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-11800 | Relation between Signed and Unsigned Stirling Numbers of the First Kind | Let $m, n \in \Z_{\ge 0}$ be positive integers.
Then:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
where:
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind
:$\map s {n, m}$ denotes a signed Stirling number of the first kind. | {{begin-eqn}}
{{eqn | l = \sum_k \map s {n, k} x^k
| r = x^{\underline n}
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | ll= \leadsto
| l = \map s {n, m}
... | Let $m, n \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
Then:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
where:
:$\ds {n \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$\map s {n, m}$ denotes a [[Definit... | {{begin-eqn}}
{{eqn | l = \sum_k \map s {n, k} x^k
| r = x^{\underline n}
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | ll= \leadsto
| l = \map s {n, m}
... | Relation between Signed and Unsigned Stirling Numbers of the First Kind | https://proofwiki.org/wiki/Relation_between_Signed_and_Unsigned_Stirling_Numbers_of_the_First_Kind | https://proofwiki.org/wiki/Relation_between_Signed_and_Unsigned_Stirling_Numbers_of_the_First_Kind | [
"Stirling Numbers"
] | [
"Definition:Positive/Integer",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Stirling Numbers of the First Kind/Signed"
] | [
"Category:Stirling Numbers"
] |
proofwiki-11801 | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1 | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}} | By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the {{... | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}} | By definition, [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
wher... | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_1 | [
"Unsigned Stirling Number of the First Kind of Number with Greater"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Coefficient of Polynomial",
"Definition:Falling Factorial",
"Definition:Expression",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Coefficient of Polynomial"
] |
proofwiki-11802 | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2 | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}} | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
:$\fo... | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Unsigned Stirling Number of the First Kind of Number with Greater}} | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P v$ be the [[Definition:Proposition|proposition]]:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from [[Unsigned Stirli... | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_2 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_2 | [
"Unsigned Stirling Number of the First Kind of Number with Greater"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Unsigned Stirling Number of the First Kind of 0",
"Definition:Kronecker Delta",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Unsigned Stirling Number of the First Kind of Number ... |
proofwiki-11803 | Stirling Number of the Second Kind of Number with Greater/Proof 1 | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}} | By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.
Both of the expressions on the {{LHS}} and {{RHS}} are polynomials ... | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}} | By definition, the [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]] are defined as the [[Definition:Coefficient|coefficients]] $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the [[Definition:Falling... | Stirling Number of the Second Kind of Number with Greater/Proof 1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_1 | [
"Stirling Number of the Second Kind of Number with Greater"
] | [] | [
"Definition:Stirling Numbers of the Second Kind",
"Definition:Coefficient",
"Definition:Falling Factorial",
"Definition:Expression",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Coefficient of Polynomial"
] |
proofwiki-11804 | Stirling Number of the Second Kind of Number with Greater/Proof 2 | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}} | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from Stirling Number of the Second Kind of 0.
So by definition of Kronecker delta:
:$\forall k \... | Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
{{:Stirling Number of the Second Kind of Number with Greater}} | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from [[Stirling Number... | Stirling Number of the Second Kind of Number with Greater/Proof 2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_2 | [
"Stirling Number of the Second Kind of Number with Greater"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Stirling Number of the Second Kind of 0",
"Definition:Kronecker Delta",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Stirling Number of the Second Kind of Number with Greater/Pro... |
proofwiki-11805 | Signed Stirling Number of the First Kind of Number with Self | :$\map s {n, n} = 1$ | From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n} = \paren {-1}^{n + n} \map s {n, n}$
We have that:
:$\paren {-1}^{n + n} = \paren {-1}^{2 n} = 1$
and so:
:$\ds {n \brack n} = \map s {n, n}$
The result follows from Unsigned Stirling Number of the First Kind of Number with... | :$\map s {n, n} = 1$ | From [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
:$\ds {n \brack n} = \paren {-1}^{n + n} \map s {n, n}$
We have that:
:$\paren {-1}^{n + n} = \paren {-1}^{2 n} = 1$
and so:
:$\ds {n \brack n} = \map s {n, n}$
The result follows from [[Unsigned Stirling Number of the First Kind of N... | Signed Stirling Number of the First Kind of Number with Self | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_Number_with_Self | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_Number_with_Self | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Relation between Signed and Unsigned Stirling Numbers of the First Kind",
"Unsigned Stirling Number of the First Kind of Number with Self"
] |
proofwiki-11806 | Unsigned Stirling Number of the First Kind of n with n-1 | :$\ds {n \brack n - 1} = \binom n 2$ | The proof proceeds by induction. | :$\ds {n \brack n - 1} = \binom n 2$ | The proof proceeds by [[Principle of Mathematical Induction|induction]]. | Unsigned Stirling Number of the First Kind of n with n-1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n_with_n-1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-11807 | Stirling Number of the Second Kind of n with n-1 | :$\ds {n \brace n - 1} = \binom n 2$ | The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n - 1} = \binom n 2$ | :$\ds {n \brace n - 1} = \binom n 2$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds {n \brace n - 1} = \binom n 2$ | Stirling Number of the Second Kind of n with n-1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n_with_n-1 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11808 | Signed Stirling Number of the First Kind of n with n-1 | :$\map s {n, n - 1} = -\dbinom n 2$ | From Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$
where $\ds {n \brack n - 1}$ denotes an unsigned Stirling number of the first kind.
We have that:
:$\paren {-1}^{n + n - 1} = \paren {-1}^{2 n - 1} = -1$
and so:
:$\ds {n \br... | :$\map s {n, n - 1} = -\dbinom n 2$ | From [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
:$\ds {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$
where $\ds {n \brack n - 1}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
We have that:
:$\paren ... | Signed Stirling Number of the First Kind of n with n-1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n_with_n-1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n_with_n-1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Relation between Signed and Unsigned Stirling Numbers of the First Kind",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Unsigned Stirling Number of the First Kind of Number with Self"
] |
proofwiki-11809 | Unsigned Stirling Number of the First Kind of n+1 with 0 | :$\ds {n + 1 \brack 0} = 0$ | We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brack k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus:
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = 0
| c = by hypothesis
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = >
... | :$\ds {n + 1 \brack 0} = 0$ | We are given that $k = 0$.
So by definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
:$\ds {n \brack k} = \delta_{n k}$
where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
Thus:
{{begin-eqn}}
{{eqn | l = n
| o = \ge
... | Unsigned Stirling Number of the First Kind of n+1 with 0 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_0 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_0 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Kronecker Delta",
"Definition:By Hypothesis"
] |
proofwiki-11810 | Stirling Number of the Second Kind of n+1 with 0 | :$\ds {n + 1 \brace 0} = 0$ | We are given that $k = 0$.
So by definition of unsigned Stirling number of the first kind:
:$\ds {n \brace k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = 0
| c = by hypothesis
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = >
... | :$\ds {n + 1 \brace 0} = 0$ | We are given that $k = 0$.
So by definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
:$\ds {n \brace k} = \delta_{n k}$
where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
Thus
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| ... | Stirling Number of the Second Kind of n+1 with 0 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_0 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_0 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Kronecker Delta",
"Definition:By Hypothesis"
] |
proofwiki-11811 | Signed Stirling Number of the First Kind of n+1 with 0 | :$\map s {n + 1, 0} = 0$ | By definition of signed Stirling number of the first kind:
:$\map s {n, k} = \delta_{n k}$
where $\delta_{n k}$ is the Kronecker delta.
Thus
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = 0
| c = by hypothesis
}}
{{eqn | ll= \leadsto
| l = n + 1
| o = >
| r = 0
| c =
}}
{{eqn | l... | :$\map s {n + 1, 0} = 0$ | By definition of [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]]:
:$\map s {n, k} = \delta_{n k}$
where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
Thus
{{begin-eqn}}
{{eqn | l = n
| o = \ge
| r = 0
| c = [[Definition:By Hypo... | Signed Stirling Number of the First Kind of n+1 with 0 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n+1_with_0 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n+1_with_0 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Signed",
"Definition:Kronecker Delta",
"Definition:By Hypothesis"
] |
proofwiki-11812 | Unsigned Stirling Number of the First Kind of n+1 with 1 | :$\ds {n + 1 \brack 1} = n!$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brack 1} = n!$ | :$\ds {n + 1 \brack 1} = n!$ | 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]]:
:$\ds {n + 1 \brack 1} = n!$ | Unsigned Stirling Number of the First Kind of n+1 with 1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11813 | Stirling Number of the Second Kind of n+1 with 1 | :$\ds {n + 1 \brace 1} = 1$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 1} = 1$ | :$\ds {n + 1 \brace 1} = 1$ | 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]]:
:$\ds {n + 1 \brace 1} = 1$ | Stirling Number of the Second Kind of n+1 with 1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_1 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11814 | Signed Stirling Number of the First Kind of n+1 with 1 | :$\map s {n + 1, 1} = \paren {-1}^n n!$ | By Relation between Signed and Unsigned Stirling Numbers of the First Kind:
:$\ds {n + 1 \brack 1} = \paren {-1}^{n + 1 + 1} \map s {n + 1, 1}$
where $\ds {n + 1 \brack 1}$ denotes an unsigned Stirling number of the first kind.
We have that:
:$\paren {-1}^{n + 1 + 1} = \paren {-1}^n$
and so:
:$\ds {n + 1 \brack 1} = \p... | :$\map s {n + 1, 1} = \paren {-1}^n n!$ | By [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
:$\ds {n + 1 \brack 1} = \paren {-1}^{n + 1 + 1} \map s {n + 1, 1}$
where $\ds {n + 1 \brack 1}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
We have that:
:$\paren {-... | Signed Stirling Number of the First Kind of n+1 with 1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n+1_with_1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_n+1_with_1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Relation between Signed and Unsigned Stirling Numbers of the First Kind",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Unsigned Stirling Number of the First Kind of Number with Self"
] |
proofwiki-11815 | Stirling Number of the Second Kind of n+1 with 2 | :$\ds {n + 1 \brace 2} = 2^n - 1$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds {n + 1 \brace 2} = 2^n - 1$ | :$\ds {n + 1 \brace 2} = 2^n - 1$ | 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]]:
:$\ds {n + 1 \brace 2} = 2^n - 1$ | Stirling Number of the Second Kind of n+1 with 2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_n+1_with_2 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11816 | Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$
where:
:$\ds {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$\dbinom k m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | 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]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} \binom k m = {n + 1 \brack m + 1}$ | Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_the_First_Kind_of_n_with_k_by_k_choose_m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_the_First_Kind_of_n_with_k_by_k_choose_m | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11817 | Semantically Equivalent Terms are Equal | Let $\tau_1, \tau_2$ be terms.
Suppose that they are semantically equivalent with respect to the empty set.
Then $\tau_1 = \tau_2$. | {{proof wanted|While an exercise, this basically requires construction of the syntactic model in order to achieve full rigour}} | Let $\tau_1, \tau_2$ be [[Definition:Term (Predicate Logic)|terms]].
Suppose that they are [[Definition:Relative Semantic Equivalence of Terms|semantically equivalent]] with respect to the [[Definition:Empty Set|empty set]].
Then $\tau_1 = \tau_2$. | {{proof wanted|While an exercise, this basically requires construction of the syntactic model in order to achieve full rigour}} | Semantically Equivalent Terms are Equal | https://proofwiki.org/wiki/Semantically_Equivalent_Terms_are_Equal | https://proofwiki.org/wiki/Semantically_Equivalent_Terms_are_Equal | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Relative Semantic Equivalence/Term",
"Definition:Empty Set"
] | [] |
proofwiki-11818 | Down Mapping is Generated by Approximating Relation | Let $L = \struct {S, \wedge, \preceq}$ be a bounded below meet-continuous meet semilattice.
Let $\map {\operatorname{Ids} } L$ be the set of all ideals in $L$.
Let $I$ be an ideal in $L$.
Let $f: S \to \map {\operatorname{Ids} } L$ be a mapping such that:
:$\forall x \in S: x \preceq \sup I \implies \map f x = \set {x ... | Define relation $\RR$ on $S$:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$
By definition of $\RR$-segment:
:$\forall x \in S: \map f x = x^\RR$
We will prove that:
:$\RR$ is an approximating relation.
Let $x \in S$.
Suppose the case holds that:
:$x \preceq \sup I$
Thus:
{{begin-eqn}}
{{eqn | l = \ma... | Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous]] [[Definition:Meet Semilattice|meet semilattice]].
Let $\map {\operatorname{Ids} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $... | Define [[Definition:Relation|relation]] $\RR$ on $S$:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$
By definition of [[Definition:Relation Segment|$\RR$-segment]]:
:$\forall x \in S: \map f x = x^\RR$
We will prove that:
:$\RR$ is an [[Definition:Approximating Relation|approximating relation]].
Le... | Down Mapping is Generated by Approximating Relation | https://proofwiki.org/wiki/Down_Mapping_is_Generated_by_Approximating_Relation | https://proofwiki.org/wiki/Down_Mapping_is_Generated_by_Approximating_Relation | [
"Meet-Continuous Lattices",
"Approximating Relations"
] | [
"Definition:Bounded Below",
"Definition:Meet-Continuous Lattice",
"Definition:Meet Semilattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Mapping",
"Definition:Lower Closure/Element",
"Definition:Auxiliary Relation",
"Definition:Ap... | [
"Definition:Relation",
"Definition:Relation Segment",
"Definition:Approximating Relation",
"Preceding iff Meet equals Less Operand",
"Supremum of Lower Closure of Element"
] |
proofwiki-11819 | Preceding is Approximating Relation | Let $\struct {S, \preceq}$ be an ordered set.
Then $\preceq$ is an approximating relation on $S$. | Let $x \in S$.
Define $\RR := \mathord\preceq$.
By definitions of lower closure of element and $\RR$-segment:
:$x^\preceq = x^\RR$
where:
:$x^\preceq$ denotes the lower closure of $x$
:$x^\RR$ denotes the $\RR$-segment of $x$
Thus by Supremum of Lower Closure of Element:
:$x = \map \sup {x^\preceq} = \map \sup {x^\RR}$... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Then $\preceq$ is an [[Definition:Approximating Relation|approximating relation]] on $S$. | Let $x \in S$.
Define $\RR := \mathord\preceq$.
By definitions of [[Definition:Lower Closure of Element|lower closure of element]] and [[Definition:Relation Segment|$\RR$-segment]]:
:$x^\preceq = x^\RR$
where:
:$x^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $x$
:$x^\RR$ denotes th... | Preceding is Approximating Relation | https://proofwiki.org/wiki/Preceding_is_Approximating_Relation | https://proofwiki.org/wiki/Preceding_is_Approximating_Relation | [
"Approximating Relations"
] | [
"Definition:Ordered Set",
"Definition:Approximating Relation"
] | [
"Definition:Lower Closure/Element",
"Definition:Relation Segment",
"Definition:Lower Closure/Element",
"Definition:Relation Segment",
"Supremum of Lower Closure of Element",
"Definition:Approximating Relation"
] |
proofwiki-11820 | Substitution Instance of Term is Term | Let $\beta, \tau$ be terms of predicate logic.
Let $x \in \operatorname {VAR}$ be a variable.
Let $\map \beta {x \gets \tau}$ be the substitution instance of $\beta$ substituting $\tau$ for $x$.
Then $\map \beta {x \gets \tau}$ is a term. | Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$.
If $\beta = y$ for some variable $y$, then:
:$\map \beta {x \gets \tau} = \begin{cases} \tau & : \text {if $y = x$} \\ y &: \text {otherwise} \end{cases}$
In either case, $\map \beta {x \gets \tau}$ is a term.
If $\beta = \m... | Let $\beta, \tau$ be [[Definition:Term (Predicate Logic)|terms of predicate logic]].
Let $x \in \operatorname {VAR}$ be a [[Definition:Variable (Logic)|variable]].
Let $\map \beta {x \gets \tau}$ be the [[Definition:Substitution Instance of Term|substitution instance of $\beta$ substituting $\tau$ for $x$]].
