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-2500 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | We are seeking to calculate the number of $r$-permutations of $S$, that is ${}^n P_r$, where $r = n$.
Hence:
{{begin-eqn}}
{{eqn | l = {}^n P_n
| r = \dfrac {n!} {\paren {n - n}!}
| c = Number of Permutations
}}
{{eqn | r = n!
| c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}} | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | We are seeking to calculate the number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]], that is ${}^n P_r$, where $r = n$.
Hence:
{{begin-eqn}}
{{eqn | l = {}^n P_n
| r = \dfrac {n!} {\paren {n - n}!}
| c = [[Number of Permutations]]
}}
{{eqn | r = n!
| c = {{Defof|Factoria... | Number of Permutations of All Elements/Proof 1 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations_of_All_Elements/Proof_1 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Definition:Permutation/Ordered Selection",
"Number of Permutations"
] |
proofwiki-2501 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | We pick the elements of $S$ in an arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we have not picked, so there are $n - 1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options f... | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | We pick the [[Definition:Element|elements]] of $S$ in an arbitrary order.
There are $n$ [[Definition:Element|elements]] of $S$, so there are $n$ options for the first [[Definition:Element|element]].
Then there are $n - 1$ [[Definition:Element|elements]] left in $S$ that we have not picked, so there are $n - 1$ option... | Number of Permutations of All Elements/Proof 2 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations_of_All_Elements/Proof_2 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Definition:Element",
"Product Rule for Counting"
] |
proofwiki-2502 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | From the definition, it can be seen that a bijection $f: S \to S$ is an $n$-permutation.
Hence, from Cardinality of Set of Bijections the number of $n$-permutations on a set of $n$ elements is:
:${}^n P_n = \dfrac {n!} {\paren {n - n}!} = n!$
{{qed}} | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | From the definition, it can be seen that a [[Definition:Bijection|bijection]] $f: S \to S$ is an [[Definition:Permutation (Ordered Selection)|$n$-permutation]].
Hence, from [[Cardinality of Set of Bijections]] the number of [[Definition:Permutation (Ordered Selection)|$n$-permutations]] on a [[Definition:Set|set]] of ... | Number of Permutations of All Elements/Proof 3 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations_of_All_Elements/Proof_3 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Definition:Bijection",
"Definition:Permutation/Ordered Selection",
"Cardinality of Set of Bijections",
"Definition:Permutation/Ordered Selection",
"Definition:Set",
"Definition:Element"
] |
proofwiki-2503 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | {{begin-eqn}}
{{eqn | l = {}^{n - 1} P_n
| r = \dfrac {n!} {\paren {n - \paren {n - 1} }!}
| c = Number of Permutations
}}
{{eqn | r = \dfrac {n!} {1!}
| c =
}}
{{eqn | r = n!
| c =
}}
{{eqn | r = {}^n P_n
| c = Number of Permutations
}}
{{end-eqn}}
{{qed}} | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | {{begin-eqn}}
{{eqn | l = {}^{n - 1} P_n
| r = \dfrac {n!} {\paren {n - \paren {n - 1} }!}
| c = [[Number of Permutations]]
}}
{{eqn | r = \dfrac {n!} {1!}
| c =
}}
{{eqn | r = n!
| c =
}}
{{eqn | r = {}^n P_n
| c = [[Number of Permutations]]
}}
{{end-eqn}}
{{qed}} | Number of Permutations of One Less/Proof 1 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations_of_One_Less/Proof_1 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Number of Permutations",
"Number of Permutations"
] |
proofwiki-2504 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | {{begin-eqn}}
{{eqn | l = {}^{n - 1} P_n
| r = n^{\underline {n - 1} }
| c = Number of Permutations: $n^{\underline {n - 1} }$ denotes Falling Factorial
}}
{{eqn | r = n!
| c = Integer to Power of Itself Less One Falling is Factorial
}}
{{eqn | r = {}^n P_n
| c = Number of Permutations
}}
{{end-... | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | {{begin-eqn}}
{{eqn | l = {}^{n - 1} P_n
| r = n^{\underline {n - 1} }
| c = [[Number of Permutations]]: $n^{\underline {n - 1} }$ denotes [[Definition:Falling Factorial|Falling Factorial]]
}}
{{eqn | r = n!
| c = [[Integer to Power of Itself Less One Falling is Factorial]]
}}
{{eqn | r = {}^n P_n
... | Number of Permutations of One Less/Proof 2 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations_of_One_Less/Proof_2 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Number of Permutations",
"Definition:Falling Factorial",
"Integer to Power of Itself Less One Falling is Factorial",
"Number of Permutations"
] |
proofwiki-2505 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | We pick the elements of $S$ in any arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we haven't picked, so there are $n-1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options for... | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | We pick the elements of $S$ in any arbitrary order.
There are $n$ elements of $S$, so there are $n$ options for the first element.
Then there are $n - 1$ elements left in $S$ that we haven't picked, so there are $n-1$ options for the second element.
Then there are $n - 2$ elements left, so there are $n - 2$ options ... | Number of Permutations/Proof 1 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations/Proof_1 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Product Rule for Counting",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Definition:Fraction/Numerator"
] |
proofwiki-2506 | Number of Permutations | Let $S$ be a set of $n$ elements.
Let $r \in \N: r \le n$.
The number of $r$-permutations of $S$ is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the falling factorial symbol, this can also be expressed:
:${}^n P_r = n^{\underline r}$ | By definition, an $r$-permutation of $S$ is an ordered selection of $r$ elements of $S$.
Also, an $r$-permutation of $S$ is {{afortiori}} an injection from a subset of $S$ into $S$.
The result is immediate from Cardinality of Set of Injections.
{{qed}} | Let $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
Let $r \in \N: r \le n$.
The number of [[Definition:Permutation (Ordered Selection)|$r$-permutations of $S$]] is:
:${}^n P_r = \dfrac {n!} {\paren {n - r}!}$
Using the [[Definition:Falling Factorial|falling factorial]] symbol, this can als... | By definition, an [[Definition:Permutation (Ordered Selection)|$r$-permutation of $S$]] is an ordered selection of $r$ [[Definition:Element|elements]] of $S$.
Also, an [[Definition:Permutation (Ordered Selection)|$r$-permutation of $S$]] is {{afortiori}} an [[Definition:Injection|injection]] from a [[Definition:Subset... | Number of Permutations/Proof 2 | https://proofwiki.org/wiki/Number_of_Permutations | https://proofwiki.org/wiki/Number_of_Permutations/Proof_2 | [
"Number of Permutations",
"Permutations (Ordered Selections)",
"Combinatorics"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Falling Factorial"
] | [
"Definition:Permutation/Ordered Selection",
"Definition:Element",
"Definition:Permutation/Ordered Selection",
"Definition:Injection",
"Definition:Subset",
"Cardinality of Set of Injections"
] |
proofwiki-2507 | Construction of Permutations | The ${}^n P_n$ permutations of $n$ objects can be generated algorithmically.
By Number of Permutations, that number is given by:
:${}^n P_n = n!$
where $n!$ denotes the factorial of $n$.
This will be demonstrated to hold. | The following is an inductive method of creating all the permutations of $n$ objects.
=== Base Case ===
There is clearly one way to arrange one object in order.
=== Inductive Hypothesis ===
We assume that we have constructed all $n!$ permutations of $n$ objects.
=== Induction Step ===
{{WLOG}}, let a set $S_n$ of $n$ o... | The ${}^n P_n$ [[Definition:Permutation (Ordered Selection)|permutations]] of $n$ [[Definition:Object|objects]] can be generated [[Definition:Algorithm|algorithmically]].
By [[Number of Permutations]], that number is given by:
:${}^n P_n = n!$
where $n!$ denotes the [[Definition:Factorial|factorial]] of $n$.
This wi... | The following is an [[Principle of Mathematical Induction|inductive]] method of creating all the [[Definition:Permutation (Ordered Selection)|permutations]] of $n$ [[Definition:Object|objects]].
=== Base Case ===
There is clearly one way to arrange one [[Definition:Object|object]] in order.
=== Inductive Hypothesi... | Construction of Permutations/Proof 1 | https://proofwiki.org/wiki/Construction_of_Permutations | https://proofwiki.org/wiki/Construction_of_Permutations/Proof_1 | [
"Construction of Permutations",
"Permutations (Ordered Selections)",
"Counting Arguments"
] | [
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Algorithm",
"Number of Permutations",
"Definition:Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Object",
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Set",
"Definition:Object",
"Definition:Permutation/Ordered Selection",
"Definition:Permutation/O... |
proofwiki-2508 | Construction of Permutations | The ${}^n P_n$ permutations of $n$ objects can be generated algorithmically.
By Number of Permutations, that number is given by:
:${}^n P_n = n!$
where $n!$ denotes the factorial of $n$.
This will be demonstrated to hold. | The following is an inductive method of creating all the permutations of $n$ objects.
=== Base Case ===
There is clearly one way to arrange one object in order.
=== Inductive Hypothesis ===
We assume that we have constructed all $n!$ permutations of $n$ objects.
=== Induction Step ===
{{WLOG}}, let a set $S_n$ of $n$ o... | The ${}^n P_n$ [[Definition:Permutation (Ordered Selection)|permutations]] of $n$ [[Definition:Object|objects]] can be generated [[Definition:Algorithm|algorithmically]].
By [[Number of Permutations]], that number is given by:
:${}^n P_n = n!$
where $n!$ denotes the [[Definition:Factorial|factorial]] of $n$.
This wi... | The following is an [[Principle of Mathematical Induction|inductive]] method of creating all the [[Definition:Permutation (Ordered Selection)|permutations]] of $n$ [[Definition:Object|objects]].
=== Base Case ===
There is clearly one way to arrange one [[Definition:Object|object]] in order.
=== Inductive Hypothesi... | Construction of Permutations/Proof 2 | https://proofwiki.org/wiki/Construction_of_Permutations | https://proofwiki.org/wiki/Construction_of_Permutations/Proof_2 | [
"Construction of Permutations",
"Permutations (Ordered Selections)",
"Counting Arguments"
] | [
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Algorithm",
"Number of Permutations",
"Definition:Factorial"
] | [
"Principle of Mathematical Induction",
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Object",
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"Definition:Set",
"Definition:Object",
"Definition:Permutation/Ordered Selection",
"Definition:Object",
"D... |
proofwiki-2509 | De Polignac's Formula | Let $n!$ be the factorial of $n$.
Let $p$ be a prime number.
Then $p^\mu$ is a divisor of $n!$, and $p^{\mu + 1}$ is not, where:
:$\ds \mu = \sum_{k \mathop > 0} \floor {\frac n {p^k} }$
where $\floor {\, \cdot \,}$ denotes the floor function. | Note that although the summation given in the statement of the theorem is given as an infinite sum, in fact it terminates after a finite number of terms (because when $p^k > n$ we have $0 < n/p^k < 1$).
From Number of Multiples less than Given Number, we have that $\floor{\dfrac n {p^k} }$ is the number of integers $m$... | Let $n!$ be the [[Definition:Factorial|factorial]] of $n$.
Let $p$ be a [[Definition:Prime Number|prime number]].
Then $p^\mu$ is a [[Definition:Divisor of Integer|divisor]] of $n!$, and $p^{\mu + 1}$ is not, where:
:$\ds \mu = \sum_{k \mathop > 0} \floor {\frac n {p^k} }$
where $\floor {\, \cdot \,}$ denotes the [[... | Note that although the [[Definition:Summation|summation]] given in the statement of the theorem is given as an [[Definition:Infinite Summation|infinite sum]], in fact it terminates after a [[Definition:Finite Set|finite number]] of terms (because when $p^k > n$ we have $0 < n/p^k < 1$).
From [[Number of Multiples less... | De Polignac's Formula | https://proofwiki.org/wiki/De_Polignac's_Formula | https://proofwiki.org/wiki/De_Polignac's_Formula | [
"Factorials",
"Prime Decompositions",
"De Polignac's Formula"
] | [
"Definition:Factorial",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Floor Function"
] | [
"Definition:Summation",
"Definition:Summation/Infinite",
"Definition:Finite Set",
"Number of Multiples less than Given Number",
"Definition:Integer",
"Definition:Integer",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-2510 | Sum of Binomial Coefficients over Lower Index | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$
$\map P 0$ is true, as this just says $\dbinom 0 0 = 1$.
This holds by definition.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $\dbinom 1 0 + \dbinom 1 1 = 2$.
This holds by Binomial Coefficie... | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$
$\map P 0$ is true, as this just says $\dbinom 0 0 = 1$.
This holds by [[Definition:Binomial Coefficient|definition]].
=== Basis for the Induction ===
$\map P 1$ is true, as this just s... | Sum of Binomial Coefficients over Lower Index/Proof 1 | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index/Proof_1 | [
"Sum of Binomial Coefficients over Lower Index",
"Binomial Coefficients"
] | [] | [
"Definition:Proposition",
"Definition:Binomial Coefficient",
"Binomial Coefficient with Zero",
"Binomial Coefficient with One",
"Binomial Coefficient with Self",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Pascal's Rule",
"Translation of I... |
proofwiki-2511 | Sum of Binomial Coefficients over Lower Index | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | Let $S$ be a set with $n$ elements.
From the definition of $r$-combination, $\ds \sum_{i \mathop = 0}^n \binom n i$ is the total number of subsets of $S$.
Hence $\ds \sum_{i \mathop = 0}^n \binom n i$ is equal to the cardinality of the power set of $S$.
Hence the result.
{{qed}} | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | Let $S$ be a [[Definition:Set|set]] with $n$ [[Definition:Element|elements]].
From the definition of [[Definition:Combination|$r$-combination]], $\ds \sum_{i \mathop = 0}^n \binom n i$ is the total number of [[Definition:Subset|subsets]] of $S$.
Hence $\ds \sum_{i \mathop = 0}^n \binom n i$ is equal to the [[Cardinal... | Sum of Binomial Coefficients over Lower Index/Proof 2 | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index/Proof_2 | [
"Sum of Binomial Coefficients over Lower Index",
"Binomial Coefficients"
] | [] | [
"Definition:Set",
"Definition:Element",
"Definition:Combination",
"Definition:Subset",
"Cardinality of Power Set of Finite Set"
] |
proofwiki-2512 | Sum of Binomial Coefficients over Lower Index | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | From the Binomial Theorem, we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = y = 1$ we get:
{{begin-eqn}}
{{eqn | l = 2^n
| r = \paren {1 + 1}^n
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \binom n i 1^{n - i} 1^i
| c =
}}
... | :$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$ | From the [[Binomial Theorem]], we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = y = 1$ we get:
{{begin-eqn}}
{{eqn | l = 2^n
| r = \paren {1 + 1}^n
| c =
}}
{{eqn | r = \sum_{i \mathop = 0}^n \binom n i 1^{n - i} 1^i
| ... | Sum of Binomial Coefficients over Lower Index/Proof 3 | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index | https://proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Lower_Index/Proof_3 | [
"Sum of Binomial Coefficients over Lower Index",
"Binomial Coefficients"
] | [] | [
"Binomial Theorem"
] |
proofwiki-2513 | Alternating Sum and Difference of Binomial Coefficients for Given n | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | For $n > 0$:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren{-1}^i \binom n i
| r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \binom n i + \paren{-1}^n \binom n n
| c =
}}
{{eqn | r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \paren{\binom {n - 1} {i - 1} + \binom {n - 1} ... | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | For $n > 0$:
{{begin-eqn}}
{{eqn | l = \sum_{i \mathop = 0}^n \paren{-1}^i \binom n i
| r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \binom n i + \paren{-1}^n \binom n n
| c =
}}
{{eqn | r = \binom n 0 + \sum_{i \mathop = 1}^{n - 1} \paren{-1}^i \paren{\binom {n - 1} {i - 1} + \binom {n - 1} ... | Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 1 | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n/Proof_1 | [
"Alternating Sum and Difference of Binomial Coefficients for Given n",
"Binomial Coefficients"
] | [] | [
"Pascal's Rule",
"Telescoping Series/Example 1",
"Binomial Coefficient with Zero",
"Binomial Coefficient with Self"
] |
proofwiki-2514 | Alternating Sum and Difference of Binomial Coefficients for Given n | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | Let $n > 0$.
From the Binomial Theorem, we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = 1, y = -1$, we get:
{{begin-eqn}}
{{eqn | l = 0
| r = \paren {1 - 1}^n
| c = which holds for all $n > 0$
}}
{{eqn | r = \sum_{i \mathop = 0}... | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | Let $n > 0$.
From the [[Binomial Theorem]], we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = 1, y = -1$, we get:
{{begin-eqn}}
{{eqn | l = 0
| r = \paren {1 - 1}^n
| c = which holds for all $n > 0$
}}
{{eqn | r = \sum_{i \ma... | Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 2 | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n/Proof_2 | [
"Alternating Sum and Difference of Binomial Coefficients for Given n",
"Binomial Coefficients"
] | [] | [
"Binomial Theorem"
] |
proofwiki-2515 | Alternating Sum and Difference of Binomial Coefficients for Given n | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | Let $n > 0$.
The assertion can be expressed:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = 0$ for all $n > 0$
as $\dbinom n i = 0$ when $i < 0$ by definition of binomial coefficient.
From Alternating Sum and Difference of r Choose k up to n we have:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom r i = \par... | :$\ds \forall n \in \Z_{\geq 0}: \sum_{i \mathop = 0}^n \paren {-1}^i \binom n i = \delta_{n 0}$ | Let $n > 0$.
The assertion can be expressed:
:$\ds \sum_{i \mathop \le n} \paren {-1}^i \binom n i = 0$ for all $n > 0$
as $\dbinom n i = 0$ when $i < 0$ by definition of [[Definition:Binomial Coefficient|binomial coefficient]].
From [[Alternating Sum and Difference of r Choose k up to n]] we have:
:$\ds \sum_{i \ma... | Alternating Sum and Difference of Binomial Coefficients for Given n/Proof 3 | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_Binomial_Coefficients_for_Given_n/Proof_3 | [
"Alternating Sum and Difference of Binomial Coefficients for Given n",
"Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient",
"Alternating Sum and Difference of r Choose k up to n",
"Definition:Binomial Coefficient"
] |
proofwiki-2516 | Symmetry Rule for Binomial Coefficients | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | Follows directly from the definition of binomial coefficient, as follows.
