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-11700 | Ceiling is between Number and One More | :$x \le \ceiling x < x + 1$
where $\ceiling x$ denotes the ceiling of $x$. | From Number is between Ceiling and One Less:
:$\ceiling x - 1 < x \le \ceiling x$
Thus by adding $1$:
:$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$
So:
:$x \le \ceiling x$
and:
:$\ceiling x < x + 1$
as required.
{{qed}} | :$x \le \ceiling x < x + 1$
where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$. | From [[Number is between Ceiling and One Less]]:
:$\ceiling x - 1 < x \le \ceiling x$
Thus by adding $1$:
:$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$
So:
:$x \le \ceiling x$
and:
:$\ceiling x < x + 1$
as required.
{{qed}} | Ceiling is between Number and One More | https://proofwiki.org/wiki/Ceiling_is_between_Number_and_One_More | https://proofwiki.org/wiki/Ceiling_is_between_Number_and_One_More | [
"Ceiling Function"
] | [
"Definition:Ceiling Function"
] | [
"Number is between Ceiling and One Less"
] |
proofwiki-11701 | Ceiling of Number plus Integer | :$\forall n \in \Z: \ceiling x + n = \ceiling {x + n}$ | {{begin-eqn}}
{{eqn | l = \ceiling {x + n} - 1
| o = <
| m = x + n
| mo= \le
| r = \ceiling {x + n}
| c = Number is between Ceiling and One Less
}}
{{eqn | ll= \leadsto
| l = \ceiling {x + n} - n - 1
| o = <
| m = x
| mo= \le
| r = \ceiling {x + n} - n
... | :$\forall n \in \Z: \ceiling x + n = \ceiling {x + n}$ | {{begin-eqn}}
{{eqn | l = \ceiling {x + n} - 1
| o = <
| m = x + n
| mo= \le
| r = \ceiling {x + n}
| c = [[Number is between Ceiling and One Less]]
}}
{{eqn | ll= \leadsto
| l = \ceiling {x + n} - n - 1
| o = <
| m = x
| mo= \le
| r = \ceiling {x + n} - n
... | Ceiling of Number plus Integer | https://proofwiki.org/wiki/Ceiling_of_Number_plus_Integer | https://proofwiki.org/wiki/Ceiling_of_Number_plus_Integer | [
"Ceiling Function"
] | [] | [
"Number is between Ceiling and One Less",
"Number is between Ceiling and One Less"
] |
proofwiki-11702 | Relation Segment of Auxiliary Relation is Ideal | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $R$ be auxiliary relation on $S$.
Let $x \in S$.
Then
:$x^R$ is ideal in $L$
where $x^R$ denotes the $R$-segment of $x$. | === Non-empty ===
By definition of auxiliary relation:
:$\tuple {\bot, x} \in R$
where $\bot$ denotes the smallest element in $L$.
By definition of relation segment:
:$\bot \in x^R$
Thus by definition:
:$x^R$ is non-empty.
{{qed|lemma}} | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $R$ be [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
Let $x \in S$.
Then
:$x^R$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$
where $x^R$ denotes the [[Defi... | === Non-empty ===
By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$\tuple {\bot, x} \in R$
where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
By definition of [[Definition:Relation Segment|relation segment]]:
:$\bot \in x^R$
Thus by definition:
:$x^R$ is [[Defin... | Relation Segment of Auxiliary Relation is Ideal | https://proofwiki.org/wiki/Relation_Segment_of_Auxiliary_Relation_is_Ideal | https://proofwiki.org/wiki/Relation_Segment_of_Auxiliary_Relation_is_Ideal | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Auxiliary Relation",
"Definition:Ideal in Ordered Set",
"Definition:Relation Segment"
] | [
"Definition:Auxiliary Relation",
"Definition:Smallest Element",
"Definition:Relation Segment",
"Definition:Non-Empty Set",
"Definition:Relation Segment",
"Definition:Auxiliary Relation",
"Definition:Relation Segment",
"Definition:Relation Segment",
"Definition:Auxiliary Relation",
"Definition:Rela... |
proofwiki-11703 | Quotient of Modulo Operation with Modulus | Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y \ne 0$.
Then:
:$0 \le \dfrac x y - \floor {\dfrac x y} = \dfrac {x \bmod y}... | From Real Number minus Floor:
:$\dfrac x y - \floor {\dfrac x y} \in \hointr 0 1$
Thus by definition of half-open real interval:
:$0 \le \dfrac x y - \floor {\dfrac x y} < 1$
Then:
{{begin-eqn}}
{{eqn | l = x \bmod y
| r = x - y \floor {\frac x y}
| c = {{Defof|Modulo Operation}}
}}
{{eqn | ll= \leadsto
... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfra... | From [[Real Number minus Floor]]:
:$\dfrac x y - \floor {\dfrac x y} \in \hointr 0 1$
Thus by definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
:$0 \le \dfrac x y - \floor {\dfrac x y} < 1$
Then:
{{begin-eqn}}
{{eqn | l = x \bmod y
| r = x - y \floor {\frac x y}
| c = {{Defof... | Quotient of Modulo Operation with Modulus | https://proofwiki.org/wiki/Quotient_of_Modulo_Operation_with_Modulus | https://proofwiki.org/wiki/Quotient_of_Modulo_Operation_with_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation",
"Definition:Floor Function"
] | [
"Real Number minus Floor",
"Definition:Real Interval/Half-Open"
] |
proofwiki-11704 | Range of Modulo Operation for Positive Modulus | Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y > 0$.
Then:
:$0 \le x \bmod y < y$ | {{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = \frac {x \bmod y} y
| mo= <
| r = 1
| c = Quotient of Modulo Operation with Modulus
}}
{{eqn | ll= \leadsto
| l = 0
| o = \le
| m = \frac {x \bmod y} y \times y
| mo= <
| r = 1 \times y
| c = Real Number Orderi... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfra... | {{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = \frac {x \bmod y} y
| mo= <
| r = 1
| c = [[Quotient of Modulo Operation with Modulus]]
}}
{{eqn | ll= \leadsto
| l = 0
| o = \le
| m = \frac {x \bmod y} y \times y
| mo= <
| r = 1 \times y
| c = [[Real Number ... | Range of Modulo Operation for Positive Modulus | https://proofwiki.org/wiki/Range_of_Modulo_Operation_for_Positive_Modulus | https://proofwiki.org/wiki/Range_of_Modulo_Operation_for_Positive_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation",
"Definition:Floor Function"
] | [
"Quotient of Modulo Operation with Modulus",
"Real Number Ordering is Compatible with Multiplication"
] |
proofwiki-11705 | Range of Modulo Operation for Negative Modulus | Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y < 0$.
Then:
:$0 \ge x \bmod y > y$ | {{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = \frac {x \bmod y} y
| mo= <
| r = 1
| c = Quotient of Modulo Operation with Modulus
}}
{{eqn | ll= \leadsto
| l = 0
| o = \ge
| m = \frac {x \bmod y} y \times y
| mo= >
| r = 1 \times y
| c = Real Number Orderi... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfra... | {{begin-eqn}}
{{eqn | l = 0
| o = \le
| m = \frac {x \bmod y} y
| mo= <
| r = 1
| c = [[Quotient of Modulo Operation with Modulus]]
}}
{{eqn | ll= \leadsto
| l = 0
| o = \ge
| m = \frac {x \bmod y} y \times y
| mo= >
| r = 1 \times y
| c = [[Real Number ... | Range of Modulo Operation for Negative Modulus | https://proofwiki.org/wiki/Range_of_Modulo_Operation_for_Negative_Modulus | https://proofwiki.org/wiki/Range_of_Modulo_Operation_for_Negative_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation",
"Definition:Floor Function"
] | [
"Quotient of Modulo Operation with Modulus",
"Real Number Ordering is Compatible with Multiplication"
] |
proofwiki-11706 | Zero is Integer Multiple of Zero | Zero is an integer multiple of zero. | We have that:
:$0 \times 0 = 0$
The result follows by definition of integer multiple.
{{qed}}
Category:Number Theory
cr4qeo16keg2mqnhx82y0t2xypme5gz | [[Definition:Zero (Number)|Zero]] is an [[Definition:Integer Multiple|integer multiple]] of [[Definition:Zero (Number)|zero]]. | We have that:
:$0 \times 0 = 0$
The result follows by definition of [[Definition:Integer Multiple|integer multiple]].
{{qed}}
[[Category:Number Theory]]
cr4qeo16keg2mqnhx82y0t2xypme5gz | Zero is Integer Multiple of Zero | https://proofwiki.org/wiki/Zero_is_Integer_Multiple_of_Zero | https://proofwiki.org/wiki/Zero_is_Integer_Multiple_of_Zero | [
"Number Theory"
] | [
"Definition:Zero (Number)",
"Definition:Integral Multiple/Real Numbers",
"Definition:Zero (Number)"
] | [
"Definition:Integral Multiple/Real Numbers",
"Category:Number Theory"
] |
proofwiki-11707 | Number minus Modulo is Integer Multiple | Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Let $y < 0$.
Then:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$. | When $y = 0$ we have:
:$x \bmod y := x$
Thus:
:$x - \paren {x \bmod y} = 0$
From Zero is Integer Multiple of Zero it follows that:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$.
Let $y \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \bmod y
| r = x - y \floor {\dfrac x y}
| c = {{Defof|Modulo Operation}... | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := \begin{cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end{cases}$
where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfra... | When $y = 0$ we have:
:$x \bmod y := x$
Thus:
:$x - \paren {x \bmod y} = 0$
From [[Zero is Integer Multiple of Zero]] it follows that:
:$x - \paren {x \bmod y}$
is an [[Definition:Integer Multiple|integer multiple]] of $y$.
Let $y \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \bmod y
| r = x - y \floor {\dfrac x... | Number minus Modulo is Integer Multiple | https://proofwiki.org/wiki/Number_minus_Modulo_is_Integer_Multiple | https://proofwiki.org/wiki/Number_minus_Modulo_is_Integer_Multiple | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation",
"Definition:Floor Function",
"Definition:Integral Multiple/Real Numbers"
] | [
"Zero is Integer Multiple of Zero",
"Definition:Integral Multiple/Real Numbers",
"Floor Function is Integer",
"Definition:Integer",
"Definition:Integral Multiple/Real Numbers"
] |
proofwiki-11708 | Modulo Operation/Examples/5 mod 3 | :$5 \bmod 3 = 2$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 3 = 1 + \dfrac 2 3$
and so:
:$\floor {\dfrac 5 3} = 1$
Thus:
:$5 \bmod 3 = 5 - 3 \times \floor {\dfrac 5 3} = 5 - 3 \times 1 = 2$
{{qed}} | :$5 \bmod 3 = 2$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac 5 3 = 1 + \dfrac 2 3$
and so:
:$\floor {\dfrac 5 3} = 1$
Thus:
:$5 \bmod 3 = 5 - 3 \times \floor {\dfrac 5 3} = 5 - 3 \times 1 = 2$
{{qed}} | Modulo Operation/Examples/5 mod 3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/5_mod_3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/5_mod_3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-11709 | Modulo Operation/Examples/18 mod 3 | :$18 \bmod 3 = 0$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} 3 = 6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} 3} = 6$
Thus:
:$18 \bmod 3 = 18 - 3 \times \floor {\dfrac {18} 3} = 18 - 3 \times 6 = 0$
{{qed}} | :$18 \bmod 3 = 0$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {18} 3 = 6 + \dfrac 0 3$
and so:
:$\floor {\dfrac {18} 3} = 6$
Thus:
:$18 \bmod 3 = 18 - 3 \times \floor {\dfrac {18} 3} = 18 - 3 \times 6 = 0$
{{qed}} | Modulo Operation/Examples/18 mod 3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/18_mod_3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/18_mod_3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-11710 | Modulo Operation/Examples/-2 mod 3 | :$-2 \bmod 3 = 1$ | By definition of modulo operation:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} 3 = -1 + \dfrac 1 3$
and so:
:$\floor {\dfrac {-2} 3} = -1$
Thus:
:$-2 \bmod 3 = -2 - 3 \times \floor {\dfrac {-2} 3} = -2 - 3 \times \paren {-1} = 1$
{{qed}} | :$-2 \bmod 3 = 1$ | By definition of [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := x - y \floor {\dfrac x y}$
for $y \ne 0$.