Then $... | Proceed by the [[Principle of Structural Induction]] on the [[Definition:Term (Predicate Logic)|definition of term]], applied to $\beta$.
If $\beta = y$ for some [[Definition:Variable (Logic)|variable]] $y$, then:
:$\map \beta {x \gets \tau} = \begin{cases} \tau & : \text {if $y = x$} \\ y &: \text {otherwise} \end{... | Substitution Instance of Term is Term | https://proofwiki.org/wiki/Substitution_Instance_of_Term_is_Term | https://proofwiki.org/wiki/Substitution_Instance_of_Term_is_Term | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Variable/Predicate Logic",
"Definition:Substitution (Formal Systems)/Term/In Term",
"Definition:Language of Predicate Logic/Formal Grammar/Term"
] | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Variable/Predicate Logic",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Principle of Structural Induction"
] |
proofwiki-11821 | Substitution Theorem for Terms | Let $\beta, \tau$ be terms.
Let $x \in \mathrm {VAR}$ be a variable.
Let $\map \beta {x \gets \tau}$ be the substitution instance of $\beta$ substituting $\tau$ for $x$.
Let $\AA$ be a structure for predicate logic.
Let $\sigma$ be an assignment for $\beta$ and $\tau$.
Suppose that:
:$\map {\operatorname{val}_\AA} \tau... | Proceed by the Principle of Structural Induction on the definition of term, applied to $\beta$.
If $\beta = y$ for some variable $y$, then:
:$\map \beta {x \gets \tau} = \begin {cases} \tau & : \text{if $y = x$} \\ y & : \text {otherwise} \end {cases}$
In the first case:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{va... | Let $\beta, \tau$ be [[Definition:Term (Predicate Logic)|terms]].
Let $x \in \mathrm {VAR}$ be a [[Definition:Variable (Logic)|variable]].
Let $\map \beta {x \gets \tau}$ be the [[Definition:Substitution Instance of Term|substitution instance of $\beta$ substituting $\tau$ for $x$]].
Let $\AA$ be a [[Definition:Stru... | Proceed by the [[Principle of Structural Induction]] on the [[Definition:Term (Predicate Logic)|definition of term]], applied to $\beta$.
If $\beta = y$ for some [[Definition:Variable (Logic)|variable]] $y$, then:
:$\map \beta {x \gets \tau} = \begin {cases} \tau & : \text{if $y = x$} \\ y & : \text {otherwise} \end... | Substitution Theorem for Terms | https://proofwiki.org/wiki/Substitution_Theorem_for_Terms | https://proofwiki.org/wiki/Substitution_Theorem_for_Terms | [
"Named Theorems",
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Variable/Predicate Logic",
"Definition:Substitution (Formal Systems)/Term/In Term",
"Definition:Structure for Predicate Logic",
"Definition:Assignment for Structure/Term",
"Definition:Value of Term under Assignment",
"Definition:E... | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Variable/Predicate Logic",
"Principle of Structural Induction"
] |
proofwiki-11822 | Substitution Instance of WFF is WFF | Let $\mathbf A$ be a WFF of predicate logic.
Let $\tau$ be a term of predicate logic.
Let $x \in \mathrm{VAR}$ be a variable.
Let $\mathbf A \left({x \gets \tau}\right)$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$.
Then $\mathbf A \left({x \gets \tau}\right)$ is a WFF. | Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$.
If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
:$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\r... | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]].
Let $x \in \mathrm{VAR}$ be a [[Definition:Variable (Logic)|variable]].
Let $\mathbf A \left({x \gets \tau}\right)$ be the [[Definition:Substitution Instan... | Proceed by the [[Principle of Structural Induction]] on the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]], applied to $\mathbf A$.
If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
:$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x ... | Substitution Instance of WFF is WFF | https://proofwiki.org/wiki/Substitution_Instance_of_WFF_is_WFF | https://proofwiki.org/wiki/Substitution_Instance_of_WFF_is_WFF | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Variable/Predicate Logic",
"Definition:Substitution (Formal Systems)/Term/In WFF",
"Definition:Language of Predicate Logic/Formal Grammar"
] | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Substitution (Formal Systems)/Term/In Term",
"Substitution Instance of Term is Term",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Language of Predicate Logic/Formal Gramm... |
proofwiki-11823 | Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
where:
:$\ds {n + 1 \brack k + 1}$ etc. denotes an unsigned Stirling number of the first kind
:$\dbinom k m$ denotes a binomial coefficient. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$
where:
:$\ds {n + 1 \brack k + 1}$ etc. denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$\dbinom k m$ denotes a [[Definition:Binomial Coefficient... | 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]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brack k + 1} \binom k m \paren {-1}^{k - m} = {n \brack m}$ | Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_the_First_Kind_of_n+1_with_k+1_by_k_choose_m_by_-1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_the_First_Kind_of_n+1_with_k+1_by_k_choose_m_by_-1^k-m | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11824 | Substitution Theorem for Well-Formed Formulas | Let $\mathbf A$ be a WFF of predicate logic.
Let $x \in \mathrm{VAR}$ be a variable.
Let $\tau$ be a term of predicate logic which is freely substitutable for $x$ in $\mathbf A$.
Let $\map {\mathbf A} {x \gets \tau}$ be the substitution instance of $\mathbf A$ substituting $\tau$ for $x$.
Let $\AA$ be a structure for p... | Proceed by the Principle of Structural Induction on the bottom-up specification of predicate logic, applied to $\mathbf A$.
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\mathbf A} {x \gets \tau} = \map p {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$
where $\map {\tau_i} {x \g... | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $x \in \mathrm{VAR}$ be a [[Definition:Variable (Logic)|variable]].
Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]] which is [[Definition:Freely Substitutable|freely substitutable]] for $x$ in $\mathbf... | Proceed by the [[Principle of Structural Induction]] on the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]], applied to $\mathbf A$.
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\mathbf A} {x \gets \tau} = \map p {\map {\tau_1} {x \gets \tau}, \... | Substitution Theorem for Well-Formed Formulas | https://proofwiki.org/wiki/Substitution_Theorem_for_Well-Formed_Formulas | https://proofwiki.org/wiki/Substitution_Theorem_for_Well-Formed_Formulas | [
"Named Theorems",
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Variable/Predicate Logic",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Freely Substitutable",
"Definition:Substitution (Formal Systems)/Term/In WFF",
"Definition:Structure for Predicate Logic",
"Definition... | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Substitution (Formal Systems)/Term/In Term",
"Substitution Theorem for Terms",
"Definition:Freely Substitutable",
"Definition:Freely Substitutable",
"Definition:Freely Substitutable",
"Definition:... |
proofwiki-11825 | Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
where:
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$
where:
:$\ds {k \brace m}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | 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]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k \brace m} \binom n k = {n + 1 \brace m + 1}$ | Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Numbers_of_the_Second_Kind_of_k_with_m_by_n_choose_k | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Numbers_of_the_Second_Kind_of_k_with_m_by_n_choose_k | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Stirling Numbers of the Second Kind",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11826 | Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
where:
:$\ds {k + 1 \brace m + 1}$ and so on denotes a Stirling number of the second kind
:$\dbinom n k$ denotes a binomial coefficient. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$
where:
:$\ds {k + 1 \brace m + 1}$ and so on denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial co... | 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]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {k + 1 \brace m + 1} \binom n k \paren {-1}^{n - k} = {n \brace m}$ | Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Numbers_of_the_Second_Kind_of_k+1_with_m+1_by_n_choose_k_by_-1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Numbers_of_the_Second_Kind_of_k+1_with_m+1_by_n_choose_k_by_-1^k-m | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Stirling Numbers of the Second Kind",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11827 | Universal Instantiation/Informal Statement | Suppose we have a universal statement:
:$\forall x: \map P x$
where $\forall$ is the universal quantifier and $\map P x$ is a propositional function.
Then we can deduce:
:$\map P {\mathbf a}$
where $\mathbf a$ is any arbitrary object we care to choose in the universe of discourse.
In natural language:
:''Suppose $P$ is... | In the language of symbolic logic:
{{begin-eqn}}
{{eqn | q = \forall x
| l = \map P x
| o =
}}
{{eqn | ll= \therefore
| l = \map P {\mathbf a}
| o =
}}
{{end-eqn}}
{{qed}} | Suppose we have a [[Definition:Universal Statement|universal statement]]:
:$\forall x: \map P x$
where $\forall$ is the [[Definition:Universal Quantifier|universal quantifier]] and $\map P x$ is a [[Definition:Propositional Function|propositional function]].
Then we can deduce:
:$\map P {\mathbf a}$
where $\mathbf a$ ... | In the language of [[Definition:Symbolic Logic|symbolic logic]]:
{{begin-eqn}}
{{eqn | q = \forall x
| l = \map P x
| o =
}}
{{eqn | ll= \therefore
| l = \map P {\mathbf a}
| o =
}}
{{end-eqn}}
{{qed}} | Universal Instantiation/Informal Statement | https://proofwiki.org/wiki/Universal_Instantiation/Informal_Statement | https://proofwiki.org/wiki/Universal_Instantiation/Informal_Statement | [
"Universal Instantiation"
] | [
"Definition:Universal Statement",
"Definition:Universal Quantifier",
"Definition:Propositional Function",
"Definition:Object",
"Definition:Universe of Discourse",
"Definition:Natural Language",
"Definition:Universe of Discourse",
"Definition:Universe of Discourse"
] | [
"Definition:Symbolic Logic"
] |
proofwiki-11828 | Universal Instantiation/Model | Let $\map {\mathbf A} x$ be a WFF of predicate logic.
Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$.
Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a tautology. | Let $\AA$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \mathop {\map {\operatorname{val}_\AA} \tau} \sqbrk \sigma$
the value of $\tau$ under $\sigma$.
From the definition of value under $\sigma$:
:$\map {\mathrm {v... | Let $\map {\mathbf A} x$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\tau$ be a [[Definition:Term (Predicate Logic)|term]] which is [[Definition:Freely Substitutable|freely substitutable]] for $x$ in $\mathbf A$.
Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a [[D... | Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure]] on a [[Definition:Set|set]] $A$, and let $\sigma$ be an [[Definition:Assignment for Formula|assignment]] for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \mathop {\map {\operatorname{val}_\AA} \tau} \sqbrk \si... | Universal Instantiation/Model | https://proofwiki.org/wiki/Universal_Instantiation/Model | https://proofwiki.org/wiki/Universal_Instantiation/Model | [
"Universal Instantiation"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Freely Substitutable",
"Definition:Tautology/Formal Semantics/Predicate Logic"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Set",
"Definition:Assignment for Structure/Formula",
"Definition:Value of Term under Assignment",
"Definition:Value of Formula under Assignment",
"Definition:Conditional/Truth Function",
"Definition:Value of Formula under Assignment",
"Substituti... |
proofwiki-11829 | Existential Generalisation/Model | Let $\map {\mathbf A} x$ be a WFF of predicate logic.
Let $\tau$ be a term which is freely substitutable for $x$ in $\mathbf A$.
Then $\map {\mathbf A} \tau \implies \exists x: \map {\mathbf A} x$ is a tautology. | Let $\AA$ be a structure on a set $A$, and let $\sigma$ be an assignment for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \map {\operatorname{val}_\AA} \tau \sqbrk \sigma$
the value of $\tau$ under $\sigma$.
From the definition of value under $\sigma$:
:$\map {\operatorname{val}_\... | Let $\map {\mathbf A} x$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\tau$ be a [[Definition:Term (Predicate Logic)|term]] which is [[Definition:Freely Substitutable|freely substitutable]] for $x$ in $\mathbf A$.
Then $\map {\mathbf A} \tau \implies \exists x: \map {\mathbf A} x$ is a [[D... | Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure]] on a [[Definition:Set|set]] $A$, and let $\sigma$ be an [[Definition:Assignment for Formula|assignment]] for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
Define:
:$a_\tau := \map {\operatorname{val}_\AA} \tau \sqbrk \sigma$
the ... | Existential Generalisation/Model | https://proofwiki.org/wiki/Existential_Generalisation/Model | https://proofwiki.org/wiki/Existential_Generalisation/Model | [
"Existential Generalisation"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Freely Substitutable",
"Definition:Tautology/Formal Semantics/Predicate Logic"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Set",
"Definition:Assignment for Structure/Formula",
"Definition:Value of Term under Assignment",
"Definition:Value of Formula under Assignment",
"Definition:Conditional/Truth Function",
"Substitution Theorem for Well-Formed Formulas",
"Definitio... |
proofwiki-11830 | Way Below is Approximating Relation | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Then $\ll$ is an approximating relation on $S$. | Let $x \in S$.
Define $\RR := \mathord \ll$.
By definitions of way below closure and $\RR$-segment:
:$x^\ll = x^\RR$
where:
:$x^\ll$ denotes the way below closure of $x$
:$x^\RR$ denotes the $\RR$-segment of $x$
By definition of continuous:
:$L$ satisfies the axiom of approximation.
Thus by the axiom of approximation:
... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Then $\ll$ is an [[Definition:Approximating Relation|approximating relation]] on $S$. | Let $x \in S$.
Define $\RR := \mathord \ll$.
By definitions of [[Definition:Way Below Closure|way below closure]] and [[Definition:Relation Segment|$\RR$-segment]]:
:$x^\ll = x^\RR$
where:
:$x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$
:$x^\RR$ denotes the [[Definition:Relation Segm... | Way Below is Approximating Relation | https://proofwiki.org/wiki/Way_Below_is_Approximating_Relation | https://proofwiki.org/wiki/Way_Below_is_Approximating_Relation | [
"Approximating Relations",
"Continuous Lattices",
"Way Below Relation"
] | [
"Definition:Bounded Below",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)",
"Definition:Approximating Relation"
] | [
"Definition:Way Below Closure",
"Definition:Relation Segment",
"Definition:Way Below Closure",
"Definition:Relation Segment",
"Definition:Continuous Ordered Set",
"Axiom:Axiom of Approximation",
"Axiom:Axiom of Approximation",
"Definition:Approximating Relation"
] |
proofwiki-11831 | Intersection of Relation Segments of Approximating Relations equals Way Below Closure | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below meet-continuous lattice.
Let $\map {\operatorname {App} } L$ be the set of all auxiliary approximating relations on $S$.
Let $x \in S$.
Then
:$\ds \bigcap \set {x^\RR: \RR \in \map {\operatorname {App} } L} = x^\ll$ | By Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element:
:$\ds \bigcap \set {I \in \operatorname {Ids}: x \preceq \sup I} = x^\ll$
where $\operatorname {Ids}$ denotes the set of all ideals in $L$.
For all $I \in \operatorname {Ids}$ define a mapping $m_I: S \to \operatorname {Ids}$:
:... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
Let $\map {\operatorname {App} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|a... | By [[Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element]]:
:$\ds \bigcap \set {I \in \operatorname {Ids}: x \preceq \sup I} = x^\ll$
where $\operatorname {Ids}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
For all $I \in \oper... | Intersection of Relation Segments of Approximating Relations equals Way Below Closure | https://proofwiki.org/wiki/Intersection_of_Relation_Segments_of_Approximating_Relations_equals_Way_Below_Closure | https://proofwiki.org/wiki/Intersection_of_Relation_Segments_of_Approximating_Relations_equals_Way_Below_Closure | [
"Approximating Relations",
"Meet-Continuous Lattices"
] | [
"Definition:Bounded Below",
"Definition:Meet-Continuous Lattice",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation"
] | [
"Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Mapping",
"Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element",
"Down Mapping is Generated by Approx... |
proofwiki-11832 | Convergent Sequence in Set of Integers/Corollary | Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence of distinct terms in the set $\Z$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is not convergent. | By Convergent Sequence in Set of Integers, $\sequence {x_n}_{n \mathop \in \N}$ is convergent {{iff}}:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
But since $\sequence {x_n}_{n \mathop \in \N}$ is a sequence of distinct terms, for all $k \ne m$, one has $x_m \ne x_k$.
The result follows.