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k = \dbinom n {n - k} = 0$
Let $0 \le k \le n$.
{{begin-eqn}}
{{eqn | l = \binom n k
| r = \frac {n!} {k! \paren {n - k}!}
| c =
}}
{{eq... | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | Follows directly from the definition of [[Definition:Binomial Coefficient|binomial coefficient]], as follows.
If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k = \dbinom n {n - k} = 0$
Let $0 \le k \le n$.
{{begin-eqn}}
{{eqn | l = \binom n k
| r = \frac {n!}... | Symmetry Rule for Binomial Coefficients/Proof 1 | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients/Proof_1 | [
"Symmetry Rule for Binomial Coefficients",
"Binomial Coefficients"
] | [] | [
"Definition:Binomial Coefficient"
] |
proofwiki-2517 | Symmetry Rule for Binomial Coefficients | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | From the definition of Cardinality of Set of Subsets, $\dbinom n k$ is the number of subsets of cardinality $k$ of a set with cardinality $n$.
Let $S$ be a set with cardinality $n$.
Let $\powerset S$ denote the power set of $S$.
Let $\AA_k \subseteq \powerset S$ be the set of elements of $\powerset S$ which have cardin... | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | From the definition of [[Cardinality of Set of Subsets]], $\dbinom n k$ is the number of [[Definition:Subset|subsets]] of [[Definition:Cardinality|cardinality]] $k$ of a [[Definition:Set|set]] with [[Definition:Cardinality|cardinality]] $n$.
Let $S$ be a [[Definition:Set|set]] with [[Definition:Cardinality|cardinality... | Symmetry Rule for Binomial Coefficients/Proof 2 | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients/Proof_2 | [
"Symmetry Rule for Binomial Coefficients",
"Binomial Coefficients"
] | [] | [
"Cardinality of Set of Subsets",
"Definition:Subset",
"Definition:Cardinality",
"Definition:Set",
"Definition:Cardinality",
"Definition:Set",
"Definition:Cardinality",
"Definition:Power Set",
"Definition:Set",
"Definition:Element",
"Definition:Cardinality",
"Definition:Relative Complement",
... |
proofwiki-2518 | Symmetry Rule for Binomial Coefficients | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | If we choose $k$ objects from $n$, then we leave $n - k$ objects.
Hence for every choice of $k$, we are also making the same choice of $n - k$.
Hence the number of choices of $k$ objects is the same as the number of choices of $n - k$ objects.
Hence the result.
{{qed}} | Let $n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k = \dbinom n {n - k}$ | If we choose $k$ [[Definition:Object|objects]] from $n$, then we leave $n - k$ [[Definition:Object|objects]].
Hence for every choice of $k$, we are also making the same choice of $n - k$.
Hence the number of choices of $k$ [[Definition:Object|objects]] is the same as the number of choices of $n - k$ [[Definition:Obje... | Symmetry Rule for Binomial Coefficients/Proof 3 | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients | https://proofwiki.org/wiki/Symmetry_Rule_for_Binomial_Coefficients/Proof_3 | [
"Symmetry Rule for Binomial Coefficients",
"Binomial Coefficients"
] | [] | [
"Definition:Object",
"Definition:Object",
"Definition:Object",
"Definition:Object"
] |
proofwiki-2519 | Factors of Binomial Coefficient | For all $r \in \R, k \in \Z$:
:$k \dbinom r k = r \dbinom {r - 1} {k - 1}$
where $\dbinom r k$ is a binomial coefficient.
Hence:
:$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$)
and:
:$\dfrac 1 r \dbinom r k = \dfrac 1 k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$ and $r \ne 0$) | If $k = 0$ then $\dbinom r k = r \dbinom {r - 1} {k - 1} = 0$ by definition.
Otherwise:
{{begin-eqn}}
{{eqn | l = k \binom r k
| r = k \frac {r^{\underline k} } {k!}
| c =
}}
{{eqn | r = k \frac {r \paren {r - 1} \paren {r - 2} \dotsm \paren {r - k + 1} } {k \paren {k - 1} \paren {k - 2} \dotsm 1}
| ... | For all $r \in \R, k \in \Z$:
:$k \dbinom r k = r \dbinom {r - 1} {k - 1}$
where $\dbinom r k$ is a [[Definition:Binomial Coefficient|binomial coefficient]].
Hence:
:$\dbinom r k = \dfrac r k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$)
and:
:$\dfrac 1 r \dbinom r k = \dfrac 1 k \dbinom {r - 1} {k - 1}$ (if $k \ne 0$ and ... | If $k = 0$ then $\dbinom r k = r \dbinom {r - 1} {k - 1} = 0$ by [[Definition:Binomial Coefficient|definition]].
Otherwise:
{{begin-eqn}}
{{eqn | l = k \binom r k
| r = k \frac {r^{\underline k} } {k!}
| c =
}}
{{eqn | r = k \frac {r \paren {r - 1} \paren {r - 2} \dotsm \paren {r - k + 1} } {k \paren {k... | Factors of Binomial Coefficient | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient | https://proofwiki.org/wiki/Factors_of_Binomial_Coefficient | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Definition:Binomial Coefficient"
] |
proofwiki-2520 | Sum of r+k Choose k up to n | :$\ds \forall n \in \Z: n \ge 0: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ | Proof by induction:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition
:$\ds \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ | :$\ds \forall n \in \Z: n \ge 0: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]
:$\ds \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ | Sum of r+k Choose k up to n | https://proofwiki.org/wiki/Sum_of_r+k_Choose_k_up_to_n | https://proofwiki.org/wiki/Sum_of_r+k_Choose_k_up_to_n | [
"Binomial Coefficients"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2521 | Negated Upper Index of Binomial Coefficient | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom r k
| r = \frac {r^{\underline k} } {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac 1 {k!} \prod_{j \mathop = 0}^{k - 1} \paren {r - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {-\pa... | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom r k
| r = \frac {r^{\underline k} } {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac 1 {k!} \prod_{j \mathop = 0}^{k - 1} \paren {r - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \frac {\paren {-1}^k} {k!} \prod_{j \mathop = 0}^{k - 1} \paren {-\pa... | Negated Upper Index of Binomial Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient | [
"Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient"
] | [] | [
"Permutation of Indices of Product"
] |
proofwiki-2522 | Negated Upper Index of Binomial Coefficient | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom {-r} k
| r = \paren {-1}^k \binom {k - \paren {-r} - 1} k
| c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \paren {-1}^k \binom {r + k - 1} k
| c=
}}
{{end-eqn}}
{{qed}} | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom {-r} k
| r = \paren {-1}^k \binom {k - \paren {-r} - 1} k
| c = [[Negated Upper Index of Binomial Coefficient]]
}}
{{eqn | r = \paren {-1}^k \binom {r + k - 1} k
| c=
}}
{{end-eqn}}
{{qed}} | Negated Upper Index of Binomial Coefficient/Corollary 1/Proof 1 | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient/Corollary_1/Proof_1 | [
"Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient"
] | [] | [
"Negated Upper Index of Binomial Coefficient"
] |
proofwiki-2523 | Negated Upper Index of Binomial Coefficient | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom {-r} k
| r = \frac {\paren {-r}^{\underline k} } {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {-r \paren {-r - 1} \paren {-r - 2} \dotsm \paren {-r - k + 1} } {k!}
| c =
}}
{{eqn | r = \paren {-1}^k \frac {\paren r \paren {r + 1} \paren {r + 2} \dots... | :$\dbinom r k = \paren {-1}^k \dbinom {k - r - 1} k$ | {{begin-eqn}}
{{eqn | l = \binom {-r} k
| r = \frac {\paren {-r}^{\underline k} } {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {-r \paren {-r - 1} \paren {-r - 2} \dotsm \paren {-r - k + 1} } {k!}
| c =
}}
{{eqn | r = \paren {-1}^k \frac {\paren r \paren {r + 1} \paren {r + 2} \dots... | Negated Upper Index of Binomial Coefficient/Corollary 1/Proof 2 | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Binomial_Coefficient/Corollary_1/Proof_2 | [
"Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient"
] | [] | [] |
proofwiki-2524 | Alternating Sum and Difference of r Choose k up to n | :$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \le n} \paren {-1}^k \binom r k
| r = \sum_{k \mathop \le n} \binom {k - r - 1} k
| c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \binom {-r + n} n
| c = Sum of r+k Choose k up to n
}}
{{eqn | r = \paren {-1}^n \binom {r - 1} n
| c = Neg... | :$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop \le n} \paren {-1}^k \binom r k
| r = \sum_{k \mathop \le n} \binom {k - r - 1} k
| c = [[Negated Upper Index of Binomial Coefficient]]
}}
{{eqn | r = \binom {-r + n} n
| c = [[Sum of r+k Choose k up to n]]
}}
{{eqn | r = \paren {-1}^n \binom {r - 1} n
|... | Alternating Sum and Difference of r Choose k up to n/Proof 1 | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_r_Choose_k_up_to_n | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_r_Choose_k_up_to_n/Proof_1 | [
"Binomial Coefficients",
"Alternating Sum and Difference of r Choose k up to n"
] | [] | [
"Negated Upper Index of Binomial Coefficient",
"Sum of r+k Choose k up to n",
"Negated Upper Index of Binomial Coefficient"
] |
proofwiki-2525 | Alternating Sum and Difference of r Choose k up to n | :$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \le 0} \paren {-1}^k \binom r k
| r = \paren {-1}^0 \binom r 0
... | :$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$ | 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]]:
:$\ds \sum_{k \mathop \le n} \paren {-1}^k \binom r k = \paren {-1}^n \binom {r - 1} n$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop... | Alternating Sum and Difference of r Choose k up to n/Proof 2 | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_r_Choose_k_up_to_n | https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_r_Choose_k_up_to_n/Proof_2 | [
"Binomial Coefficients",
"Alternating Sum and Difference of r Choose k up to n"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Binomial Coefficient with Zero",
"Binomial Coefficient with Zero",
"Binomial Coefficient with Zero",
"Binomial Coefficient with One",
"Binomial Coefficient with One",
"Definition:Basis for the Induction",
"Definition:Induction Hypothe... |
proofwiki-2526 | Product of r Choose m with m Choose k | :$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$ | === Integral Index ===
Let $r \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \binom r m \binom m k
| r = \frac {r^{\underline m} } {m!} \frac {m^{\underline k} } {k!}
| c =
}}
{{eqn | r = \frac {r! m!} {m! \paren {r - m}! k! \paren {m - k}!}
| c =
}}
{{eqn | r = \frac {r! \paren {r - k}!} {k! \paren {r - ... | :$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$ | === Integral Index ===
Let $r \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \binom r m \binom m k
| r = \frac {r^{\underline m} } {m!} \frac {m^{\underline k} } {k!}
| c =
}}
{{eqn | r = \frac {r! m!} {m! \paren {r - m}! k! \paren {m - k}!}
| c =
}}
{{eqn | r = \frac {r! \paren {r - k}!} {k! \paren {... | Product of r Choose m with m Choose k/Proof 1 | https://proofwiki.org/wiki/Product_of_r_Choose_m_with_m_Choose_k | https://proofwiki.org/wiki/Product_of_r_Choose_m_with_m_Choose_k/Proof_1 | [
"Binomial Coefficients",
"Product of r Choose m with m Choose k"
] | [] | [
"Definition:Polynomial"
] |
proofwiki-2527 | Product of r Choose m with m Choose k | :$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$ | Consider the trinomial coefficient:
:$\dbinom r {k, m - k, r - m}$
We use Multinomial Coefficient expressed as Product of Binomial Coefficients:
:$\dbinom {k_1 + k_2 + k_3} {k_1, k_2, k_3} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2}$
and substitute as appropriate for $k_1, k_2, k_3$.
We have:
{{be... | :$\dbinom r m \dbinom m k = \dbinom r k \dbinom {r - k} {m - k}$ | Consider the [[Definition:Trinomial Coefficient|trinomial coefficient]]:
:$\dbinom r {k, m - k, r - m}$
We use [[Multinomial Coefficient expressed as Product of Binomial Coefficients]]:
:$\dbinom {k_1 + k_2 + k_3} {k_1, k_2, k_3} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2}$
and substitute as app... | Product of r Choose m with m Choose k/Proof 2 | https://proofwiki.org/wiki/Product_of_r_Choose_m_with_m_Choose_k | https://proofwiki.org/wiki/Product_of_r_Choose_m_with_m_Choose_k/Proof_2 | [
"Binomial Coefficients",
"Product of r Choose m with m Choose k"
] | [] | [
"Definition:Multinomial Coefficient/Trinomial",
"Multinomial Coefficient expressed as Product of Binomial Coefficients",
"Multinomial Coefficient expressed as Product of Binomial Coefficients",
"Multinomial Coefficient expressed as Product of Binomial Coefficients"
] |
proofwiki-2528 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From the Chu-Vandermonde Identity:
{{:Chu-Vandermonde Identity}}
The result follows on setting $r = e$, $s = \pi$ and $n = 2$.
{{qed}} | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From the [[Chu-Vandermonde Identity]]:
{{:Chu-Vandermonde Identity}}
The result follows on setting $r = e$, $s = \pi$ and $n = 2$.
{{qed}} | Chu-Vandermonde Identity/Examples/2 from e + pi/Proof 1 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Examples/2_from_e_+_pi/Proof_1 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [
"Chu-Vandermonde Identity"
] |
proofwiki-2529 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \binom {e + \pi} 2
| r = \dfrac {\paren {e + \pi} \times \paren {e + \pi - 1} } {2 \times 1}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {e + \pi}^2 - \paren {e + \pi} }
| c = multiplying out
}}
{{eqn | r = \dfrac 1 2 \paren {e^2 + 2e\pi + \p... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \binom {e + \pi} 2
| r = \dfrac {\paren {e + \pi} \times \paren {e + \pi - 1} } {2 \times 1}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {e + \pi}^2 - \paren {e + \pi} }
| c = multiplying out
}}
{{eqn | r = \dfrac 1 2 \paren {e^2 + 2e\pi + \p... | Chu-Vandermonde Identity/Examples/2 from e + pi/Proof 2 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Examples/2_from_e_+_pi/Proof_2 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [] |
proofwiki-2530 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From the Chu-Vandermonde Identity:
{{:Chu-Vandermonde Identity}}
The result follows on setting $r = 4$, $s = 5$ and $n = 3$.
{{qed}} | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From the [[Chu-Vandermonde Identity]]:
{{:Chu-Vandermonde Identity}}
The result follows on setting $r = 4$, $s = 5$ and $n = 3$.
{{qed}} | Chu-Vandermonde Identity/Examples/3 from 4 + 5/Proof 1 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Examples/3_from_4_+_5/Proof_1 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [
"Chu-Vandermonde Identity"
] |
proofwiki-2531 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \binom 9 3
| r = \binom {4 + 5} 3
| c =
}}
{{eqn | r = \dfrac {9!} {3! \times 6!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac {9 \times 8 \times 7} {3 \times 2 \times 1}
| c = {{Defof|Factorial}}
}}
{{eqn | r = 84
| c =
}}
{{end-eqn}}
{{qed|lemma}... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \binom 9 3
| r = \binom {4 + 5} 3
| c =
}}
{{eqn | r = \dfrac {9!} {3! \times 6!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \dfrac {9 \times 8 \times 7} {3 \times 2 \times 1}
| c = {{Defof|Factorial}}
}}
{{eqn | r = 84
| c =
}}
{{end-eqn}}
{{qed|lemma}... | Chu-Vandermonde Identity/Examples/3 from 4 + 5/Proof 2 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Examples/3_from_4_+_5/Proof_2 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [] |
proofwiki-2532 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^{r + s} \binom {r + s} n x^n
| r = \paren {1 + x}^{r + s}
| c = Binomial Theorem for Integral Index
}}
{{eqn | r = \paren {1 + x}^r \paren {1 + x}^s
| c = Exponent Combination Laws
}}
{{eqn | r = \sum_{k \mathop = 0}^r \binom r k x^k \sum_{k \mathop = 0}^... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^{r + s} \binom {r + s} n x^n
| r = \paren {1 + x}^{r + s}
| c = [[Binomial Theorem for Integral Index]]
}}
{{eqn | r = \paren {1 + x}^r \paren {1 + x}^s
| c = [[Exponent Combination Laws]]
}}
{{eqn | r = \sum_{k \mathop = 0}^r \binom r k x^k \sum_{k \math... | Chu-Vandermonde Identity/Proof 1 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Proof_1 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [
"Binomial Theorem/Integral Index",
"Exponent Combination Laws",
"Binomial Theorem/Integral Index",
"Product of Absolutely Convergent Series"
] |
proofwiki-2533 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | The '''Chu-Vandermonde Identity''' is a special case of Gauss's Hypergeometric Theorem:
:$\map { {}_2F_1} {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
when:
:$\map \Re {c - a - b} \gt 0$
where:
:$\map { {}_2F_1} {a, b; c; 1}$ is the hypergeometric series: $\... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | The '''[[Chu-Vandermonde Identity]]''' is a special case of [[Gauss's Hypergeometric Theorem]]:
:$\map { {}_2F_1} {a, b; c; 1} = \dfrac {\map \Gamma c \map \Gamma {c - a - b} } {\map \Gamma {c - a} \map \Gamma {c - b} }$
when:
:$\map \Re {c - a - b} \gt 0$
where:
:$\map { {}_2F_1} {a, b; c; 1}$ is the [[Definition:Hyp... | Chu-Vandermonde Identity/Proof 2 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Proof_2 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [
"Chu-Vandermonde Identity",
"Gauss's Hypergeometric Theorem",
"Definition:Hypergeometric Series",
"Definition:Rising Factorial",
"Definition:Gamma Function",
"Rising Factorial in terms of Falling Factorial of Negative",
"Falling Factorial as Quotient of Factorials",
"Rising Factorial as Quotient of Fa... |
proofwiki-2534 | Chu-Vandermonde Identity | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \paren{n - k}} {n - k} \dfrac r {r - t k}$:
{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk}}
where $r, s, t \in \R, n \in \Z$.