We have:
:$\dfrac {-2} 3 = -1 + \dfrac 1 3$
and so:
:$\floor {\dfrac {-2} 3} = -1$
Thus:
:$-2 \bmod 3 = -2 - 3 \times \floor {\dfrac {-2} 3} = -2 - 3 \times \paren {-1} = 1$
{{qed}} | Modulo Operation/Examples/-2 mod 3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-2_mod_3 | https://proofwiki.org/wiki/Modulo_Operation/Examples/-2_mod_3 | [
"Examples of Modulo Operation"
] | [] | [
"Definition:Modulo Operation"
] |
proofwiki-11711 | Number is Divisor iff Modulo is Zero | Let $x, y \in \R$ be real numbers.
Let $x \bmod y$ denote the modulo operation:
:$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$
where $\floor {\dfrac x y}$ denotes the floor of $\dfrac x y$.
Then $x \bmod y = 0$ {{iff}} $x$ is an integer multiple of $y$. | === Sufficient Condition ===
Let $x \bmod y = 0$.
From Number minus Modulo is Integer Multiple:
:$x - \paren {x \bmod y}$
is an integer multiple of $y$.
As $x \bmod y = 0$ it follows that $x - 0 = x$ is an integer multiple of $y$.
{{qed|lemma}} | Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
:$x \bmod y := \begin {cases}
x - y \floor {\dfrac x y} & : y \ne 0 \\
x & : y = 0
\end {cases}$
where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\df... | === Sufficient Condition ===
Let $x \bmod y = 0$.
From [[Number minus Modulo is Integer Multiple]]:
:$x - \paren {x \bmod y}$
is an [[Definition:Integer Multiple|integer multiple]] of $y$.
As $x \bmod y = 0$ it follows that $x - 0 = x$ is an [[Definition:Integer Multiple|integer multiple]] of $y$.
{{qed|lemma}} | Number is Divisor iff Modulo is Zero | https://proofwiki.org/wiki/Number_is_Divisor_iff_Modulo_is_Zero | https://proofwiki.org/wiki/Number_is_Divisor_iff_Modulo_is_Zero | [
"Modulo Arithmetic"
] | [
"Definition:Real Number",
"Definition:Modulo Operation",
"Definition:Floor Function",
"Definition:Integral Multiple/Real Numbers"
] | [
"Number minus Modulo is Integer Multiple",
"Definition:Integral Multiple/Real Numbers",
"Definition:Integral Multiple/Real Numbers",
"Definition:Integral Multiple/Real Numbers"
] |
proofwiki-11712 | Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure | Let $R = \struct {S, \preceq}$ be an ordered set.
Let $\map {\it Ids} R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\it Ids} R, \precsim}$ be an ordered set
where $\precsim \mathop = \subseteq \restriction_{\map {\it Ids} R \times \map {\it Ids} R}$
Let:
:$M = \struct {F, \preccurlyeq}$
where:
:$F = \left... | === Reflexivity ===
Let $f \in F$.
By definition of reflexivity:
:$\forall x \in S: \map f x \precsim \map f x$
Thus by definition of ordering on mappings:
:$f \preccurlyeq f$
{{qed|lemma}} | Let $R = \struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\map {\it Ids} R$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
Let $L = \struct {\map {\it Ids} R, \precsim}$ be an [[Definition:Ordered Set|ordered set]]
where $\precsim \mathop = \sub... | === Reflexivity ===
Let $f \in F$.
By definition of [[Definition:Reflexivity|reflexivity]]:
:$\forall x \in S: \map f x \precsim \map f x$
Thus by definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
:$f \preccurlyeq f$
{{qed|lemma}} | Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure | https://proofwiki.org/wiki/Correctness_of_Definition_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure | https://proofwiki.org/wiki/Correctness_of_Definition_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Increasing/Mapping",
"Definition:Ordering on Mappings",
"Definition:Lower Closure/Element",
"Definition:Ordered Set"
] | [
"Definition:Reflexivity",
"Definition:Ordering on Mappings",
"Definition:Ordering on Mappings",
"Definition:Ordering on Mappings",
"Definition:Ordering on Mappings"
] |
proofwiki-11713 | Number less than Integer iff Floor less than Integer | :$\floor x < n \iff x < n$ | === Necessary Condition ===
Let $x < n$.
By definition of the floor of $x$:
:$\floor x \le x$
Hence:
:$\floor x < n$
{{qed|lemma}} | :$\floor x < n \iff x < n$ | === Necessary Condition ===
Let $x < n$.
By definition of the [[Definition:Floor Function|floor]] of $x$:
:$\floor x \le x$
Hence:
:$\floor x < n$
{{qed|lemma}} | Number less than Integer iff Floor less than Integer | https://proofwiki.org/wiki/Number_less_than_Integer_iff_Floor_less_than_Integer | https://proofwiki.org/wiki/Number_less_than_Integer_iff_Floor_less_than_Integer | [
"Floor Function"
] | [] | [
"Definition:Floor Function"
] |
proofwiki-11714 | Number not less than Integer iff Floor not less than Integer | :$x \ge n \iff \floor x \ge n$ | === Necessary Condition ===
Let $\floor x \ge n$.
By definition of the floor of $x$:
:$x \ge \floor x$
Hence:
:$x \ge n$
{{qed|lemma}} | :$x \ge n \iff \floor x \ge n$ | === Necessary Condition ===
Let $\floor x \ge n$.
By definition of the [[Definition:Floor Function|floor]] of $x$:
:$x \ge \floor x$
Hence:
:$x \ge n$
{{qed|lemma}} | Number not less than Integer iff Floor not less than Integer | https://proofwiki.org/wiki/Number_not_less_than_Integer_iff_Floor_not_less_than_Integer | https://proofwiki.org/wiki/Number_not_less_than_Integer_iff_Floor_not_less_than_Integer | [
"Floor Function"
] | [] | [
"Definition:Floor Function",
"Definition:Floor Function"
] |
proofwiki-11715 | Number not greater than Integer iff Ceiling not greater than Integer | :$\ceiling x \le n \iff x \le n$ | === Necessary Condition ===
Let $\ceiling x \le n$.
By Number is between Ceiling and One Less:
:$x \le \ceiling x$
Hence:
:$x \le n$
{{qed|lemma}} | :$\ceiling x \le n \iff x \le n$ | === Necessary Condition ===
Let $\ceiling x \le n$.
By [[Number is between Ceiling and One Less]]:
:$x \le \ceiling x$
Hence:
:$x \le n$
{{qed|lemma}} | Number not greater than Integer iff Ceiling not greater than Integer | https://proofwiki.org/wiki/Number_not_greater_than_Integer_iff_Ceiling_not_greater_than_Integer | https://proofwiki.org/wiki/Number_not_greater_than_Integer_iff_Ceiling_not_greater_than_Integer | [
"Ceiling Function"
] | [] | [
"Number is between Ceiling and One Less",
"Number is between Ceiling and One Less"
] |
proofwiki-11716 | Number greater than Integer iff Ceiling greater than Integer | :$\ceiling x > n \iff x > n$ | === Necessary Condition ===
Let $x > n$.
By Number is between Ceiling and One Less:
:$\ceiling x \ge x$
Hence:
:$\ceiling x > n$
{{qed|lemma}} | :$\ceiling x > n \iff x > n$ | === Necessary Condition ===
Let $x > n$.
By [[Number is between Ceiling and One Less]]:
:$\ceiling x \ge x$
Hence:
:$\ceiling x > n$
{{qed|lemma}} | Number greater than Integer iff Ceiling greater than Integer | https://proofwiki.org/wiki/Number_greater_than_Integer_iff_Ceiling_greater_than_Integer | https://proofwiki.org/wiki/Number_greater_than_Integer_iff_Ceiling_greater_than_Integer | [
"Ceiling Function"
] | [] | [
"Number is between Ceiling and One Less",
"Number is between Ceiling and One Less"
] |
proofwiki-11717 | Integer equals Floor iff between Number and One Less | :$\floor x = n \iff x - 1 < n \le x$ | === Necessary Condition ===
Let $x - 1 < n \le x$.
From $n \le x$, we have by Number not less than Integer iff Floor not less than Integer:
:$n \le \floor x$
From $x - 1 < n$:
:$x < n + 1$
Hence by Number less than Integer iff Floor less than Integer:
:$\floor x < n + 1$
We have that:
:$\forall m, n \in \Z: m \le n \if... | :$\floor x = n \iff x - 1 < n \le x$ | === Necessary Condition ===
Let $x - 1 < n \le x$.
From $n \le x$, we have by [[Number not less than Integer iff Floor not less than Integer]]:
:$n \le \floor x$
From $x - 1 < n$:
:$x < n + 1$
Hence by [[Number less than Integer iff Floor less than Integer]]:
:$\floor x < n + 1$
We have that:
:$\forall m, n \in \Z... | Integer equals Floor iff between Number and One Less | https://proofwiki.org/wiki/Integer_equals_Floor_iff_between_Number_and_One_Less | https://proofwiki.org/wiki/Integer_equals_Floor_iff_between_Number_and_One_Less | [
"Floor Function"
] | [] | [
"Number not less than Integer iff Floor not less than Integer",
"Number less than Integer iff Floor less than Integer",
"Number not less than Integer iff Floor not less than Integer"
] |
proofwiki-11718 | Integer equals Floor iff Number between Integer and One More | :$\floor x = n \iff n \le x < n + 1$ | === Necessary Condition ===
Let $n \le x < n + 1$.
From Number not less than Integer iff Floor not less than Integer:
:$n \le x \implies n \le \floor x$
By definition of the floor of $x$:
:$\floor x \le x$
and so by hypothesis:
:$\floor x < n + 1$
We have that:
:$\forall m, n \in \Z: m \le n \iff m < n + 1$
and so:
:$\... | :$\floor x = n \iff n \le x < n + 1$ | === Necessary Condition ===
Let $n \le x < n + 1$.
From [[Number not less than Integer iff Floor not less than Integer]]:
:$n \le x \implies n \le \floor x$
By definition of the [[Definition:Floor Function|floor]] of $x$:
:$\floor x \le x$
and so [[Definition:By Hypothesis|by hypothesis]]:
:$\floor x < n + 1$
We ... | Integer equals Floor iff Number between Integer and One More | https://proofwiki.org/wiki/Integer_equals_Floor_iff_Number_between_Integer_and_One_More | https://proofwiki.org/wiki/Integer_equals_Floor_iff_Number_between_Integer_and_One_More | [
"Floor Function"
] | [] | [
"Number not less than Integer iff Floor not less than Integer",
"Definition:Floor Function",
"Definition:By Hypothesis",
"Definition:Floor Function",
"Definition:By Hypothesis",
"Definition:Floor Function",
"Definition:By Hypothesis"
] |
proofwiki-11719 | Integer equals Ceiling iff between Number and One More | :$\ceiling x = n \iff x \le n < x + 1$ | === Necessary Condition ===
Let $x \le n < x + 1$.
From $x \le n$, we have by Number not greater than Integer iff Ceiling not greater than Integer:
:$\ceiling x \le n$
From $n < x + 1$:
:$n - 1 < x$
Hence by Number greater than Integer iff Ceiling greater than Integer:
:$n - 1 < \ceiling x$
We have that:
:$\forall m, n... | :$\ceiling x = n \iff x \le n < x + 1$ | === Necessary Condition ===
Let $x \le n < x + 1$.
From $x \le n$, we have by [[Number not greater than Integer iff Ceiling not greater than Integer]]:
:$\ceiling x \le n$
From $n < x + 1$:
:$n - 1 < x$
Hence by [[Number greater than Integer iff Ceiling greater than Integer]]:
:$n - 1 < \ceiling x$
We have that:
:... | Integer equals Ceiling iff between Number and One More | https://proofwiki.org/wiki/Integer_equals_Ceiling_iff_between_Number_and_One_More | https://proofwiki.org/wiki/Integer_equals_Ceiling_iff_between_Number_and_One_More | [
"Ceiling Function"
] | [] | [
"Number not greater than Integer iff Ceiling not greater than Integer",
"Number greater than Integer iff Ceiling greater than Integer",
"Number not greater than Integer iff Ceiling not greater than Integer"
] |
proofwiki-11720 | Integer equals Ceiling iff Number between Integer and One Less | :$\ceiling x = n \iff n - 1 < x \le n$ | === Necessary Condition ===
Let $n - 1 < x \le n$.