{{qed}}
Category:Se... | Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] in the set $\Z$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is not [[Definition:Convergent Sequence (Topology)|convergent]]. | By [[Convergent Sequence in Set of Integers]], $\sequence {x_n}_{n \mathop \in \N}$ is [[Definition:Convergent Sequence (Topology)|convergent]] {{iff}}:
:$\exists k \in \N: \forall m \in \N: m > k: x_m = x_k$
But since $\sequence {x_n}_{n \mathop \in \N}$ is a [[Definition:Sequence of Distinct Terms|sequence of disti... | Convergent Sequence in Set of Integers/Corollary | https://proofwiki.org/wiki/Convergent_Sequence_in_Set_of_Integers/Corollary | https://proofwiki.org/wiki/Convergent_Sequence_in_Set_of_Integers/Corollary | [
"Sequences",
"Convergence"
] | [
"Definition:Sequence of Distinct Terms",
"Definition:Convergent Sequence/Topology"
] | [
"Convergent Sequence in Set of Integers",
"Definition:Convergent Sequence/Topology",
"Definition:Sequence of Distinct Terms",
"Category:Sequences",
"Category:Convergence"
] |
proofwiki-11833 | Limit of Monotone Real Function/Increasing | Let $f$ be a real function which is increasing and bounded above on the open interval $\openint a b$.
Let the supremum of $f$ on $\openint a b$ be $L$.
Then:
:$\ds \lim_{x \mathop \to b^-} \map f x = L$
where $\ds \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$. | Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$.
That is, that $L - \epsilon < \map f x < L + \epsilon$.
As $L$ is an upper bound for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically happens.
Since $L - \epsilon$ is ''n... | Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Increasing Real Function|increasing]] and [[Definition:Bounded Above Mapping|bounded above]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
Let the [[Definition:Supremum of Real-Valued Function|supremum]] of $f$ on $\... | Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$.
That is, that $L - \epsilon < \map f x < L + \epsilon$.
As $L$ is an [[Definition:Upper Bound of Set|upper bound]] for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically... | Limit of Monotone Real Function/Increasing | https://proofwiki.org/wiki/Limit_of_Monotone_Real_Function/Increasing | https://proofwiki.org/wiki/Limit_of_Monotone_Real_Function/Increasing | [
"Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Increasing/Real Function",
"Definition:Bounded Above Mapping",
"Definition:Real Interval/Open",
"Definition:Supremum of Mapping/Real-Valued Function",
"Definition:Limit of Real Function/Left"
] | [
"Definition:Upper Bound of Set",
"Definition:Increasing/Real Function"
] |
proofwiki-11834 | Limit of Monotone Real Function/Decreasing | Let $f$ be a real function which is decreasing and bounded below on the open interval $\openint a b$.
Let the infimum of $f$ on $\openint a b$ be $l$.
Then:
:$\ds \lim_{x \mathop \to b^-} \map f x = l$
where $\ds \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$. | Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that:
:$\forall x: b - \delta < x < b: \size {\map f x - l} < \epsilon$
That is:
:$l - \epsilon < \map f x < l + \epsilon$
As $l$ is a lower bound for $f$ on $\openint a b$:
:$l - \epsilon < \map f x$
automatically happens
Because $l + \epsilon$ is ''not'... | Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Decreasing Real Function|decreasing]] and [[Definition:Bounded Below Mapping|bounded below]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
Let the [[Definition:Infimum of Real-Valued Function|infimum]] of $f$ on $\op... | Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that:
:$\forall x: b - \delta < x < b: \size {\map f x - l} < \epsilon$
That is:
:$l - \epsilon < \map f x < l + \epsilon$
As $l$ is a [[Definition:Lower Bound of Set|lower bound]] for $f$ on $\openint a b$:
:$l - \epsilon < \map f x$
automatically hap... | Limit of Monotone Real Function/Decreasing | https://proofwiki.org/wiki/Limit_of_Monotone_Real_Function/Decreasing | https://proofwiki.org/wiki/Limit_of_Monotone_Real_Function/Decreasing | [
"Limits of Real Functions"
] | [
"Definition:Real Function",
"Definition:Decreasing/Real Function",
"Definition:Bounded Below Mapping",
"Definition:Real Interval/Open",
"Definition:Infimum of Mapping/Real-Valued Function",
"Definition:Limit of Real Function/Left"
] | [
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
"Definition:Decreasing/Real Function"
] |
proofwiki-11835 | Sum over k of m choose k by -1^m-k by k to the n | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
where:
:$\dbinom m k$ denotes a binomial coefficient
:$\ds {n \brace m}$ etc. denotes a Stirling number of the second kind
:$m!$ denotes a factorial. | The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
where:
:$\dbinom m k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$\ds {n \brace m}$ etc. denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
:$m!$ denotes a... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$ | Sum over k of m choose k by -1^m-k by k to the n | https://proofwiki.org/wiki/Sum_over_k_of_m_choose_k_by_-1^m-k_by_k_to_the_n | https://proofwiki.org/wiki/Sum_over_k_of_m_choose_k_by_-1^m-k_by_k_to_the_n | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Stirling Numbers of the Second Kind",
"Definition:Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11836 | Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
:$\ds {m + k \brace k}$ denotes a Stirling number of the second kind
:$\ds {n \brack n - m}$ denotes an unsigned Stirling number ... | The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} ... | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote [[Definition:Binomial Coefficient|binomial coefficients]]
:$\ds {m + k \brace k}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling ... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
$\map P 0$ is the case:
... | Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k | https://proofwiki.org/wiki/Sum_over_k_of_m-n_choose_m+k_by_m+n_choose_n+k_by_Stirling_Number_of_the_Second_Kind_of_m+k_with_k | https://proofwiki.org/wiki/Sum_over_k_of_m-n_choose_m+k_by_m+n_choose_n+k_by_Stirling_Number_of_the_Second_Kind_of_m+k_with_k | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Stirling Numbers of the Second Kind",
"Definition:Stirling Numbers of the First Kind/Unsigned"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Stirling Number of the Second Kind of Number with Self",
"Binomial Coefficient with Self",
"Binomial Coefficient with Zero",
"Unsigned Stirling Number of the First Kind of Number with Self",
"Principle of Mathematical Induction"
] |
proofwiki-11837 | Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote binomial coefficients
:$\ds {m + k \brack k}$ denotes an unsigned Stirling number of the first kind
:$\ds {n \brace n - m}$ denotes a Stirling number o... | The proof proceeds by induction on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall n \in \Z_{\ge 0}: \ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
where:
:$\dbinom {m - n} {m + k}$ etc. denote [[Definition:Binomial Coefficient|binomial coefficients]]
:$\ds {m + k \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall n \in \Z_{\ge 0}: \ds \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$ | Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k | https://proofwiki.org/wiki/Sum_over_k_of_m-n_choose_m+k_by_m+n_choose_n+k_by_Unsigned_Stirling_Number_of_the_First_Kind_of_m+k_with_k | https://proofwiki.org/wiki/Sum_over_k_of_m-n_choose_m+k_by_m+n_choose_n+k_by_Unsigned_Stirling_Number_of_the_First_Kind_of_m+k_with_k | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11838 | Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
where:
:$\dbinom n m$ denotes a binomial coefficient
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\ds {n + 1 \brace k + 1}$ denotes a Stirling number of the second kind. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$
where:
:$\dbinom n m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$\ds {k \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the fi... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n + 1 \brace k + 1} {k \brack m} \paren {-1}^{k - m} = \binom n m$ | Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Number_of_the_Second_Kind_of_n+1_with_k+1_by_Unsigned_Stirling_Number_of_the_First_Kind_of_k_with_m_by_-1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Stirling_Number_of_the_Second_Kind_of_n+1_with_k+1_by_Unsigned_Stirling_Number_of_the_First_Kind_of_k_with_m_by_-1^k-m | [
"Stirling Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11839 | Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
where:
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$ n!$ denotes a factorial. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
where:
:$\ds {k \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$ n!$ denotes a [[Definition:Factorial|factorial]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$ | Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial | https://proofwiki.org/wiki/Sum_over_k_to_n_of_Unsigned_Stirling_Number_of_the_First_Kind_of_k_with_m_by_n_factorial_over_k_factorial | https://proofwiki.org/wiki/Sum_over_k_to_n_of_Unsigned_Stirling_Number_of_the_First_Kind_of_k_with_m_by_n_factorial_over_k_factorial | [
"Stirling Numbers",
"Factorials"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11840 | Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below meet-continuous lattice.
Let $\operatorname {Ids}$ be the set of all ideals in $L$.
Let for all $I \in \operatorname {Ids}$: $m_I: S \to \operatorname {Ids}$ be a mapping:
:$\forall x \in S: x \preceq \sup I \implies \map {m_I} x = \set {x \wedge i: i \in ... | {{begin-eqn}}
{{eqn | l = \set {\map {m_I} x: I \in \operatorname {Ids} }
| r = \set {\map {m_I} x: I \in \operatorname {Ids} \land x \preceq \sup I} \cup \set {\map {m_I} x: I \in \operatorname {Ids} \land x \npreceq \sup I}
}}
{{eqn | r = \set {\set {x \wedge i: i \in I}: I \in \operatorname {Ids} \land x \prec... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
Let $\operatorname {Ids}$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
Let for all $I \in \operatorname {Ids}$... | {{begin-eqn}}
{{eqn | l = \set {\map {m_I} x: I \in \operatorname {Ids} }
| r = \set {\map {m_I} x: I \in \operatorname {Ids} \land x \preceq \sup I} \cup \set {\map {m_I} x: I \in \operatorname {Ids} \land x \npreceq \sup I}
}}
{{eqn | r = \set {\set {x \wedge i: i \in I}: I \in \operatorname {Ids} \land x \prec... | Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element | https://proofwiki.org/wiki/Intersection_of_Applications_of_Down_Mappings_at_Element_equals_Way_Below_Closure_of_Element | https://proofwiki.org/wiki/Intersection_of_Applications_of_Down_Mappings_at_Element_equals_Way_Below_Closure_of_Element | [
"Way Below Relation",
"Meet-Continuous Lattices"
] | [
"Definition:Bounded Below",
"Definition:Meet-Continuous Lattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Mapping",
"Definition:Lower Closure/Element",
"Definition:Way Below Closure"
] | [
"Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal",
"Intersection of Singleton",
"Set Intersection is Self-Distributive",
"Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element",
"Intersection with Subset is Subset",
"Way Below implies ... |
proofwiki-11841 | Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
where $\ds {k \brace m}$ etc. denotes a Stirling number of the second kind. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$
where $\ds {k \brace m}$ etc. denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {n + 1 \brace m + 1}$ | Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k | https://proofwiki.org/wiki/Sum_over_k_to_n_of_Stirling_Number_of_the_Second_Kind_of_k_with_m_by_m+1^n-k | https://proofwiki.org/wiki/Sum_over_k_to_n_of_Stirling_Number_of_the_Second_Kind_of_k_with_m_by_m+1^n-k | [
"Stirling Numbers"
] | [
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11842 | Zero Choose Zero | :$\dbinom 0 0 = 1$ | By Zero Choose n:
:$\dbinom 0 n = \delta_{0 n}$
where:
:$\dbinom 0 n$ denotes a binomial coefficient
:$\delta_{0 n}$ denotes the Kronecker delta.
Hence directly:
:$\dbinom 0 0 = 1$ | :$\dbinom 0 0 = 1$ | By [[Zero Choose n]]:
:$\dbinom 0 n = \delta_{0 n}$
where:
:$\dbinom 0 n$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$\delta_{0 n}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
Hence directly:
:$\dbinom 0 0 = 1$ | Zero Choose Zero/Proof 1 | https://proofwiki.org/wiki/Zero_Choose_Zero | https://proofwiki.org/wiki/Zero_Choose_Zero/Proof_1 | [
"Examples of Binomial Coefficients",
"Zero Choose Zero"
] | [] | [
"Zero Choose n",
"Definition:Binomial Coefficient",
"Definition:Kronecker Delta"
] |
proofwiki-11843 | Zero Choose Zero | :$\dbinom 0 0 = 1$ | By Binomial Coefficient with Zero:
:$\forall r \in \R: \dbinom r 0 = 1$
Hence directly:
:$\dbinom 0 0 = 1$ | :$\dbinom 0 0 = 1$ | By [[Binomial Coefficient with Zero]]:
:$\forall r \in \R: \dbinom r 0 = 1$
Hence directly:
:$\dbinom 0 0 = 1$ | Zero Choose Zero/Proof 2 | https://proofwiki.org/wiki/Zero_Choose_Zero | https://proofwiki.org/wiki/Zero_Choose_Zero/Proof_2 | [
"Examples of Binomial Coefficients",
"Zero Choose Zero"
] | [] | [
"Binomial Coefficient with Zero"
] |
proofwiki-11844 | Binomial Coefficient/Examples/Number of Bridge Hands | The total number $N$ of possible different hands for a game of [https://en.wikipedia.org/wiki/Contract_bridge bridge] is:
:$N = \dfrac {52!} {13! \, 39!} = 635 \ 013 \ 559 \ 600$ | The total number of cards in a standard deck is $52$.
The number of cards in a single bridge hand is $13$.
Thus $N$ is equal to the number of ways $13$ things can be chosen from $52$.
Thus:
{{begin-eqn}}
{{eqn | l = N
| r = \dbinom {52} {23}
| c = Cardinality of Set of Subsets
}}
{{eqn | r = \frac {52!} {13... | The total number $N$ of possible different hands for a game of [https://en.wikipedia.org/wiki/Contract_bridge bridge] is:
:$N = \dfrac {52!} {13! \, 39!} = 635 \ 013 \ 559 \ 600$ | The total number of cards in a standard deck is $52$.
The number of cards in a single bridge hand is $13$.
Thus $N$ is equal to the number of ways $13$ things can be chosen from $52$.
Thus:
{{begin-eqn}}
{{eqn | l = N
| r = \dbinom {52} {23}
| c = [[Cardinality of Set of Subsets]]
}}
{{eqn | r = \frac {5... | Binomial Coefficient/Examples/Number of Bridge Hands | https://proofwiki.org/wiki/Binomial_Coefficient/Examples/Number_of_Bridge_Hands | https://proofwiki.org/wiki/Binomial_Coefficient/Examples/Number_of_Bridge_Hands | [
"Examples of Binomial Coefficients"
] | [] | [
"Cardinality of Set of Subsets"
] |
proofwiki-11845 | Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal | Let $\struct {S, \preceq}$ be a meet semilattice.
Let $I$ be an ideal in $\struct {S, \preceq}$.
Let $x \in S$.
Then:
:$\paren {x^\preceq} \cap I = \set {x \wedge i: i \in I}$
where $x^\preceq$ denotes the lower closure of $x$. | === First Inclusion ===
Let $z \in \paren {x^\preceq} \cap I$
By definition of intersection:
:$z \in x^\preceq$ and $z \in I$
By definition of lower closure of element:
:$z \preceq x$
By Preceding iff Meet equals Less Operand:
:$x \wedge z = z$
Thus:
:$z \in \set {x \wedge i: i \in I}$
{{qed|lemma}} | Let $\struct {S, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $I$ be an [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$.
Let $x \in S$.
Then:
:$\paren {x^\preceq} \cap I = \set {x \wedge i: i \in I}$
where $x^\preceq$ denotes the [[Definition:Lower Closure of Element|lowe... | === First Inclusion ===
Let $z \in \paren {x^\preceq} \cap I$
By definition of [[Definition:Set Intersection|intersection]]:
:$z \in x^\preceq$ and $z \in I$
By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
:$z \preceq x$
By [[Preceding iff Meet equals Less Operand]]:
:$x \wedge z ... | Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal | https://proofwiki.org/wiki/Intersection_of_Lower_Closure_of_Element_with_Ideal_equals_Meet_of_Element_and_Ideal | https://proofwiki.org/wiki/Intersection_of_Lower_Closure_of_Element_with_Ideal_equals_Meet_of_Element_and_Ideal | [
"Lower Closures"
] | [
"Definition:Meet Semilattice",
"Definition:Ideal in Ordered Set",
"Definition:Lower Closure/Element"
] | [
"Definition:Set Intersection",
"Definition:Lower Closure/Element",
"Preceding iff Meet equals Less Operand",
"Definition:Lower Closure/Element",
"Definition:Set Intersection"
] |
proofwiki-11846 | Condition on Lower Coefficient for Binomial Coefficient to be Maximum | Let $n, k \in \Z_{\ge 0}$ be positive integers.