Setting $t = 0$:
:$\ds \sum_{k \mathop \ge 0} \binom r k \binom s {n - k} = \binom {r + s} n$
which is the result ... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom r k \binom s {n - k}
| r = \binom r 0 \binom s n + \binom r 1 \binom s {n - 1} + \binom r 2 \binom s {n - 2} + \cdots + \binom r n \binom s 0
| c =
}}
{{eqn | r = \binom {r + s} n
| c =
}}
{{end-eqn}} | From [[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 $\dbinom {r - t k} k \dbinom {s - t \paren{n - k}} {n - k} \dfrac r {r - t k}$]]:
{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk}}
where $r, s, t \in \R, n \in \Z$.
Setting $t = 0$:
:$\ds \sum_{k \mathop \... | Chu-Vandermonde Identity/Proof 4 | https://proofwiki.org/wiki/Chu-Vandermonde_Identity | https://proofwiki.org/wiki/Chu-Vandermonde_Identity/Proof_4 | [
"Chu-Vandermonde Identity",
"Binomial Coefficients"
] | [] | [
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk"
] |
proofwiki-2535 | Lucas' Theorem | Let $p$ be a prime number.
Let $n, k \in \Z_{\ge 0}$.
Then:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$n \bmod p$ denotes the modulo operation
:$\floor \cdot$ denotes the floor function. | First we show that:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
Consider $\dbinom n k$ as the fraction:
:$\dfrac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}$
This can be expressed as:
:$(1): \quad \... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n, k \in \Z_{\ge 0}$.
Then:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
where:
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$n \bmod p$ denotes the [[Definit... | First we show that:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
Consider $\dbinom n k$ as the fraction:
:$\dfrac {n \paren {n - 1} \paren {n - 2} \cdots \paren {n - k + 1} } {k \paren {k - 1} \paren {k - 2} \cdots 1}$
This can be expressed as:
:$(1): \qu... | Lucas' Theorem | https://proofwiki.org/wiki/Lucas'_Theorem | https://proofwiki.org/wiki/Lucas'_Theorem | [
"Binomial Coefficients",
"Number Theory",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient",
"Definition:Modulo Operation",
"Definition:Floor Function"
] | [
"Division Theorem",
"Definition:Fraction/Denominator",
"Definition:Divisor (Algebra)/Integer",
"Division Theorem",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Common Factor Cancelling in Congruence",
"Definition:Congruence (Number Theory)"
] |
proofwiki-2536 | Equivalence of Definitions of Congruence | {{TFAE|def = Congruence (Number Theory)|view = Congruence|context = Number Theory}}
Let $z \in \R$. | Let $x_1, x_2, z \in \R$.
Let $x_1 \equiv x_2 \pmod z$ as defined by an equivalence relation.
That is, let $\RR_z$ be the relation on the set of all $x, y \in \R$:
:$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x = y + k z}$
Let $\tuple {x_1, x_2} \in \RR_z$.
Then by definition, $\exists k \in \Z: x_... | {{TFAE|def = Congruence (Number Theory)|view = Congruence|context = Number Theory}}
Let $z \in \R$. | Let $x_1, x_2, z \in \R$.
Let $x_1 \equiv x_2 \pmod z$ as defined by an [[Definition:Congruence (Number Theory)|equivalence relation]].
That is, let $\RR_z$ be the [[Definition:Relation|relation]] on the [[Definition:Set|set]] of all $x, y \in \R$:
:$\RR_z = \set {\tuple {x, y} \in \R \times \R: \exists k \in \Z: x ... | Equivalence of Definitions of Congruence | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Congruence | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Congruence | [
"Modulo Arithmetic"
] | [] | [
"Definition:Congruence (Number Theory)",
"Definition:Relation",
"Definition:Set",
"Definition:Modulo Operation",
"Definition:Congruence (Number Theory)/Modulo Operation",
"Definition:Congruence (Number Theory)/Modulo Operation",
"Definition:Congruence (Number Theory)/Integer Multiple",
"Definition:Mod... |
proofwiki-2537 | Congruence Modulo Real Number is Equivalence Relation | For all $z \in \R$, congruence modulo $z$ is an equivalence relation. | Checking in turn each of the criteria for equivalence: | For all $z \in \R$, [[Definition:Congruence (Number Theory)|congruence modulo $z$]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Congruence Modulo Real Number is Equivalence Relation | https://proofwiki.org/wiki/Congruence_Modulo_Real_Number_is_Equivalence_Relation | https://proofwiki.org/wiki/Congruence_Modulo_Real_Number_is_Equivalence_Relation | [
"Modulo Arithmetic",
"Examples of Equivalence Relations"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-2538 | Congruence Modulo Zero is Diagonal Relation | Congruence modulo zero is the diagonal relation.
That is:
:$x \equiv y \pmod 0 \iff x = y$ | Follows directly from the definition of congruence:
:$x \equiv y \pmod z \iff x \bmod z = y \bmod z$
When $z = 0$ we have by definition:
:$x \bmod 0 := x$
And so $x \bmod 0 = y \bmod 0 \iff x = y$.
Hence the result.
{{qed}}
Category:Modulo Arithmetic
cekrtgnb3larl48i6x3esxn6kisvbjy | [[Definition:Congruence (Number Theory)|Congruence modulo zero]] is the [[Definition:Diagonal Relation|diagonal relation]].
That is:
:$x \equiv y \pmod 0 \iff x = y$ | Follows directly from the definition of [[Definition:Congruence (Number Theory)#Definition by Modulo Operation|congruence]]:
:$x \equiv y \pmod z \iff x \bmod z = y \bmod z$
When $z = 0$ we have by [[Definition:Modulo Operation|definition]]:
:$x \bmod 0 := x$
And so $x \bmod 0 = y \bmod 0 \iff x = y$.
Hence the r... | Congruence Modulo Zero is Diagonal Relation | https://proofwiki.org/wiki/Congruence_Modulo_Zero_is_Diagonal_Relation | https://proofwiki.org/wiki/Congruence_Modulo_Zero_is_Diagonal_Relation | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Diagonal Relation"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Modulo Operation",
"Category:Modulo Arithmetic"
] |
proofwiki-2539 | Filter Basis Generates Filter | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\BB \subset \powerset S$.
Then:
:$\FF = \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a filter on $S$
{{iff}}:
:$(1): \quad \forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$
:$(2): \quad \O \notin \BB, \BB \ne \O$
That i... | Assume first that $\FF$ is a filter on $S$.
Then $S \in \FF$ and thus $\BB \ne \O$.
Because $\O \notin \FF$ we have that $\O \notin \BB$, since $\BB \subseteq \FF$.
Let $V_1, V_2 \in \BB$.
Then:
:$V_1, V_2 \in \FF$
Because $\FF$ is a filter:
:$V := V_1 \cap V_2 \in \FF$
The definition of $\FF$ implies therefore that th... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\BB \subset \powerset S$.
Then:
:$\FF = \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a [[Definition:Filter on Set|filter]] on $S$
{{iff}}:
:$(1): \quad \forall V_1, V_2 \in \BB: \exists U \in ... | Assume first that $\FF$ is a [[Definition:Filter on Set|filter]] on $S$.
Then $S \in \FF$ and thus $\BB \ne \O$.
Because $\O \notin \FF$ we have that $\O \notin \BB$, since $\BB \subseteq \FF$.
Let $V_1, V_2 \in \BB$.
Then:
:$V_1, V_2 \in \FF$
Because $\FF$ is a [[Definition:Filter on Set|filter]]:
:$V := V_1 \ca... | Filter Basis Generates Filter | https://proofwiki.org/wiki/Filter_Basis_Generates_Filter | https://proofwiki.org/wiki/Filter_Basis_Generates_Filter | [
"Filter Bases",
"Generated Filters"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Filter on Set",
"Definition:Filter Basis",
"Definition:Filter on Set",
"Definition:Filter Basis/Generated Filter"
] | [
"Definition:Filter on Set",
"Definition:Filter on Set",
"Definition:Filter on Set",
"Definition:Filter on Set"
] |
proofwiki-2540 | Adherent Point of Filter iff Superfilter Converges | Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on $S$.
Let $x \in S$.
Then $x$ is a adherent point of $\FF$ {{iff}} there exists a superfilter $\FF'$ of $\FF$ on $S$ which converges to $x$. | Let $x$ be a adherent point of $\FF$.
Define:
:$\BB := \set {F \cap U : F \in \FF \text{ and } U \text{ is a neighborhood of } x}$
Then $\BB$ is filter basis by definition.
Let $\FF'$ be the corresponding generated filter.
By construction we have $\FF \subseteq \FF'$ and $U \in \FF'$ for every neighborhood $U$ of $x$.
... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
Let $x \in S$.
Then $x$ is a [[Definition:Adherent Point of Filter|adherent point]] of $\FF$ {{iff}} there exists a [[Definition:Superfilter|superfilter]] $\FF'$ of $\FF$ o... | Let $x$ be a [[Definition:Adherent Point of Filter|adherent point]] of $\FF$.
Define:
:$\BB := \set {F \cap U : F \in \FF \text{ and } U \text{ is a neighborhood of } x}$
Then $\BB$ is [[Definition:Filter Basis|filter basis]] by definition.
Let $\FF'$ be the corresponding [[Definition:Generated Filter|generated filt... | Adherent Point of Filter iff Superfilter Converges | https://proofwiki.org/wiki/Adherent_Point_of_Filter_iff_Superfilter_Converges | https://proofwiki.org/wiki/Adherent_Point_of_Filter_iff_Superfilter_Converges | [
"Adherent Points of Filters",
"Filter Theory"
] | [
"Definition:Topological Space",
"Definition:Filter on Set",
"Definition:Adherent Point/Filter",
"Definition:Superfilter on Set",
"Definition:Convergent Filter"
] | [
"Definition:Adherent Point/Filter",
"Definition:Filter Basis",
"Definition:Filter Basis/Generated Filter",
"Definition:Neighborhood (Topology)",
"Definition:Convergent Filter",
"Definition:Filter on Set",
"Definition:Convergent Filter",
"Definition:Neighborhood (Topology)",
"Definition:Neighborhood ... |
proofwiki-2541 | Ultrafilter Lemma | Let $S$ be a set.
Every filter on $S$ is contained in an ultrafilter on $S$. | Let $\Omega$ be the set of filters on $S$.
From Subset Relation is Ordering, the subset relation makes $\struct {\Omega, \subseteq}$ a partially ordered set.
Let $C \subseteq \Omega$ be a non-empty chain.
Then $\bigcup C$ is again a filter on $S$.
Thus $\bigcup C$ is an upper bound of $C$.
Indeed, if $A, B \in \bigcup ... | Let $S$ be a [[Definition:Set|set]].
Every [[Definition:Filter on Set|filter]] on $S$ is contained in an [[Definition:Ultrafilter on Set|ultrafilter]] on $S$. | Let $\Omega$ be the [[Definition:Set|set]] of [[Definition:Filter on Set|filters]] on $S$.
From [[Subset Relation is Ordering]], the [[Definition:Subset Relation|subset relation]] makes $\struct {\Omega, \subseteq}$ a [[Definition:Partially Ordered Set|partially ordered set]].
Let $C \subseteq \Omega$ be a [[Definiti... | Ultrafilter Lemma/Proof 1 | https://proofwiki.org/wiki/Ultrafilter_Lemma | https://proofwiki.org/wiki/Ultrafilter_Lemma/Proof_1 | [
"Ultrafilter Lemma",
"Set Theory",
"Filter Theory",
"Named Theorems"
] | [
"Definition:Set",
"Definition:Filter on Set",
"Definition:Ultrafilter on Set"
] | [
"Definition:Set",
"Definition:Filter on Set",
"Subset Relation is Ordering",
"Definition:Subset Relation",
"Definition:Partially Ordered Set",
"Definition:Non-Empty Set",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Filter on Set",
"Definition:Upper Bound of Set",
"Definition:Fil... |
proofwiki-2542 | Ultrafilter Lemma | Let $S$ be a set.
Every filter on $S$ is contained in an ultrafilter on $S$. | Let $\FF$ be a filter on $S$.
Let the power set of $S$ be ordered by inclusion.
Then, by Filter on Set is Proper Filter, $\FF$ is a proper filter on $\powerset S$.
By Filter is Ideal in Dual Ordered Set, $\FF$ is an ideal in the dual of $\powerset S$.
By Singleton of Bottom is Ideal, $\set \O$ is an ideal in $\powerset... | Let $S$ be a [[Definition:Set|set]].
Every [[Definition:Filter on Set|filter]] on $S$ is contained in an [[Definition:Ultrafilter on Set|ultrafilter]] on $S$. | Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
Let the [[Definition:Power Set|power set]] of $S$ be [[Definition:Set Ordered by Inclusion|ordered by inclusion]].
Then, by [[Filter on Set is Proper Filter]], $\FF$ is a [[Definition:Proper Filter|proper filter]] on $\powerset S$.
By [[Filter is Ideal in Du... | Ultrafilter Lemma/Proof 2 | https://proofwiki.org/wiki/Ultrafilter_Lemma | https://proofwiki.org/wiki/Ultrafilter_Lemma/Proof_2 | [
"Ultrafilter Lemma",
"Set Theory",
"Filter Theory",
"Named Theorems"
] | [
"Definition:Set",
"Definition:Filter on Set",
"Definition:Ultrafilter on Set"
] | [
"Definition:Filter on Set",
"Definition:Power Set",
"Definition:Set Ordered by Subset Relation",
"Filter on Set is Proper Filter",
"Definition:Filter/Proper Filter",
"Filter is Ideal in Dual Ordered Set",
"Definition:Ideal",
"Definition:Dual Ordering",
"Singleton of Bottom is Ideal",
"Definition:I... |
proofwiki-2543 | Generated Topology is a Topology | Let $X$ be a set.
Let $\SS \subseteq \powerset X$, where $\powerset X$ is the power set of $X$.
Let $\TT_\SS$ be the generated topology for $\SS$.
Then $\TT_\SS$ is a topology on $X$. | We show that $\SS^* = \set {\bigcap S: S \subseteq \SS \text{ finite} }$ (cf. the definition of the generated topology) is a basis.
By definition of synthetic basis, we need to prove:
:$(1): \quad X = \bigcup \SS^*$
:$(2): \quad$ For any $U_1, U_2 \in \SS^*$ and $x \in U_1 \cap U_2$ there is a $U \in \SS^*$ such that $... | Let $X$ be a [[Definition:Set|set]].
Let $\SS \subseteq \powerset X$, where $\powerset X$ is the [[Definition:Power Set|power set]] of $X$.
Let $\TT_\SS$ be the [[Definition:Generated Topology|generated topology]] for $\SS$.
Then $\TT_\SS$ is a [[Definition:Topology|topology]] on $X$. | We show that $\SS^* = \set {\bigcap S: S \subseteq \SS \text{ finite} }$ (cf. the definition of the [[Definition:Generated Topology|generated topology]]) is a [[Definition:Basis (Topology)|basis]].
By definition of [[Definition:Synthetic Basis|synthetic basis]], we need to prove:
:$(1): \quad X = \bigcup \SS^*$
:$(2):... | Generated Topology is a Topology | https://proofwiki.org/wiki/Generated_Topology_is_a_Topology | https://proofwiki.org/wiki/Generated_Topology_is_a_Topology | [
"Topology"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Generated Topology",
"Definition:Topology"
] | [
"Definition:Generated Topology",
"Definition:Basis (Topology)",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Finite Set",
"Definition:Finite Set",
"Category:Topology"
] |
proofwiki-2544 | Finite Fourier Series | Let $\map a n$ be any finite periodic real function on $\Z$ with period $b$.
Let $\xi = e^{2 \pi i/ b}$ be the first $b$th root of unity.
Then:
:$\ds \map a n = \sum_{k \mathop = 0}^{b - 1} \map {a_*} k \xi^{n k}$
where:
:$\ds \map {a_*} n = \frac 1 b \sum_{k \mathop = 0}^{b - 1} \map a k \xi^{-n k}$ | Since $a$ has period $b$, we have:
:$\map a {n + b} = \map a n$
So if we define:
:$\ds \map F z = \sum_{n \mathop \ge 0} \map a n z^n$
we have:
{{begin-eqn}}
{{eqn | l = \map F z)
| r = \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^b \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^{2 b} \paren {\s... | Let $\map a n$ be any finite [[Definition:Periodic Real Function|periodic real function]] on $\Z$ with [[Definition:Period of Periodic Real Function|period]] $b$.
Let $\xi = e^{2 \pi i/ b}$ be the [[Definition:First Complex Root of Unity|first $b$th root of unity]].
Then:
:$\ds \map a n = \sum_{k \mathop = 0}^{b - 1... | Since $a$ has [[Definition:Period of Periodic Real Function|period]] $b$, we have:
:$\map a {n + b} = \map a n$
So if we define:
:$\ds \map F z = \sum_{n \mathop \ge 0} \map a n z^n$
we have:
{{begin-eqn}}
{{eqn | l = \map F z)
| r = \paren {\sum_{k \mathop = 0}^{b - 1} \map a k z^k} + z^b \paren {\sum_{k \ma... | Finite Fourier Series | https://proofwiki.org/wiki/Finite_Fourier_Series | https://proofwiki.org/wiki/Finite_Fourier_Series | [
"Fourier Analysis"
] | [
"Definition:Periodic Function/Real",
"Definition:Periodic Real Function/Period",
"Definition:Root of Unity/Complex/First"
] | [
"Definition:Periodic Real Function/Period",
"Definition:Polynomial",
"Definition:Partial Fractions Expansion",
"Category:Fourier Analysis"
] |
proofwiki-2545 | Product Distributes over Modulo Operation | Let $x, y, z \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then:
:$z \paren {x \bmod y} = \paren {z x} \bmod \paren {z y}$ | By definition of modulo operation:
:<nowiki>$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$</nowiki>
If $y = 0$ we have that:
:$z \paren {x \bmod 0} = z x = \paren {z x} \bmod \paren {z 0}$
If $y \ne 0$ we have that:
{{begin-eqn}}
{{eqn | l = z \paren {x \bmod y}
| ... | Let $x, y, z \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
Then:
:$z \paren {x \bmod y} = \paren {z x} \bmod \paren {z y}$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:<nowiki>$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$</nowiki>
If $y = 0$ we have that:
:$z \paren {x \bmod 0} = z x = \paren {z x} \bmod \paren {z 0}$
If $y \ne 0$ we have that:
{{begin-eqn}}
{{eq... | Product Distributes over Modulo Operation | https://proofwiki.org/wiki/Product_Distributes_over_Modulo_Operation | https://proofwiki.org/wiki/Product_Distributes_over_Modulo_Operation | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation"
] | [
"Definition:Modulo Operation"
] |
proofwiki-2546 | Congruence by Product of Moduli | Let $a, b, m \in \Z$.