From Number not greater than Integer iff Ceiling not greater than Integer:
:$x \le n \implies \ceiling x \le n$
From Number is between Ceiling and One Less:
:$x \le \ceiling x$
and so:
:$n - 1 < \ceiling x$
We have that:
:$\forall m, n \in \Z: m - 1 < n \iff m \le n$
a... | :$\ceiling x = n \iff n - 1 < x \le n$ | === Necessary Condition ===
Let $n - 1 < x \le n$.
From [[Number not greater than Integer iff Ceiling not greater than Integer]]:
:$x \le n \implies \ceiling x \le n$
From [[Number is between Ceiling and One Less]]:
:$x \le \ceiling x$
and so:
:$n - 1 < \ceiling x$
We have that:
:$\forall m, n \in \Z: m - 1 < n \... | Integer equals Ceiling iff Number between Integer and One Less | https://proofwiki.org/wiki/Integer_equals_Ceiling_iff_Number_between_Integer_and_One_Less | https://proofwiki.org/wiki/Integer_equals_Ceiling_iff_Number_between_Integer_and_One_Less | [
"Ceiling Function"
] | [] | [
"Number not greater than Integer iff Ceiling not greater than Integer",
"Number is between Ceiling and One Less",
"Number is between Ceiling and One Less",
"Number is between Ceiling and One Less"
] |
proofwiki-11721 | Floor of Root of Floor equals Floor of Root | :$\ds \floor {\sqrt {\floor x} } = \floor {\sqrt x}$ | The square root is defined on the interval $\hointr 0 \to$.
We have that Square Root is Strictly Increasing.
From Continuity of Root Function, the square root is continuous.
Hence the conditions are fulfilled for for McEliece's Theorem (Integer Functions) to be applied:
:$\forall x \in A: \paren {\map f x \in \Z \impli... | :$\ds \floor {\sqrt {\floor x} } = \floor {\sqrt x}$ | The [[Definition:Square Root|square root]] is defined on the [[Definition:Real Interval|interval]] $\hointr 0 \to$.
We have that [[Square Root is Strictly Increasing]].
From [[Continuity of Root Function]], the [[Definition:Square Root|square root]] is [[Definition:Continuous Real Function|continuous]].
Hence the co... | Floor of Root of Floor equals Floor of Root/Proof 2 | https://proofwiki.org/wiki/Floor_of_Root_of_Floor_equals_Floor_of_Root | https://proofwiki.org/wiki/Floor_of_Root_of_Floor_equals_Floor_of_Root/Proof_2 | [
"Floor Function",
"Floor of Root of Floor equals Floor of Root"
] | [] | [
"Definition:Square Root",
"Definition:Real Interval",
"Square Root is Strictly Increasing",
"Continuity of Root Function",
"Definition:Square Root",
"Definition:Continuous Real Function",
"McEliece's Theorem (Integer Functions)",
"Definition:Square Root",
"Integer Multiplication is Closed"
] |
proofwiki-11722 | Ceiling of Root of Ceiling equals Ceiling of Root | :$\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$ | The square root is defined on the interval $\hointr 0 \to$.
We have that Square Root is Strictly Increasing.
From Continuity of Root Function, the square root is continuous.
Hence the conditions are fulfilled for for McEliece's Theorem (Integer Functions) to be applied:
:$\forall x \in A: \paren {\map f x \in \Z \impli... | :$\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$ | The [[Definition:Square Root|square root]] is defined on the [[Definition:Real Interval|interval]] $\hointr 0 \to$.
We have that [[Square Root is Strictly Increasing]].
From [[Continuity of Root Function]], the [[Definition:Square Root|square root]] is [[Definition:Continuous Real Function|continuous]].
Hence the co... | Ceiling of Root of Ceiling equals Ceiling of Root/Proof 2 | https://proofwiki.org/wiki/Ceiling_of_Root_of_Ceiling_equals_Ceiling_of_Root | https://proofwiki.org/wiki/Ceiling_of_Root_of_Ceiling_equals_Ceiling_of_Root/Proof_2 | [
"Ceiling Function",
"Ceiling of Root of Ceiling equals Ceiling of Root"
] | [] | [
"Definition:Square Root",
"Definition:Real Interval",
"Square Root is Strictly Increasing",
"Continuity of Root Function",
"Definition:Square Root",
"Definition:Continuous Real Function",
"McEliece's Theorem (Integer Functions)",
"Definition:Square Root",
"Integer Multiplication is Closed"
] |
proofwiki-11723 | Congruence by Divisor of Modulus/Integer Modulus | Let $r, s \in \Z$ be integers.
Let $a, b \in \Z$ such that $a$ is congruent modulo $r s$ to $b$, that is:
:$a \equiv b \pmod {r s}$
Then:
:$a \equiv b \pmod r$
and:
:$a \equiv b \pmod s$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod {r s}
| c =
}}
{{eqn | ll= \leadsto
| l = a - b
| r = q r s
| c = {{Defof|Congruence Modulo Integer}}
}}
{{eqn | ll= \leadsto
| l = a - b
| r = \paren {q r} s
| c =
}}
{{eqn | lo= \land
| l =... | Let $r, s \in \Z$ be [[Definition:Integer|integers]].
Let $a, b \in \Z$ such that $a$ is [[Definition:Congruence Modulo Integer|congruent modulo $r s$]] to $b$, that is:
:$a \equiv b \pmod {r s}$
Then:
:$a \equiv b \pmod r$
and:
:$a \equiv b \pmod s$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod {r s}
| c =
}}
{{eqn | ll= \leadsto
| l = a - b
| r = q r s
| c = {{Defof|Congruence Modulo Integer}}
}}
{{eqn | ll= \leadsto
| l = a - b
| r = \paren {q r} s
| c =
}}
{{eqn | lo= \land
| l =... | Congruence by Divisor of Modulus/Integer Modulus | https://proofwiki.org/wiki/Congruence_by_Divisor_of_Modulus/Integer_Modulus | https://proofwiki.org/wiki/Congruence_by_Divisor_of_Modulus/Integer_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Integer",
"Definition:Congruence (Number Theory)/Integers"
] | [
"Definition:Integer",
"Definition:Integer"
] |
proofwiki-11724 | Modulo Addition is Well-Defined/Real Modulus | Let $z \in \R$ be a real number.
{{Explain|What if z is zero?}}
Let:
:$a \equiv b \pmod z$
and:
:$x \equiv y \pmod z$
where $a, b, x, y \in \R$.
Then:
: $a + x \equiv b + y \pmod z$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | l = c
| o = \equiv
| r = d
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadsto
| l = a \bmod z
| r = b \bmod z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | l = x ... | Let $z \in \R$ be a [[Definition:Real Number|real number]].
{{Explain|What if z is zero?}}
Let:
:$a \equiv b \pmod z$
and:
:$x \equiv y \pmod z$
where $a, b, x, y \in \R$.
Then:
: $a + x \equiv b + y \pmod z$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | l = c
| o = \equiv
| r = d
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadsto
| l = a \bmod z
| r = b \bmod z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | l = x ... | Modulo Addition is Well-Defined/Real Modulus | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined/Real_Modulus | https://proofwiki.org/wiki/Modulo_Addition_is_Well-Defined/Real_Modulus | [
"Modulo Addition",
"Modulo Addition is Well-Defined"
] | [
"Definition:Real Number"
] | [] |
proofwiki-11725 | Modulo Multiplication is Well-Defined/Warning | Let $z \in \R$ be a real number.
Let:
:$a \equiv b \pmod z$
and:
:$x \equiv y \pmod z$
where $a, b, x, y \in \R$.
Then it does '''not''' necessarily hold that:
:$a x \equiv b y \pmod z$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
| c =
}}
{{eqn | l = x
| o = \equiv
| r = y
| rr= \pmod m
| c =
}}
{{eqn | ll= \leadsto
| l = a \bmod m
| r = b \bmod m
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | l = ... | Let $z \in \R$ be a [[Definition:Real Number|real number]].
Let:
:$a \equiv b \pmod z$
and:
:$x \equiv y \pmod z$
where $a, b, x, y \in \R$.
Then it does '''not''' necessarily hold that:
:$a x \equiv b y \pmod z$ | {{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod m
| c =
}}
{{eqn | l = x
| o = \equiv
| r = y
| rr= \pmod m
| c =
}}
{{eqn | ll= \leadsto
| l = a \bmod m
| r = b \bmod m
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | l = ... | Modulo Multiplication is Well-Defined/Warning | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined/Warning | https://proofwiki.org/wiki/Modulo_Multiplication_is_Well-Defined/Warning | [
"Modulo Multiplication"
] | [
"Definition:Real Number"
] | [
"Integer Multiplication Distributes over Addition",
"Definition:Integer",
"Definition:Integer"
] |
proofwiki-11726 | Congruence by Product of Moduli/Real Modulus | Let $a, b, z \in \R$.
Let $a \equiv b \pmod z$ denote that $a$ is congruent to $b$ modulo $z$.
Then $\forall y \in \R, y \ne 0$:
:$a \equiv b \pmod z \iff y a \equiv y b \pmod {y z}$ | Let $y \in \R: y \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \bmod z
| r = b \bmod z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | ll= \leadstoandfrom
| l = y \paren {a \bmod ... | Let $a, b, z \in \R$.
Let $a \equiv b \pmod z$ denote that [[Definition:Congruence (Number Theory)|$a$ is congruent to $b$ modulo $z$]].
Then $\forall y \in \R, y \ne 0$:
:$a \equiv b \pmod z \iff y a \equiv y b \pmod {y z}$ | Let $y \in \R: y \ne 0$.
Then:
{{begin-eqn}}
{{eqn | l = a
| o = \equiv
| r = b
| rr= \pmod z
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = a \bmod z
| r = b \bmod z
| c = {{Defof|Congruence (Number Theory)|Congruence}}
}}
{{eqn | ll= \leadstoandfrom
| l = y \paren {a \bmo... | Congruence by Product of Moduli/Real Modulus | https://proofwiki.org/wiki/Congruence_by_Product_of_Moduli/Real_Modulus | https://proofwiki.org/wiki/Congruence_by_Product_of_Moduli/Real_Modulus | [
"Modulo Arithmetic"
] | [
"Definition:Congruence (Number Theory)"
] | [
"Product Distributes over Modulo Operation"
] |
proofwiki-11727 | Segment of Auxiliary Relation is Subset of Lower Closure | Let $\struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $R$ be auxiliary relation on $S$.
Let $x \in S$.
Then
:$x^R \subseteq x^\preceq$
where
:$x^R$ denotes the $R$-segment of $x$,
:$x^\preceq$ denotes the lower closure of $x$. | Let $a \in x^R$.
By definition of $R$-segment of $x$:
:$\tuple {a, x} \in R$
By definition of auxiliary relation:
:$a \preceq x$
Thus by definition of lower closure of element:
:$a \in x^\preceq$
{{qed}} | Let $\struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $R$ be [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
Let $x \in S$.
Then
:$x^R \subseteq x^\preceq$
where
:$x^R$ denotes the [[Definition:Relation Segment|$R$-segment... | Let $a \in x^R$.
By definition of [[Definition:Relation Segment|$R$-segment]] of $x$:
:$\tuple {a, x} \in R$
By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$a \preceq x$
Thus by definition of [[Definition:Lower Closure of Element|lower closure of element]]:
:$a \in x^\preceq$
{{qed}} | Segment of Auxiliary Relation is Subset of Lower Closure | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_is_Subset_of_Lower_Closure | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_is_Subset_of_Lower_Closure | [
"Auxiliary Relations",
"Lower Closures"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Auxiliary Relation",
"Definition:Relation Segment",
"Definition:Lower Closure/Element"
] | [
"Definition:Relation Segment",
"Definition:Auxiliary Relation",
"Definition:Lower Closure/Element"
] |
proofwiki-11728 | Power Function is Completely Multiplicative/Integers | Let $c \in \Z$ be an integer.