Let $\dbinom n k$ denote the binomial coefficient of $n$ choose $k$.
Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a maximum is:
:$k_\max = \floor {\dfrac n 2}$
and also:
:$k_\max = \ceiling {\dfrac n 2}$ | From Condition for Increasing Binomial Coefficients:
:$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$
Let $n$ be an even integer, such that:
:$n = 2 r$
Then:
:$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
and:
:$\forall k < r: \dbinom n r > \dbinom n k$
From Symmetry Rule for Binomia... | Let $n, k \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
Let $\dbinom n k$ denote the [[Definition:Binomial Coefficient|binomial coefficient]] of $n$ choose $k$.
Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a [[Definition:Maximum Element|maximum]] is:
:$k_\max = \f... | From [[Condition for Increasing Binomial Coefficients]]:
:$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$
Let $n$ be an [[Definition:Even Integer|even integer]], such that:
:$n = 2 r$
Then:
:$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
and:
:$\forall k < r: \dbinom n r > \dbinom ... | Condition on Lower Coefficient for Binomial Coefficient to be Maximum | https://proofwiki.org/wiki/Condition_on_Lower_Coefficient_for_Binomial_Coefficient_to_be_Maximum | https://proofwiki.org/wiki/Condition_on_Lower_Coefficient_for_Binomial_Coefficient_to_be_Maximum | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient",
"Definition:Greatest Element"
] | [
"Condition for Increasing Binomial Coefficients",
"Definition:Even Integer",
"Symmetry Rule for Binomial Coefficients",
"Definition:Odd Integer",
"Symmetry Rule for Binomial Coefficients"
] |
proofwiki-11847 | Row in Pascal's Triangle forms Palindromic Sequence | Each of the rows of Pascal's triangle forms a palindromic sequence. | The $n$th row of Pascal's triangle consists of the finite sequence:
:$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$
By the Symmetry Rule for Binomial Coefficients:
:$\dbinom n m = \dbinom n {n - m}$
Hence we can write the $n$th row in reverse order:
{{begin-eqn}}
{{eq... | Each of the [[Definition:Row of Pascal's Triangle|rows]] of [[Definition:Pascal's Triangle|Pascal's triangle]] forms a [[Definition:Palindrome|palindromic]] [[Definition:Sequence|sequence]]. | The $n$th [[Definition:Row of Pascal's Triangle|row]] of [[Definition:Pascal's Triangle|Pascal's triangle]] consists of the [[Definition:Finite Sequence|finite sequence]]:
:$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$
By the [[Symmetry Rule for Binomial Coefficien... | Row in Pascal's Triangle forms Palindromic Sequence | https://proofwiki.org/wiki/Row_in_Pascal's_Triangle_forms_Palindromic_Sequence | https://proofwiki.org/wiki/Row_in_Pascal's_Triangle_forms_Palindromic_Sequence | [
"Pascal's Triangle"
] | [
"Definition:Pascal's Triangle/Row",
"Definition:Pascal's Triangle",
"Definition:Palindrome",
"Definition:Sequence"
] | [
"Definition:Pascal's Triangle/Row",
"Definition:Pascal's Triangle",
"Definition:Finite Sequence",
"Symmetry Rule for Binomial Coefficients",
"Definition:Pascal's Triangle/Row",
"Definition:Finite Sequence"
] |
proofwiki-11848 | N Choose Negative Number is Zero | Let $n \in \Z$ be an integer.
Let $k \in \Z_{<0}$ be a (strictly) negative integer.
Then:
:$\dbinom n k = 0$ | From Pascal's Rule we have:
:$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$
Thus it is sufficient to prove that:
:$\forall n \in \Z: \dbinom n {-1} = 0$
So:
{{begin-eqn}}
{{eqn | l = \dbinom n {-1}
| r = \dbinom {n + 1} 0 - \dbinom n 0
| c = Pascal's Rule
}}
{{eqn | r = 1 - 1
... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Let $k \in \Z_{<0}$ be a [[Definition:Strictly Negative Integer|(strictly) negative integer]].
Then:
:$\dbinom n k = 0$ | From [[Pascal's Rule]] we have:
:$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$
Thus it is sufficient to prove that:
:$\forall n \in \Z: \dbinom n {-1} = 0$
So:
{{begin-eqn}}
{{eqn | l = \dbinom n {-1}
| r = \dbinom {n + 1} 0 - \dbinom n 0
| c = [[Pascal's Rule]]
}}
{{eqn | r ... | N Choose Negative Number is Zero | https://proofwiki.org/wiki/N_Choose_Negative_Number_is_Zero | https://proofwiki.org/wiki/N_Choose_Negative_Number_is_Zero | [
"Binomial Coefficients"
] | [
"Definition:Integer",
"Definition:Strictly Negative/Integer"
] | [
"Pascal's Rule",
"Pascal's Rule",
"Binomial Coefficient with Zero",
"Category:Binomial Coefficients"
] |
proofwiki-11849 | Kummer's Theorem | Let $p$ be a prime number.
Let $a, b \in \Z_{\ge 0}$.
Let:
:$p^n \divides \dbinom {a + b} b$
but
:$p^{n + 1} \nmid \dbinom {a + b} b$
where:
:$\divides$ denotes divisibility
:$\nmid$ denotes non-divisibility
:$\dbinom {a + b} b$ denotes a binomial coefficient.
Then $n$ equals the number of carries that occur when $a$ i... | {{ProofWanted|We need to establish the classical algorithms and build up the knowledge of exactly what a carry is before tackling this delightful little number.}}
{{Namedfor|Ernst Eduard Kummer|cat = Kummer}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $a, b \in \Z_{\ge 0}$.
Let:
:$p^n \divides \dbinom {a + b} b$
but
:$p^{n + 1} \nmid \dbinom {a + b} b$
where:
:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]]
:$\nmid$ denotes non-[[Definition:Divisor of Integer|divisibility]]
:$\dbinom {a ... | {{ProofWanted|We need to establish the [[Definition:Classical Algorithm|classical algorithms]] and build up the knowledge of exactly what a [[Definition:Carry Digit|carry]] is before tackling this delightful little number.}}
{{Namedfor|Ernst Eduard Kummer|cat = Kummer}} | Kummer's Theorem | https://proofwiki.org/wiki/Kummer's_Theorem | https://proofwiki.org/wiki/Kummer's_Theorem | [
"Binomial Coefficients",
"Number Bases",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Binomial Coefficient",
"Definition:Carry Digit",
"Definition:Classical Algorithm/Addition",
"Definition:Number Base"
] | [
"Definition:Classical Algorithm",
"Definition:Carry Digit"
] |
proofwiki-11850 | Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1 | Let $n \in \Z_{\ge 0}$ be a positive integer.
Then the non-zero elements of the $n$th row of Pascal's triangle are all odd {{iff}}:
:$n = 2^m - 1$
for some $m \in \Z_{\ge 0}$.
{{:Definition:Pascal's Triangle}}
As can be seen, the entries in rows $0, 1, 3, 7$ are all odd. | The statement:
:$\dbinom n k$ is odd
is equivalent to:
:$\dbinom n k \equiv 1 \pmod p$
{{Corollary|Lucas' Theorem}} gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$ and $p$ is prime
:the representations of $n$ and $k$ to the base $p$ are given by:
::$n =... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then the non-zero elements of the $n$th [[Definition:Row of Pascal's Triangle|row]] of [[Definition:Pascal's Triangle|Pascal's triangle]] are all [[Definition:Odd Integer|odd]] {{iff}}:
:$n = 2^m - 1$
for some $m \in \Z_{\ge 0}$.
{{:Definit... | The statement:
:$\dbinom n k$ is [[Definition:Odd Integer|odd]]
is equivalent to:
:$\dbinom n k \equiv 1 \pmod p$
{{Corollary|Lucas' Theorem}} gives:
:$\ds \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
where:
:$n, k \in \Z_{\ge 0}$ and $p$ is [[Definition:Prime Number|prime]]
:the [[Definiti... | Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1 | https://proofwiki.org/wiki/Numbers_in_Row_of_Pascal's_Triangle_all_Odd_iff_Row_number_2^n_-_1 | https://proofwiki.org/wiki/Numbers_in_Row_of_Pascal's_Triangle_all_Odd_iff_Row_number_2^n_-_1 | [
"Pascal's Triangle"
] | [
"Definition:Positive/Integer",
"Definition:Pascal's Triangle/Row",
"Definition:Pascal's Triangle",
"Definition:Odd Integer",
"Definition:Odd Integer"
] | [
"Definition:Odd Integer",
"Definition:Prime Number",
"Definition:Number Base",
"Definition:Digit",
"Definition:Biconditional/Semantics of Biconditional/Necessary and Sufficient",
"Definition:Digit"
] |
proofwiki-11851 | Sum of Sequence of Fourth Powers | :$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$ | From Faulhaber's Formula:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the Bernoulli numb... | :$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$ | From [[Faulhaber's Formula]]:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 1}^n i^p
| r = 1^p + 2^p + \cdots + n^p
| c =
}}
{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
| c =
}}
{{end-eqn}}
where $B_k$ are the [[Defini... | Sum of Sequence of Fourth Powers/Proof using Bernoulli Numbers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Fourth_Powers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Fourth_Powers/Proof_using_Bernoulli_Numbers | [
"Sum of Sequence of Fourth Powers",
"Sums of Sequences",
"Fourth Powers"
] | [] | [
"Faulhaber's Formula",
"Definition:Bernoulli Numbers"
] |
proofwiki-11852 | Sum of Sequence of Fourth Powers | :$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$ | By definition of binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom k 4
| r = \frac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4 !}
| c =
}}
{{eqn | l = \binom k 3
| r = \frac {k \paren {k - 1} \paren {k - 2} } {3 !}
| c =
}}
{{eqn | l = \binom k 2
| r = \frac {k \paren {k - ... | :$\ds \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$ | By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
{{begin-eqn}}
{{eqn | l = \binom k 4
| r = \frac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4 !}
| c =
}}
{{eqn | l = \binom k 3
| r = \frac {k \paren {k - 1} \paren {k - 2} } {3 !}
| c =
}}
{{eqn | l = \binom ... | Sum of Sequence of Fourth Powers/Proof using Binomial Coefficients | https://proofwiki.org/wiki/Sum_of_Sequence_of_Fourth_Powers | https://proofwiki.org/wiki/Sum_of_Sequence_of_Fourth_Powers/Proof_using_Binomial_Coefficients | [
"Sum of Sequence of Fourth Powers",
"Sums of Sequences",
"Fourth Powers"
] | [] | [
"Definition:Binomial Coefficient",
"Sum of Binomial Coefficients over Upper Index"
] |
proofwiki-11853 | Complete Lattice is Bounded | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Then
:$L$ is bounded. | By definition of complete lattice:
:$\O$ admits a supremum and an infimum.
By Infimum of Empty Set is Greatest Element:
:$\forall x \in S: x \preceq \inf \O$
Thus by definition:
:$L$ is bounded above.
By Supremum of Empty Set is Smallest Element:
:$\forall x \in S: \sup \O \preceq x$
Thus by definition:
:$L$ is bounded... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Then
:$L$ is [[Definition:Bounded Set|bounded]]. | By definition of [[Definition:Complete Lattice|complete lattice]]:
:$\O$ admits a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
By [[Infimum of Empty Set is Greatest Element]]:
:$\forall x \in S: x \preceq \inf \O$
Thus by definition:
:$L$ is [[Definition:Bounded Above Set|boun... | Complete Lattice is Bounded | https://proofwiki.org/wiki/Complete_Lattice_is_Bounded | https://proofwiki.org/wiki/Complete_Lattice_is_Bounded | [
"Bounded Lattices",
"Complete Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Bounded Set"
] | [
"Definition:Complete Lattice",
"Definition:Supremum of Set",
"Definition:Infimum of Set",
"Infimum of Empty Set is Greatest Element",
"Definition:Bounded Above Set",
"Supremum of Empty Set is Smallest Element",
"Definition:Bounded Below Set",
"Definition:Bounded Set"
] |
proofwiki-11854 | Constant Function is Primitive Recursive/General Case | The constant function of $k$ variables: $f_c^k: \N^k \to \N$, defined as:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = c$ where $c \in \N$
is primitive recursive. | For $k \ge 1$, let $f_c^k$ be the constant function of $k$ variables with value $c$.
We know from Constant Function is Primitive Recursive that $f_c^1$ is primitive recursive.
Now:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = \map {f_c^1} {n_1} = \map {f_c^1} {\map {\pr_1^k} {n_1, n_2, \ldots, n_k} }$
where $\pr_1^k$ is a ... | The [[Definition:Constant Mapping|constant function]] of $k$ variables: $f_c^k: \N^k \to \N$, defined as:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = c$ where $c \in \N$
is [[Definition:Primitive Recursive Function|primitive recursive]]. | For $k \ge 1$, let $f_c^k$ be the [[Definition:Constant Mapping|constant function]] of $k$ variables with value $c$.
We know from [[Constant Function is Primitive Recursive]] that $f_c^1$ is [[Definition:Primitive Recursive Function|primitive recursive]].
Now:
:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = \map {f_c^1} {n... | Constant Function is Primitive Recursive/General Case | https://proofwiki.org/wiki/Constant_Function_is_Primitive_Recursive/General_Case | https://proofwiki.org/wiki/Constant_Function_is_Primitive_Recursive/General_Case | [
"Primitive Recursive Functions",
"Constant Mappings"
] | [
"Definition:Constant Mapping",
"Definition:Primitive Recursive/Function"
] | [
"Definition:Constant Mapping",
"Constant Function is Primitive Recursive",
"Definition:Primitive Recursive/Function",
"Definition:Basic Primitive Recursive Function/Projection Function",
"Definition:Basic Primitive Recursive Function",
"Definition:Primitive Recursive/Function",
"Definition:Basic Primiti... |
proofwiki-11855 | Value of Multiplicative Function at One | Let $f: \N \to \C$ be a multiplicative function.
If $f$ is not identically zero, then $\map f 1 = 1$. | If $f$ is not identically zero, then:
:$\exists m \in \Z: \map f m \ne 0$
Then:
:$\map f m = \map f {1 \times m} = \map f 1 \, \map f m$
Hence $\map f 1 = 1$.
{{qed}}
Category:Multiplicative Functions
3yg93q06vyrng6lsj376wzn4zqpmn45 | Let $f: \N \to \C$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
If $f$ is not identically zero, then $\map f 1 = 1$. | If $f$ is not identically zero, then:
:$\exists m \in \Z: \map f m \ne 0$
Then:
:$\map f m = \map f {1 \times m} = \map f 1 \, \map f m$
Hence $\map f 1 = 1$.
{{qed}}
[[Category:Multiplicative Functions]]
3yg93q06vyrng6lsj376wzn4zqpmn45 | Value of Multiplicative Function at One | https://proofwiki.org/wiki/Value_of_Multiplicative_Function_at_One | https://proofwiki.org/wiki/Value_of_Multiplicative_Function_at_One | [
"Multiplicative Functions"
] | [
"Definition:Multiplicative Arithmetic Function"
] | [
"Category:Multiplicative Functions"
] |
proofwiki-11856 | Value of Multiplicative Function is Product of Values of Prime Power Factors | Let $f: \N \to \C$ be a multiplicative function.
Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the prime decomposition of $n$.
Then:
:$\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$ | We have:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
We also have:
:$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$
So:
:$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$
It is a simple inductive process to show that $\map f n = \map f {p_1^{k_1} } \, \map f {p_2... | Let $f: \N \to \C$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the [[Definition:Prime Decomposition|prime decomposition]] of $n$.