Let $a \equiv b \pmod m$ denote that $a$ is congruent to $b$ modulo $m$.
Then $\forall n \in \Z, n \ne 0$:
:$a \equiv b \pmod m \iff a n \equiv b n \pmod {m n}$ | Let $n \in \Z: n \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \bmod m
| r = b \bmod m
| c = {{Defof|Congruence Modulo Integer}}
}}
{{eqn | ll= \leadstoandfrom
| l = n \paren {a \bmod n}
| r... | Let $a, b, m \in \Z$.
Let $a \equiv b \pmod m$ denote that [[Definition:Congruence Modulo Integer|$a$ is congruent to $b$ modulo $m$]].
Then $\forall n \in \Z, n \ne 0$:
:$a \equiv b \pmod m \iff a n \equiv b n \pmod {m n}$ | Let $n \in \Z: n \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \bmod m
| r = b \bmod m
| c = {{Defof|Congruence Modulo Integer}}
}}
{{eqn | ll= \leadstoandfrom
| l = n \paren {a \bmod n}
|... | Congruence by Product of Moduli | https://proofwiki.org/wiki/Congruence_by_Product_of_Moduli | https://proofwiki.org/wiki/Congruence_by_Product_of_Moduli | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)/Integers"
] | [
"Product Distributes over Modulo Operation"
] |
proofwiki-2547 | Definite Integral of Partial Derivative | Let $\map f {x, y}$ and $\map {\dfrac {\partial f} {\partial x} } {x, y}$ be continuous functions of $x$ and $y$ on $D = \closedint {x_1} {x_2} \times \closedint a b$.
Then:
:$\ds \frac \d {\d x} \int_a^b \map f {x, y} \rd y = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$
for $x \in \closedint {x_1} {x... | From Leibniz's Integral Rule:
:$\ds \frac \d {\d y} \int_{\map a y}^{\map b y} \map f {x, y} \rd x = \map f {y, \map b y} \frac {\d b} {\d y} - \map f {y, \map a y} \frac {\d a} {\d y} + \int_{\map a y}^{\map b y} \frac {\partial} {\partial y} \map f {x, y} \rd x$
where $\map a y$, $\map b y$ are continuously different... | Let $\map f {x, y}$ and $\map {\dfrac {\partial f} {\partial x} } {x, y}$ be [[Definition:Continuous Real Function|continuous functions]] of $x$ and $y$ on $D = \closedint {x_1} {x_2} \times \closedint a b$.
Then:
:$\ds \frac \d {\d x} \int_a^b \map f {x, y} \rd y = \int_a^b \map {\frac {\partial f} {\partial x} } {x,... | From [[Leibniz's Integral Rule]]:
:$\ds \frac \d {\d y} \int_{\map a y}^{\map b y} \map f {x, y} \rd x = \map f {y, \map b y} \frac {\d b} {\d y} - \map f {y, \map a y} \frac {\d a} {\d y} + \int_{\map a y}^{\map b y} \frac {\partial} {\partial y} \map f {x, y} \rd x$
where $\map a y$, $\map b y$ are [[Definition:Con... | Definite Integral of Partial Derivative/Proof 1 | https://proofwiki.org/wiki/Definite_Integral_of_Partial_Derivative | https://proofwiki.org/wiki/Definite_Integral_of_Partial_Derivative/Proof_1 | [
"Partial Differentiation",
"Definite Integrals",
"Definite Integral of Partial Derivative"
] | [
"Definition:Continuous Real Function"
] | [
"Leibniz's Integral Rule",
"Definition:Continuously Differentiable/Real-Valued Function/Open Set",
"Definition:Constant"
] |
proofwiki-2548 | Definite Integral of Partial Derivative | Let $\map f {x, y}$ and $\map {\dfrac {\partial f} {\partial x} } {x, y}$ be continuous functions of $x$ and $y$ on $D = \closedint {x_1} {x_2} \times \closedint a b$.
Then:
:$\ds \frac \d {\d x} \int_a^b \map f {x, y} \rd y = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$
for $x \in \closedint {x_1} {x... | Define $\ds \map G x = \int_a^b \map f {x, y} \rd y$.
The continuity of $f$ ensures that $G$ exists.
Then by linearity of the integral:
:$\dfrac {\Delta G} {\Delta x} = \dfrac {\map G {x + \Delta x} - \map G x} {\Delta x} = \ds \int_a^b \frac {\map f {x + \Delta x, y} - \map f {x, y} } {\Delta x} \rd y$
We want to find... | Let $\map f {x, y}$ and $\map {\dfrac {\partial f} {\partial x} } {x, y}$ be [[Definition:Continuous Real Function|continuous functions]] of $x$ and $y$ on $D = \closedint {x_1} {x_2} \times \closedint a b$.
Then:
:$\ds \frac \d {\d x} \int_a^b \map f {x, y} \rd y = \int_a^b \map {\frac {\partial f} {\partial x} } {x,... | Define $\ds \map G x = \int_a^b \map f {x, y} \rd y$.
The [[Definition:Continuous Real Function|continuity]] of $f$ ensures that $G$ exists.
Then by linearity of the integral:
:$\dfrac {\Delta G} {\Delta x} = \dfrac {\map G {x + \Delta x} - \map G x} {\Delta x} = \ds \int_a^b \frac {\map f {x + \Delta x, y} - \map f ... | Definite Integral of Partial Derivative/Proof 2 | https://proofwiki.org/wiki/Definite_Integral_of_Partial_Derivative | https://proofwiki.org/wiki/Definite_Integral_of_Partial_Derivative/Proof_2 | [
"Partial Differentiation",
"Definite Integrals",
"Definite Integral of Partial Derivative"
] | [
"Definition:Continuous Real Function"
] | [
"Definition:Continuous Real Function",
"Mean Value Theorem",
"Definition:Compact Space/Real Analysis",
"Definition:Uniform Continuity/Real Function"
] |
proofwiki-2549 | Law of Inverses (Modulo Arithmetic) | Let $m, n \in \Z$.
Then:
:$\exists n' \in \Z: n n' \equiv d \pmod m$
where $d = \gcd \set {m, n}$. | We have that $d = \gcd \set {m, n}$.
So:
{{begin-eqn}}
{{eqn | q = \exists a, b \in \Z
| l = a m + b n
| r = d
| c = Bézout's Identity
}}
{{eqn | ll= \leadsto
| l = a m
| r = d - b n
| c =
}}
{{eqn | ll= \leadsto
| l = d - b n
| o = \equiv
| r = 0
| rr= \pmod... | Let $m, n \in \Z$.
Then:
:$\exists n' \in \Z: n n' \equiv d \pmod m$
where $d = \gcd \set {m, n}$. | We have that $d = \gcd \set {m, n}$.
So:
{{begin-eqn}}
{{eqn | q = \exists a, b \in \Z
| l = a m + b n
| r = d
| c = [[Bézout's Identity]]
}}
{{eqn | ll= \leadsto
| l = a m
| r = d - b n
| c =
}}
{{eqn | ll= \leadsto
| l = d - b n
| o = \equiv
| r = 0
| rr= ... | Law of Inverses (Modulo Arithmetic) | https://proofwiki.org/wiki/Law_of_Inverses_(Modulo_Arithmetic) | https://proofwiki.org/wiki/Law_of_Inverses_(Modulo_Arithmetic) | [
"Modulo Arithmetic",
"Named Theorems"
] | [] | [
"Bézout's Identity",
"Definition:Modulo Addition",
"Category:Modulo Arithmetic",
"Category:Named Theorems"
] |
proofwiki-2550 | Common Factor Cancelling in Congruence | Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is congruent modulo $m$ to $b$.
Then:
:$x \equiv y \pmod {m / d}$
where $d = \gcd \set {a, m}$. | We have that $d = \gcd \set {a, m}$.
From Law of Inverses (Modulo Arithmetic), we have:
: $\exists a' \in \Z: a a' \equiv d \pmod m$
Hence:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
}}
{{eqn | ll= \leadsto
| l = a a'
| o = \equiv
| r = b a'
| rr= \pmod m
... | Let $a, b, x, y, m \in \Z$.
Let:
:$a x \equiv b y \pmod m$ and $a \equiv b \pmod m$
where $a \equiv b \pmod m$ denotes that $a$ is [[Definition:Congruence Modulo Integer|congruent modulo $m$]] to $b$.
Then:
:$x \equiv y \pmod {m / d}$
where $d = \gcd \set {a, m}$. | We have that $d = \gcd \set {a, m}$.
From [[Law of Inverses (Modulo Arithmetic)]], we have:
: $\exists a' \in \Z: a a' \equiv d \pmod m$
Hence:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
}}
{{eqn | ll= \leadsto
| l = a a'
| o = \equiv
| r = b a'
| rr= \p... | Common Factor Cancelling in Congruence | https://proofwiki.org/wiki/Common_Factor_Cancelling_in_Congruence | https://proofwiki.org/wiki/Common_Factor_Cancelling_in_Congruence | [
"Modulo Arithmetic",
"Common Factor Cancelling in Congruence"
] | [
"Definition:Congruence (Number Theory)/Integers"
] | [
"Law of Inverses (Modulo Arithmetic)",
"Congruence by Product of Moduli"
] |
proofwiki-2551 | Congruence by Factors of Modulo | Let $a, b \in \Z$.
Let $r$ and $s$ be coprime integers.
Then:
:$a \equiv b \pmod {r s}$ {{iff}} $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is congruent modulo $r$ to $b$. | === Necessary Condition ===
This is proved in Congruence by Divisor of Modulus.
Note that for this result it is not required that $r \perp s$.
{{qed|lemma}} | Let $a, b \in \Z$.
Let $r$ and $s$ be [[Definition:Coprime Integers|coprime integers]].
Then:
:$a \equiv b \pmod {r s}$ {{iff}} $a \equiv b \pmod r$ and $a \equiv b \pmod s$
where $a \equiv b \pmod r$ denotes that $a$ is [[Definition:Congruence (Number Theory)|congruent modulo $r$]] to $b$. | === Necessary Condition ===
This is proved in [[Congruence by Divisor of Modulus/Integer Modulus|Congruence by Divisor of Modulus]].
Note that for this result it is not required that $r \perp s$.
{{qed|lemma}} | Congruence by Factors of Modulo | https://proofwiki.org/wiki/Congruence_by_Factors_of_Modulo | https://proofwiki.org/wiki/Congruence_by_Factors_of_Modulo | [
"Congruence by Factors of Modulo",
"Modulo Arithmetic"
] | [
"Definition:Coprime/Integers",
"Definition:Congruence (Number Theory)"
] | [
"Congruence by Divisor of Modulus/Integer Modulus"
] |
proofwiki-2552 | Sum of Floors not greater than Floor of Sum | Let $\floor x$ denote the floor function.
Then:
:$\floor x + \floor y \le \floor {x + y}$
The equality holds:
:$\floor x + \floor y = \floor {x + y}$
{{iff}}:
:$x \bmod 1 + y \bmod 1 < 1$
where $x \bmod 1$ denotes the modulo operation. | From the definition of the modulo operation, we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which:
{{begin-eqn}}
{{eqn | l = \floor {x + y}
| r = \floor {\floor x + \paren {x \bmod 1} + \floor y + \paren {y \bmod 1} }
| c =
}}
{{eqn | r = \floor x + \floor y + \floor {\paren {x \bmod 1} + \paren {... | Let $\floor x$ denote the [[Definition:Floor Function|floor function]].
Then:
:$\floor x + \floor y \le \floor {x + y}$
The equality holds:
:$\floor x + \floor y = \floor {x + y}$
{{iff}}:
:$x \bmod 1 + y \bmod 1 < 1$
where $x \bmod 1$ denotes the [[Definition:Modulo Operation|modulo operation]]. | From the definition of the [[Definition:Modulo Operation|modulo operation]], we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which:
{{begin-eqn}}
{{eqn | l = \floor {x + y}
| r = \floor {\floor x + \paren {x \bmod 1} + \floor y + \paren {y \bmod 1} }
| c =
}}
{{eqn | r = \floor x + \floor y + \fl... | Sum of Floors not greater than Floor of Sum | https://proofwiki.org/wiki/Sum_of_Floors_not_greater_than_Floor_of_Sum | https://proofwiki.org/wiki/Sum_of_Floors_not_greater_than_Floor_of_Sum | [
"Floor Function"
] | [
"Definition:Floor Function",
"Definition:Modulo Operation"
] | [
"Definition:Modulo Operation",
"Floor of Number plus Integer"
] |
proofwiki-2553 | Sum of Ceilings not less than Ceiling of Sum | Let $\ceiling x$ be the ceiling function.
Then:
:$\ceiling x + \ceiling y \ge \ceiling {x + y}$
The equality holds:
:$\ceiling x + \ceiling y = \ceiling {x + y}$
{{iff}} either:
:$x \in \Z$ or $y \in \Z$
or:
:$x \bmod 1 + y \bmod 1 > 1$
where $x \bmod 1$ denotes the modulo operation. | From the definition of the modulo operation, we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which we obtain:
:$x = \ceiling x - \sqbrk {x \notin \Z} + \paren {x \bmod 1}$
where $\sqbrk {x \notin \Z}$ uses Iverson's convention.
{{begin-eqn}}
{{eqn | l = \ceiling {x + y}
| r = \ceiling {\floor x + \paren {... | Let $\ceiling x$ be the [[Definition:Ceiling Function|ceiling function]].
Then:
:$\ceiling x + \ceiling y \ge \ceiling {x + y}$
The equality holds:
:$\ceiling x + \ceiling y = \ceiling {x + y}$
{{iff}} either:
:$x \in \Z$ or $y \in \Z$
or:
:$x \bmod 1 + y \bmod 1 > 1$
where $x \bmod 1$ denotes the [[Definition:Modu... | From the definition of the [[Definition:Modulo Operation|modulo operation]], we have that:
:$x = \floor x + \paren {x \bmod 1}$
from which we obtain:
:$x = \ceiling x - \sqbrk {x \notin \Z} + \paren {x \bmod 1}$
where $\sqbrk {x \notin \Z}$ uses [[Definition:Iverson's Convention|Iverson's convention]].
{{begin-eqn}... | Sum of Ceilings not less than Ceiling of Sum | https://proofwiki.org/wiki/Sum_of_Ceilings_not_less_than_Ceiling_of_Sum | https://proofwiki.org/wiki/Sum_of_Ceilings_not_less_than_Ceiling_of_Sum | [
"Ceiling Function"
] | [
"Definition:Ceiling Function",
"Definition:Modulo Operation"
] | [
"Definition:Modulo Operation",
"Definition:Iverson's Convention",
"Ceiling of Number plus Integer"
] |
proofwiki-2554 | Integer to Power of p-1 over 2 Modulo p | Let $a \in \Z$.
Let $p$ be an odd prime.
Let $b = a^{\frac {\paren {p - 1} } 2}$.
Then one of the following cases holds:
:$b \bmod p = 0$
which happens exactly when $a \equiv 0 \pmod p$, or:
:$b \bmod p = 1$
or:
:$b \bmod p = p - 1$
where:
:$b \bmod p$ denotes the modulo operation
:$x \equiv y \pmod p$ denotes that $x$... | By definition of congruence modulo $p$:
:$\forall x, y \in \R: x \equiv y \pmod p \iff x \bmod p = y \bmod p$
We have that:
:$b = a^{\frac{\paren {p - 1} } 2}$
and so:
:$b^2 = a^{p - 1}$
Let $a \equiv 0 \pmod p$.
Then by definition of congruence modulo $p$:
:$p \divides a$
and so:
:$p \divides a^{\frac{\paren {p - 1} ... | Let $a \in \Z$.
Let $p$ be an [[Definition:Odd Prime|odd prime]].
Let $b = a^{\frac {\paren {p - 1} } 2}$.
Then one of the following cases holds:
:$b \bmod p = 0$
which happens exactly when $a \equiv 0 \pmod p$, or:
:$b \bmod p = 1$
or:
:$b \bmod p = p - 1$
where:
:$b \bmod p$ denotes the [[Definition:Modulo Opera... | By definition of [[Definition:Congruence (Number Theory)|congruence modulo $p$]]:
:$\forall x, y \in \R: x \equiv y \pmod p \iff x \bmod p = y \bmod p$
We have that:
:$b = a^{\frac{\paren {p - 1} } 2}$
and so:
:$b^2 = a^{p - 1}$
Let $a \equiv 0 \pmod p$.
Then by definition of [[Definition:Congruence (Number Theor... | Integer to Power of p-1 over 2 Modulo p | https://proofwiki.org/wiki/Integer_to_Power_of_p-1_over_2_Modulo_p | https://proofwiki.org/wiki/Integer_to_Power_of_p-1_over_2_Modulo_p | [
"Modulo Arithmetic",
"Number Theory"
] | [
"Definition:Odd Prime",
"Definition:Modulo Operation",
"Definition:Congruence (Number Theory)"
] | [
"Definition:Congruence (Number Theory)",
"Definition:Congruence (Number Theory)",
"Definition:Divisor (Algebra)/Integer",
"Definition:Congruence (Number Theory)",
"Fermat's Little Theorem",
"Difference of Two Squares",
"Modulo Subtraction is Well-Defined",
"Definition:Odd Prime",
"Definition:Contrad... |
proofwiki-2555 | Image Filter is Filter | Let $X, Y$ be sets.
Let $\powerset X$ and $\powerset Y$ be the power sets of $X$ and $Y$ respectively.
Let $f: X \to Y$ be a mapping.
Let $\FF \subset \powerset X$ be a filter on $X$.
Then the image filter of $\FF$ {{WRT}} $f$:
:$f \sqbrk \FF := \set {U \subseteq Y: f^{-1} \sqbrk U \in \FF}$
is a filter on $Y$. | From the definition of a filter we have to prove four things:
: $(1): \quad f \sqbrk \FF \subset \powerset Y$
: $(2): \quad Y \in f \sqbrk \FF, \O \notin f \sqbrk \FF$
: $(3): \quad U, V \in f \sqbrk \FF \implies U \cap V \in f \sqbrk \FF$
: $(4): \quad U \in f \sqbrk \FF, U \subseteq V \subseteq Y \implies V \in f \sq... | Let $X, Y$ be [[Definition:Set|sets]].