Let $f_c: \Z \to \Z$ be the mapping defined as:
:$\forall n \in \Z: \map {f_c} n = n^c$
Then $f_c$ is completely multiplicative. | Let $r, s \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f_c} {r s}
| r = \paren {r s}^c
| c =
}}
{{eqn | r = r^c s^c
| c = Product of Powers
}}
{{eqn | r = \map {f_c} r \map {f_c} s
| c =
}}
{{end-eqn}}
{{Qed}} | Let $c \in \Z$ be an [[Definition:Integer|integer]].
Let $f_c: \Z \to \Z$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall n \in \Z: \map {f_c} n = n^c$
Then $f_c$ is [[Definition:Completely Multiplicative Function|completely multiplicative]]. | Let $r, s \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f_c} {r s}
| r = \paren {r s}^c
| c =
}}
{{eqn | r = r^c s^c
| c = [[Product of Powers]]
}}
{{eqn | r = \map {f_c} r \map {f_c} s
| c =
}}
{{end-eqn}}
{{Qed}} | Power Function is Completely Multiplicative/Integers | https://proofwiki.org/wiki/Power_Function_is_Completely_Multiplicative/Integers | https://proofwiki.org/wiki/Power_Function_is_Completely_Multiplicative/Integers | [
"Completely Multiplicative Functions",
"Number Theory"
] | [
"Definition:Integer",
"Definition:Mapping",
"Definition:Completely Multiplicative Function"
] | [
"Exponent Combination Laws/Product of Powers"
] |
proofwiki-11729 | Preceding implies Inclusion of Segments of Auxiliary Relation | Let $\left({S, \vee, \preceq}\right)$ be a bounded below join semilattice.
Let $R$ be an auxiliary relation on $S$.
Let $x, y \in S$ such that
:$x \preceq y$
Then
:$x^R \subseteq y^R$
where $x^R$ denotes the $R$-segment of $x$. | Let $a \in x^R$.
By definition of $R$-segment of $x$:
:$\left({a, x}\right) \in R$
By definition of reflexivity:
:$a \preceq a$
By definition of auxiliary relation:
:$\left({a, y}\right) \in R$
Thus by definition of $R$-segment of $y$:
:$a \in y^R$
{{qed}} | Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $R$ be an [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
Let $x, y \in S$ such that
:$x \preceq y$
Then
:$x^R \subseteq y^R$
where $x^R$ denotes the [[Definition:... | Let $a \in x^R$.
By definition of [[Definition:Relation Segment|$R$-segment]] of $x$:
:$\left({a, x}\right) \in R$
By definition of [[Definition:Reflexivity|reflexivity]]:
:$a \preceq a$
By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
:$\left({a, y}\right) \in R$
Thus by definition of [[Defin... | Preceding implies Inclusion of Segments of Auxiliary Relation | https://proofwiki.org/wiki/Preceding_implies_Inclusion_of_Segments_of_Auxiliary_Relation | https://proofwiki.org/wiki/Preceding_implies_Inclusion_of_Segments_of_Auxiliary_Relation | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Auxiliary Relation",
"Definition:Relation Segment"
] | [
"Definition:Relation Segment",
"Definition:Reflexivity",
"Definition:Auxiliary Relation",
"Definition:Relation Segment"
] |
proofwiki-11730 | Characteristic Function of Square-Free Integers is Multiplicative | Let $S \subseteq \Z$ be the set of positive integers defined as:
:$S := \set {n \in \Z: \forall k \in \Z_{>1}: k^2 \nmid n}$
That is, let $S$ be the set of all square-free positive integers.
Let $\chi_S: \N \to \Z$ denote the characteristic function of $S$:
:$\forall n \in \Z: \map {\chi_S} n = \sqbrk {n \in S}$
where ... | Let $r, s \in \Z$ such that $r \perp s$. | Let $S \subseteq \Z$ be the [[Definition:Set|set]] of [[Definition:Positive Integer|positive integers]] defined as:
:$S := \set {n \in \Z: \forall k \in \Z_{>1}: k^2 \nmid n}$
That is, let $S$ be the [[Definition:Set|set]] of all [[Definition:Square-Free Integer|square-free]] [[Definition:Positive Integer|positive int... | Let $r, s \in \Z$ such that $r \perp s$. | Characteristic Function of Square-Free Integers is Multiplicative | https://proofwiki.org/wiki/Characteristic_Function_of_Square-Free_Integers_is_Multiplicative | https://proofwiki.org/wiki/Characteristic_Function_of_Square-Free_Integers_is_Multiplicative | [
"Multiplicative Functions",
"Number Theory",
"Square-Free Integers"
] | [
"Definition:Set",
"Definition:Positive/Integer",
"Definition:Set",
"Definition:Square-Free Integer",
"Definition:Positive/Integer",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Iverson's Convention",
"Definition:Multiplicative Arithmetic Function"
] | [] |
proofwiki-11731 | Segment of Auxiliary Relation Mapping is Increasing | Let $R = \left({S, \preceq}\right)$ be an ordered set.
Let ${\it Ids}\left({R}\right)$ be the set of all ideals in $R$.
Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an ordered set
where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$
Let $r$ b... | $f$ is well-defined because by Relation Segment of Auxiliary Relation is Ideal:
:$\forall x \in S: x^r$ is ideal in $L$
Let $x, y \in S$ such that
:$x \preceq y$
By Preceding implies Inclusion of Segments of Auxiliary Relation:
:$x^r \subseteq y^r$
Thus by definitions of $\precsim$ and $f$:
:$f\left({x}\right) \precsim... | Let $R = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let ${\it Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
w... | $f$ is well-defined because by [[Relation Segment of Auxiliary Relation is Ideal]]:
:$\forall x \in S: x^r$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$
Let $x, y \in S$ such that
:$x \preceq y$
By [[Preceding implies Inclusion of Segments of Auxiliary Relation]]:
:$x^r \subseteq y^r$
Thus by definitions of $... | Segment of Auxiliary Relation Mapping is Increasing | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_Mapping_is_Increasing | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_Mapping_is_Increasing | [
"Auxiliary Relations"
] | [
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Auxiliary Relation",
"Definition:Mapping",
"Definition:Relation Segment",
"Definition:Increasing/Mapping"
] | [
"Relation Segment of Auxiliary Relation is Ideal",
"Definition:Ideal in Ordered Set",
"Preceding implies Inclusion of Segments of Auxiliary Relation",
"Definition:Increasing/Mapping"
] |
proofwiki-11732 | Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function | Let $c \in \R$ be a constant.
Let $f: \N \to \R$ denotes the mapping defined as:
:$\forall n \in \N: \map f n = c^k$
where $k$ is number of distinct primes that divide $n$.
Then $f$ is multiplicative. | Let $r, s \in \Z$ such that $r \perp s$.
Let $r$ be composed of $p$ distinct primes:
:$r_1, r_2, \ldots r_p$
Thus:
:$\map f r = c^p$
Let $s$ be composed of $q$ distinct primes:
:$s_1, s_2, \ldots s_q$
Thus:
:$\map f s = c^q$
As $r \perp s$, all the $r_k$ and $s_k$ are distinct.
Thus $r s$ is composed of:
:the $p$ disti... | Let $c \in \R$ be a [[Definition:Constant|constant]].
Let $f: \N \to \R$ denotes the [[Definition:Mapping|mapping]] defined as:
:$\forall n \in \N: \map f n = c^k$
where $k$ is [[Definition:Number|number]] of [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]] that [[Definition:Divisor of Integer|divid... | Let $r, s \in \Z$ such that $r \perp s$.
Let $r$ be composed of $p$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]]:
:$r_1, r_2, \ldots r_p$
Thus:
:$\map f r = c^p$
Let $s$ be composed of $q$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]]:
:$s_1, s_2, \ldots s_q$
Thus:
:$\ma... | Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function | https://proofwiki.org/wiki/Constant_to_Power_of_Number_of_Distinct_Prime_Divisors_is_Multiplicative_Function | https://proofwiki.org/wiki/Constant_to_Power_of_Number_of_Distinct_Prime_Divisors_is_Multiplicative_Function | [
"Multiplicative Functions",
"Number Theory"
] | [
"Definition:Constant",
"Definition:Mapping",
"Definition:Number",
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Multiplicative Arithmetic Function"
] | [
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Distinct",
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Distinct",
"Definition:Prime Number",
"Definition:Distinct",
"Definition:Prime Number",
"Exponent Combination... |
proofwiki-11733 | Product of Multiplicative Functions is Multiplicative | Let $f: \N \to \C$ and $g: \N \to \C$ be multiplicative functions.
Then their pointwise product:
:$f \times g: \Z \to \Z: \forall s \in S: \map {\paren {f \times g} } s := \map f s \times \map g s$
is also multiplicative. | Let $f$ and $g$ be multiplicative.
Let $m \perp n$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f \times g} {m \times n}
| r = \map f {m \times n} \times \map g {m \times n}
| c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}}
}}
{{eqn | r = \map f m \times \map f n \times \map g m \times \map g n... | Let $f: \N \to \C$ and $g: \N \to \C$ be [[Definition:Multiplicative Arithmetic Function|multiplicative functions]].
Then their [[Definition:Pointwise Multiplication of Integer-Valued Functions|pointwise product]]:
:$f \times g: \Z \to \Z: \forall s \in S: \map {\paren {f \times g} } s := \map f s \times \map g s$
is ... | Let $f$ and $g$ be [[Definition:Multiplicative Arithmetic Function|multiplicative]].
Let $m \perp n$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f \times g} {m \times n}
| r = \map f {m \times n} \times \map g {m \times n}
| c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}}
}}
{{eqn | r = \map... | Product of Multiplicative Functions is Multiplicative | https://proofwiki.org/wiki/Product_of_Multiplicative_Functions_is_Multiplicative | https://proofwiki.org/wiki/Product_of_Multiplicative_Functions_is_Multiplicative | [
"Multiplicative Functions"
] | [
"Definition:Multiplicative Arithmetic Function",
"Definition:Pointwise Multiplication of Integer-Valued Functions",
"Definition:Multiplicative Arithmetic Function"
] | [
"Definition:Multiplicative Arithmetic Function",
"Integer Multiplication is Commutative",
"Definition:Multiplicative Arithmetic Function"
] |
proofwiki-11734 | Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure | Let $R = \struct {S, \preceq}$ be an ordered set.
Let $\map {\mathit {Ids} } R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\mathit {Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq {\restriction_{\map {\mathit {Ids} } R \times \map {\mathit {Ids} } R} }$.
Let $r$ be an auxiliary ... | === Condition $(1)$ ===
By Segment of Auxiliary Relation Mapping is Increasing:
:$f$ is an increasing mapping.
By Segment of Auxiliary Relation is Subset of Lower Closure:
:$\forall x \in S: x^r \subseteq x^\preceq$
By definition of $f$:
:$\forall x \in S: \map f x \subseteq x^\preceq$
Thus
:$f \in F$
{{qed|lemma}} | Let $R = \struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\map {\mathit {Ids} } R$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
Let $L = \struct {\map {\mathit {Ids} } R, \precsim}$ be an [[Definition:Ordered Set|ordered set]] where $\precsim ... | === Condition $(1)$ ===
By [[Segment of Auxiliary Relation Mapping is Increasing]]:
:$f$ is an [[Definition:Increasing Mapping|increasing mapping]].
By [[Segment of Auxiliary Relation is Subset of Lower Closure]]:
:$\forall x \in S: x^r \subseteq x^\preceq$
By definition of $f$:
:$\forall x \in S: \map f x \subseteq... | Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_Mapping_is_Element_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure | https://proofwiki.org/wiki/Segment_of_Auxiliary_Relation_Mapping_is_Element_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure | [
"Auxiliary Relations"
] | [
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Auxiliary Relation",
"Definition:Increasing Mappings Satisfying Inclusion in Lower Closure",
"Definition:Mapping",
"Definition:Relation Segment",
"Definition:Smallest Element"... | [
"Segment of Auxiliary Relation Mapping is Increasing",
"Definition:Increasing/Mapping",
"Segment of Auxiliary Relation is Subset of Lower Closure",
"Definition:Increasing/Mapping",
"Segment of Auxiliary Relation is Subset of Lower Closure",
"Definition:Increasing/Mapping"
] |
proofwiki-11735 | Singleton of Bottom is Ideal | Let $\struct {S, \preceq}$ be a bounded below ordered set.