Then:
:$\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$ | We have:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
We also have:
:$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$
So:
:$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$
It is a simple [[Principle of Mathematical Induction|inductive process]] to show that $\... | Value of Multiplicative Function is Product of Values of Prime Power Factors | https://proofwiki.org/wiki/Value_of_Multiplicative_Function_is_Product_of_Values_of_Prime_Power_Factors | https://proofwiki.org/wiki/Value_of_Multiplicative_Function_is_Product_of_Values_of_Prime_Power_Factors | [
"Multiplicative Functions"
] | [
"Definition:Multiplicative Arithmetic Function",
"Definition:Prime Decomposition"
] | [
"Principle of Mathematical Induction",
"Category:Multiplicative Functions"
] |
proofwiki-11857 | -1^n by -n choose k-1 equals -1^k by -k choose n-1 | :$\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$ | {{begin-eqn}}
{{eqn | l = \paren {-1}^n \dbinom {-n} {k - 1}
| r = \paren {-1}^n \paren {\paren {-1}^{-n - \paren {k - 1} } \binom {-\paren {k - 1 + 1} } {-n - \paren {k - 1} } }
| c = Moving Top Index to Bottom in Binomial Coefficient
}}
{{eqn | r = \paren {-1}^{- \paren {k - 1} } \binom {-k} {-\paren {n -... | :$\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$ | {{begin-eqn}}
{{eqn | l = \paren {-1}^n \dbinom {-n} {k - 1}
| r = \paren {-1}^n \paren {\paren {-1}^{-n - \paren {k - 1} } \binom {-\paren {k - 1 + 1} } {-n - \paren {k - 1} } }
| c = [[Moving Top Index to Bottom in Binomial Coefficient]]
}}
{{eqn | r = \paren {-1}^{- \paren {k - 1} } \binom {-k} {-\paren ... | -1^n by -n choose k-1 equals -1^k by -k choose n-1 | https://proofwiki.org/wiki/-1^n_by_-n_choose_k-1_equals_-1^k_by_-k_choose_n-1 | https://proofwiki.org/wiki/-1^n_by_-n_choose_k-1_equals_-1^k_by_-k_choose_n-1 | [
"Binomial Coefficients"
] | [] | [
"Negated Upper Index of Binomial Coefficient/Corollary 2",
"Symmetry Rule for Binomial Coefficients"
] |
proofwiki-11858 | One Choose n | :$\dbinom 1 n = \begin{cases} 1 & : n \in \set {0, 1} \\ 0 & : \text {otherwise} \end{cases}$ | By definition of binomial coefficient:
:<nowiki>$\dbinom m n = \begin{cases}
\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\
& \\
0 & : \text { otherwise } \end{cases}$</nowiki>
Thus when $n > 1$:
:$\dbinom 1 n = 0$
and when $n < 0$:
:$\dbinom 1 n = 0$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom 1 0
| r = \dfra... | :$\dbinom 1 n = \begin{cases} 1 & : n \in \set {0, 1} \\ 0 & : \text {otherwise} \end{cases}$ | By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
:<nowiki>$\dbinom m n = \begin{cases}
\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\
& \\
0 & : \text { otherwise } \end{cases}$</nowiki>
Thus when $n > 1$:
:$\dbinom 1 n = 0$
and when $n < 0$:
:$\dbinom 1 n = 0$
Then:
{{begin-eqn}}
{... | One Choose n | https://proofwiki.org/wiki/One_Choose_n | https://proofwiki.org/wiki/One_Choose_n | [
"Examples of Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient",
"Symmetry Rule for Binomial Coefficients",
"Category:Examples of Binomial Coefficients"
] |
proofwiki-11859 | Separable Discrete Space is Countable | Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $T$ be separable.
Then $S$ is countable. | Let $T$ be separable.
By Space is Separable iff Density not greater than Aleph Zero:
:$\map d T \le \aleph_0$
where:
:$\map d T$ denotes the density of $T$
:$\aleph$ denotes the aleph mapping.
By definition of density:
:$\exists A \subseteq S: A$ is everywhere dense and $\map d T = \card A$
where $\card A$ denotes the ... | Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Space|discrete]] [[Definition:Topological Space|topological space]].
Let $T$ be [[Definition:Separable Space|separable]].
Then $S$ is [[Definition:Countable Set|countable]]. | Let $T$ be [[Definition:Separable Space|separable]].
By [[Space is Separable iff Density not greater than Aleph Zero]]:
:$\map d T \le \aleph_0$
where:
:$\map d T$ denotes the [[Definition:Density of Topological Space|density]] of $T$
:$\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]].
By definition of... | Separable Discrete Space is Countable/Proof 1 | https://proofwiki.org/wiki/Separable_Discrete_Space_is_Countable | https://proofwiki.org/wiki/Separable_Discrete_Space_is_Countable/Proof_1 | [
"Separable Discrete Space is Countable",
"Discrete Topologies",
"Separable Spaces",
"Countable Sets"
] | [
"Definition:Discrete Topology",
"Definition:Topological Space",
"Definition:Separable Space",
"Definition:Countable Set"
] | [
"Definition:Separable Space",
"Space is Separable iff Density not greater than Aleph Zero",
"Definition:Density of Topological Space",
"Definition:Aleph Mapping",
"Definition:Density of Topological Space",
"Definition:Everywhere Dense",
"Definition:Cardinality",
"Definition:Everywhere Dense",
"Defin... |
proofwiki-11860 | Separable Discrete Space is Countable | Let $T = \struct {S, \tau}$ be a discrete topological space.
Let $T$ be separable.
Then $S$ is countable. | Let $T$ be separable.
{{AimForCont}} $S$ is uncountable.
Then by Uncountable Discrete Space is not Separable, $T$ is not separable.
Hence the result by Proof by Contradiction.
{{qed}} | Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Space|discrete]] [[Definition:Topological Space|topological space]].
Let $T$ be [[Definition:Separable Space|separable]].
Then $S$ is [[Definition:Countable Set|countable]]. | Let $T$ be [[Definition:Separable Space|separable]].
{{AimForCont}} $S$ is [[Definition:Uncountable Set|uncountable]].
Then by [[Uncountable Discrete Space is not Separable]], $T$ is not [[Definition:Separable Space|separable]].
Hence the result by [[Proof by Contradiction]].
{{qed}} | Separable Discrete Space is Countable/Proof 2 | https://proofwiki.org/wiki/Separable_Discrete_Space_is_Countable | https://proofwiki.org/wiki/Separable_Discrete_Space_is_Countable/Proof_2 | [
"Separable Discrete Space is Countable",
"Discrete Topologies",
"Separable Spaces",
"Countable Sets"
] | [
"Definition:Discrete Topology",
"Definition:Topological Space",
"Definition:Separable Space",
"Definition:Countable Set"
] | [
"Definition:Separable Space",
"Definition:Uncountable/Set",
"Uncountable Discrete Space is not Separable",
"Definition:Separable Space",
"Proof by Contradiction"
] |
proofwiki-11861 | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction | Let $r, s, t \in \R, n \in \Z$.
Consider the equation:
:$\ds (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
where $\dbinom {r - t k} k$ etc. are binomial coefficients.
Then equation $(1)$ holds for the special case where $s = n - 1... | Substituting $n - 1 - r + n t$ for $s$ in the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \binom {r + s - t n} n
| r = \binom {r + \paren {n - 1 - r + n t} - t n} n
| c =
}}
{{eqn | r = \binom {r + n - 1 - r + n t - t n} n
| c =
}}
{{eqn | r = \binom {n - 1} n
| c =
}}
{{end-eqn}}
Substituting $n - 1 ... | Let $r, s, t \in \R, n \in \Z$.
Consider the equation:
:$\ds (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
where $\dbinom {r - t k} k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]].
Then equation $(1)$ hol... | Substituting $n - 1 - r + n t$ for $s$ in the {{RHS}}:
{{begin-eqn}}
{{eqn | l = \binom {r + s - t n} n
| r = \binom {r + \paren {n - 1 - r + n t} - t n} n
| c =
}}
{{eqn | r = \binom {r + n - 1 - r + n t - t n} n
| c =
}}
{{eqn | r = \binom {n - 1} n
| c =
}}
{{end-eqn}}
Substituting $n -... | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_1/Basis_for_the_Induction | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_1/Basis_for_the_Induction | [
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk"
] | [
"Definition:Binomial Coefficient"
] | [
"Definition:Polynomial",
"Definition:Degree of Polynomial"
] |
proofwiki-11862 | Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Then:
:$L$ is continuous
{{iff}}:
:$L$ is meet-continuous and there exists the smallest auxiliary approximating relation on $S$
That is:
:$L$ is continuous
{{iff}}:
:$L$ is meet-continuous and there exists an auxiliary approximating relation $\RR$ on $... | === Sufficient Condition ===
Let $L$ be continuous.
Thus by Continuous Lattice is Meet-Continuous:
:$L$ is meet-continuous.
Thus by Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:
:$\ll$ is auxiliary approximating relation on $S$.
Thus by Continuous Lattice iff Auxiliary Approximating ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Then:
:$L$ is [[Definition:Continuous Ordered Set|continuous]]
{{iff}}:
:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] and there exists the [[Definition:Smallest Element|smallest]] [[Definition:Auxiliar... | === Sufficient Condition ===
Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
Thus by [[Continuous Lattice is Meet-Continuous]]:
:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
Thus by [[Way Below is Approximating Relation]] and [[Way Below Relation is Auxiliary Relation]]:
:$\ll$ is [[De... | Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation | https://proofwiki.org/wiki/Continuous_iff_Meet-Continuous_and_There_Exists_Smallest_Auxiliary_Approximating_Relation | https://proofwiki.org/wiki/Continuous_iff_Meet-Continuous_and_There_Exists_Smallest_Auxiliary_Approximating_Relation | [
"Approximating Relations",
"Continuous Lattices",
"Meet-Continuous Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Continuous Ordered Set",
"Definition:Meet-Continuous Lattice",
"Definition:Smallest Element",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation",
"Definition:Continuous Ordered Set",
"Definition:Meet-Continuous Lattice",
"Definition:Auxili... | [
"Definition:Continuous Ordered Set",
"Continuous Lattice is Meet-Continuous",
"Definition:Meet-Continuous Lattice",
"Way Below is Approximating Relation",
"Way Below Relation is Auxiliary Relation",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation",
"Continuous Lattice iff Auxiliary... |
proofwiki-11863 | Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $\map {\operatorname{App} } L$ be the set of all auxiliary approximating relation on $S$.
Then
:$L$ is continuous
{{iff}}
:$\forall \RR \in \map {\operatorname{App} } L: \ll \subseteq \RR$ and $\ll$ is an approximating relation | === Sufficient Condition ===
Let $L$ be continuous.
Let $\RR \in \map {\operatorname{App} } L$
Let $\tuple {a, b} \in S \times S$ such that:
:$a \ll b$
By definition of way below closure:
:$a \in b^\ll$
where $b^\ll$ denotes the way below closure of $b$.
By Complete Lattice is Bounded and Continuous Lattice is Meet-Con... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $\map {\operatorname{App} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
Then
:$L$ is [[Definitio... | === Sufficient Condition ===
Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
Let $\RR \in \map {\operatorname{App} } L$
Let $\tuple {a, b} \in S \times S$ such that:
:$a \ll b$
By definition of [[Definition:Way Below Closure|way below closure]]:
:$a \in b^\ll$
where $b^\ll$ denotes the [[Definition:Way... | Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation | https://proofwiki.org/wiki/Continuous_Lattice_iff_Auxiliary_Approximating_Relation_is_Superset_of_Way_Below_Relation | https://proofwiki.org/wiki/Continuous_Lattice_iff_Auxiliary_Approximating_Relation_is_Superset_of_Way_Below_Relation | [
"Approximating Relations",
"Continuous Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation",
"Definition:Continuous Ordered Set",
"Definition:Approximating Relation"
] | [
"Definition:Continuous Ordered Set",
"Definition:Way Below Closure",
"Definition:Way Below Closure",
"Complete Lattice is Bounded",
"Continuous Lattice is Meet-Continuous",
"Definition:Bounded Below Set",
"Definition:Meet-Continuous Lattice",
"Intersection of Relation Segments of Approximating Relatio... |
proofwiki-11864 | Completely Additive Function is Additive | Let $f: \N \to \C$ be a completely additive function.
Then $f$ is also additive. | Let $m, n$ be coprime integers.
Then in particular, $m, n \in \Z$.
Hence, since $f$ is completely additive:
:$f \left({m \times n}\right) = f \left({m}\right) + f \left({n}\right)$
and $f$ is additive, as desired.
{{qed}}
Category:Number Theory
Category:Ring Theory
okn9cbt94w16hdr1f8wcq1ox7gqmhnb | Let $f: \N \to \C$ be a [[Definition:Completely Additive Function|completely additive function]].
Then $f$ is also [[Definition:Additive Arithmetic Function|additive]]. | Let $m, n$ be [[Definition:Coprime Integers|coprime integers]].
Then in particular, $m, n \in \Z$.
Hence, since $f$ is [[Definition:Completely Additive Function|completely additive]]:
:$f \left({m \times n}\right) = f \left({m}\right) + f \left({n}\right)$
and $f$ is [[Definition:Additive Arithmetic Function|additi... | Completely Additive Function is Additive | https://proofwiki.org/wiki/Completely_Additive_Function_is_Additive | https://proofwiki.org/wiki/Completely_Additive_Function_is_Additive | [
"Number Theory",
"Ring Theory"
] | [
"Definition:Completely Additive Function",
"Definition:Additive Arithmetic Function"
] | [
"Definition:Coprime/Integers",
"Definition:Completely Additive Function",
"Definition:Additive Arithmetic Function",
"Category:Number Theory",
"Category:Ring Theory"
] |
proofwiki-11865 | Real Logarithm is Completely Additive | Let $\log_b: \R_{>0} \to \R$ be the real logarithm to base $b$.
Then $\log_b$ is completely additive. | From Sum of General Logarithms:
:$\log_b x + \log_b y = \log_b \left({x y}\right)$
The result follows from the definition of complete additivity.
{{qed}}
Category:Logarithms
Category:Number Theory
8wryrdf91qouep0q87julzna5febp3i | Let $\log_b: \R_{>0} \to \R$ be the [[Definition:Real General Logarithm|real logarithm]] to [[Definition:Base of Logarithm|base]] $b$.
Then $\log_b$ is [[Definition:Completely Additive Function|completely additive]]. | From [[Sum of General Logarithms]]:
:$\log_b x + \log_b y = \log_b \left({x y}\right)$
The result follows from the definition of [[Definition:Completely Additive Function|complete additivity]].
{{qed}}
[[Category:Logarithms]]
[[Category:Number Theory]]
8wryrdf91qouep0q87julzna5febp3i | Real Logarithm is Completely Additive | https://proofwiki.org/wiki/Real_Logarithm_is_Completely_Additive | https://proofwiki.org/wiki/Real_Logarithm_is_Completely_Additive | [
"Logarithms",
"Number Theory"
] | [
"Definition:General Logarithm/Positive Real",
"Definition:Logarithm/Base",
"Definition:Completely Additive Function"
] | [
"Sum of Logarithms/General Logarithm",
"Definition:Completely Additive Function",
"Category:Logarithms",
"Category:Number Theory"
] |
proofwiki-11866 | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma | Let this hold for $\tuple {r, s, t, n}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
and also for $\tuple {r, s - t, t, n - 1}$.
Then it also holds for $\tuple {r, s + 1, t, n}$. | Evaluating the equation for $\tuple {r, s - t, t, n - 1}$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {s - t} - t \paren {\paren {n - 1} - k} } {\paren {n - 1} - k} \frac r {r - t k}
| r = \binom {r + \paren {s - t} - t \paren {n - 1} } {n - 1}
| c =
}}
{{eqn | ll= \... | Let this hold for $\tuple {r, s, t, n}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
and also for $\tuple {r, s - t, t, n - 1}$.