Let $\powerset X$ and $\powerset Y$ be the [[Definition:Power Set|power sets]] of $X$ and $Y$ respectively.
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $\FF \subset \powerset X$ be a [[Definition:Filter on Set|filter]] on $X$.
Then the [[Definition:Image Filter|... | From the definition of a [[Definition:Filter on Set|filter]] we have to prove four things:
: $(1): \quad f \sqbrk \FF \subset \powerset Y$
: $(2): \quad Y \in f \sqbrk \FF, \O \notin f \sqbrk \FF$
: $(3): \quad U, V \in f \sqbrk \FF \implies U \cap V \in f \sqbrk \FF$
: $(4): \quad U \in f \sqbrk \FF, U \subseteq V \su... | Image Filter is Filter | https://proofwiki.org/wiki/Image_Filter_is_Filter | https://proofwiki.org/wiki/Image_Filter_is_Filter | [
"Set Theory",
"Filter Theory"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Mapping",
"Definition:Filter on Set",
"Definition:Image Filter",
"Definition:Filter on Set"
] | [
"Definition:Filter on Set",
"Preimage of Intersection under Mapping",
"Preimage of Subset is Subset of Preimage",
"Definition:Filter on Set",
"Category:Set Theory",
"Category:Filter Theory"
] |
proofwiki-2556 | Equivalence of Definitions of Continuous Mapping between Topological Spaces/Point | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.
Let $x \in S_1$.
{{TFAE|def = Continuous at Point of Topological Space|view = continuity at a point of a topological space}}
=== Definition using Open Sets ===
{{Definition:C... | === Definition using Open Sets implies Definition using Neighborhoods ===
Let $f$ be a continuous mapping defined using Open Sets.
Then by definition:
:For every open set $U_2$ of $T_2$ such that $\map f x \in U_2$, there exists an open set $U_1$ of $T_1$ such that $x \in U_1$ and $f \sqbrk {U_1} \subseteq U_2$.
Let $N... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]] from $S_1$ to $S_2$.
Let $x \in S_1$.
{{TFAE|def = Continuous at Point of Topological Space|view = continuity at a point of a topologica... | === Definition using Open Sets implies Definition using Neighborhoods ===
Let $f$ be a [[Definition:Continuous Mapping (Topology)/Point/Open Sets|continuous mapping defined using Open Sets]].
Then by definition:
:For every [[Definition:Open Set (Topology)|open set]] $U_2$ of $T_2$ such that $\map f x \in U_2$, there ... | Equivalence of Definitions of Continuous Mapping between Topological Spaces/Point | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Continuous_Mapping_between_Topological_Spaces/Point | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Continuous_Mapping_between_Topological_Spaces/Point | [
"Equivalence of Definitions of Continuous Mapping between Topological Spaces",
"Continuous Mappings (Topology)",
"Filter Theory"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Point/Open Sets",
"Definition:Continuous Mapping (Topology)/Point/Neighborhoods",
"Definition:Continuous Mapping (Topology)/Point/Neighborhood Inverse",
"Definition:Continuous Mapping (Topology)/Point/Filters"
... | [
"Definition:Continuous Mapping (Topology)/Point/Open Sets",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Neighborhood (Topology)/Point",
"Definition:Neighborhood (Topology)/Point",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topol... |
proofwiki-2557 | Filter on Product Space Converges to Point iff Projections Converge to Projections of Point | Let $\family{X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces where $I$ is an arbitrary index set.
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Let $\FF$ be a filter on $X$.
Let $x \in X$.
Let $\pr_i: X \to X_i$ denote the projection from $X$ onto $X_i$.
The... | === Necessary Condition ===
Let $\FF$ converge to $x$.
Let $i \in I$.
From Projection from Product Topology is Continuous, $\pr_i : X \to X_i$ is continuous.
Thus $\map {\pr_i} \FF$ converges to $\map {\pr_i} x$ as claimed.
{{qed|lemma}} | Let $\family{X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
Let $\ds X := \prod_{i \mathop \in I} X_i$ be the corresponding [[Defini... | === Necessary Condition ===
Let $\FF$ [[Definition:Convergent Filter|converge]] to $x$.
Let $i \in I$.
From [[Projection from Product Topology is Continuous]], $\pr_i : X \to X_i$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
Thus $\map {\pr_i} \FF$ [[Definition:Convergent Filter|converges]... | Filter on Product Space Converges to Point iff Projections Converge to Projections of Point | https://proofwiki.org/wiki/Filter_on_Product_Space_Converges_to_Point_iff_Projections_Converge_to_Projections_of_Point | https://proofwiki.org/wiki/Filter_on_Product_Space_Converges_to_Point_iff_Projections_Converge_to_Projections_of_Point | [
"Product Topology",
"Filter Theory",
"Projections"
] | [
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Product Space (Topology)",
"Definition:Filter on Set",
"Definition:Projection (Mapping Theory)",
"Definition:Convergent Filter",
"Definition:Image Filter"
] | [
"Definition:Convergent Filter",
"Projection from Product Topology is Continuous",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Convergent Filter"
] |
proofwiki-2558 | Image of Ultrafilter is Ultrafilter | Let $X, Y$ be sets.
Let $f: X \to Y$ be a mapping.
Let $\FF$ be an ultrafilter on $X$.
Then the image filter $f \sqbrk \FF$ is an ultrafilter on $Y$. | From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.
Let $\GG$ be a filter on $Y$ such that $f \sqbrk \FF \subseteq \GG$.
We have to show that $f \sqbrk \FF = \GG$.
{{AimForCont}} there exists $U \in \GG$ such that $U \notin f \sqbrk \FF$.
By the definition of $f \sqbrk \FF$ this implies that $f^{-1} \sq... | Let $X, Y$ be [[Definition:Set|sets]].
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $X$.
Then the [[Definition:Image Filter|image filter]] $f \sqbrk \FF$ is an [[Definition:Ultrafilter on Set|ultrafilter]] on $Y$. | From [[Image Filter is Filter]], we have that $\FF$ is a [[Definition:Filter on Set|filter]] on $Y$.
Let $\GG$ be a [[Definition:Filter on Set|filter]] on $Y$ such that $f \sqbrk \FF \subseteq \GG$.
We have to show that $f \sqbrk \FF = \GG$.
{{AimForCont}} there exists $U \in \GG$ such that $U \notin f \sqbrk \FF$.... | Image of Ultrafilter is Ultrafilter/Proof 1 | https://proofwiki.org/wiki/Image_of_Ultrafilter_is_Ultrafilter | https://proofwiki.org/wiki/Image_of_Ultrafilter_is_Ultrafilter/Proof_1 | [
"Image of Ultrafilter is Ultrafilter",
"Ultrafilters on Sets"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Ultrafilter on Set",
"Definition:Image Filter",
"Definition:Ultrafilter on Set"
] | [
"Image Filter is Filter",
"Definition:Filter on Set",
"Definition:Filter on Set",
"Definition:Ultrafilter on Set/Definition 3",
"Definition:Filter on Set",
"Definition:Contradiction",
"Definition:Ultrafilter on Set"
] |
proofwiki-2559 | Image of Ultrafilter is Ultrafilter | Let $X, Y$ be sets.
Let $f: X \to Y$ be a mapping.
Let $\FF$ be an ultrafilter on $X$.
Then the image filter $f \sqbrk \FF$ is an ultrafilter on $Y$. | From Image Filter is Filter, we have that $\FF$ is a filter on $Y$.
In view of definition of ultrafilter, we need to show that:
:for every $A \subseteq Y$, either $A \in f \sqbrk \FF$ or $\relcomp Y A \in f \sqbrk \FF$
Let $A \subseteq Y$ be arbitrary.
Suppose $A \not \in f \sqbrk \FF$.
Then, by definition of image fil... | Let $X, Y$ be [[Definition:Set|sets]].
Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
Let $\FF$ be an [[Definition:Ultrafilter on Set|ultrafilter]] on $X$.
Then the [[Definition:Image Filter|image filter]] $f \sqbrk \FF$ is an [[Definition:Ultrafilter on Set|ultrafilter]] on $Y$. | From [[Image Filter is Filter]], we have that $\FF$ is a [[Definition:Filter on Set|filter]] on $Y$.
In view of definition of [[Definition:Ultrafilter on Set/Definition 3|ultrafilter]], we need to show that:
:for every $A \subseteq Y$, either $A \in f \sqbrk \FF$ or $\relcomp Y A \in f \sqbrk \FF$
Let $A \subseteq Y... | Image of Ultrafilter is Ultrafilter/Proof 2 | https://proofwiki.org/wiki/Image_of_Ultrafilter_is_Ultrafilter | https://proofwiki.org/wiki/Image_of_Ultrafilter_is_Ultrafilter/Proof_2 | [
"Image of Ultrafilter is Ultrafilter",
"Ultrafilters on Sets"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Ultrafilter on Set",
"Definition:Image Filter",
"Definition:Ultrafilter on Set"
] | [
"Image Filter is Filter",
"Definition:Filter on Set",
"Definition:Ultrafilter on Set/Definition 3",
"Definition:Image Filter",
"Definition:Ultrafilter on Set/Definition 3",
"Complement of Preimage equals Preimage of Complement",
"Definition:Image Filter"
] |
proofwiki-2560 | Tychonoff's Theorem | Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact. | First assume that $X$ is compact.
From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.
It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
By Equivalence of Definitions of Compact Topological... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]].
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Pr... | First assume that $X$ is [[Definition:Compact Topological Space|compact]].
From [[Projection from Product Topology is Continuous]], the [[Definition:Projection (Mapping Theory)|projections]] $\pr_i : X \to X_i$ are [[Definition:Continuous Mapping (Topology)|continuous]].
It follows from [[Continuous Image of Compact... | Tychonoff's Theorem/Proof 1 | https://proofwiki.org/wiki/Tychonoff's_Theorem | https://proofwiki.org/wiki/Tychonoff's_Theorem/Proof_1 | [
"Tychonoff's Theorem",
"Compact Topological Spaces",
"Product Spaces"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Product Space (Topology)",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space"
] | [
"Definition:Compact Topological Space",
"Projection from Product Topology is Continuous",
"Definition:Projection (Mapping Theory)",
"Definition:Continuous Mapping (Topology)",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space",
... |
proofwiki-2561 | Tychonoff's Theorem | Let $I$ be an indexing set.
Let $\family {X_i}_{i \mathop \in I}$ be an indexed family of non-empty topological spaces.
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding product space.
Then $X$ is compact {{iff}} each $X_i$ is compact. | First assume that $X$ is compact.
From Projection from Product Topology is Continuous, the projections $\pr_i : X \to X_i$ are continuous.
It follows from Continuous Image of Compact Space is Compact that the $X_i$ are compact.
Assume now that each $X_i$ is compact.
By Compact Space satisfies Finite Intersection Axiom... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {X_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Topological Space|topological spaces]].
Let $\ds X = \prod_{i \mathop \in I} X_i$ be the corresponding [[Definition:Pr... | First assume that $X$ is [[Definition:Compact Topological Space|compact]].
From [[Projection from Product Topology is Continuous]], the [[Definition:Projection (Mapping Theory)|projections]] $\pr_i : X \to X_i$ are [[Definition:Continuous Mapping (Topology)|continuous]].
It follows from [[Continuous Image of Compact... | Tychonoff's Theorem/Proof 2 | https://proofwiki.org/wiki/Tychonoff's_Theorem | https://proofwiki.org/wiki/Tychonoff's_Theorem/Proof_2 | [
"Tychonoff's Theorem",
"Compact Topological Spaces",
"Product Spaces"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Non-Empty Set",
"Definition:Topological Space",
"Definition:Product Space (Topology)",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space"
] | [
"Definition:Compact Topological Space",
"Projection from Product Topology is Continuous",
"Definition:Projection (Mapping Theory)",
"Definition:Continuous Mapping (Topology)",
"Continuous Image of Compact Space is Compact",
"Definition:Compact Topological Space",
"Definition:Compact Topological Space",
... |
proofwiki-2562 | Schur's Inequality | Let $x, y, z \in \R_{\ge 0}$ be positive real numbers.
Let $t \in \R, t > 0$ be a (strictly) positive real number.
Then:
:$x^t \paren {x - y} \paren {x - z} + y^t \paren {y - z} \paren {y - x} + z^t \paren {z - x} \paren {z - y} \ge 0$
The equality holds {{iff}} either:
: $x = y = z$
: Two of them are equal and the oth... | We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.
{{WLOG}}, we can assume that $x \ge y \ge z \ge 0$.
Consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
We see that every term in the above is non-negative. So, directly:... | Let $x, y, z \in \R_{\ge 0}$ be [[Definition:Positive Real Number|positive real numbers]].
Let $t \in \R, t > 0$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Then:
:$x^t \paren {x - y} \paren {x - z} + y^t \paren {y - z} \paren {y - x} + z^t \paren {z - x} \paren {z - y} \ge 0$
... | We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.
{{WLOG}}, we can assume that $x \ge y \ge z \ge 0$.
Consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
We see that every term in the above is [[Definition:Non-Negat... | Schur's Inequality | https://proofwiki.org/wiki/Schur's_Inequality | https://proofwiki.org/wiki/Schur's_Inequality | [
"Algebra"
] | [
"Definition:Positive/Real Number",
"Definition:Strictly Positive/Real Number",
"Definition:Positive/Integer",
"Definition:Even Integer"
] | [
"Definition:Positive/Real Number",
"Definition:Positive/Real Number",
"Definition:Positive/Integer",
"Definition:Even Integer",
"Dirichlet's Box Principle/Corollary",
"Definition:Sign of Number",
"Definition:Positive/Number",
"Definition:Positive/Real Number",
"Definition:Positive/Real Number",
"E... |
proofwiki-2563 | Wallis's Product | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | From the Reduction Formula for Integral of Power of Sine, we have:
:$\ds (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} x \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\pi / 2} \sin^n x \rd x$
As $\cos \dfrac \pi 2 = 0$ from Shape of Cosine Function, we have... | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | From the [[Reduction Formula for Integral of Power of Sine]], we have:
:$\ds (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} x \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$
Let $I_n$ be defined as:
:$\ds I_n = \int_0^{\pi / 2} \sin^n x \rd x$
As $\cos \dfrac \pi 2 = 0$ from [[Shape of Cosine Function... | Wallis's Product/Original Proof | https://proofwiki.org/wiki/Wallis's_Product | https://proofwiki.org/wiki/Wallis's_Product/Original_Proof | [
"Wallis's Product",
"Examples of Infinite Products",
"Formulas for Pi"
] | [] | [
"Reduction Formula for Integral of Power of Sine",
"Shape of Cosine Function",
"Definition:Even Integer",
"Definition:Odd Integer",
"Shape of Sine Function",
"Relative Sizes of Definite Integrals",
"Squeeze Theorem"
] |
proofwiki-2564 | Wallis's Product | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | Into Euler Formula for Sine Function:
{{begin-eqn}}
{{eqn | l = \dfrac {\sin x} x
| r = \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots
| c =
}}
{{eqn | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
| c =
}... | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | Into [[Euler Formula for Sine Function]]:
{{begin-eqn}}
{{eqn | l = \dfrac {\sin x} x
| r = \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots
| c =
}}
{{eqn | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
| c... | Wallis's Product/Proof 1 | https://proofwiki.org/wiki/Wallis's_Product | https://proofwiki.org/wiki/Wallis's_Product/Proof_1 | [
"Wallis's Product",
"Examples of Infinite Products",
"Formulas for Pi"
] | [] | [
"Euler Formula for Sine Function",
"Sine of Half-Integer Multiple of Pi"
] |
proofwiki-2565 | Wallis's Product | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \prod_{n \mathop = 1}^\infty \frac {\paren {2 n - 0} \paren {2 n - 0} \times \dfrac 1 2 \times \dfrac 1 2} {\paren {2 n - 1} \paren {2 n + 1} \times \dfrac 1 2 \times \dfrac 1 2}
| c = multiplying ... | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots
| c =
}}
{{eqn | r = \frac \pi 2
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1}
| r = \prod_{n \mathop = 1}^\infty \frac {\paren {2 n - 0} \paren {2 n - 0} \times \dfrac 1 2 \times \dfrac 1 2} {\paren {2 n - 1} \paren {2 n + 1} \times \dfrac 1 2 \times \dfrac 1 2}
| c = multiplying ... | Wallis's Product/Proof 3 | https://proofwiki.org/wiki/Wallis's_Product | https://proofwiki.org/wiki/Wallis's_Product/Proof_3 | [
"Wallis's Product",
"Examples of Infinite Products",
"Formulas for Pi"
] | [] | [
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta",
"Gamma Function of One Half",
"Gamma Difference Equation",
"Gamma Function Extends Factorial"
] |
proofwiki-2566 | Number of Regions in Plane Defined by Given Number of Lines | The maximum number $L_n$ of regions in the plane that can be defined by $n$ straight lines in the plane is:
:$L_n = \dfrac {n \paren {n + 1} } 2 + 1$
{{OEIS|A000124}} | === Setting up a Recurrence Rule ===
First we consider the plane with no lines at all.
This has one region, so $L_0 = 1$.
Now when we have one line, we divide the plane into two regions, so $L_1 = 2$.
Now consider the $n$th line.
This increase the number of regions by $k$ iff it splits $k$ of the old regions.
It can sp... | The maximum number $L_n$ of [[Definition:Region of Plane|regions in the plane]] that can be defined by $n$ [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]] is:
:$L_n = \dfrac {n \paren {n + 1} } 2 + 1$
{{OEIS|A000124}} | === Setting up a Recurrence Rule ===
First we consider the plane with no lines at all.
This has one [[Definition:Region of Plane|region]], so $L_0 = 1$.
Now when we have one line, we divide the plane into two regions, so $L_1 = 2$.
Now consider the $n$th line.