Then
:$\set \bot$ is an ideal in $\struct {S, \preceq}$
where $\bot$ denotes the smallest element in $S$. | === Non-empty ===
By definition of singleton:
:$\bot \in \set \bot$
By definition:
:$\set \bot$ is a non-empty set.
{{qed|lemma}} | Let $\struct {S, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Ordered Set|ordered set]].
Then
:$\set \bot$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$
where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $S$. | === Non-empty ===
By definition of [[Definition:Singleton|singleton]]:
:$\bot \in \set \bot$
By definition:
:$\set \bot$ is a [[Definition:Non-Empty Set|non-empty set]].
{{qed|lemma}} | Singleton of Bottom is Ideal | https://proofwiki.org/wiki/Singleton_of_Bottom_is_Ideal | https://proofwiki.org/wiki/Singleton_of_Bottom_is_Ideal | [
"Order Theory"
] | [
"Definition:Bounded Below",
"Definition:Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Smallest Element"
] | [
"Definition:Singleton",
"Definition:Non-Empty Set",
"Definition:Singleton",
"Definition:Singleton"
] |
proofwiki-11736 | Number of Digits in Factorial | Let $n!$ denote the factorial of $n$.
The number of digits in $n!$ is approximately:
:$1 + \floor {\dfrac 1 2 \paren {\log_{10} 2 + \log_{10} \pi} + \dfrac 1 2 \log_{10} n + n \paren {\log_{10} n - \log_{10} e} }$
when $n!$ is shown in decimal notation.
This evaluates to:
:$1 + \floor {\paren {n + \dfrac 1 2} \log_{10}... | From Stirling's Formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
from which the result can be calculated.
To count the number of digits:
{{begin-eqn}}
{{eqn | l = \log_{10} n!
| o = \sim
| r = \log_{10} \paren {\sqrt {2 \pi n} \paren {\dfrac n e}^n}
| c = Stirling's Formula
}}
{{eqn | r = \lo... | Let $n!$ denote the [[Definition:Factorial|factorial]] of $n$.
The number of [[Definition:Digit|digits]] in $n!$ is approximately:
:$1 + \floor {\dfrac 1 2 \paren {\log_{10} 2 + \log_{10} \pi} + \dfrac 1 2 \log_{10} n + n \paren {\log_{10} n - \log_{10} e} }$
when $n!$ is shown in [[Definition:Decimal Notation|decimal... | From [[Stirling's Formula]]:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
from which the result can be calculated.
To count the number of [[Definition:Digit|digits]]:
{{begin-eqn}}
{{eqn | l = \log_{10} n!
| o = \sim
| r = \log_{10} \paren {\sqrt {2 \pi n} \paren {\dfrac n e}^n}
| c = [[Stirling... | Number of Digits in Factorial | https://proofwiki.org/wiki/Number_of_Digits_in_Factorial | https://proofwiki.org/wiki/Number_of_Digits_in_Factorial | [
"Factorials"
] | [
"Definition:Factorial",
"Definition:Digit",
"Definition:Decimal Notation"
] | [
"Stirling's Formula",
"Definition:Digit",
"Stirling's Formula",
"Sum of Logarithms",
"Logarithm of Power",
"Sum of Logarithms",
"Difference of Logarithms",
"Common Logarithm/Examples/2",
"Common Logarithm/Examples/pi",
"Common Logarithm/Examples/e",
"Number of Digits in Number",
"Category:Fact... |
proofwiki-11737 | Rising Factorial in terms of Falling Factorial | :$x^{\overline n} = \paren {x + n - 1}^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x^{\overline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = x \paren {x + 1} \cdots \paren {x + n - 1}
| c = {{Defof|Continued Product}}
}}
{{eqn | r = \paren {x + n - 1} \paren {x + n - 2} \cdots \paren {x + 1} x
|... | :$x^{\overline n} = \paren {x + n - 1}^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x^{\overline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = x \paren {x + 1} \cdots \paren {x + n - 1}
| c = {{Defof|Continued Product}}
}}
{{eqn | r = \paren {x + n - 1} \paren {x + n - 2} \cdots \paren {x + 1} x
|... | Rising Factorial in terms of Falling Factorial | https://proofwiki.org/wiki/Rising_Factorial_in_terms_of_Falling_Factorial | https://proofwiki.org/wiki/Rising_Factorial_in_terms_of_Falling_Factorial | [
"Falling Factorials",
"Rising Factorials"
] | [] | [] |
proofwiki-11738 | Rising Factorial in terms of Falling Factorial of Negative | :$x^{\overline k} = \paren {-1}^k \paren {-x}^{\underline k}$ | {{begin-eqn}}
{{eqn | l = \paren {-x}^{\underline k}
| r = \prod_{j \mathop = 0}^{k - 1} \paren {-x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \paren {x + j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \prod_{j \mathop = 0}^{k - ... | :$x^{\overline k} = \paren {-1}^k \paren {-x}^{\underline k}$ | {{begin-eqn}}
{{eqn | l = \paren {-x}^{\underline k}
| r = \prod_{j \mathop = 0}^{k - 1} \paren {-x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \paren {x + j}
| c =
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \prod_{j \mathop = 0}^{k - ... | Rising Factorial in terms of Falling Factorial of Negative | https://proofwiki.org/wiki/Rising_Factorial_in_terms_of_Falling_Factorial_of_Negative | https://proofwiki.org/wiki/Rising_Factorial_in_terms_of_Falling_Factorial_of_Negative | [
"Falling Factorials",
"Rising Factorials"
] | [] | [
"Product of Products"
] |
proofwiki-11739 | Falling Factorial as Quotient of Factorials | :$x^{\underline n} = \dfrac {x!} {\paren {x - n}!} = \dfrac {\map \Gamma {x + 1} } {\map \Gamma {x - n + 1} }$ | {{begin-eqn}}
{{eqn | l = x^{\underline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = x \paren {x - 1} \paren {x - 2} \dotsm \paren {x - n + 1}
| c =
}}
{{eqn | r = \dfrac {x!} {\paren {x - n}!}
| c = {{Defof|Factorial}}
}}
{{eqn | r = \d... | :$x^{\underline n} = \dfrac {x!} {\paren {x - n}!} = \dfrac {\map \Gamma {x + 1} } {\map \Gamma {x - n + 1} }$ | {{begin-eqn}}
{{eqn | l = x^{\underline n}
| r = \prod_{j \mathop = 0}^{n - 1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = x \paren {x - 1} \paren {x - 2} \dotsm \paren {x - n + 1}
| c =
}}
{{eqn | r = \dfrac {x!} {\paren {x - n}!}
| c = {{Defof|Factorial}}
}}
{{eqn | r = \d... | Falling Factorial as Quotient of Factorials | https://proofwiki.org/wiki/Falling_Factorial_as_Quotient_of_Factorials | https://proofwiki.org/wiki/Falling_Factorial_as_Quotient_of_Factorials | [
"Falling Factorials",
"Factorials"
] | [] | [] |
proofwiki-11740 | One to Integer Rising is Integer Factorial | :$1^{\overline n} = n!$ | {{begin-eqn}}
{{eqn | l = 1^{\overline n}
| r = \dfrac {\paren {1 + n - 1}!} {\paren {1 - 1}!}
| c = Rising Factorial as Quotient of Factorials
}}
{{eqn | r = \dfrac {n!} {0!}
| c =
}}
{{eqn | r = n!
| c = Factorial of Zero
}}
{{end-eqn}}
{{qed}}
Category:Rising Factorials
Category:Factorials
4... | :$1^{\overline n} = n!$ | {{begin-eqn}}
{{eqn | l = 1^{\overline n}
| r = \dfrac {\paren {1 + n - 1}!} {\paren {1 - 1}!}
| c = [[Rising Factorial as Quotient of Factorials]]
}}
{{eqn | r = \dfrac {n!} {0!}
| c =
}}
{{eqn | r = n!
| c = [[Factorial of Zero]]
}}
{{end-eqn}}
{{qed}}
[[Category:Rising Factorials]]
[[Catego... | One to Integer Rising is Integer Factorial | https://proofwiki.org/wiki/One_to_Integer_Rising_is_Integer_Factorial | https://proofwiki.org/wiki/One_to_Integer_Rising_is_Integer_Factorial | [
"Rising Factorials",
"Factorials"
] | [] | [
"Rising Factorial as Quotient of Factorials",
"Factorial/Examples/0",
"Category:Rising Factorials",
"Category:Factorials"
] |
proofwiki-11741 | Number to Power of One Rising is Itself | :$x^{\overline 1} = x$ | {{begin-eqn}}
{{eqn | l = x^{\overline 1}
| r = \dfrac {\map \Gamma {x + 1} } {\map \Gamma x}
| c = Rising Factorial as Quotient of Factorials
}}
{{eqn | r = x
| c = Gamma Difference Equation
}}
{{end-eqn}}
{{qed}}
Category:Rising Factorials
sa1f6ul61j0vgsfk8waanaiyu8o7pg1 | :$x^{\overline 1} = x$ | {{begin-eqn}}
{{eqn | l = x^{\overline 1}
| r = \dfrac {\map \Gamma {x + 1} } {\map \Gamma x}
| c = [[Rising Factorial as Quotient of Factorials]]
}}
{{eqn | r = x
| c = [[Gamma Difference Equation]]
}}
{{end-eqn}}
{{qed}}
[[Category:Rising Factorials]]
sa1f6ul61j0vgsfk8waanaiyu8o7pg1 | Number to Power of One Rising is Itself | https://proofwiki.org/wiki/Number_to_Power_of_One_Rising_is_Itself | https://proofwiki.org/wiki/Number_to_Power_of_One_Rising_is_Itself | [
"Rising Factorials"
] | [] | [
"Rising Factorial as Quotient of Factorials",
"Gamma Difference Equation",
"Category:Rising Factorials"
] |
proofwiki-11742 | Integer to Power of Itself Falling is Factorial | :$n^{\underline n} = n!$ | {{begin-eqn}}
{{eqn | l = n^{\underline n}
| r = \dfrac {n!} {\paren {n - n}!}
| c = Falling Factorial as Quotient of Factorials
}}
{{eqn | r = \dfrac {n!} {0!}
| c =
}}
{{eqn | r = n!
| c = Factorial of Zero
}}
{{end-eqn}}
{{qed}}
Category:Falling Factorials
opk3w0uaoprxmpgvd9rzvqako48cfx9 | :$n^{\underline n} = n!$ | {{begin-eqn}}
{{eqn | l = n^{\underline n}
| r = \dfrac {n!} {\paren {n - n}!}
| c = [[Falling Factorial as Quotient of Factorials]]
}}
{{eqn | r = \dfrac {n!} {0!}
| c =
}}
{{eqn | r = n!
