Way Below if Operands are Way Below",
"Bottom is Way Below Any Element",
"Definition:Smallest Element",
"Definition:Auxiliary Relation"
] |
proofwiki-11698 | Exponential Sequence is Eventually Strictly Positive | Let $\sequence {E_n}$ be the sequence of real functions $E_n: \R \to \R$ defined as:
:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for each $x \in \R$ and for sufficiently large $n \in \N$, $\map {E_n} x$ is positive.
That is:
:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x > 0$... | Fix $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = \ceiling {\size x}
}}
{{eqn | ll= \leadsto
| l = n
| o = >
| r = -x
| c = Real Number is between Ceiling Functions and Negative of Absolute Value
}}
{{eqn | ll= \leadsto
| l = 1
| o = >
| r = \frac {-... | Let $\sequence {E_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $E_n: \R \to \R$ defined as:
:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
Then, for each $x \in \R$ and for [[Definition:Sufficiently Large|sufficiently large]] $n \in \N$, $\map {E_n} x$ is [[Definition:Positi... | Fix $x \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = \ceiling {\size x}
}}
{{eqn | ll= \leadsto
| l = n
| o = >
| r = -x
| c = [[Real Number is between Ceiling Functions]] and [[Negative of Absolute Value]]
}}
{{eqn | ll= \leadsto
| l = 1
| o = >
| r ... | Exponential Sequence is Eventually Strictly Positive | https://proofwiki.org/wiki/Exponential_Sequence_is_Eventually_Strictly_Positive | https://proofwiki.org/wiki/Exponential_Sequence_is_Eventually_Strictly_Positive | [
"Exponential Function"
] | [
"Definition:Sequence",
"Definition:Real Function",
"Definition:Sufficiently Large",
"Definition:Positive/Real Number",
"Definition: Ceiling Function"
] | [
"Real Number is between Ceiling Functions",
"Negative of Absolute Value",
"Power of Strictly Positive Real Number is Strictly Positive/Positive Integer",
"Category:Exponential Function"
] |
proofwiki-11699 | Number is between Ceiling and One Less | :$\ceiling x - 1 < x \le \ceiling x$ | By definition of ceiling of $x$:
:$\forall x \in \R: \ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
By definition of infimum:
:$\ceiling x \ge x$
Also by definition of infimum:
:$\ceiling x - 1 \not \ge x$
as $\ceiling x$ is the smallest integer with that property.
That is:
:$x > \ceiling x - 1$
Hence the result.
... | :$\ceiling x - 1 < x \le \ceiling x$ | By definition of [[Definition:Ceiling Function|ceiling]] of $x$:
:$\forall x \in \R: \ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
By definition of [[Definition:Infimum of Subset of Real Numbers|infimum]]:
:$\ceiling x \ge x$
Also by definition of [[Definition:Infimum of Subset of Real Numbers|infimum]]:
:$\ce... | Number is between Ceiling and One Less | https://proofwiki.org/wiki/Number_is_between_Ceiling_and_One_Less | https://proofwiki.org/wiki/Number_is_between_Ceiling_and_One_Less | [
"Ceiling Function"
] | [] | [
"Definition:Ceiling Function",
"Definition:Infimum of Set/Real Numbers",
"Definition:Infimum of Set/Real Numbers",
"Definition:Integer",
"Category:Ceiling Function"
] |
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