This increase the number of regions by $k$ iff it spl... | Number of Regions in Plane Defined by Given Number of Lines | https://proofwiki.org/wiki/Number_of_Regions_in_Plane_Defined_by_Given_Number_of_Lines | https://proofwiki.org/wiki/Number_of_Regions_in_Plane_Defined_by_Given_Number_of_Lines | [
"Discrete Mathematics",
"Number of Regions in Plane Defined by Given Number of Lines"
] | [
"Definition:Region/Plane",
"Definition:Line/Straight Line",
"Definition:Plane Surface/The Plane"
] | [
"Definition:Region/Plane",
"Definition:Line/Straight Line",
"Definition:Intersection (Geometry)",
"Definition:Point"
] |
proofwiki-2567 | Forward Difference of Falling Factorial | Let $f: \R \to \R$ be a real function.
Let $\Delta$ denote the forward difference operator.
Let $x^{\underline m}$ be the $m$th falling factorial of $x$
Then:
:$\map \Delta {x^{\underline m} } = m x^{\underline {m - 1} }$ | From the definitions:
{{begin-eqn}}
{{eqn | l = \map \Delta {x^{\underline m} }
| r = \paren {x + 1}^{\underline m} - x^{\underline m}
| c = {{Defof|Forward Difference Operator}}
}}
{{eqn | r = \prod_{k \mathop = 0}^{m - 1} \paren {x + 1 - k} - \prod_{k \mathop = 0}^{m - 1} \paren {x - k}
| c = {{Defo... | Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
Let $\Delta$ denote the [[Definition:Forward Difference Operator|forward difference operator]].
Let $x^{\underline m}$ be the [[Definition:Falling Factorial|$m$th falling factorial]] of $x$
Then:
:$\map \Delta {x^{\underline m} } = m x^{\underline... | From the definitions:
{{begin-eqn}}
{{eqn | l = \map \Delta {x^{\underline m} }
| r = \paren {x + 1}^{\underline m} - x^{\underline m}
| c = {{Defof|Forward Difference Operator}}
}}
{{eqn | r = \prod_{k \mathop = 0}^{m - 1} \paren {x + 1 - k} - \prod_{k \mathop = 0}^{m - 1} \paren {x - k}
| c = {{Def... | Forward Difference of Falling Factorial | https://proofwiki.org/wiki/Forward_Difference_of_Falling_Factorial | https://proofwiki.org/wiki/Forward_Difference_of_Falling_Factorial | [
"Finite Calculus"
] | [
"Definition:Real Function",
"Definition:Finite Difference Operator/Forward Difference",
"Definition:Falling Factorial"
] | [
"Translation of Index Variable of Product",
"Category:Finite Calculus"
] |
proofwiki-2568 | Chen's Theorem | Every sufficiently large even integer is the sum of either:
:$(1): \quad$ two primes, or
:$(2): \quad$ a prime and a semiprime. | {{ProofWanted}}
{{Namedfor|Chen Jingrun|cat = Chen}} | Every [[Definition:Sufficiently Large|sufficiently large]] [[Definition:Even Integer|even integer]] is the sum of either:
:$(1): \quad$ two [[Definition:Prime Number|primes]], or
:$(2): \quad$ a [[Definition:Prime Number|prime]] and a [[Definition:Semiprime Number|semiprime]]. | {{ProofWanted}}
{{Namedfor|Chen Jingrun|cat = Chen}} | Chen's Theorem | https://proofwiki.org/wiki/Chen's_Theorem | https://proofwiki.org/wiki/Chen's_Theorem | [
"Prime Numbers",
"Semiprimes",
"Number Theory"
] | [
"Definition:Sufficiently Large",
"Definition:Even Integer",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Semiprime Number"
] | [] |
proofwiki-2569 | Binomial Coefficient involving Prime | Let $p$ be a prime number.
Let $\dbinom n p$ be a binomial coefficient.
Then:
:$\dbinom n p \equiv \floor {\dfrac n p} \pmod p$
where:
:$\floor {\dfrac n p}$
denotes the floor function. | Follows directly from Lucas' Theorem:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
where $k = p$.
Then:
:$k \bmod p = 0$
and so by Binomial Coefficient with Zero:
:$\dbinom {n \bmod p} {k \bmod p} = 1$
Also:
:$\floor {k / p} = 1$
and by Binomial Coefficient w... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $\dbinom n p$ be a [[Definition:Binomial Coefficient|binomial coefficient]].
Then:
:$\dbinom n p \equiv \floor {\dfrac n p} \pmod p$
where:
:$\floor {\dfrac n p}$
denotes the [[Definition:Floor Function|floor function]]. | Follows directly from [[Lucas' Theorem]]:
:$\dbinom n k \equiv \dbinom {\floor {n / p} } {\floor {k / p} } \dbinom {n \bmod p} {k \bmod p} \pmod p$
where $k = p$.
Then:
:$k \bmod p = 0$
and so by [[Binomial Coefficient with Zero]]:
:$\dbinom {n \bmod p} {k \bmod p} = 1$
Also:
:$\floor {k / p} = 1$
and by [[Binom... | Binomial Coefficient involving Prime | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_involving_Prime | [
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient",
"Definition:Floor Function"
] | [
"Lucas' Theorem",
"Binomial Coefficient with Zero",
"Binomial Coefficient with One"
] |
proofwiki-2570 | Binomial Coefficient of Prime Minus One Modulo Prime | Let $p$ be a prime number.
Then:
:$0 \le k \le p - 1 \implies \dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$
where $\dbinom {p - 1} k$ denotes a binomial coefficient. | From Binomial Coefficient of Prime, we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From Pascal's Rule:
:$\dbinom {p-1} k + \dbinom {p - 1} {k - 1} = \dbinom p k \equiv 0 \pmod p$
This certainly holds for $k = 1$, and so we have:
:$\dbinom {p - 1} 1 + \dbinom {p - 1} 0 = \dbinom p 1 \equiv 0 \pmod p$... | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$0 \le k \le p - 1 \implies \dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$
where $\dbinom {p - 1} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Binomial Coefficient of Prime]], we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From [[Pascal's Rule]]:
:$\dbinom {p-1} k + \dbinom {p - 1} {k - 1} = \dbinom p k \equiv 0 \pmod p$
This certainly holds for $k = 1$, and so we have:
:$\dbinom {p - 1} 1 + \dbinom {p - 1} 0 = \dbinom p 1 \equiv ... | Binomial Coefficient of Prime Minus One Modulo Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime_Minus_One_Modulo_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime_Minus_One_Modulo_Prime | [
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient"
] | [
"Binomial Coefficient of Prime",
"Pascal's Rule",
"Binomial Coefficient with Zero"
] |
proofwiki-2571 | Binomial Coefficient of Prime Plus One Modulo Prime | Let $p$ be a prime number.
Then:
:$2 \le k \le p - 1 \implies \dbinom {p + 1} k \equiv 0 \pmod p$
where $\dbinom {p + 1} k$ denotes a binomial coefficient. | From Binomial Coefficient of Prime, we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From Pascal's Rule:
:$\dbinom {p + 1} k = \dbinom p {k - 1} + \dbinom p k$
The result follows immediately.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Then:
:$2 \le k \le p - 1 \implies \dbinom {p + 1} k \equiv 0 \pmod p$
where $\dbinom {p + 1} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Binomial Coefficient of Prime]], we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From [[Pascal's Rule]]:
:$\dbinom {p + 1} k = \dbinom p {k - 1} + \dbinom p k$
The result follows immediately.
{{qed}} | Binomial Coefficient of Prime Plus One Modulo Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime_Plus_One_Modulo_Prime | https://proofwiki.org/wiki/Binomial_Coefficient_of_Prime_Plus_One_Modulo_Prime | [
"Binomial Coefficients"
] | [
"Definition:Prime Number",
"Definition:Binomial Coefficient"
] | [
"Binomial Coefficient of Prime",
"Pascal's Rule"
] |
proofwiki-2572 | Set of Prime Numbers is Infinite | The number of primes is infinite. | {{AimForCont}} there exists a greatest prime number $N$.
Hence, let $S$ denote the finite set of all prime numbers.
Euclid's Theorem states that:
:For any finite set of prime numbers, there exists a prime number not in that set.
Hence there exists $N'$ not in $S$.
If $N' < N$ then $N \in S$.
Hence:
:$N' > N$
which cont... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | {{AimForCont}} there exists a [[Definition:Greatest Element|greatest]] [[Definition:Prime Number|prime number]] $N$.
Hence, let $S$ denote the [[Definition:Finite Set|finite set]] of all [[Definition:Prime Number|prime numbers]].
[[Euclid's Theorem]] states that:
:For any [[Definition:Finite Set|finite set]] of [[De... | Set of Prime Numbers is Infinite/Proof 1 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_1 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Greatest Element",
"Definition:Prime Number",
"Definition:Finite Set",
"Definition:Prime Number",
"Euclid's Theorem",
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Prime Number",
"Definition:Set",
"Definition:Contradiction",
"Definition:Greatest Element",
"Definit... |
proofwiki-2573 | Set of Prime Numbers is Infinite | The number of primes is infinite. | Define a topology on the integers $\Z$ by declaring a subset $U \subseteq \Z$ to be an open set {{iff}} it is either:
:the empty set $\O$
or:
:a union of sequences $\map S {a, b}$ such that $a \ne 0$, where:
::$\map S {a, b} = \set {a n + b: n \in \Z} = a \Z + b$
In other words, $U$ is open {{iff}} every $x \in U$ admi... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | Define a [[Definition:Topology|topology]] on the [[Definition:Integer|integers]] $\Z$ by declaring a [[Definition:Subset|subset]] $U \subseteq \Z$ to be an [[Definition:Open Set (Topology)|open set]] {{iff}} it is either:
:the [[Definition:Empty Set|empty set]] $\O$
or:
:a [[Definition:Set Union|union]] of [[Definition... | Set of Prime Numbers is Infinite/Proof 2 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_2 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Topology",
"Definition:Integer",
"Definition:Subset",
"Definition:Open Set/Topology",
"Definition:Empty Set",
"Definition:Set Union",
"Definition:Sequence",
"Definition:Open Set/Topology",
"Definition:Zero (Number)",
"Definition:Integer",
"Axiom:Open Set Axioms",
"Definition:Set Un... |
proofwiki-2574 | Set of Prime Numbers is Infinite | The number of primes is infinite. | {{AimForCont}} that there are only $N$ prime numbers.
Let the set of all primes be:
:$\Bbb P = \set {p_1, p_2, \ldots, p_N}$
By the Fundamental Theorem of Arithmetic, every integer $k > 1$ can be expressed in the form:
:$k = {p_1}^{a_1} {p_2}^{a_2} \dotsm {p_N}^{a_N}$
Let $n > 1$ be fixed.
Let $a$ be the largest expone... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | {{AimForCont}} that there are only $N$ [[Definition:Prime Number|prime numbers]].
Let the [[Definition:Set|set]] of all [[Definition:Prime Number|primes]] be:
:$\Bbb P = \set {p_1, p_2, \ldots, p_N}$
By the [[Fundamental Theorem of Arithmetic]], every [[Definition:Integer|integer]] $k > 1$ can be expressed in the for... | Set of Prime Numbers is Infinite/Proof 3 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_3 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Prime Number",
"Definition:Set",
"Definition:Prime Number",
"Fundamental Theorem of Arithmetic",
"Definition:Integer",
"Definition:Power (Algebra)/Exponent",
"Definition:Prime Decomposition",
"Definition:Positive/Integer",
"Sum of Infinite Geometric Sequence",
"Definition:Contradiction... |
proofwiki-2575 | Set of Prime Numbers is Infinite | The number of primes is infinite. | {{AimForCont}} there exist only a finite number of primes.
From Sum of Reciprocals of Powers as Euler Product:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
When $z = 1$ this gives:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 n = \prod_p \frac 1 {1 - 1 / p}$
{{mistake|You cannot choose $z{{=}}1... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | {{AimForCont}} there exist only a [[Definition:Finite Set|finite]] number of [[Definition:Prime Number|primes]].
From [[Sum of Reciprocals of Powers as Euler Product]]:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
When $z = 1$ this gives:
:$\ds \sum_{n \mathop \ge 1} \dfrac 1 n = \prod... | Set of Prime Numbers is Infinite/Proof 4 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_4 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Finite Set",
"Definition:Prime Number",
"Sum of Reciprocals of Powers as Euler Product",
"Sum of Reciprocals of Powers as Euler Product",
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Finite Set",
"Harmonic Series is Divergent",
"Definition:Divergent Series",
"Definit... |
proofwiki-2576 | Set of Prime Numbers is Infinite | The number of primes is infinite. | {{AimForCont}} there exist $n$ prime numbers.
Consider the Fermat number $F_n$.
From Goldbach's Theorem, $F_n$ is coprime to each of $F_0$ to $F_{n - 1}$.
Therefore there must be a prime number which is a divisor of $F_n$ which is not a divisor of any of $F_0$ to $F_n$.
But, again from Goldbach's Theorem, each of $F_0$... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | {{AimForCont}} there exist $n$ [[Definition:Prime Number|prime numbers]].
Consider the [[Definition:Fermat Number|Fermat number]] $F_n$.
From [[Goldbach's Theorem]], $F_n$ is [[Definition:Coprime Integers|coprime]] to each of $F_0$ to $F_{n - 1}$.
Therefore there must be a [[Definition:Prime Number|prime number]] wh... | Set of Prime Numbers is Infinite/Proof 5 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_5 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Prime Number",
"Definition:Fermat Number",
"Goldbach's Theorem",
"Definition:Coprime/Integers",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Goldbach's Theorem",
"Definition:Coprime/Integers",
"Definition:Fermat Number",
... |
proofwiki-2577 | Set of Prime Numbers is Infinite | The number of primes is infinite. | Let $p_1, p_2, \ldots, p_j$ be the first $j$ primes.
For each real $x$ and natural number $j$, let:
:$\map {M_j} x = \set {n \in \N \mid n \le x, \, n \text { is not divisible by any prime } p \text { with } p > p_j}$
Define:
:$\map {N_j} x = \# \map {M_j} x$
Let $n \in \map {M_j} x$ for some $x$, $j$.
We can write:
... | The number of [[Definition:Prime Number|primes]] is [[Definition:Infinite Set|infinite]]. | Let $p_1, p_2, \ldots, p_j$ be the first $j$ [[Definition:Prime Number|primes]].
For each [[Definition:Real Number|real]] $x$ and [[Definition:Natural Number|natural number]] $j$, let:
:$\map {M_j} x = \set {n \in \N \mid n \le x, \, n \text { is not divisible by any prime } p \text { with } p > p_j}$
Define:
:$\map ... | Set of Prime Numbers is Infinite/Proof 6 | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite | https://proofwiki.org/wiki/Set_of_Prime_Numbers_is_Infinite/Proof_6 | [
"Euclid's Theorem"
] | [
"Definition:Prime Number",
"Definition:Infinite Set"
] | [
"Definition:Prime Number",
"Definition:Real Number",
"Definition:Natural Numbers",
"Definition:Integer",
"Definition:Square-Free Integer",
"Fundamental Theorem of Arithmetic",
"Definition:Finite Set",
"Definition:Prime Number",
"Definition:Contradiction",
"Definition:Infinite Set",
"Definition:P... |
proofwiki-2578 | Factorial Divisible by Binary Root | Let $n \in \Z: n \ge 1$.
Let $n$ be expressed in binary notation:
:$n = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}$
where $e_1 > e_2 > \cdots > e_r \ge 0$.
Let $n!$ be the factorial of $n$.
Then $n!$ is divisible by $2^{n - r}$, but not by $2^{n - r + 1}$. | A direct application of Factorial Divisible by Prime Power.
{{qed}}
Category:Factorials
Category:Binary Notation
lhbklcp3l1qd7gzyj76tln0qu2wmrl6 | Let $n \in \Z: n \ge 1$.
Let $n$ be expressed in [[Definition:Binary Notation|binary notation]]:
:$n = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}$
where $e_1 > e_2 > \cdots > e_r \ge 0$.
Let $n!$ be the [[Definition:Factorial|factorial]] of $n$.
Then $n!$ is [[Definition:Divisor of Integer|divisible]] by $2^{n - r}$, but... | A direct application of [[Factorial Divisible by Prime Power]].
{{qed}}
[[Category:Factorials]]
[[Category:Binary Notation]]
lhbklcp3l1qd7gzyj76tln0qu2wmrl6 | Factorial Divisible by Binary Root | https://proofwiki.org/wiki/Factorial_Divisible_by_Binary_Root | https://proofwiki.org/wiki/Factorial_Divisible_by_Binary_Root | [
"Factorials",
"Binary Notation"
] | [
"Definition:Binary Notation",
"Definition:Factorial",
"Definition:Divisor (Algebra)/Integer"
] | [
"Factorial Divisible by Prime Power",
"Category:Factorials",
"Category:Binary Notation"
] |
proofwiki-2579 | Quotient and Remainder to Number Base | Let $n \in \Z: n > 0$ be an integer.
Let $n$ be expressed in base $b$:
:$\ds n = \sum_{j \mathop = 0}^m {r_j b^j}$
that is:
:$n = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Then:
:$\ds \floor {\frac n b} = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1}_b$
:$n \bmod b = r_0$
where:
:$\floor {\, \cdot \,}$ denotes the floor fun... | From the Quotient-Remainder Theorem, we have:
:$\exists q, r \in \Z: n = q b + r$
where $0 \le b < r$.
We have that:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m {r_j b^j}
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^m {r_j b^j} + r_0
| c =
}}
{{eqn | r = b \sum_{j \mathop = 1}^m {r_j b^{j-... | Let $n \in \Z: n > 0$ be an [[Definition:Integer|integer]].
Let $n$ be expressed in [[Definition:Number Base|base $b$]]:
:$\ds n = \sum_{j \mathop = 0}^m {r_j b^j}$
that is:
:$n = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
Then:
:$\ds \floor {\frac n b} = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1}_b$
:$n \bmod b = r_0$... | From the [[Quotient-Remainder Theorem]], we have:
:$\exists q, r \in \Z: n = q b + r$
where $0 \le b < r$.