| c = [[Factorial of Zero]]
}}
{{end-eqn}}
{{qed}}
[[Category:Falling Factorials]]
opk3w0uaoprxmpgvd9rzvq... | Integer to Power of Itself Falling is Factorial | https://proofwiki.org/wiki/Integer_to_Power_of_Itself_Falling_is_Factorial | https://proofwiki.org/wiki/Integer_to_Power_of_Itself_Falling_is_Factorial | [
"Falling Factorials"
] | [] | [
"Falling Factorial as Quotient of Factorials",
"Factorial/Examples/0",
"Category:Falling Factorials"
] |
proofwiki-11743 | Number to Power of One Falling is Itself | :$x^{\underline 1} = x$ | {{begin-eqn}}
{{eqn | l = x^{\underline 1}
| r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x - 1 + 1}\right)}
| c = Falling Factorial as Quotient of Factorials
}}
{{eqn | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x}\right)}
| c =
}}
{{eqn | r = x
| c = Gamma Difference E... | :$x^{\underline 1} = x$ | {{begin-eqn}}
{{eqn | l = x^{\underline 1}
| r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x - 1 + 1}\right)}
| c = [[Falling Factorial as Quotient of Factorials]]
}}
{{eqn | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x}\right)}
| c =
}}
{{eqn | r = x
| c = [[Gamma Differ... | Number to Power of One Falling is Itself | https://proofwiki.org/wiki/Number_to_Power_of_One_Falling_is_Itself | https://proofwiki.org/wiki/Number_to_Power_of_One_Falling_is_Itself | [
"Falling Factorials"
] | [] | [
"Falling Factorial as Quotient of Factorials",
"Gamma Difference Equation",
"Category:Falling Factorials"
] |
proofwiki-11744 | Number of Permutations of One Less | :${}^{n - 1} P_n = {}^n P_n$
where ${}^k P_n$ denotes the number of ordered selections of $k$ objects from $n$. | {{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}} | :${}^{n - 1} P_n = {}^n P_n$
where ${}^k P_n$ denotes the number of [[Definition:Permutation (Ordered Selection)|ordered selections of $k$ objects from $n$]]. | {{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_of_One_Less | https://proofwiki.org/wiki/Number_of_Permutations_of_One_Less/Proof_1 | [
"Number of Permutations of One Less",
"Permutations (Ordered Selections)"
] | [
"Definition:Permutation/Ordered Selection"
] | [
"Number of Permutations",
"Number of Permutations"
] |
proofwiki-11745 | Number of Permutations of One Less | :${}^{n - 1} P_n = {}^n P_n$
where ${}^k P_n$ denotes the number of ordered selections of $k$ objects from $n$. | {{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-... | :${}^{n - 1} P_n = {}^n P_n$
where ${}^k P_n$ denotes the number of [[Definition:Permutation (Ordered Selection)|ordered selections of $k$ objects from $n$]]. | {{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_of_One_Less | https://proofwiki.org/wiki/Number_of_Permutations_of_One_Less/Proof_2 | [
"Number of Permutations of One Less",
"Permutations (Ordered Selections)"
] | [
"Definition:Permutation/Ordered Selection"
] | [
"Number of Permutations",
"Definition:Falling Factorial",
"Integer to Power of Itself Less One Falling is Factorial",
"Number of Permutations"
] |
proofwiki-11746 | Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation | Let $R = \struct {S, \preceq}$ be a bounded below join semilattice.
Let $\map {\operatorname{Ids} } R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\map {\operatorname{Ids} } R \times \map {\operatorname{Id... | Define relation $\RR$ on $S$:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$
We will prove that:
:$(1): \quad \forall x, y \in S: \tuple {x, y} \in \RR \implies x \preceq y$
Let $x, y \in S$ such that:
:$\tuple {x, y} \in \RR$
By definitions of $\RR$ and $F$:
:$x \in \map f y \subseteq y^\preceq$
By d... | Let $R = \struct {S, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\operatorname{Ids} } R$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$... | Define [[Definition:Relation|relation]] $\RR$ on $S$:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff x \in \map f y$
We will prove that:
:$(1): \quad \forall x, y \in S: \tuple {x, y} \in \RR \implies x \preceq y$
Let $x, y \in S$ such that:
:$\tuple {x, y} \in \RR$
By definitions of $\RR$ and $F$:
:$x \in \map f ... | Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation | https://proofwiki.org/wiki/Element_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure_is_Generated_by_Auxiliary_Relation | https://proofwiki.org/wiki/Element_of_Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure_is_Generated_by_Auxiliary_Relation | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Increasing Mappings Satisfying Inclusion in Lower Closure",
"Definition:Auxiliary Relation",
"Definition:Relation Segment"
] | [
"Definition:Relation",
"Definition:Subset",
"Definition:Lower Closure/Element",
"Definition:Increasing/Mapping",
"Definition:Subset",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Lower Section",
"Definition:Lower Section",
"Definition:Ideal in Ordered Set",
"... |
proofwiki-11747 | Binomial Coefficient with Two/Corollary | :$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$
where $T_n$ is the $n$th triangular number. | From the definition of binomial coefficient:
:$\dbinom n 2 = \dfrac {n!} {2! \paren {n - 2}!}$
The result follows directly from the definition of the factorial:
:$2! = 1 \times 2$
{{qed}} | :$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$
where $T_n$ is the [[Definition:Triangular Number|$n$th triangular number]]. | From the definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
:$\dbinom n 2 = \dfrac {n!} {2! \paren {n - 2}!}$
The result follows directly from the definition of the [[Definition:Factorial|factorial]]:
:$2! = 1 \times 2$
{{qed}} | Binomial Coefficient with Two/Corollary | https://proofwiki.org/wiki/Binomial_Coefficient_with_Two/Corollary | https://proofwiki.org/wiki/Binomial_Coefficient_with_Two/Corollary | [
"Examples of Binomial Coefficients",
"Triangular Numbers"
] | [
"Definition:Triangular Number"
] | [
"Definition:Binomial Coefficient",
"Definition:Factorial"
] |
proofwiki-11748 | Increasing Mappings Satisfying Inclusion in Lower Closure is Isomorphic to Auxiliary Relations | Let $R = \struct {S, \preceq}$ be a bounded below join semilattice.
Let $\map {\operatorname{Ids} } R$ be the set of all ideals in $R$.
Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\map {\operatorname{Ids} } R \times \map {\operatorname{Id... | By Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure define $G: \map {\operatorname{Aux} } R \to F$:
:$\forall \RR \in \map {\operatorname{Aux} } R: \map G \RR = \paren {S \ni x \mapsto x^\RR}$
We will prove by Order Isomorphism is Surjective Order Embedding t... | Let $R = \struct {S, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\operatorname{Ids} } R$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$... | By [[Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure]] define $G: \map {\operatorname{Aux} } R \to F$:
:$\forall \RR \in \map {\operatorname{Aux} } R: \map G \RR = \paren {S \ni x \mapsto x^\RR}$
We will prove by [[Order Isomorphism is Surjective Order Embe... | Increasing Mappings Satisfying Inclusion in Lower Closure is Isomorphic to Auxiliary Relations | https://proofwiki.org/wiki/Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure_is_Isomorphic_to_Auxiliary_Relations | https://proofwiki.org/wiki/Increasing_Mappings_Satisfying_Inclusion_in_Lower_Closure_is_Isomorphic_to_Auxiliary_Relations | [
"Auxiliary Relations"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Increasing Mappings Satisfying Inclusion in Lower Closure",
"Definition:Set of Sets",
"Definition:Auxiliary Relation",
"Definition:Ordered Set... | [
"Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure",
"Order Isomorphism is Surjective Order Embedding",
"Definition:Order Isomorphism"
] |
proofwiki-11749 | Binomial Theorem/Abel's Generalisation | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | By admitting $y = \paren {x + y} - x$, we have that:
:$\paren {x + y}^n = \paren {x + \paren {x + y} - x}^n$
Expanding the {{RHS}} in powers of $\paren {x + y}$:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}
| c =
}}
{{eqn | r = \sum_k \binom n k x \... | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | By admitting $y = \paren {x + y} - x$, we have that:
:$\paren {x + y}^n = \paren {x + \paren {x + y} - x}^n$
Expanding the {{RHS}} in powers of $\paren {x + y}$:
{{begin-eqn}}
{{eqn | o =
| r = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}
| c =
}}
{{eqn | r = \sum_k \binom n k ... | Binomial Theorem/Abel's Generalisation/Proof 1 | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/Proof_1 | [
"Binomial Coefficients",
"Binomial Theorem"
] | [] | [
"Binomial Theorem/Abel's Generalisation/x+y = 0",
"Binomial Theorem"
] |
proofwiki-11750 | Binomial Theorem/Abel's Generalisation | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | From this formula:
:$(1): \quad \ds \sum_{k \mathop \in \Z} \binom n k x \paren {x + k z}^{k - 1} y \paren {y + \paren {n - k} z}^{n - k - 1} = \paren {x + y} \paren {x + y + n z}^{n - 1}$
The given formula:
:$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$
can then be tran... | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | From [[Link required to result in Knuth: exercise 2.3.4.4-29|this formula]]:
:$(1): \quad \ds \sum_{k \mathop \in \Z} \binom n k x \paren {x + k z}^{k - 1} y \paren {y + \paren {n - k} z}^{n - k - 1} = \paren {x + y} \paren {x + y + n z}^{n - 1}$
The given formula:
:$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {... | Binomial Theorem/Abel's Generalisation/Proof 2 | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/Proof_2 | [
"Binomial Coefficients",
"Binomial Theorem"
] | [] | [
"Link required to result in Knuth: exercise 2.3.4.4-29"
] |
proofwiki-11751 | Binomial Theorem/Abel's Generalisation | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | From Hurwitz's Generalisation of Binomial Theorem:
:$(1): \quad \paren {x + y}^n = \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
Setting $z = z_1 = z_2 = \cdots z_n$ we... | :$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$ | From [[Hurwitz's Generalisation of Binomial Theorem]]:
:$(1): \quad \paren {x + y}^n = \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
Setting $z = z_1 = z_2 = \cdots z_... | Binomial Theorem/Abel's Generalisation/Proof 3 | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation | https://proofwiki.org/wiki/Binomial_Theorem/Abel's_Generalisation/Proof_3 | [
"Binomial Coefficients",
"Binomial Theorem"
] | [] | [
"Binomial Theorem/Hurwitz's Generalisation",
"Symmetry Rule for Binomial Coefficients"
] |
proofwiki-11752 | Sum over k of r Choose m+k by s Choose n+k | Let $s \in \R, r \in \Z_{\ge 0}, m, n \in \Z$.
Then:
:$\ds \sum_k \binom r {m + k} \binom s {n + k} = \binom {r + s} {r - m + n}$ | {{begin-eqn}}
{{eqn | l = \sum_k \binom r {m + k} \binom s {n + k}
| r = \sum_k \binom r {r - m - k} \binom s {s - n - k}
| c = Symmetry Rule for Binomial Coefficients
}}
{{eqn | r = \sum_k \binom r k \binom s {s - n - \left({r - m - k}\right)}
| c = Change of Index Variable of Summation
}}
{{eqn | r ... | Let $s \in \R, r \in \Z_{\ge 0}, m, n \in \Z$.
Then:
:$\ds \sum_k \binom r {m + k} \binom s {n + k} = \binom {r + s} {r - m + n}$ | {{begin-eqn}}
{{eqn | l = \sum_k \binom r {m + k} \binom s {n + k}
| r = \sum_k \binom r {r - m - k} \binom s {s - n - k}
| c = [[Symmetry Rule for Binomial Coefficients]]
}}
{{eqn | r = \sum_k \binom r k \binom s {s - n - \left({r - m - k}\right)}
| c = [[Change of Index Variable of Summation]]
}}
{{... | Sum over k of r Choose m+k by s Choose n+k/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_m+k_by_s_Choose_n+k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_m+k_by_s_Choose_n+k/Proof_1 | [
"Binomial Coefficients"
] | [] | [
"Symmetry Rule for Binomial Coefficients",
"Change of Index Variable of Summation",
"Symmetry Rule for Binomial Coefficients",
"Chu-Vandermonde Identity"
] |
proofwiki-11753 | Sum over k of r Choose k by s+k Choose n by -1^r-k | Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
Then:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$ | The proof proceeds by induction on $r$.
For all $r \in \Z_{>0}$, let $\map P r$ be the proposition:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_k \binom 0 k \binom {s + k} n \paren {-1}^{0 - k}
... | Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
Then:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$ | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $r$.
For all $r \in \Z_{>0}$, let $\map P r$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_k \binom r k \binom {s + k} n \paren {-1}^{r - k} = \binom s {n - r}$
=== Basis for the Induction ===
$\map P 0$ is the case:
{{begin-eqn... | Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s+k_Choose_n_by_-1^r-k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s+k_Choose_n_by_-1^r-k/Proof_1 | [
"Binomial Coefficients",
"Sum over k of r Choose k by s+k Choose n by -1^r-k"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Zero Choose n",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Pascal's Rule",
"Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1",
"Pascal's Rule",
"Translation of In... |
proofwiki-11754 | Sum over k of r-k Choose m by s+k Choose n | Let $m, n, r, s \in \Z_{\ge 0}$ such that $n \ge s$.
Then:
:$\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$ | {{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom {-\left({n + 1}\right)} {s + k - m} \left({-1}\right)^{r - m + s - m}
| c = Moving Top Index to Bottom in Binomial Co... | Let $m, n, r, s \in \Z_{\ge 0}$ such that $n \ge s$.