Then it also holds for $\tuple {r, s + 1, t, n}$. | Evaluating the equation for $\tuple {r, s - t, t, n - 1}$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {s - t} - t \paren {\paren {n - 1} - k} } {\paren {n - 1} - k} \frac r {r - t k}
| r = \binom {r + \paren {s - t} - t \paren {n - 1} } {n - 1}
| c =
}}
{{eqn | ll= ... | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_1/Lemma | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_1/Lemma | [
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk"
] | [] | [
"Pascal's Rule"
] |
proofwiki-11867 | Sum over k of r-kt choose k by r over r-kt by z^k | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $\map {A_n} {x, t}$ be the polynomial of degree $n$ defined as:
:$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
for $x \ne n t$.
Let $z = x^{t + 1} - x^t$.
Then:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
for sufficiently small $z$. | From Sum over $k$ of $\paren {-1}^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$ and renaming variables:
:$\ds \sum_j \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$
where $\delta_{k 0}$ is the Kronecker delta.
Thus:
:$\ds \sum_{j, k} \paren {-1}^j \dbinom k j \dbi... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $\map {A_n} {x, t}$ be the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree $n$]] defined as:
:$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
for $x \ne n t$.
Let $z = x^{... | From [[Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt|Sum over $k$ of $\paren {-1}^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$]] and renaming variables:
:$\ds \sum_j \paren {-1}^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$
where $\delta_{k 0}$ is the [[... | Sum over k of r-kt choose k by r over r-kt by z^k | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_r_over_r-kt_by_z^k | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_r_over_r-kt_by_z^k | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Sufficiently Small"
] | [
"Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt",
"Definition:Kronecker Delta",
"Product of r Choose m with m Choose k",
"Binomial Theorem"
] |
proofwiki-11868 | Not Preceding implies Approximating Relation and not Preceding | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, y \in S$ such that
:$x \npreceq y$
Let $\RR$ be an approximating relation on $S$.
Then
:$\exists u \in S: \tuple {u, x} \in \RR \land u \npreceq y$ | By definition of approximating relation:
:$x = \map \sup {x^\RR}$
By definition of supremum:
:$y$ is upper bound for $x^\RR \implies x \preceq y$
By definition of upper bound:
:$\exists u \in S: u \in x^\RR \land u \npreceq y$
Thus by definition of $\RR$-segment:
:$\exists u \in S: \tuple {u, x} \in \RR \land u \nprece... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $x, y \in S$ such that
:$x \npreceq y$
Let $\RR$ be an [[Definition:Approximating Relation|approximating relation]] on $S$.
Then
:$\exists u \in S: \tuple {u, x} \in \RR \land u \npreceq y$ | By definition of [[Definition:Approximating Relation|approximating relation]]:
:$x = \map \sup {x^\RR}$
By definition of [[Definition:Supremum of Set|supremum]]:
:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\RR \implies x \preceq y$
By definition of [[Definition:Upper Bound of Set|upper bound]]:
:$\ex... | Not Preceding implies Approximating Relation and not Preceding | https://proofwiki.org/wiki/Not_Preceding_implies_Approximating_Relation_and_not_Preceding | https://proofwiki.org/wiki/Not_Preceding_implies_Approximating_Relation_and_not_Preceding | [
"Approximating Relations",
"Complete Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Approximating Relation"
] | [
"Definition:Approximating Relation",
"Definition:Supremum of Set",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Relation Segment"
] |
proofwiki-11869 | Auxiliary Approximating Relation has Quasi Interpolation Property | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, z \in S$.
Let $\RR$ be an auxiliary approximating relation on $S$ such that
:$\tuple {x, z} \in \RR \land x \ne z$
Then:
:$\exists y \in S: x \preceq y \land \tuple {y, z} \in \RR \land x \ne y$ | By definition of auxiliary relation:
:$x \preceq z$
By definition of $\prec$:
:$x \prec z$
By definition of antisymmetry:
:$z \nprec x$
Then:
:$z \npreceq x$
By Not Preceding implies Approximating Relation and not Preceding:
:$\exists u \in S: \tuple {u, z} \land u \npreceq x$
Define $y = x \vee u$.
Thus by Join Succee... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $x, z \in S$.
Let $\RR$ be an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$ such that
:$\tuple {x, z} \in \RR \land x \ne z$
Then:
:$\exists y \in... | By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$x \preceq z$
By definition of $\prec$:
:$x \prec z$
By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
:$z \nprec x$
Then:
:$z \npreceq x$
By [[Not Preceding implies Approximating Relation and not Preceding]]:
:$\exists u \in... | Auxiliary Approximating Relation has Quasi Interpolation Property | https://proofwiki.org/wiki/Auxiliary_Approximating_Relation_has_Quasi_Interpolation_Property | https://proofwiki.org/wiki/Auxiliary_Approximating_Relation_has_Quasi_Interpolation_Property | [
"Approximating Relations",
"Complete Lattices",
"Auxiliary Relations"
] | [
"Definition:Complete Lattice",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation"
] | [
"Definition:Auxiliary Relation",
"Definition:Antisymmetric Relation",
"Not Preceding implies Approximating Relation and not Preceding",
"Join Succeeds Operands",
"Definition:Auxiliary Relation",
"Preceding iff Join equals Larger Operand"
] |
proofwiki-11870 | Preceding iff Join equals Larger Operand | Let $\left({S, \preceq}\right)$ be a join semilattice.
Let $x, y \in S$.
Then
:$x \preceq y$ {{iff}} $x \vee y = y$ | === Sufficient Condition ===
Let
:$x \preceq y$
By definition of join:
:$x \vee y = \sup \left\{ {x, y}\right\}$
By definitions of upper bound and reflexivity:
:$y$ is upper bound for $\left\{ {x, y}\right\}$
and
:$\forall z \in S: z$ is upper bound for $\left\{ {x, y}\right\} \implies y \preceq z$
Thus by definition o... | Let $\left({S, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
Let $x, y \in S$.
Then
:$x \preceq y$ {{iff}} $x \vee y = y$ | === Sufficient Condition ===
Let
:$x \preceq y$
By definition of [[Definition:Join (Order Theory)|join]]:
:$x \vee y = \sup \left\{ {x, y}\right\}$
By definitions of [[Definition:Upper Bound of Set|upper bound]] and [[Definition:Reflexivity|reflexivity]]:
:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $\l... | Preceding iff Join equals Larger Operand | https://proofwiki.org/wiki/Preceding_iff_Join_equals_Larger_Operand | https://proofwiki.org/wiki/Preceding_iff_Join_equals_Larger_Operand | [
"Join and Meet",
"Join and Meet Semilattices"
] | [
"Definition:Join Semilattice"
] | [
"Definition:Join (Order Theory)",
"Definition:Upper Bound of Set",
"Definition:Reflexivity",
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set"
] |
proofwiki-11871 | Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt | :$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
where $\delta_{n 0}$ is the Kronecker delta. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^k \dbinom 0 k \dbinom {r - k ... | :$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
where $\delta_{n 0}$ is the [[Definition:Kronecker Delta|Kronecker delta]]. | 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]]:
:$\ds \sum_k \paren {-1}^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{... | Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt | https://proofwiki.org/wiki/Sum_over_k_of_-1^k_by_n_choose_k_by_r-kt_choose_n_by_r_over_r-kt | https://proofwiki.org/wiki/Sum_over_k_of_-1^k_by_n_choose_k_by_r-kt_choose_n_by_r_over_r-kt | [
"Binomial Coefficients"
] | [
"Definition:Kronecker Delta"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Zero Choose n",
"Binomial Coefficient with Zero",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Principle of Mathematical Induction"
] |
proofwiki-11872 | Sum over k of r-kt choose k by z^k | Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
where $\dbinom {r - t k} k$ denotes a binomial coefficient. | From Sum over $k$ of $\dbinom r k$ by $\dbinom {s - k t} r$ by $\paren {-1}^k$ and renaming variables:
:$\ds \sum_j \paren {-1}^j \binom k j \binom {r - j t} k = t^k$
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{j, k} \binom k j \binom {r - j t} k \paren {-1}^j
| r = \sum_{k \mathop \ge 0} t^k
| c = when $k < 0$ ... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
where $\dbinom {r - t k} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Sum over k of r Choose k by s-kt Choose r by -1^k|Sum over $k$ of $\dbinom r k$ by $\dbinom {s - k t} r$ by $\paren {-1}^k$]] and renaming variables:
:$\ds \sum_j \paren {-1}^j \binom k j \binom {r - j t} k = t^k$
Thus:
{{begin-eqn}}
{{eqn | l = \sum_{j, k} \binom k j \binom {r - j t} k \paren {-1}^j
|... | Sum over k of r-kt choose k by z^k/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_z^k | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_z^k/Proof_1 | [
"Binomial Coefficients",
"Sum over k of r-kt choose k by z^k"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient"
] | [
"Sum over k of r Choose k by s-kt Choose r by -1^k",
"Sum of Infinite Geometric Sequence",
"Product of r Choose m with m Choose k"
] |
proofwiki-11873 | Sum over k of r-kt choose k by z^k | Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
where $\dbinom {r - t k} k$ denotes a binomial coefficient. | From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:
:$\ds (1): \quad \sum_k \map {A_k} {r, t} z^k = x^r$
where:
:$\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
::$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
:for $x \ne n t$
:$z = x^{t + 1} - x^t$.
Differentiati... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then:
:$\ds \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\paren {t + 1} x - t}$
where $\dbinom {r - t k} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
:$\ds (1): \quad \sum_k \map {A_k} {r, t} z^k = x^r$
where:
:$\map {A_n} {x, t}$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree ... | Sum over k of r-kt choose k by z^k/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_z^k | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_choose_k_by_z^k/Proof_2 | [
"Binomial Coefficients",
"Sum over k of r-kt choose k by z^k"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient"
] | [
"Sum over k of r-kt choose k by r over r-kt by z^k",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Differentiation",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Power Rule for Derivatives",
"Derivative of Inverse Function"
] |
proofwiki-11874 | Sum over k of r+tk choose k by s-tk choose n-k | Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$
where $\dbinom {r + t k} k$ and so on denotes a binomial coefficient. | Let $\map f {r, s, t, n}$ be the function defined as:
:$\ds \map f {r, s, t, n} := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$
We have:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}
| c =
}}
{{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then:
:$\ds \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$
where $\dbinom {r + t k} k$ and so on denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | Let $\map f {r, s, t, n}$ be the [[Definition:Real-Valued Function|function]] defined as:
:$\ds \map f {r, s, t, n} := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$
We have:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}
| c =
}}
{{eqn | r = \sum_k \dbinom {... | Sum over k of r+tk choose k by s-tk choose n-k | https://proofwiki.org/wiki/Sum_over_k_of_r+tk_choose_k_by_s-tk_choose_n-k | https://proofwiki.org/wiki/Sum_over_k_of_r+tk_choose_k_by_s-tk_choose_n-k | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Binomial Coefficient"
] | [
"Definition:Real-Valued Function"
] |
proofwiki-11875 | Auxiliary Approximating Relation has Interpolation Property | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x, z \in S$ such that:
:$x \ll z \land x \ne z$
{{explain|$\ll$}}
Let $\RR$ be an auxiliary approximating relation on $S$.
Then
:$\exists y \in S: \tuple {x, y} \in \RR \land \tuple {y, z} \in \RR \land x \ne y$ | Define $I := \set {u \in S: \exists y \in S: \tuple {u, y} \in \RR \land \tuple {y, z} \in \RR}$
By definition of auxiliary relation:
:$\tuple {\bot, \bot} \in \RR$ and $\tuple {\bot, z} \in \RR$
where $\bot$ denotes the smallest element in $L$.
Then
:$\bot \in I$
By definition:
:$I$ is a non-empty set.
We will prove t... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $x, z \in S$ such that:
:$x \ll z \land x \ne z$
{{explain|$\ll$}}
Let $\RR$ be an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
Then
:$\exists ... | Define $I := \set {u \in S: \exists y \in S: \tuple {u, y} \in \RR \land \tuple {y, z} \in \RR}$
By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$\tuple {\bot, \bot} \in \RR$ and $\tuple {\bot, z} \in \RR$
where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
Then
:... | Auxiliary Approximating Relation has Interpolation Property | https://proofwiki.org/wiki/Auxiliary_Approximating_Relation_has_Interpolation_Property | https://proofwiki.org/wiki/Auxiliary_Approximating_Relation_has_Interpolation_Property | [
"Approximating Relations",
"Complete Lattices",
"Auxiliary Relations"
] | [
"Definition:Complete Lattice",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation"
] | [
"Definition:Auxiliary Relation",
"Definition:Smallest Element",
"Definition:Non-Empty Set",
"Definition:Lower Section",
"Definition:Auxiliary Relation",
"Definition:Reflexivity",
"Definition:Directed Subset",
"Auxiliary Relation is Congruent",
"Definition:Auxiliary Relation",
"Join Succeeds Operan... |
proofwiki-11876 | Way Below has Strong Interpolation Property | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Let $x, z \in S$ such that
:$x \ll z \land x \ne z$
Then
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$ | By Way Below is Approximating Relation and Way Below Relation is Auxiliary Relation:
:$\ll$ is an auxiliary approximating relation on $S$.
Thus by Auxiliary Approximating Relation has Interpolation Property:
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
{{qed}} | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Let $x, z \in S$ such that
:$x \ll z \land x \ne z$
Then
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$ | By [[Way Below is Approximating Relation]] and [[Way Below Relation is Auxiliary Relation]]:
:$\ll$ is an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
Thus by [[Auxiliary Approximating Relation has Interpolation Property]]:
:$\exists y \in S: x \ll y ... | Way Below has Strong Interpolation Property | https://proofwiki.org/wiki/Way_Below_has_Strong_Interpolation_Property | https://proofwiki.org/wiki/Way_Below_has_Strong_Interpolation_Property | [
"Approximating Relations",
"Continuous Lattices",
"Way Below Relation"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)"
] | [
"Way Below is Approximating Relation",
"Way Below Relation is Auxiliary Relation",
"Definition:Auxiliary Relation",
"Definition:Approximating Relation",
"Auxiliary Approximating Relation has Interpolation Property"
] |
proofwiki-11877 | Way Below has Interpolation Property | Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.