We have that:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m {r_j b^j}
| c =
}}
{{eqn | r = \sum_{j \mathop = 1}^m {r_j b^j} + r_0
| c =
}}
{{eqn | r = b \sum_{j \mathop = 1}^m {r... | Quotient and Remainder to Number Base | https://proofwiki.org/wiki/Quotient_and_Remainder_to_Number_Base | https://proofwiki.org/wiki/Quotient_and_Remainder_to_Number_Base | [
"Number Theory"
] | [
"Definition:Integer",
"Definition:Number Base",
"Definition:Floor Function",
"Definition:Modulo Operation"
] | [
"Division Theorem",
"Definition:Modulo Operation"
] |
proofwiki-2580 | Factorial Divisible by Prime Power | Let $n \in \Z: n \ge 1$.
Let $p$ be a prime number.
Let $n$ be expressed in base $p$ notation:
:$\ds n = \sum_{j \mathop = 0}^m r_j p^j$
where $0 \le r_j < p$.
Let $n!$ be the factorial of $n$.
Let $p^\mu$ be the largest power of $p$ which divides $n!$, that is:
:$p^\mu \divides n!$
:$p^{\mu + 1} \nmid n!$
Then:
:$\mu ... | If $p > n$, then $\map {s_p} n = n$ and we have that:
:$\dfrac {n - \map {s_p} n} {p - 1} = 0$
which ties in with the fact that $\floor {\dfrac n p} = 0$.
Hence the result holds for $p > n$.
So, let $p \le n$.
From De Polignac's Formula, we have that:
{{begin-eqn}}
{{eqn | l = \mu
| r = \sum_{k \mathop > 0} \floo... | Let $n \in \Z: n \ge 1$.
Let $p$ be a [[Definition:Prime Number|prime number]].
Let $n$ be expressed in [[Definition:Number Base|base $p$ notation]]:
:$\ds n = \sum_{j \mathop = 0}^m r_j p^j$
where $0 \le r_j < p$.
Let $n!$ be the [[Definition:Factorial|factorial]] of $n$.
Let $p^\mu$ be the largest [[Definition:P... | If $p > n$, then $\map {s_p} n = n$ and we have that:
:$\dfrac {n - \map {s_p} n} {p - 1} = 0$
which ties in with the fact that $\floor {\dfrac n p} = 0$.
Hence the result holds for $p > n$.
So, let $p \le n$.
From [[De Polignac's Formula]], we have that:
{{begin-eqn}}
{{eqn | l = \mu
| r = \sum_{k \mathop >... | Factorial Divisible by Prime Power | https://proofwiki.org/wiki/Factorial_Divisible_by_Prime_Power | https://proofwiki.org/wiki/Factorial_Divisible_by_Prime_Power | [
"Factorials",
"Prime Numbers"
] | [
"Definition:Prime Number",
"Definition:Number Base",
"Definition:Factorial",
"Definition:Power (Algebra)",
"Definition:Digit Sum",
"Definition:Number Base"
] | [
"De Polignac's Formula",
"Quotient and Remainder to Number Base/General Result",
"Sum of Geometric Sequence"
] |
proofwiki-2581 | Wilson's Theorem/Corollary 2 | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime factor of $n!$ with multiplicity $\mu$.
Let $n$ be expressed in a base $p$ representation as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m a_j p^j
| c = where $0 \le a_j < p$
}}
{{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + ... | Proof by induction:
Let $\map P n$ be the proposition:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$
where $p, a_0, \dots, a_k, \mu$ are as defined above.
=== Basis for the Induction ===
For $n = 1$:
:$a_0 = 1, \mu = 0$
$\map P 1$ reduces to:
:$\dfrac {1!} {p^0} \equiv \paren {-1}^0 1! \pm... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $p$ be a [[Definition:Prime Factor|prime factor]] of $n!$ with [[Definition:Multiplicity of Prime Factor|multiplicity]] $\mu$.
Let $n$ be expressed in a [[Definition:Number Base|base $p$ representation]] as:
{{begin-eq... | Proof by [[Principle of Mathematical Induction|induction]]:
Let $\map P n$ be the proposition:
:$\dfrac {n!} {p^\mu} \equiv \paren {-1}^\mu a_0! a_1! \dotsm a_k! \pmod p$
where $p, a_0, \dots, a_k, \mu$ are as defined above.
=== Basis for the Induction ===
For $n = 1$:
:$a_0 = 1, \mu = 0$
$\map P 1$ reduces to:
:... | Wilson's Theorem/Corollary 2/Proof 1 | https://proofwiki.org/wiki/Wilson's_Theorem/Corollary_2 | https://proofwiki.org/wiki/Wilson's_Theorem/Corollary_2/Proof_1 | [
"Factorials",
"Prime Numbers",
"Modulo Arithmetic",
"Wilson's Theorem"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Prime Factor",
"Definition:Prime Decomposition/Multiplicity",
"Definition:Number Base"
] | [
"Principle of Mathematical Induction",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Number Base",
"Definition:Induction Step",
"Definition:Power (Algebra)",
"Wilson's Theorem/Corollary 2/Proof 1",
"Wilson's Theorem",
"Factorial/Examples/0",
"Principle of Mat... |
proofwiki-2582 | Wilson's Theorem/Corollary 2 | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $p$ be a prime factor of $n!$ with multiplicity $\mu$.
Let $n$ be expressed in a base $p$ representation as:
{{begin-eqn}}
{{eqn | l = n
| r = \sum_{j \mathop = 0}^m a_j p^j
| c = where $0 \le a_j < p$
}}
{{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + ... | Consider the numbers of $\set {1, 2, \ldots, n}$ which are not multiples of $p$.
There are $\floor {\dfrac n p}$ complete sets of $p - 1$ such consecutive elements of $\set {1, 2, \ldots, n}$.
Each one of these has a product which is congruent to $-1 \pmod p$ by Wilson's Theorem.
There are also $a_0$ left over which ar... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $p$ be a [[Definition:Prime Factor|prime factor]] of $n!$ with [[Definition:Multiplicity of Prime Factor|multiplicity]] $\mu$.
Let $n$ be expressed in a [[Definition:Number Base|base $p$ representation]] as:
{{begin-eq... | Consider the numbers of $\set {1, 2, \ldots, n}$ which are not [[Definition:Integer Multiple|multiples]] of $p$.
There are $\floor {\dfrac n p}$ complete sets of $p - 1$ such consecutive elements of $\set {1, 2, \ldots, n}$.
Each one of these has a product which is [[Definition:Congruence Modulo Integer|congruent]] t... | Wilson's Theorem/Corollary 2/Proof 2 | https://proofwiki.org/wiki/Wilson's_Theorem/Corollary_2 | https://proofwiki.org/wiki/Wilson's_Theorem/Corollary_2/Proof_2 | [
"Factorials",
"Prime Numbers",
"Modulo Arithmetic",
"Wilson's Theorem"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Prime Factor",
"Definition:Prime Decomposition/Multiplicity",
"Definition:Number Base"
] | [
"Definition:Integral Multiple/Real Numbers",
"Definition:Congruence (Number Theory)/Integers",
"Wilson's Theorem",
"Definition:Congruence (Number Theory)/Integers",
"Definition:Divisor (Algebra)/Integer",
"Definition:Integral Multiple/Real Numbers",
"Definition:Divisor (Algebra)/Integer",
"Definition:... |
proofwiki-2583 | Negative Number is Congruent to Modulus minus Number | :$\forall m, n \in \Z: -m \equiv n - m \pmod n$
where $\bmod n$ denotes congruence modulo $n$. | Let $-m = r + k n$.
Then $-m + n = r + \paren {k + 1} n$
and the result follows directly by definition. | :$\forall m, n \in \Z: -m \equiv n - m \pmod n$
where $\bmod n$ denotes [[Definition:Congruence Modulo Integer|congruence modulo $n$]]. | Let $-m = r + k n$.
Then $-m + n = r + \paren {k + 1} n$
and the result follows directly by [[Definition:Congruence Modulo Integer|definition]]. | Negative Number is Congruent to Modulus minus Number | https://proofwiki.org/wiki/Negative_Number_is_Congruent_to_Modulus_minus_Number | https://proofwiki.org/wiki/Negative_Number_is_Congruent_to_Modulus_minus_Number | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)/Integers"
] | [
"Definition:Congruence (Number Theory)/Integers"
] |
proofwiki-2584 | Gaussian Integers form Subring of Complex Numbers | The ring of Gaussian integers:
:$\struct {\Z \sqbrk i, +, \times}$
forms a subring of the set of complex numbers $\C$. | From Complex Numbers form Field, $\C$ forms a field.
By definition, a field is a ring.
Thus it is possible to use the Subring Test.
We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.
Let $a + b i, c + d i \in \Z \sqbrk i$.
Then we have $-\paren {c + d i} = -c - d i$, and so:
{{begin-eq... | The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
:$\struct {\Z \sqbrk i, +, \times}$
forms a [[Definition:Subring|subring]] of the set of [[Definition:Complex Number|complex numbers]] $\C$. | From [[Complex Numbers form Field]], $\C$ forms a [[Definition:Field (Abstract Algebra)|field]].
By definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Ring (Abstract Algebra)|ring]].
Thus it is possible to use the [[Subring Test]].
We note that $\Z \sqbrk i$ is not [[Definition:Empty Set|... | Gaussian Integers form Subring of Complex Numbers | https://proofwiki.org/wiki/Gaussian_Integers_form_Subring_of_Complex_Numbers | https://proofwiki.org/wiki/Gaussian_Integers_form_Subring_of_Complex_Numbers | [
"Subrings",
"Gaussian Integers"
] | [
"Definition:Ring of Gaussian Integers",
"Definition:Subring",
"Definition:Complex Number"
] | [
"Complex Numbers form Field",
"Definition:Field (Abstract Algebra)",
"Definition:Field (Abstract Algebra)",
"Definition:Ring (Abstract Algebra)",
"Subring Test",
"Definition:Empty Set",
"Integers form Integral Domain",
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)",
"Definition... |
proofwiki-2585 | Gaussian Rationals form Number Field | The set of Gaussian rationals $\Q \sqbrk i$, under the operations of complex addition and complex multiplication, forms a number field. | By definition, a number field is a subfield of the field of complex numbers $\C$.
Recall the definition of the Gaussian rationals:
:$\Q \sqbrk i = \set {z \in \C: z = a + b i: a, b \in \Q}$
From Complex Numbers form Field, $\C$ forms a field.
Thus it is possible to use the Subfield Test.
$\Q \sqbrk i$ is not empty, as ... | The [[Definition:Set|set]] of [[Definition:Gaussian Rational|Gaussian rationals]] $\Q \sqbrk i$, under the operations of [[Definition:Complex Addition|complex addition]] and [[Definition:Complex Multiplication|complex multiplication]], forms a [[Definition:Number Field|number field]]. | By definition, a [[Definition:Number Field|number field]] is a [[Definition:Subfield|subfield]] of the [[Definition:Field of Complex Numbers|field of complex numbers]] $\C$.
Recall the definition of the [[Definition:Gaussian Rational|Gaussian rationals]]:
:$\Q \sqbrk i = \set {z \in \C: z = a + b i: a, b \in \Q}$
Fr... | Gaussian Rationals form Number Field | https://proofwiki.org/wiki/Gaussian_Rationals_form_Number_Field | https://proofwiki.org/wiki/Gaussian_Rationals_form_Number_Field | [
"Number Fields",
"Gaussian Rationals"
] | [
"Definition:Set",
"Definition:Gaussian Rational",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers",
"Definition:Number Field"
] | [
"Definition:Number Field",
"Definition:Subfield",
"Definition:Field of Complex Numbers",
"Definition:Gaussian Rational",
"Complex Numbers form Field",
"Definition:Field (Abstract Algebra)",
"Subfield Test",
"Definition:Empty Set",
"Rational Numbers form Field",
"Definition:Multiplication/Complex N... |
proofwiki-2586 | Complex Conjugation is Field Automorphism of Complex Numbers | Consider the field of complex numbers $\C$.
The operation of complex conjugation:
:$\forall z \in \C: z \mapsto \overline z$
is a field automorphism of $\C$. | Let:
:$z_1 = x_1 + i y_1$
and:
:$z_2 = x_2 + i y_2$.
Let us define the mapping $\phi: \C \to \C$ defined as:
:$\forall z \in \C: \map \phi z = \overline z$
We check that $\phi$ has the morphism property:
By Sum of Complex Conjugates:
:$\map \phi {z_1 + z_2} = \map \phi {z_1} + \map \phi {z_2}$
By Product of Complex Con... | Consider the [[Definition:Field of Complex Numbers|field of complex numbers]] $\C$.
The operation of [[Definition:Complex Conjugate|complex conjugation]]:
:$\forall z \in \C: z \mapsto \overline z$
is a [[Definition:Field Automorphism|field automorphism]] of $\C$. | Let:
:$z_1 = x_1 + i y_1$
and:
:$z_2 = x_2 + i y_2$.
Let us define the [[Definition:Mapping|mapping]] $\phi: \C \to \C$ defined as:
:$\forall z \in \C: \map \phi z = \overline z$
We check that $\phi$ has the [[Definition:Morphism Property|morphism property]]:
By [[Sum of Complex Conjugates]]:
:$\map \phi {z_1 + z_2... | Complex Conjugation is Field Automorphism of Complex Numbers | https://proofwiki.org/wiki/Complex_Conjugation_is_Field_Automorphism_of_Complex_Numbers | https://proofwiki.org/wiki/Complex_Conjugation_is_Field_Automorphism_of_Complex_Numbers | [
"Complex Conjugates",
"Examples of Field Automorphisms"
] | [
"Definition:Field of Complex Numbers",
"Definition:Complex Conjugate",
"Definition:Field Automorphism"
] | [
"Definition:Mapping",
"Definition:Morphism Property",
"Sum of Complex Conjugates",
"Product of Complex Conjugates",
"Definition:Morphism Property",
"Definition:Addition/Complex Numbers",
"Definition:Multiplication/Complex Numbers",
"Definition:Complex Conjugate",
"Definition:Field Homomorphism",
"... |
proofwiki-2587 | Tschirnhaus Transformation yields Depressed Polynomial | Let $\map f x$ be a polynomial of order $n$:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$
Then the Tschirnhaus transformation:
$y = x + \dfrac {a_{n - 1} } {n a_n}$
converts $f$ into a depressed polynomial:
:$b_n y^n + b_{n - 1} y^{n - 1} + \cdots + b_1 y + b_0$
where $b_{n - 1} = 0$. | Substituting $y = x + \dfrac {a_{n - 1} } {n a_n}$ gives us:
:$x = y - \dfrac {a_{n - 1} } {n a_n}$
By the Binomial Theorem:
:$a_n x^n = a_n \paren {y^n - \dfrac {a_{n - 1} } {a_n} y^{n - 1} + \map {f'_{n - 2} } y}$
where $\map {f'_{n - 2} } y$ is a polynomial in $y$ of order $n - 2$.
Now we note that:
:$a_{n - 1} x^{n... | Let $\map f x$ be a [[Definition:Polynomial|polynomial]] of order $n$:
:$a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$
Then the [[Definition:Tschirnhaus Transformation|Tschirnhaus transformation]]:
$y = x + \dfrac {a_{n - 1} } {n a_n}$
converts $f$ into a [[Definition:Depressed Polynomial|depressed polynomia... | Substituting $y = x + \dfrac {a_{n - 1} } {n a_n}$ gives us:
:$x = y - \dfrac {a_{n - 1} } {n a_n}$
By the [[Binomial Theorem]]:
:$a_n x^n = a_n \paren {y^n - \dfrac {a_{n - 1} } {a_n} y^{n - 1} + \map {f'_{n - 2} } y}$
where $\map {f'_{n - 2} } y$ is a [[Definition:Polynomial|polynomial]] in $y$ of order $n - 2$.
No... | Tschirnhaus Transformation yields Depressed Polynomial | https://proofwiki.org/wiki/Tschirnhaus_Transformation_yields_Depressed_Polynomial | https://proofwiki.org/wiki/Tschirnhaus_Transformation_yields_Depressed_Polynomial | [
"Polynomial Theory",
"Tschirnhaus Transformations"
] | [
"Definition:Polynomial",
"Definition:Tschirnhaus Transformation",
"Definition:Depressed Polynomial"
] | [
"Binomial Theorem",
"Definition:Polynomial",
"Definition:Polynomial",
"Category:Polynomial Theory",
"Category:Tschirnhaus Transformations"
] |
proofwiki-2588 | Ring of Sets is Semiring of Sets | Let $\RR$ be a ring of sets.
Then $\RR$ is also a semiring of sets. | {{Explain|It is not clear which of the definitions in Definition:Ring of Sets is being used. In any case, this proof seems to just address axiom (3) in Definition:Semiring of Sets. An explanation (even if trivial) of why (1) and (2) hold is missing. Especially because the explanation depends on which definition of ring... | Let $\RR$ be a [[Definition:Ring of Sets|ring of sets]].
Then $\RR$ is also a [[Definition:Semiring of Sets|semiring of sets]]. | {{Explain|It is not clear which of the definitions in [[Definition:Ring of Sets]] is being used. In any case, this proof seems to just address axiom (3) in [[Definition:Semiring of Sets]]. An explanation (even if trivial) of why (1) and (2) hold is missing. Especially because the explanation depends on which definition... | Ring of Sets is Semiring of Sets | https://proofwiki.org/wiki/Ring_of_Sets_is_Semiring_of_Sets | https://proofwiki.org/wiki/Ring_of_Sets_is_Semiring_of_Sets | [
"Rings of Sets",
"Semirings of Sets"
] | [
"Definition:Ring of Sets",
"Definition:Semiring of Sets"
] | [
"Definition:Ring of Sets",
"Definition:Semiring of Sets",
"Definition:Set Difference",
"Definition:Relative Complement",
"Union with Relative Complement",
"Ring of Sets Closed under Various Operations",
"Set Difference Intersection with Second Set is Empty Set",
"Definition:Set Partition/Finite Expans... |
proofwiki-2589 | Set of All Real Intervals is Semiring of Sets | Let $\mathbb S$ be the set of all real intervals.
Then $\mathbb S$ is a semiring of sets, but is '''not''' a ring of sets. | Consider the types of real interval that exist.
In the following, $a, b \in \R$ are real numbers.