Then:
:$\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$ | {{begin-eqn}}
{{eqn | o =
| r = \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom {-\left({n + 1}\right)} {s + k - m} \left({-1}\right)^{r - m + s - m}
| c = [[Moving Top Index to Bottom in Binomial ... | Sum over k of r-k Choose m by s+k Choose n/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r-k_Choose_m_by_s+k_Choose_n | https://proofwiki.org/wiki/Sum_over_k_of_r-k_Choose_m_by_s+k_Choose_n/Proof_1 | [
"Binomial Coefficients"
] | [] | [
"Negated Upper Index of Binomial Coefficient/Corollary 2",
"Chu-Vandermonde Identity",
"Negated Upper Index of Binomial Coefficient/Corollary 2"
] |
proofwiki-11755 | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk | Let $r, s, t \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ | For all $n \in \Z$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
Let the {{LHS}} of this equation be denoted $\tuple {r, s, t, n}$.
Let $n = 0$.
Then:
{{begin-eqn}}
{{eqn | o =
| r = \tuple {r... | Let $r, s, t \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ | For all $n \in \Z$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
Let the {{LHS}} of this equation be denoted $\tuple {r, s, t, n}$.
Let $n = 0$.
Then:
{{begin-eqn}}
{{e... | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_1 | [
"Binomial Coefficients",
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk"
] | [] | [
"Definition:Proposition",
"N Choose Negative Number is Zero",
"Binomial Coefficient with Zero",
"Binomial Coefficient with Zero",
"N Choose Negative Number is Zero",
"N Choose Negative Number is Zero",
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Sum over k of r-tk Choose... |
proofwiki-11756 | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk | Let $r, s, t \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ | From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
where:
:$\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
::$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
:for $x \ne n t$
:$z = x^{t + 1} - x^t$.
From Sum over $k$ of $\d... | Let $r, s, t \in \R, n \in \Z$.
Then:
:$\ds \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$ | From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
where:
:$\map {A_n} {x, t}$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree $n$]] defin... | Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk | https://proofwiki.org/wiki/Sum_over_k_of_r-tk_Choose_k_by_s-t(n-k)_Choose_n-k_by_r_over_r-tk/Proof_2 | [
"Binomial Coefficients",
"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 r-kt choose k by r over r-kt by z^k",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Sum over k of r-kt choose k by z^k"
] |
proofwiki-11757 | Sum over k of r Choose k by s Choose k by k | :$\ds \sum_k \binom r k \binom s k k = \binom {r + s - 1} {r - 1} s$ | {{begin-eqn}}
{{eqn | l = \sum_k \binom r k \binom s k k
| r = \sum_k \binom r k \binom {s - 1} {k - 1} s
| c = Factors of Binomial Coefficient
}}
{{eqn | r = s \sum_k \binom r k \binom {s - 1} {k - 1}
| c = as $s$ is constant
}}
{{eqn | r = s \binom {r + s - 1} {r - 1}
| c = Sum over $k$ of $\d... | :$\ds \sum_k \binom r k \binom s k k = \binom {r + s - 1} {r - 1} s$ | {{begin-eqn}}
{{eqn | l = \sum_k \binom r k \binom s k k
| r = \sum_k \binom r k \binom {s - 1} {k - 1} s
| c = [[Factors of Binomial Coefficient]]
}}
{{eqn | r = s \sum_k \binom r k \binom {s - 1} {k - 1}
| c = as $s$ is [[Definition:Constant|constant]]
}}
{{eqn | r = s \binom {r + s - 1} {r - 1}
... | Sum over k of r Choose k by s Choose k by k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s_Choose_k_by_k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s_Choose_k_by_k | [
"Binomial Coefficients"
] | [] | [
"Factors of Binomial Coefficient",
"Definition:Constant",
"Sum over k of r Choose m+k by s Choose n+k"
] |
proofwiki-11758 | Relation Segment is Increasing | Let $S$ be a set.
Let $\RR, \QQ$ be relations on $S$ such that
:$\RR \subseteq \QQ$
Let $x \in S$.
Then
:$x^\RR \subseteq x^\QQ$
where $x^\RR$ denotes the $\RR$-segment of $x$. | Let $y \in x^\RR$.
By definition of $\RR$-segment:
:$\tuple {y, x} \in \RR$
By definition of subset:
:$\tuple {y, x} \in \QQ$
Thus by definition of $\QQ$-segment:
:$y \in x^\QQ$
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $\RR, \QQ$ be [[Definition:Relation|relations]] on $S$ such that
:$\RR \subseteq \QQ$
Let $x \in S$.
Then
:$x^\RR \subseteq x^\QQ$
where $x^\RR$ denotes the [[Definition:Relation Segment|$\RR$-segment]] of $x$. | Let $y \in x^\RR$.
By definition of [[Definition:Relation Segment|$\RR$-segment]]:
:$\tuple {y, x} \in \RR$
By definition of [[Definition:Subset|subset]]:
:$\tuple {y, x} \in \QQ$
Thus by definition of [[Definition:Relation Segment|$\QQ$-segment]]:
:$y \in x^\QQ$
{{qed}} | Relation Segment is Increasing | https://proofwiki.org/wiki/Relation_Segment_is_Increasing | https://proofwiki.org/wiki/Relation_Segment_is_Increasing | [
"Relation Theory"
] | [
"Definition:Set",
"Definition:Relation",
"Definition:Relation Segment"
] | [
"Definition:Relation Segment",
"Definition:Subset",
"Definition:Relation Segment"
] |
proofwiki-11759 | Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1 | :$\ds \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$ | {{begin-eqn}}
{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
| o =
| c =
}}
{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
| c = Product of $\dbinom r m$ with $\dbinom m k$
}}
{{eqn | r = \sum_k \binom {n + k} k \binom {n ... | :$\ds \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$ | {{begin-eqn}}
{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
| o =
| c =
}}
{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
| c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
}}
{{eq... | Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_2_k_by_2_k_Choose_k_by_-1^k_over_k+1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_2_k_by_2_k_Choose_k_by_-1^k_over_k+1/Proof_1 | [
"Binomial Coefficients",
"Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1"
] | [] | [
"Product of r Choose m with m Choose k",
"Factors of Binomial Coefficient",
"Symmetry Rule for Binomial Coefficients",
"Translation of Index Variable of Summation",
"Sum over k of r Choose k by s+k Choose n by -1^r-k"
] |
proofwiki-11760 | Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1 | :$\ds \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$ | {{begin-eqn}}
{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
| o =
| c =
}}
{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
| c = Product of $\dbinom r m$ with $\dbinom m k$
}}
{{eqn | r = \sum_k \binom {n + k} k \binom {n ... | :$\ds \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$ | {{begin-eqn}}
{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
| o =
| c =
}}
{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
| c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
}}
{{eq... | Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_2_k_by_2_k_Choose_k_by_-1^k_over_k+1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_2_k_by_2_k_Choose_k_by_-1^k_over_k+1/Proof_2 | [
"Binomial Coefficients",
"Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1"
] | [] | [
"Product of r Choose m with m Choose k",
"Factors of Binomial Coefficient",
"Negated Upper Index of Binomial Coefficient",
"Sum over k of r Choose m+k by s Choose n+k"
] |
proofwiki-11761 | Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1 | :$\ds \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \binom {n - 1} {m - 1}$ | From Sum over $k$ of $\dbinom {r - k} m \dbinom {s + k} n$:
:$\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$
Setting $r = n + k - 1$, $m = 2 k$, $s = 0$, $n = m - 1$ and using $j$ as the index variable, this gives us:
:$\ds \sum_{j \mathop = 0}^{n + k - 1} \binom {n + k -... | :$\ds \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \binom {n - 1} {m - 1}$ | From [[Sum over k of r-k Choose m by s+k Choose n|Sum over $k$ of $\dbinom {r - k} m \dbinom {s + k} n$]]:
:$\ds \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$
Setting $r = n + k - 1$, $m = 2 k$, $s = 0$, $n = m - 1$ and using $j$ as the [[Definition:Index Variable of Summa... | Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_m+2k_by_2k_Choose_k_by_-1^k_over_k+1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_m+2k_by_2k_Choose_k_by_-1^k_over_k+1/Proof_1 | [
"Binomial Coefficients",
"Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1"
] | [] | [
"Sum over k of r-k Choose m by s+k Choose n",
"Definition:Summation/Index Variable",
"Definition:Binomial Coefficient",
"Definition:Summation",
"Definition:Summation",
"Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1",
"Definition:Kronecker Delta"
] |
proofwiki-11762 | Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1 | :$\ds \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \binom {n - 1} {m - 1}$ | Let:
:$\ds S := \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1}$
Then:
{{begin-eqn}}
{{eqn | l = \binom {2 k + 1} {k + 1}
| r = \binom {2 k} k \frac {2 k + 1} {k + 1}
| c = Factors of Binomial Coefficient
}}
{{eqn | ll= \leadsto
| l = \binom {2 k} k
| r = \binom {2 k... | :$\ds \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \binom {n - 1} {m - 1}$ | Let:
:$\ds S := \sum_k \binom {n + k} {m + 2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1}$
Then:
{{begin-eqn}}
{{eqn | l = \binom {2 k + 1} {k + 1}
| r = \binom {2 k} k \frac {2 k + 1} {k + 1}
| c = [[Factors of Binomial Coefficient]]
}}
{{eqn | ll= \leadsto
| l = \binom {2 k} k
| r = \binom... | Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_m+2k_by_2k_Choose_k_by_-1^k_over_k+1 | https://proofwiki.org/wiki/Sum_over_k_of_n+k_Choose_m+2k_by_2k_Choose_k_by_-1^k_over_k+1/Proof_2 | [
"Binomial Coefficients",
"Sum over k of n+k Choose m+2k by 2k Choose k by -1^k over k+1"
] | [] | [
"Factors of Binomial Coefficient",
"Symmetry Rule for Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient/Corollary 2",
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk",
"Negated Upper Index of Binomial Coefficient",
"Symmetry Rule for Binomial Coefficients"
] |
proofwiki-11763 | Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $x \in S$.
Then:
:$\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} = x^\ll$
where $\mathit {Ids}$ denotes the set of all ideals in $L$. | By Supremum of Lower Closure of Element:
:$\map \sup {x^\preceq} = x$
By Lower Closure of Element is Ideal:
:$x^\preceq \in \mathit {Ids}$
Then by definition of reflexivity:
:$x^\preceq \in \set {I \in \mathit {Ids}: x \preceq \sup I}$
We will prove that:
:$\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} \subs... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $x \in S$.
Then:
:$\ds \bigcap \set {I \in \mathit {Ids}: x \preceq \sup I} = x^\ll$
where $\mathit {Ids}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$. | By [[Supremum of Lower Closure of Element]]:
:$\map \sup {x^\preceq} = x$
By [[Lower Closure of Element is Ideal]]:
:$x^\preceq \in \mathit {Ids}$
Then by definition of [[Definition:Reflexivity|reflexivity]]:
:$x^\preceq \in \set {I \in \mathit {Ids}: x \preceq \sup I}$
We will prove that:
:$\ds \bigcap \set {I \in ... | Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element | https://proofwiki.org/wiki/Intersection_of_Ideals_with_Suprema_Succeed_Element_equals_Way_Below_Closure_of_Element | https://proofwiki.org/wiki/Intersection_of_Ideals_with_Suprema_Succeed_Element_equals_Way_Below_Closure_of_Element | [
"Way Below Relation"
] | [
"Definition:Complete Lattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set"
] | [
"Supremum of Lower Closure of Element",
"Lower Closure of Element is Ideal",
"Definition:Reflexivity",
"Definition:Set Intersection/Set of Sets",
"Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal",
"Definition:Way Below Closure",
"Definition:Set Equality... |
proofwiki-11764 | Way Below Closure is Ideal in Bounded Below Join Semilattice | Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $x \in S$.
Then
:$x^\ll$ is ideal in $L$. | By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$x^\ll$ is a non-empty directed set.
Let $y \in x^\ll, z \in S$ such that
:$z \preceq y$
By definition of way below closure:
:$y \ll x$
By definition of reflexivity:
:$x \preceq x$
By Preceding and Way Below implies Way Below:
:$z \ll x$
Thus by defin... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
Let $x \in S$.
Then
:$x^\ll$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$. | By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
:$x^\ll$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Directed Subset|directed set]].