Let $x, z \in S$ such that
:$x \ll z$
Then
:$\exists y \in S: x \ll y \land y \ll z$ | Case $x \ne z$:
By Way Below has Strong Interpolation Property:
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
Thus
:$\exists y \in S: x \ll y \land y \ll z$
Case $x = z$:
Define $y = x$
Thus
:$x \ll y \land y \ll z$
{{qed}} | Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Let $x, z \in S$ such that
:$x \ll z$
Then
:$\exists y \in S: x \ll y \land y \ll z$ | Case $x \ne z$:
By [[Way Below has Strong Interpolation Property]]:
:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
Thus
:$\exists y \in S: x \ll y \land y \ll z$
Case $x = z$:
Define $y = x$
Thus
:$x \ll y \land y \ll z$
{{qed}} | Way Below has Interpolation Property | https://proofwiki.org/wiki/Way_Below_has_Interpolation_Property | https://proofwiki.org/wiki/Way_Below_has_Interpolation_Property | [
"Continuous Lattices",
"Way Below Relation"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)"
] | [
"Way Below has Strong Interpolation Property"
] |
proofwiki-11878 | Left Coset Equals Subgroup iff Element in Subgroup | :$x H = H \iff x \in H$ | {{begin-eqn}}
{{eqn | l = x H
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x H
| r = e H
| c = Left Coset by Identity: $e H = H$
}}
{{eqn | ll= \leadstoandfrom
| l = x e^{-1}
| o = \in
| r = H
| c = Left Cosets are Equal iff Product with Inverse in Subgroup
}... | :$x H = H \iff x \in H$ | {{begin-eqn}}
{{eqn | l = x H
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x H
| r = e H
| c = [[Left Coset by Identity]]: $e H = H$
}}
{{eqn | ll= \leadstoandfrom
| l = x e^{-1}
| o = \in
| r = H
| c = [[Left Cosets are Equal iff Product with Inverse in Subg... | Left Coset Equals Subgroup iff Element in Subgroup | https://proofwiki.org/wiki/Left_Coset_Equals_Subgroup_iff_Element_in_Subgroup | https://proofwiki.org/wiki/Left_Coset_Equals_Subgroup_iff_Element_in_Subgroup | [
"Cosets"
] | [] | [
"Left Coset by Identity",
"Left Cosets are Equal iff Product with Inverse in Subgroup",
"Definition:Group"
] |
proofwiki-11879 | Right Coset Equals Subgroup iff Element in Subgroup | :$H x = H \iff x \in H$ | {{begin-eqn}}
{{eqn | l = H x
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = H x
| r = H e
| c = Right Coset by Identity: $H = H e$
}}
{{eqn | ll= \leadstoandfrom
| l = x e^{-1}
| o = \in
| r = H
| c = Right Cosets are Equal iff Product with Inverse in Subgroup... | :$H x = H \iff x \in H$ | {{begin-eqn}}
{{eqn | l = H x
| r = H
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = H x
| r = H e
| c = [[Right Coset by Identity]]: $H = H e$
}}
{{eqn | ll= \leadstoandfrom
| l = x e^{-1}
| o = \in
| r = H
| c = [[Right Cosets are Equal iff Product with Inverse in Su... | Right Coset Equals Subgroup iff Element in Subgroup | https://proofwiki.org/wiki/Right_Coset_Equals_Subgroup_iff_Element_in_Subgroup | https://proofwiki.org/wiki/Right_Coset_Equals_Subgroup_iff_Element_in_Subgroup | [
"Cosets"
] | [] | [
"Right Coset by Identity",
"Right Cosets are Equal iff Product with Inverse in Subgroup",
"Definition:Group"
] |
proofwiki-11880 | Elements in Same Left Coset iff Product with Inverse in Subgroup | : $x, y$ are in the same left coset of $H$ {{iff}} $x^{-1} y \in H$. | Let: $x H$ denote the left coset of $H$ by $x$
First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the left cosets of $G$ form a partition of $G$. | : $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$ {{iff}} $x^{-1} y \in H$. | Let: $x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$
First we note that, from [[Congruence Class Modulo Subgroup is Coset]], we have that the [[Definition:Left Coset|left cosets]] of $G$ form a [[Definition:Partition (Set Theory)|partition]] of $G$. | Elements in Same Left Coset iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Elements_in_Same_Left_Coset_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Elements_in_Same_Left_Coset_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Left Coset"
] | [
"Definition:Coset/Left Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Coset/Left Coset",
"Definition:Set Partition",
"Definition:Coset/Left Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Coset/Left Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:C... |
proofwiki-11881 | Elements in Same Right Coset iff Product with Inverse in Subgroup | : $x, y$ are in the same right coset of $H$ {{iff}} $x y^{-1} \in H$ | Let $H x$ denote the right coset of $H$ by $x$.
First we note that, from Congruence Class Modulo Subgroup is Coset, we have that the right cosets of $G$ form a partition of $G$. | : $x, y$ are in the same [[Definition:Right Coset|right coset]] of $H$ {{iff}} $x y^{-1} \in H$ | Let $H x$ denote the [[Definition:Right Coset|right coset]] of $H$ by $x$.
First we note that, from [[Congruence Class Modulo Subgroup is Coset]], we have that the [[Definition:Right Coset|right cosets]] of $G$ form a [[Definition:Partition (Set Theory)|partition]] of $G$. | Elements in Same Right Coset iff Product with Inverse in Subgroup | https://proofwiki.org/wiki/Elements_in_Same_Right_Coset_iff_Product_with_Inverse_in_Subgroup | https://proofwiki.org/wiki/Elements_in_Same_Right_Coset_iff_Product_with_Inverse_in_Subgroup | [
"Cosets"
] | [
"Definition:Coset/Right Coset"
] | [
"Definition:Coset/Right Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Coset/Right Coset",
"Definition:Set Partition",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Coset/Right Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Coset/Right Coset"
] |
proofwiki-11882 | Sum over k of m+r+s Choose k by n+r-s Choose n-k by r+k Choose m+n | Let $m, n \in \Z_{\ge 0}$.
Then:
:$\ds \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n} = \binom r m \binom s n$ | {{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n}
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \paren { \sum_{j \mathop \in \Z} \binom r {m + n - j} \binom k j}
| c = Chu-Vand... | Let $m, n \in \Z_{\ge 0}$.
Then:
:$\ds \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n} = \binom r m \binom s n$ | {{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n}
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \paren { \sum_{j \mathop \in \Z} \binom r {m + n - j} \binom k j}
| c = [[Chu-Va... | Sum over k of m+r+s Choose k by n+r-s Choose n-k by r+k Choose m+n | https://proofwiki.org/wiki/Sum_over_k_of_m+r+s_Choose_k_by_n+r-s_Choose_n-k_by_r+k_Choose_m+n | https://proofwiki.org/wiki/Sum_over_k_of_m+r+s_Choose_k_by_n+r-s_Choose_n-k_by_r+k_Choose_m+n | [
"Binomial Coefficients"
] | [] | [
"Chu-Vandermonde Identity",
"Product of r Choose m with m Choose k",
"Chu-Vandermonde Identity",
"Symmetry Rule for Binomial Coefficients",
"Product of r Choose m with m Choose k",
"Chu-Vandermonde Identity"
] |
proofwiki-11883 | Sum over k of Unsigned Stirling Numbers of First Kind by x^k | :$\ds \sum_k {n \brack k} x^k = x^{\overline n}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$x^{\overline n}$ denotes $x$ to the $n$ rising. | {{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
| r = x^{\underline n}
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | ll= \leadsto
| l = \sum_k \paren {-1}^{n - k} {n \brack k} \paren {-x}^k
| r = \paren {-x}^{\underline n}
| c = putting $-x$ f... | :$\ds \sum_k {n \brack k} x^k = x^{\overline n}$
where:
:$\ds {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$x^{\overline n}$ denotes [[Definition:Rising Factorial|$x$ to the $n$ rising]]. | {{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
| r = x^{\underline n}
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | ll= \leadsto
| l = \sum_k \paren {-1}^{n - k} {n \brack k} \paren {-x}^k
| r = \paren {-x}^{\underline n}
| c = putting $-x$ f... | Sum over k of Unsigned Stirling Numbers of First Kind by x^k | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_First_Kind_by_x^k | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Numbers_of_First_Kind_by_x^k | [
"Stirling Numbers",
"Rising Factorials",
"Examples of Power Series"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Rising Factorial"
] | [
"Rising Factorial in terms of Falling Factorial of Negative"
] |
proofwiki-11884 | Capelli's Sum | :$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$x^{\overline k}$ denotes $x$ to the $k$ rising. | The proof proceeds by induction on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \paren {x + y}^{\overline 1}
| r = x... | :$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$x^{\overline k}$ denotes [[Definition:Rising Factorial|$x$ to the $k$ rising]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
=== Basis for the Induction ===
$\map P 1$ is the ca... | Capelli's Sum | https://proofwiki.org/wiki/Capelli's_Sum | https://proofwiki.org/wiki/Capelli's_Sum | [
"Rising Factorials"
] | [
"Definition:Binomial Coefficient",
"Definition:Rising Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Number to Power of One Rising is Itself",
"Number to Power of Zero Rising is One",
"One Choose n",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Capelli's Sum",
"Translation o... |
proofwiki-11885 | Torelli's Sum | :$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x \paren {x - k z + 1}^{\overline {k - 1} } \paren {y + k z}^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$x^{\overline k}$ denotes $x$ to the $k$ rising. | From Rising Factorial as Factorial by Binomial Coefficient:
:$\paren {x + y}^{\overline n} = n! \dbinom {x + y + n - 1} n$
Recall Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \paren {n - k} } {n - k}$ by $\dfrac r {r - t k}$:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n ... | :$\ds \paren {x + y}^{\overline n} = \sum_k \binom n k x \paren {x - k z + 1}^{\overline {k - 1} } \paren {y + k z}^{\overline {n - k} }$
where:
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$x^{\overline k}$ denotes [[Definition:Rising Factorial|$x$ to the $k$ rising]]. | From [[Rising Factorial as Factorial by Binomial Coefficient]]:
:$\paren {x + y}^{\overline n} = n! \dbinom {x + y + n - 1} n$
Recall [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk|Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \paren {n - k} } {n - k}$ by $\dfrac r {r - t k}$]]:
:$\d... | Torelli's Sum | https://proofwiki.org/wiki/Torelli's_Sum | https://proofwiki.org/wiki/Torelli's_Sum | [
"Rising Factorials"
] | [
"Definition:Binomial Coefficient",
"Definition:Rising Factorial"
] | [
"Rising Factorial as Factorial by Binomial Coefficient",
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk",
"Rising Factorial as Factorial by Binomial Coefficient",
"Rising Factorial as Factorial by Binomial Coefficient"
] |
proofwiki-11886 | Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element | Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded below continuous lattice.
Let $x, y$ be elements of $S$.
Then
:$x \ll y$
{{iff}}
:for every directed subset $D$ of $S$ such that $y \preceq \sup D$
::there exists an element $d$ of $D$: $x \ll d$ | === Sufficient Condition ===
Let $x \ll y$.
Let $D$ be a directed subset of $S$ such that
:$y \preceq \sup D$
By Way Below has Interpolation Property:
:$\exists x' \in S: x \ll x' \land x' \ll y$
By definition of way below relation:
:$\exists d \in D: x' \preceq d$
Thus by Preceding and Way Below implies Way Below and ... | Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Let $x, y$ be [[Definition:Element|elements]] of $S$.
Then
:$x \ll y$
{{iff}}
:for every [[Definition:Directed Subset|d... | === Sufficient Condition ===
Let $x \ll y$.
Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
:$y \preceq \sup D$
By [[Way Below has Interpolation Property]]:
:$\exists x' \in S: x \ll x' \land x' \ll y$
By definition of [[Definition:Element is Way Below|way below relation]]:
:$\exists d ... | Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element | https://proofwiki.org/wiki/Way_Below_iff_Second_Operand_Preceding_Supremum_of_Directed_Set_There_Exists_Element_of_Directed_Set_First_Operand_Way_Below_Element | https://proofwiki.org/wiki/Way_Below_iff_Second_Operand_Preceding_Supremum_of_Directed_Set_There_Exists_Element_of_Directed_Set_First_Operand_Way_Below_Element | [
"Way Below Relation",
"Continuous Lattices"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)",
"Definition:Element",
"Definition:Directed Subset",
"Definition:Element"
] | [
"Definition:Directed Subset",
"Way Below has Interpolation Property",
"Definition:Element is Way Below",
"Preceding and Way Below implies Way Below",
"Definition:Reflexivity",
"Definition:Directed Subset",
"Definition:Element",
"Definition:Directed Subset",
"Definition:Element",
"Definition:Elemen... |
proofwiki-11887 | Continuous iff Way Below Closure is Ideal and Element Precedes Supremum | Let $L = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.
Then
:$L$ is continuous
{{iff}}
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and:
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
where $x^\ll$ denotes the way below closure of $x$. | === Sufficient Condition ===
Let $L$ be continuous.
Let $x \in S$.
By definition of continuous:
:$x^\ll$ is directed.
By Way Below Closure is Lower Section:
:$x^\ll$ is a lower section.
Thus by definition:
:$x^\ll$ is an ideal in $L$.
By definition of continuous:
:$L$ satisfies the axiom of approximation.
By the axiom ... | Let $L = \struct {S, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
Then
:$L$ is [[Definition:Continuous Ordered Set|continuous]]
{{iff}}
:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \map \sup {x^\ll}$ ... | === Sufficient Condition ===
Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
Let $x \in S$.
By definition of [[Definition:Continuous Ordered Set|continuous]]:
:$x^\ll$ is [[Definition:Directed Subset|directed]].
By [[Way Below Closure is Lower Section]]:
:$x^\ll$ is a [[Definition:Lower Section|lower s... | Continuous iff Way Below Closure is Ideal and Element Precedes Supremum | https://proofwiki.org/wiki/Continuous_iff_Way_Below_Closure_is_Ideal_and_Element_Precedes_Supremum | https://proofwiki.org/wiki/Continuous_iff_Way_Below_Closure_is_Ideal_and_Element_Precedes_Supremum | [
"Continuous Lattices",
"Way Below Relation",
"Up-Complete Semilattices"
] | [
"Definition:Up-Complete",
"Definition:Meet Semilattice",
"Definition:Continuous Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Way Below Closure"
] | [
"Definition:Continuous Ordered Set",
"Definition:Continuous Ordered Set",
"Definition:Directed Subset",
"Way Below Closure is Lower Section",
"Definition:Lower Section",
"Definition:Ideal in Ordered Set",
"Definition:Continuous Ordered Set",
"Axiom:Axiom of Approximation",
"Axiom:Axiom of Approximat... |
proofwiki-11888 | Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed | Let $\OO^n$ and $\CC^n$ be the collections of open and closed subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
where $\sigma$ denotes generated $\sigma$-algebra. | Recall that a closed set is by definition the relative complement of an open set.
Hence Sigma-Algebra Generated by Complements of Generators applies to yield:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
{{qed}} | Let $\OO^n$ and $\CC^n$ be the collections of [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
where $\si... | Recall that a [[Definition:Closed Set (Topology)|closed set]] is by definition the [[Definition:Relative Complement|relative complement]] of an [[Definition:Open Set (Topology)|open set]].
Hence [[Sigma-Algebra Generated by Complements of Generators]] applies to yield:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
{{q... | Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Open_equals_Closed | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Open_equals_Closed | [
"Characterization of Euclidean Borel Sigma-Algebra"
] | [
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Subset",
"Definition:Euclidean Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Definition:Closed Set/Topology",
"Definition:Relative Complement",
"Definition:Open Set/Topology",
"Sigma-Algebra Generated by Complements of Generators"
] |
proofwiki-11889 | Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact | Let $\CC^n$ and $\KK^n$ be the collections of closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\CC^n} = \map \sigma {\KK^n}$
where $\sigma$ denotes generated $\sigma$-algebra. | By the Heine-Borel Theorem:
:$\KK^n \subseteq \CC^n$.
Thus from Generated Sigma-Algebra Preserves Subset:
:$\map \sigma {\KK^n} \subseteq \map \sigma {\CC^n}$.
Next, let $\map {B_n^-} {\mathbf 0}$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$ for all $n \in \N$.
Observe that:
:$\R^n = \ds \bigcup_{n \ma... | Let $\CC^n$ and $\KK^n$ be the collections of [[Definition:Closed Set (Topology)|closed]] and [[Definition:Compact (Real Analysis)|compact]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\CC^n} = \map \sigma {\KK^n}$
wh... | By the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]]:
:$\KK^n \subseteq \CC^n$.
Thus from [[Generated Sigma-Algebra Preserves Subset]]:
:$\map \sigma {\KK^n} \subseteq \map \sigma {\CC^n}$.
Next, let $\map {B_n^-} {\mathbf 0}$ be the [[Definition:Closed Ball|closed ball]] of radius $n$ around $\mathbf ... | Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Closed_equals_Compact | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Closed_equals_Compact | [
"Characterization of Euclidean Borel Sigma-Algebra"
] | [
"Definition:Closed Set/Topology",
"Definition:Compact Space/Real Analysis",
"Definition:Subset",
"Definition:Euclidean Space",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Heine-Borel Theorem/Euclidean Space",
"Generated Sigma-Algebra Preserves Subset",
"Definition:Closed Ball",
"Definition:Closed Set/Topology",
"Intersection with Subset is Subset",
"Intersection Distributes over Union",
"Intersection of Closed Sets is Closed/Topology",
"Definition:Closed Set/Topology"... |
proofwiki-11890 | Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle | Let $\OO^n$ be the collection of open subsets of the Euclidean space $\struct {\R^n, \tau}$.
Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Then:
:$\map \sigma {\OO^n} = \map \sigma {\JJ_{ho}^n}$
where $\sigma$ denotes generated $\sigma$-algebra. | Let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$.