There are:
* Closed intervals:
** $\closedint a b = \set {x \in \R: a \le x \le b}$
* Open intervals:
** $\openint a b = \set {x \in \R: a < x < b}$
* Half-open intervals:
** $\hointr a b = \set {x \in \R: a \le x < b}$
**... | Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Real Interval|real intervals]].
Then $\mathbb S$ is a [[Definition:Semiring of Sets|semiring of sets]], but is '''not''' a [[Definition:Ring of Sets|ring of sets]]. | Consider the types of [[Definition:Real Interval|real interval]] that exist.
In the following, $a, b \in \R$ are [[Definition:Real Number|real numbers]].
There are:
* [[Definition:Closed Real Interval|Closed intervals]]:
** $\closedint a b = \set {x \in \R: a \le x \le b}$
* [[Definition:Open Real Interval|Open inte... | Set of All Real Intervals is Semiring of Sets | https://proofwiki.org/wiki/Set_of_All_Real_Intervals_is_Semiring_of_Sets | https://proofwiki.org/wiki/Set_of_All_Real_Intervals_is_Semiring_of_Sets | [
"Semirings of Sets",
"Analysis"
] | [
"Definition:Set",
"Definition:Real Interval",
"Definition:Semiring of Sets",
"Definition:Ring of Sets"
] | [
"Definition:Real Interval",
"Definition:Real Number",
"Definition:Real Interval/Closed",
"Definition:Real Interval/Open",
"Definition:Real Interval/Half-Open",
"Definition:Real Interval/Unbounded Closed",
"Definition:Real Interval/Unbounded Open",
"Definition:Real Interval/Empty",
"Definition:Real I... |
proofwiki-2590 | Ring of Sets Generated by Semiring | Let $\SS$ be a semiring of sets.
Let $\map \RR \SS$ be the minimal ring generated by $\SS$.
Let $\LL$ be the system of sets $A$ with the finite expansions:
:$\ds A = \bigcup_{k \mathop = 1}^n A_k$
with respect to the sets $A_k \in \SS$.
Then $\LL = \map \RR \SS$. | First we need to show that $\LL$ is a ring of sets.
Let $A, B \in \LL$.
Then by definition of $\LL$, they have expansions:
{{begin-eqn}}
{{eqn | l = A
| r = \bigcup_{i \mathop = 1}^m A_i
| c = where $A_i \in \SS$
}}
{{eqn | l = B
| r = \bigcup_{j \mathop = 1}^n B_j
| c = where $B_j \in \SS$
}}
{... | Let $\SS$ be a [[Definition:Semiring of Sets|semiring of sets]].
Let $\map \RR \SS$ be the [[Minimal Ring Generated by System of Sets|minimal ring generated by $\SS$]].
Let $\LL$ be the [[Definition:System of Sets|system of sets]] $A$ with the [[Definition:Finite Expansion|finite expansions]]:
:$\ds A = \bigcup_{k \m... | First we need to show that $\LL$ is a [[Definition:Ring of Sets|ring of sets]].
Let $A, B \in \LL$.
Then by definition of $\LL$, they have expansions:
{{begin-eqn}}
{{eqn | l = A
| r = \bigcup_{i \mathop = 1}^m A_i
| c = where $A_i \in \SS$
}}
{{eqn | l = B
| r = \bigcup_{j \mathop = 1}^n B_j
... | Ring of Sets Generated by Semiring | https://proofwiki.org/wiki/Ring_of_Sets_Generated_by_Semiring | https://proofwiki.org/wiki/Ring_of_Sets_Generated_by_Semiring | [
"Rings of Sets",
"Semirings of Sets"
] | [
"Definition:Semiring of Sets",
"Minimal Ring Generated by System of Sets",
"Definition:Set of Sets",
"Definition:Set Partition/Finite Expansion"
] | [
"Definition:Ring of Sets",
"Definition:Semiring of Sets",
"Pairwise Disjoint Subsets in Semiring Part of Partition",
"Definition:Set Partition/Finite Expansion",
"Definition:Set Partition/Finite Expansion",
"Definition:Ring of Sets",
"Category:Rings of Sets",
"Category:Semirings of Sets"
] |
proofwiki-2591 | Ring of Sets is Closed under Finite Intersection | Let $\RR$ be a ring of sets.
Let $A_1, A_2, \ldots, A_n \in \RR$.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$ | Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$
$\map P 1$ is true, as this just says $A_1 \in \RR$. | Let $\RR$ be a [[Definition:Ring of Sets|ring of sets]].
Let $A_1, A_2, \ldots, A_n \in \RR$.
Then:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \bigcap_{j \mathop = 1}^n A_j \in \RR$
$\map P 1$ is true, as this just says $A_1 \in \RR$. | Ring of Sets is Closed under Finite Intersection | https://proofwiki.org/wiki/Ring_of_Sets_is_Closed_under_Finite_Intersection | https://proofwiki.org/wiki/Ring_of_Sets_is_Closed_under_Finite_Intersection | [
"Rings of Sets"
] | [
"Definition:Ring of Sets"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2592 | Sigma-Algebra is Delta-Algebra | A $\sigma$-algebra is also a $\delta$-algebra. | Let $\SS$ be a $\sigma$-algebra whose unit is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a countably infinite collection of elements of $\SS$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \mathbb U \setminus A_i
| o = \in
| r = \SS
| c = $\SS$ is closed under relative complement with $\mathbb U$
... | A [[Definition:Sigma-Algebra|$\sigma$-algebra]] is also a [[Definition:Delta-Algebra|$\delta$-algebra]]. | Let $\SS$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] whose [[Definition:Unit of System of Sets|unit]] is $\mathbb U$.
Let $A_1, A_2, \ldots$ be a [[Definition:Countable|countably infinite]] collection of [[Definition:Element|elements]] of $\SS$.
Then:
{{begin-eqn}}
{{eqn | q = \forall i
| l = \mathbb ... | Sigma-Algebra is Delta-Algebra | https://proofwiki.org/wiki/Sigma-Algebra_is_Delta-Algebra | https://proofwiki.org/wiki/Sigma-Algebra_is_Delta-Algebra | [
"Sigma-Algebras"
] | [
"Definition:Sigma-Algebra",
"Definition:Delta-Algebra"
] | [
"Definition:Sigma-Algebra",
"Definition:Unit of System of Sets",
"Definition:Countable Set",
"Definition:Element",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Relative Complement",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Countable Set",
"... |
proofwiki-2593 | Addition Law of Probability | Let $\Pr$ be a probability measure on an event space $\Sigma$.
Let $A, B \in \Sigma$.
The probability of the occurrence of the union of $A$ and $B$ can be evaluated as:
:$\map \Pr {A \cup B} = \map \Pr A + \map \Pr B - \map \Pr {A \cap B}$
That is, the probability of either event occurring equals the sum of their indiv... | By definition, a probability measure is a measure.
Hence, again by definition, it is a countably additive function.
By Measure is Finitely Additive Function, we have that $\Pr$ is an additive function.
So Additive Function is Strongly Additive can be applied directly.
{{qed}} | Let $\Pr$ be a [[Definition:Probability Measure|probability measure]] on an [[Definition:Event Space|event space]] $\Sigma$.
Let $A, B \in \Sigma$.
The [[Definition:Probability|probability]] of the [[Definition:Occurrence of Event|occurrence]] of the [[Definition:Union of Events|union]] of $A$ and $B$ can be evaluat... | By definition, a [[Definition:Probability Measure|probability measure]] is a [[Definition:Measure (Measure Theory)|measure]].
Hence, again by definition, it is a [[Definition:Countably Additive Function|countably additive function]].
By [[Measure is Finitely Additive Function]], we have that $\Pr$ is an [[Definition:... | Addition Law of Probability/Proof 1 | https://proofwiki.org/wiki/Addition_Law_of_Probability | https://proofwiki.org/wiki/Addition_Law_of_Probability/Proof_1 | [
"Addition Law of Probability",
"Unions of Events",
"Intersections of Events",
"Probability Theory",
"Named Theorems"
] | [
"Definition:Probability Measure",
"Definition:Event Space",
"Definition:Probability",
"Definition:Event/Occurrence",
"Definition:Event/Occurrence/Union",
"Definition:Probability",
"Definition:Event",
"Definition:Event/Occurrence",
"Definition:Addition/Real Numbers",
"Definition:Probability",
"De... | [
"Definition:Probability Measure",
"Definition:Measure (Measure Theory)",
"Definition:Countably Additive Function",
"Measure is Finitely Additive Function",
"Definition:Additive Function (Measure Theory)",
"Additive Function is Strongly Additive"
] |
proofwiki-2594 | Addition Law of Probability | Let $\Pr$ be a probability measure on an event space $\Sigma$.
Let $A, B \in \Sigma$.
The probability of the occurrence of the union of $A$ and $B$ can be evaluated as:
:$\map \Pr {A \cup B} = \map \Pr A + \map \Pr B - \map \Pr {A \cap B}$
That is, the probability of either event occurring equals the sum of their indiv... | From Set Difference and Intersection form Partition:
:$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
:$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.
So, by the definition of probability measure:
:$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
:$\map \P... | Let $\Pr$ be a [[Definition:Probability Measure|probability measure]] on an [[Definition:Event Space|event space]] $\Sigma$.
Let $A, B \in \Sigma$.
The [[Definition:Probability|probability]] of the [[Definition:Occurrence of Event|occurrence]] of the [[Definition:Union of Events|union]] of $A$ and $B$ can be evaluat... | From [[Set Difference and Intersection form Partition]]:
:$A$ is the [[Definition:Set Union|union]] of the two [[Definition:Disjoint Sets|disjoint sets]] $A \setminus B$ and $A \cap B$
:$B$ is the [[Definition:Set Union|union]] of the two [[Definition:Disjoint Sets|disjoint sets]] $B \setminus A$ and $A \cap B$.
So,... | Addition Law of Probability/Proof 2 | https://proofwiki.org/wiki/Addition_Law_of_Probability | https://proofwiki.org/wiki/Addition_Law_of_Probability/Proof_2 | [
"Addition Law of Probability",
"Unions of Events",
"Intersections of Events",
"Probability Theory",
"Named Theorems"
] | [
"Definition:Probability Measure",
"Definition:Event Space",
"Definition:Probability",
"Definition:Event/Occurrence",
"Definition:Event/Occurrence/Union",
"Definition:Probability",
"Definition:Event",
"Definition:Event/Occurrence",
"Definition:Addition/Real Numbers",
"Definition:Probability",
"De... | [
"Set Difference and Intersection form Partition",
"Definition:Set Union",
"Definition:Disjoint Sets",
"Definition:Set Union",
"Definition:Disjoint Sets",
"Definition:Probability Measure",
"Set Difference is Disjoint with Reverse",
"Set Difference and Intersection form Partition/Corollary 1"
] |
proofwiki-2595 | Cardinality is Additive Function | Let $S$ be a finite set.
Let $\powerset S$ be the power set of $S$.
The function $C: \powerset S \to \R$, where $C$ is defined as the cardinality of a set, is an additive function. | We have that $\powerset S$ is an algebra of sets.
{{ProofWanted}}
Category:Set Systems
r3r1nsnaic0xb44ofi9q9och3iiqjos | Let $S$ be a [[Definition:Finite|finite set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
The function $C: \powerset S \to \R$, where $C$ is defined as the [[Definition:Cardinality|cardinality]] of a set, is an [[Definition:Additive Function (Measure Theory)|additive function]]. | We have that [[Power Set is Algebra of Sets|$\powerset S$ is an algebra of sets]].
{{ProofWanted}}
[[Category:Set Systems]]
r3r1nsnaic0xb44ofi9q9och3iiqjos | Cardinality is Additive Function | https://proofwiki.org/wiki/Cardinality_is_Additive_Function | https://proofwiki.org/wiki/Cardinality_is_Additive_Function | [
"Set Systems"
] | [
"Definition:Finite",
"Definition:Power Set",
"Definition:Cardinality",
"Definition:Additive Function (Measure Theory)"
] | [
"Power Set is Algebra of Sets",
"Category:Set Systems"
] |
proofwiki-2596 | Finite Union of Sets in Additive Function | Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline {\R}$ be an additive function.
Let $A_1, A_2, \ldots, A_n$ be any finite collection of pairwise disjoint elements of $\AA$.
Then:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
That is, for any collection of pairwise disjoin... | Proof by induction:
In the below, we assume that $A_1, A_2, \ldots$ are all pairwise disjoint elements of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} = \sum_{i \mathop = 1}^n \map f {A_i}$
$\map P 1$ is the case:
:$\ds \map f {\bigcup_{i \mathop = 1}^1... | Let $\AA$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \AA \to \overline {\R}$ be an [[Definition:Additive Function (Measure Theory)|additive function]].
Let $A_1, A_2, \ldots, A_n$ be any [[Definition:Finite Set|finite]] collection of [[Definition:Pairwise Disjoint|pairwise disjoint]] elements of $\... | Proof by [[Principle of Mathematical Induction|induction]]:
In the below, we assume that $A_1, A_2, \ldots$ are all [[Definition:Pairwise Disjoint|pairwise disjoint]] elements of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i}... | Finite Union of Sets in Additive Function | https://proofwiki.org/wiki/Finite_Union_of_Sets_in_Additive_Function | https://proofwiki.org/wiki/Finite_Union_of_Sets_in_Additive_Function | [
"Algebras of Sets",
"Additive Functions",
"Set Union",
"Proofs by Induction"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)",
"Definition:Finite Set",
"Definition:Pairwise Disjoint",
"Definition:Pairwise Disjoint"
] | [
"Principle of Mathematical Induction",
"Definition:Pairwise Disjoint",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2597 | Measure is Finitely Additive Function | Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\mu: \Sigma \to \overline {\R}$ be a measure on $\Sigma$.
Then $\mu$ is finitely additive. | From the definition of a measure, $\mu$ is countably additive.
From Countably Additive Function also Finitely Additive, $\mu$ is finitely additive.
{{qed}} | Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on a [[Definition:Set|set]] $X$.
Let $\mu: \Sigma \to \overline {\R}$ be a [[Definition:Measure (Measure Theory)|measure]] on $\Sigma$.
Then $\mu$ is [[Definition:Finitely Additive Function|finitely additive]]. | From the definition of a [[Definition:Measure (Measure Theory)|measure]], $\mu$ is [[Definition:Countably Additive Function|countably additive]].
From [[Countably Additive Function also Finitely Additive]], $\mu$ is [[Definition:Finitely Additive Function|finitely additive]].
{{qed}} | Measure is Finitely Additive Function | https://proofwiki.org/wiki/Measure_is_Finitely_Additive_Function | https://proofwiki.org/wiki/Measure_is_Finitely_Additive_Function | [
"Finitely Additive Functions",
"Measures"
] | [
"Definition:Sigma-Algebra",
"Definition:Set",
"Definition:Measure (Measure Theory)",
"Definition:Additive Function (Measure Theory)"
] | [
"Definition:Measure (Measure Theory)",
"Definition:Countably Additive Function",
"Countably Additive Function also Finitely Additive",
"Definition:Additive Function (Measure Theory)"
] |
proofwiki-2598 | Finite Union of Sets in Subadditive Function | Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be a subadditive function.
Let $A_1, A_2, \ldots, A_n$ be any finite collection of elements of $\AA$.
Then:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
That is, for any finite collection of elements of $\AA$, $f$ of... | Proof by induction:
In the below, let $A_1, A_2, \ldots$ all be elements of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}$
$\map P 1$ is trivially true, as this just says $\map f {A_1} \le \map f {A_1}$. | Let $\AA$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \AA \to \overline \R$ be a [[Definition:Subadditive Function (Measure Theory)|subadditive function]].
Let $A_1, A_2, \ldots, A_n$ be any [[Definition:Finite Set|finite collection]] of [[Definition:Element|elements]] of $\AA$.
Then:
:$\ds \map f ... | Proof by [[Principle of Mathematical Induction|induction]]:
In the below, let $A_1, A_2, \ldots$ all be [[Definition:Element|elements]] of $\AA$.
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map f {\bigcup_{i \mathop = 1}^n A_i} \le \sum_{i \mathop = 1}^n \map f {A_i}... | Finite Union of Sets in Subadditive Function | https://proofwiki.org/wiki/Finite_Union_of_Sets_in_Subadditive_Function | https://proofwiki.org/wiki/Finite_Union_of_Sets_in_Subadditive_Function | [
"Algebras of Sets",
"Subadditive Functions",
"Set Union",
"Proofs by Induction"
] | [
"Definition:Algebra of Sets",
"Definition:Subadditive Function (Measure Theory)",
"Definition:Finite Set",
"Definition:Element",
"Definition:Finite Set",
"Definition:Element",
"Definition:Set Union",
"Definition:Element"
] | [
"Principle of Mathematical Induction",
"Definition:Element",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-2599 | Additive Function on Empty Set is Zero | Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be an additive function on $\AA$.
Then $\map f \O = 0$. | From Properties of Algebras of Sets:
:$\O \in \AA$
Let $X \in \AA$.
Then:
{{begin-eqn}}
{{eqn | l = X \cap \O
| r = \O
| c = Intersection with Empty Set
}}
{{eqn | ll= \leadsto
| l = \map f X + \map f \O
| r = \map f {X \cup \O}
| c = {{Defof|Additive Function (Measure Theory)}}
}}
{{eqn |... | Let $\AA$ be an [[Definition:Algebra of Sets|algebra of sets]].
Let $f: \AA \to \overline \R$ be an [[Definition:Additive Function (Measure Theory)|additive function]] on $\AA$.
Then $\map f \O = 0$. | From [[Properties of Algebras of Sets]]:
:$\O \in \AA$
Let $X \in \AA$.
Then:
{{begin-eqn}}
{{eqn | l = X \cap \O
| r = \O
| c = [[Intersection with Empty Set]]
}}
{{eqn | ll= \leadsto
| l = \map f X + \map f \O
| r = \map f {X \cup \O}
| c = {{Defof|Additive Function (Measure Theory)}... | Additive Function on Empty Set is Zero | https://proofwiki.org/wiki/Additive_Function_on_Empty_Set_is_Zero | https://proofwiki.org/wiki/Additive_Function_on_Empty_Set_is_Zero | [
"Set Systems"
] | [
"Definition:Algebra of Sets",
"Definition:Additive Function (Measure Theory)"
] | [
"Properties of Algebras of Sets",
"Intersection with Empty Set",
"Union with Empty Set",
"Category:Set Systems"
] |
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