Then:
:$\horectr {\mathbf a} {\mathbf b} = \paren {\openint \gets {\mathbf b} } \cap \horectr {\mathbf a} \to$
provides a way of writing this half-open $n$-rectangle as an intersection of an open and a closed set.
By Characterization of Euclidean Borel Sigma-Algebr... | Let $\OO^n$ be the collection of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\struct {\R^n, \tau}$.
Let $\JJ_{ho}^n$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$.
Then:
:$\map \sigma {\OO^n} = ... | Let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$.
Then:
:$\horectr {\mathbf a} {\mathbf b} = \paren {\openint \gets {\mathbf b} } \cap \horectr {\mathbf a} \to$
provides a way of writing this [[Definition:Half-Open Rectangle|half-open $n$-rectangle]] as an [[Definition:Set Intersection|intersection]] of an [[De... | Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Open_equals_Rectangle | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Open_equals_Rectangle | [
"Characterization of Euclidean Borel Sigma-Algebra"
] | [
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Euclidean Space",
"Definition:Half-Open Rectangle",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Definition:Half-Open Rectangle",
"Definition:Set Intersection",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed",
"Sigma-Algebra Closed under Finite Intersection",
"Definition:Sigma-Algebra Generated by Collection o... |
proofwiki-11891 | Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle | Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Let $\JJ^n_{ho, \text{rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.
Then:
:$\map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text{rat} } }$
where $\sigma$ denotes generated $\sigma$-algebra. | From Generated Sigma-Algebra Preserves Subset:
:$\map \sigma {\JJ_{ho, \text{rat} }^n} \subseteq \map \sigma {\JJ_{ho}^n}$
For the converse, it will suffice to show:
:$\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho, \text{rat} }^n}$
by definition of generated $\sigma$-algebra.
So, let $\horectr {\mathbf a} {\mathbf b}$ be a... | Let $\JJ_{ho}^n$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$.
Let $\JJ^n_{ho, \text{rat} }$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$ with [[Definition:Rational Number|rational]] endpoints.
Then:
:$\map \sigma {\JJ_{ho}^n} = \ma... | From [[Generated Sigma-Algebra Preserves Subset]]:
:$\map \sigma {\JJ_{ho, \text{rat} }^n} \subseteq \map \sigma {\JJ_{ho}^n}$
For the converse, it will suffice to show:
:$\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho, \text{rat} }^n}$
by definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|gener... | Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Rectangle_equals_Rational_Rectangle | https://proofwiki.org/wiki/Characterization_of_Euclidean_Borel_Sigma-Algebra/Rectangle_equals_Rational_Rectangle | [
"Characterization of Euclidean Borel Sigma-Algebra"
] | [
"Definition:Half-Open Rectangle",
"Definition:Half-Open Rectangle",
"Definition:Rational Number",
"Definition:Sigma-Algebra Generated by Collection of Subsets"
] | [
"Generated Sigma-Algebra Preserves Subset",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Half-Open Rectangle",
"Definition:Sequence",
"Definition:Limit of Sequence",
"Definition:Sequence",
"Definition:Componentwise Ordering",
"Definition:Componentwise Ordering",
"Sigma-... |
proofwiki-11892 | Rising Factorial as Factorial by Binomial Coefficient | Let $x \in \R$ be a real number.
Let $n \in \Z_{\ge 0}$ be a positive integer.
:$x^{\overline n} = n! \dbinom {x + n - 1} n$
where:
:$x^{\overline n}$ denotes $x$ to the $n$ rising
:$n!$ denotes the factorial of $n$
:$\dbinom {x + n - 1} n$ denotes a binomial coefficient. | By definition of $x$ to the $n$ rising:
:$x^{\overline n} = x \paren {x + 1} \cdots \paren {x + n - 1}$
By definition of the binomial coefficient of a real number:
:$\dbinom {x + n - 1} n = \dfrac {\paren {x + n - 1} \paren {x + n - 2} \cdots x} {n!}$
Hence the result.
{{qed}}
Category:Binomial Coefficients
Category:Fa... | Let $x \in \R$ be a [[Definition:Real Number|real number]].
Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
:$x^{\overline n} = n! \dbinom {x + n - 1} n$
where:
:$x^{\overline n}$ denotes [[Definition:Rising Factorial|$x$ to the $n$ rising]]
:$n!$ denotes the [[Definition:Factorial|factor... | By definition of [[Definition:Rising Factorial|$x$ to the $n$ rising]]:
:$x^{\overline n} = x \paren {x + 1} \cdots \paren {x + n - 1}$
By definition of the [[Definition:Binomial Coefficient/Real Numbers|binomial coefficient of a real number]]:
:$\dbinom {x + n - 1} n = \dfrac {\paren {x + n - 1} \paren {x + n - 2} \c... | Rising Factorial as Factorial by Binomial Coefficient | https://proofwiki.org/wiki/Rising_Factorial_as_Factorial_by_Binomial_Coefficient | https://proofwiki.org/wiki/Rising_Factorial_as_Factorial_by_Binomial_Coefficient | [
"Binomial Coefficients",
"Factorials",
"Rising Factorials"
] | [
"Definition:Real Number",
"Definition:Positive/Integer",
"Definition:Rising Factorial",
"Definition:Factorial",
"Definition:Binomial Coefficient"
] | [
"Definition:Rising Factorial",
"Definition:Binomial Coefficient/Real Numbers",
"Category:Binomial Coefficients",
"Category:Factorials",
"Category:Rising Factorials"
] |
proofwiki-11893 | Ramus's Identity | Let $k, m, n \in \Z_{\ge 0}$ be positive integers such that $0 \le k < m$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop \ge 0} \binom n {j m + k}
| r = \dbinom n k + \dbinom n {m + k} + \dbinom n {2 m + k} + \cdots
| c =
}}
{{eqn | r = \dfrac 1 m \sum_{0 \mathop \le j \mathop < m} \paren {2 \cos \dfra... | Let $\omega := e^{2 \pi i / m}$.
Then by the Binomial Theorem:
:$(1): \quad \ds \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k} = \sum_t \sum_{0 \mathop \le j \mathop < m} \binom n t \omega^{j \paren {t - k} }$
By Sum of Geometric Sequence:
:$\ds \sum_{0 \mathop \ge j \mathop < m} \omega^{j \pa... | Let $k, m, n \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]] such that $0 \le k < m$.
Then:
{{begin-eqn}}
{{eqn | l = \sum_{j \mathop \ge 0} \binom n {j m + k}
| r = \dbinom n k + \dbinom n {m + k} + \dbinom n {2 m + k} + \cdots
| c =
}}
{{eqn | r = \dfrac 1 m \sum_{0 \mathop \le ... | Let $\omega := e^{2 \pi i / m}$.
Then by the [[Binomial Theorem]]:
:$(1): \quad \ds \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k} = \sum_t \sum_{0 \mathop \le j \mathop < m} \binom n t \omega^{j \paren {t - k} }$
By [[Sum of Geometric Sequence]]:
:$\ds \sum_{0 \mathop \ge j \mathop < m} \... | Ramus's Identity | https://proofwiki.org/wiki/Ramus's_Identity | https://proofwiki.org/wiki/Ramus's_Identity | [
"Binomial Coefficients",
"Ramus's Identity"
] | [
"Definition:Positive/Integer"
] | [
"Binomial Theorem",
"Sum of Geometric Sequence",
"Sum of Geometric Sequence",
"Definition:Iverson's Convention",
"Definition:Summation",
"Definition:Summation",
"Definition:Complex Number/Wholly Real",
"Definition:Complex Number/Real Part"
] |
proofwiki-11894 | Summation over Lower Index of Unsigned Stirling Numbers of the First Kind | Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\ds \sum_k {n \brack k} = n!$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$n!$ denotes the factorial of $n$. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P N$ be the proposition:
:$\ds \sum_k {n \brack k} = n!$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k {0 \brack k}
| r = \sum_k \delta_{0 k}
| c = Unsigned Stirling Number of the First Kind of 0
}}
{{eqn | r = 1
... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\ds \sum_k {n \brack k} = n!$
where:
:$\ds {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$n!$ denotes the [[Definition:Factorial|factorial of $n$]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P N$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_k {n \brack k} = n!$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k {0 \brack k}
| r = \sum_k \delta_{0 k}
| c =... | Summation over Lower Index of Unsigned Stirling Numbers of the First Kind | https://proofwiki.org/wiki/Summation_over_Lower_Index_of_Unsigned_Stirling_Numbers_of_the_First_Kind | https://proofwiki.org/wiki/Summation_over_Lower_Index_of_Unsigned_Stirling_Numbers_of_the_First_Kind | [
"Stirling Numbers",
"Factorials"
] | [
"Definition:Positive/Integer",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Unsigned Stirling Number of the First Kind of 0",
"Principle of Mathematical Induction"
] |
proofwiki-11895 | Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs | Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\delta_{n 0}$ denotes the Kronecker delta. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^k {0 \brack k}
| r = \sum_k \delta_{0 k}
| c = Unsigned Stirling ... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Then:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
where:
:$\ds {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$\delta_{n 0}$ denotes... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \paren {-1}^k {0 ... | Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs | https://proofwiki.org/wiki/Summation_over_Lower_Index_of_Unsigned_Stirling_Numbers_of_the_First_Kind_with_Alternating_Signs | https://proofwiki.org/wiki/Summation_over_Lower_Index_of_Unsigned_Stirling_Numbers_of_the_First_Kind_with_Alternating_Signs | [
"Stirling Numbers",
"Factorials"
] | [
"Definition:Positive/Integer",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Kronecker Delta"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Unsigned Stirling Number of the First Kind of 0",
"Unsigned Stirling Number of the First Kind of 1",
"Principle of Mathematical Induction"
] |
proofwiki-11896 | Beta Function of Real Number with 1 | :$\map \Beta {x, 1} = \map \Beta {1, x} = \dfrac 1 x$ | By definition:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, 1}
| r = \int_0^1 t^{x - 1} \paren {1 - t}^{1 - 1} \rd t
| c = {{Defof|Beta Function}}
}}
{{eqn | r = \int_0^1 t^{x - 1} \rd t
| c =
}}
{{eqn | r = \int_0^1 t^{x - 1} \rd t
| c =
}}
{{eqn | r = \intlimits {\frac {t^x} x} 0 1
| c =
... | :$\map \Beta {x, 1} = \map \Beta {1, x} = \dfrac 1 x$ | By definition:
{{begin-eqn}}
{{eqn | l = \map \Beta {x, 1}
| r = \int_0^1 t^{x - 1} \paren {1 - t}^{1 - 1} \rd t
| c = {{Defof|Beta Function}}
}}
{{eqn | r = \int_0^1 t^{x - 1} \rd t
| c =
}}
{{eqn | r = \int_0^1 t^{x - 1} \rd t
| c =
}}
{{eqn | r = \intlimits {\frac {t^x} x} 0 1
| c = ... | Beta Function of Real Number with 1 | https://proofwiki.org/wiki/Beta_Function_of_Real_Number_with_1 | https://proofwiki.org/wiki/Beta_Function_of_Real_Number_with_1 | [
"Beta Function"
] | [] | [
"Integration by Substitution"
] |
proofwiki-11897 | Way Below Closure is Lower Section | Let $L = \struct {S, \vee, \preceq}$ be an ordered set.
Let $x \in S$.
Then
:$x^\ll$ is a lower section. | Let $y \in x^\ll, z \in S$ such that:
:$z \preceq y$
By definition of way below closure:
:$y \ll x$
By definition of reflexivity:
:$x \preceq x$
By Preceding and Way Below implies Way Below:
:$z \ll x$
Thus by definition of way below closure:
:$z \in x^\ll$
Thus by definition:
:$x^\ll$ is a lower section.
{{qed}} | Let $L = \struct {S, \vee, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x \in S$.
Then
:$x^\ll$ is a [[Definition:Lower Section|lower section]]. | Let $y \in x^\ll, z \in S$ such that:
:$z \preceq y$
By definition of [[Definition:Way Below Closure|way below closure]]:
:$y \ll x$
By definition of [[Definition:Reflexivity|reflexivity]]:
:$x \preceq x$
By [[Preceding and Way Below implies Way Below]]:
:$z \ll x$
Thus by definition of [[Definition:Way Below Closu... | Way Below Closure is Lower Section | https://proofwiki.org/wiki/Way_Below_Closure_is_Lower_Section | https://proofwiki.org/wiki/Way_Below_Closure_is_Lower_Section | [
"Join and Meet Semilattices",
"Way Below Relation",
"Lower Sections"
] | [
"Definition:Ordered Set",
"Definition:Lower Section"
] | [
"Definition:Way Below Closure",
"Definition:Reflexivity",
"Preceding and Way Below implies Way Below",
"Definition:Way Below Closure",
"Definition:Lower Section"
] |
proofwiki-11898 | Continuous iff For Every Element There Exists Ideal Element Precedes Supremum | Let $L = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.
Then
:$L$ is continuous
{{iff}}
:for every element $x$ of $S$ there exists ideal $I$ in $L$:
::$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$ | === Sufficient Condition ===
Let $L$ be continuous.
By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:
:$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and
::for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
where $x^\ll$ denotes the way bel... | Let $L = \struct {S, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
Then
:$L$ is [[Definition:Continuous Ordered Set|continuous]]
{{iff}}
:for every [[Definition:Element|element]] $x$ of $S$ there exists [[Definition:Ideal in Ordered Set|ideal]] $I$ in ... | === Sufficient Condition ===
Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
By [[Continuous iff Way Below Closure is Ideal and Element Precedes Supremum]]:
:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \map \sup {x^\ll}$ and
::for every [[Definition:Idea... | Continuous iff For Every Element There Exists Ideal Element Precedes Supremum | https://proofwiki.org/wiki/Continuous_iff_For_Every_Element_There_Exists_Ideal_Element_Precedes_Supremum | https://proofwiki.org/wiki/Continuous_iff_For_Every_Element_There_Exists_Ideal_Element_Precedes_Supremum | [
"Continuous Lattices",
"Up-Complete Semilattices"
] | [
"Definition:Up-Complete",
"Definition:Meet Semilattice",
"Definition:Continuous Ordered Set",
"Definition:Element",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set"
] | [
"Definition:Continuous Ordered Set",
"Continuous iff Way Below Closure is Ideal and Element Precedes Supremum",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Way Below Closure",
"Definition:Element",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Se... |
proofwiki-11899 | Beta Function of x+1 with y plus Beta Function of x with y+1 | :$\map \Beta {x + 1, y} + \map \Beta {x, y + 1} = \map \Beta {x, y}$ | By definition:
{{begin-eqn}}
{{eqn | l = \map \Beta {x + 1, y} + \map \Beta {x, y + 1}
| r = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t + \int_0^1 t^{x - 1} \paren {1 - t}^y \rd t
| c = {{Defof|Beta Function}}
}}
{{eqn | r = \int_0^1 \paren {t^x \paren {1 - t}^{y - 1} + t^{x - 1} \paren {1 - t}^y} \rd t
}}
{... | :$\map \Beta {x + 1, y} + \map \Beta {x, y + 1} = \map \Beta {x, y}$ | By definition:
{{begin-eqn}}
{{eqn | l = \map \Beta {x + 1, y} + \map \Beta {x, y + 1}
| r = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t + \int_0^1 t^{x - 1} \paren {1 - t}^y \rd t
| c = {{Defof|Beta Function}}
}}
{{eqn | r = \int_0^1 \paren {t^x \paren {1 - t}^{y - 1} + t^{x - 1} \paren {1 - t}^y} \rd t
}}
... | Beta Function of x+1 with y plus Beta Function of x with y+1 | https://proofwiki.org/wiki/Beta_Function_of_x+1_with_y_plus_Beta_Function_of_x_with_y+1 | https://proofwiki.org/wiki/Beta_Function_of_x+1_with_y_plus_Beta_Function_of_x_with_y+1 | [
"Beta Function"
] | [] | [] |
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