Let $y \in x^\ll, z \in S$ such that
:$z \preceq y$
By definition of [[Definition:Way Below Closure|way below closure]]:
:$y \ll x$
By definition o... | Way Below Closure is Ideal in Bounded Below Join Semilattice | https://proofwiki.org/wiki/Way_Below_Closure_is_Ideal_in_Bounded_Below_Join_Semilattice | https://proofwiki.org/wiki/Way_Below_Closure_is_Ideal_in_Bounded_Below_Join_Semilattice | [
"Join and Meet Semilattices",
"Way Below Relation"
] | [
"Definition:Bounded Below",
"Definition:Join Semilattice",
"Definition:Ideal in Ordered Set"
] | [
"Way Below Closure is Directed in Bounded Below Join Semilattice",
"Definition:Non-Empty Set",
"Definition:Directed Subset",
"Definition:Way Below Closure",
"Definition:Reflexivity",
"Preceding and Way Below implies Way Below",
"Definition:Way Below Closure",
"Definition:Lower Section",
"Definition:... |
proofwiki-11765 | Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t | For $n \in \Z_{\ge 0}$:
:$\ds \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t} = \map {A_n} {r + s, t}$
where $\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
:$\map {A_n} {x, t} = \dbinom {x - n t} n \dfrac x {x - n t}$
where $x \ne n t$. | Let:
:$\ds S = \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t}$
Both sides of the statement of the theorem are polynomials in $r$, $s$ and $t$.
Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined.
By replacing the polynomials $A_n$ with their binomial coefficient ... | For $n \in \Z_{\ge 0}$:
:$\ds \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t} = \map {A_n} {r + s, t}$
where $\map {A_n} {x, t}$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$ defined as:
:$\map {A_n} {x, t} = \dbinom {x - n t} n \dfrac x {x - n t}$
w... | Let:
:$\ds S = \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t}$
Both sides of the statement of the theorem are [[Definition:Polynomial over Real Numbers|polynomials]] in $r$, $s$ and $t$.
Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined.
By replacing the [[D... | Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_Choose_k_by_r_over_r-kt_by_s-(n-k)t_Choose_n-k_by_s_over_s-(n-k)t | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_Choose_k_by_r_over_r-kt_by_s-(n-k)t_Choose_n-k_by_s_over_s-(n-k)t/Proof_1 | [
"Binomial Coefficients",
"Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Polynomial/Real Numbers",
"Definition:Binomial Coefficient",
"Definition:Partial Fractions Expansion",
"Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk"
] |
proofwiki-11766 | Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t | For $n \in \Z_{\ge 0}$:
:$\ds \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t} = \map {A_n} {r + s, t}$
where $\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
:$\map {A_n} {x, t} = \dbinom {x - n t} n \dfrac x {x - n t}$
where $x \ne n t$. | From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
and:
:$\ds \sum_k \map {A_k} {s, t} z^k = x^s$
Hence:
{{begin-eqn}}
{{eqn | l = x^{r + s}
| r = \sum_k \map {A_k} {r, t} z^k \sum_k \map {A_k} {s, t} z^k
| c =
}}
{{eqn | r = \sum_k \map {... | For $n \in \Z_{\ge 0}$:
:$\ds \sum_k \map {A_k} {r, t} \map {A_{n - k} } {s, t} = \map {A_n} {r + s, t}$
where $\map {A_n} {x, t}$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$ defined as:
:$\map {A_n} {x, t} = \dbinom {x - n t} n \dfrac x {x - n t}$
w... | From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
:$\ds \sum_k \map {A_k} {r, t} z^k = x^r$
and:
:$\ds \sum_k \map {A_k} {s, t} z^k = x^s$
Hence:
{{begin-eqn}}
{{eqn | l = x^{r + s}
| r = \sum_k \map {A_k} {r, t} z^k \sum_k \ma... | Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_Choose_k_by_r_over_r-kt_by_s-(n-k)t_Choose_n-k_by_s_over_s-(n-k)t | https://proofwiki.org/wiki/Sum_over_k_of_r-kt_Choose_k_by_r_over_r-kt_by_s-(n-k)t_Choose_n-k_by_s_over_s-(n-k)t/Proof_2 | [
"Binomial Coefficients",
"Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] | [
"Sum over k of r-kt choose k by r over r-kt by z^k"
] |
proofwiki-11767 | Sum over k of r Choose k by -1^r-k by Polynomial | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$. | From the {{Corollary|Sum over k of r Choose k by s+k Choose n by -1^r-k|disp = Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$}}:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
where $\delta_{n r}$ denotes the Kronecker delta.
Thus when $n \ne r$:
:$\ds \sum_k \binom r k \bin... | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ of [[Definition:Degree of Polynomial|degree]] $r$. | From the {{Corollary|Sum over k of r Choose k by s+k Choose n by -1^r-k|disp = Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$}}:
:$\ds \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
where $\delta_{n r}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
Thus when $n \n... | Sum over k of r Choose k by -1^r-k by Polynomial/Proof 1 | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_-1^r-k_by_Polynomial | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_-1^r-k_by_Polynomial/Proof_1 | [
"Binomial Coefficients",
"Sum over k of r Choose k by -1^r-k by Polynomial"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] | [
"Definition:Kronecker Delta",
"Definition:Coefficient of Polynomial",
"Definition:Polynomial/Real Numbers",
"Definition:Summation",
"Definition:Binomial Coefficient",
"Definition:Polynomial/Real Numbers",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Degree of ... |
proofwiki-11768 | Sum over k of r Choose k by -1^r-k by Polynomial | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$. | From Summation of Powers over Product of Differences:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n ... | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ of [[Definition:Degree of Polynomial|degree]] $r$. | From [[Summation of Powers over Product of Differences]]:
:$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end {pmatrix} = \begin {cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1... | Sum over k of r Choose k by -1^r-k by Polynomial/Proof 2 | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_-1^r-k_by_Polynomial | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_-1^r-k_by_Polynomial/Proof_2 | [
"Binomial Coefficients",
"Sum over k of r Choose k by -1^r-k by Polynomial"
] | [
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] | [
"Summation of Powers over Product of Differences"
] |
proofwiki-11769 | Sum over k of r Choose k by s-kt Choose r by -1^k | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \binom {s - k t} r \paren {-1}^k = t^r$
where $\dbinom r k$ etc. are binomial coefficients. | From Sum over $k$ of $\dbinom r k \paren {-1}^k$ by Polynomial:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
{{proof wanted}} | Let $r \in \Z_{\ge 0}$.
Then:
:$\ds \sum_k \binom r k \binom {s - k t} r \paren {-1}^k = t^r$
where $\dbinom r k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]]. | From [[Sum over k of r Choose k by -1^r-k by Polynomial|Sum over $k$ of $\dbinom r k \paren {-1}^k$ by Polynomial]]:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ of [[Defi... | Sum over k of r Choose k by s-kt Choose r by -1^k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s-kt_Choose_r_by_-1^k | https://proofwiki.org/wiki/Sum_over_k_of_r_Choose_k_by_s-kt_Choose_r_by_-1^k | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Sum over k of r Choose k by -1^r-k by Polynomial",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial"
] |
proofwiki-11770 | Sum over k of -2 Choose k | :$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
where:
:$\dbinom {-2} k$ is a binomial coefficient
:$\ceiling x$ denotes the ceiling of $x$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom {-2} k
| r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k - \paren {-2} - 1} k
| c = Negated Upper Index of Binomial Coefficient
}}
{{eqn | r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k + 1} k
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n \pare... | :$\ds \sum_{k \mathop = 0}^n \binom {-2} k = \paren {-1}^n \ceiling {\dfrac {n + 1} 2}$
where:
:$\dbinom {-2} k$ is a [[Definition:Binomial Coefficient|binomial coefficient]]
:$\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom {-2} k
| r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k - \paren {-2} - 1} k
| c = [[Negated Upper Index of Binomial Coefficient]]
}}
{{eqn | r = \sum_{k \mathop = 0}^n \paren {-1} \binom {k + 1} k
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n \... | Sum over k of -2 Choose k | https://proofwiki.org/wiki/Sum_over_k_of_-2_Choose_k | https://proofwiki.org/wiki/Sum_over_k_of_-2_Choose_k | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Ceiling Function"
] | [
"Negated Upper Index of Binomial Coefficient",
"Binomial Coefficient with Self minus One",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Closed Form for Triangular Numbers",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Closed Fo... |
proofwiki-11771 | Sum over k of m Choose k by k minus m over 2 | :$\ds \sum_{k \mathop = 0}^n \binom m k \paren {k - \dfrac m 2} = -\dfrac m 2 \binom {m - 1} n$
where $\dbinom m k$ etc. are binomial coefficients. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom m k \paren {k - \dfrac m 2}
| r = \sum_{k \mathop = 0}^n k \binom m k - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n m \binom {m - 1} {k - 1} - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
| c = Factor... | :$\ds \sum_{k \mathop = 0}^n \binom m k \paren {k - \dfrac m 2} = -\dfrac m 2 \binom {m - 1} n$
where $\dbinom m k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^n \binom m k \paren {k - \dfrac m 2}
| r = \sum_{k \mathop = 0}^n k \binom m k - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
| c =
}}
{{eqn | r = \sum_{k \mathop = 0}^n m \binom {m - 1} {k - 1} - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
| c = [[Fact... | Sum over k of m Choose k by k minus m over 2 | https://proofwiki.org/wiki/Sum_over_k_of_m_Choose_k_by_k_minus_m_over_2 | https://proofwiki.org/wiki/Sum_over_k_of_m_Choose_k_by_k_minus_m_over_2 | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Factors of Binomial Coefficient",
"Pascal's Rule",
"Definition:Telescoping Series"
] |
proofwiki-11772 | Binomial Coefficient is instance of Gaussian Binomial Coefficient | Let $\dbinom r m_q$ denote the Gaussian binomial coefficient:
Then:
:$\ds \lim_{q \mathop \to 1^-} \dbinom r m_q = \dbinom r m$
where $\dbinom r m$ denotes the conventional binomial coefficient. | We have by definition of Gaussian binomial coefficient:
:$\ds \dbinom r m_q = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }$
Consider a typical factor of this continued product:
{{begin-eqn}}
{{eqn | l = \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }
| r = \dfrac {\paren {1 - q^{r - k} } / \par... | Let $\dbinom r m_q$ denote the [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
Then:
:$\ds \lim_{q \mathop \to 1^-} \dbinom r m_q = \dbinom r m$
where $\dbinom r m$ denotes the conventional [[Definition:Binomial Coefficient|binomial coefficient]]. | We have by definition of [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
:$\ds \dbinom r m_q = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }$
Consider a typical [[Definition:Factor (Algebra)|factor]] of this [[Definition:Continued Product|continued product]]:
{{be... | Binomial Coefficient is instance of Gaussian Binomial Coefficient | https://proofwiki.org/wiki/Binomial_Coefficient_is_instance_of_Gaussian_Binomial_Coefficient | https://proofwiki.org/wiki/Binomial_Coefficient_is_instance_of_Gaussian_Binomial_Coefficient | [
"Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient",
"Definition:Binomial Coefficient"
] | [
"Definition:Gaussian Binomial Coefficient",
"Definition:Divisor (Algebra)/Integer",
"Definition:Continued Product",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Sum of Geometric Sequence",
"Translation of Index Variable of Product"
] |
proofwiki-11773 | Multinomial Coefficient expressed as Product of Binomial Coefficients | :$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
where:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a multinomial coefficient
:$\dbinom {k_... | The proof proceeds by induction.
For all $m \in \Z_{> 1}$, let $\map P m$ be the proposition:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$ | :$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
where:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a [[Definition:Multinomial Coefficient... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $m \in \Z_{> 1}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cd... | Multinomial Coefficient expressed as Product of Binomial Coefficients | https://proofwiki.org/wiki/Multinomial_Coefficient_expressed_as_Product_of_Binomial_Coefficients | https://proofwiki.org/wiki/Multinomial_Coefficient_expressed_as_Product_of_Binomial_Coefficients | [
"Binomial Coefficients",
"Multinomial Coefficients"
] | [
"Definition:Multinomial Coefficient",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-11774 | Equivalence of Definitions of Semantic Equivalence for Predicate Logic | Let $\mathbf A, \mathbf B$ be WFFs of predicate logic.
{{TFAE|def = Semantic Equivalence for Predicate Logic|view = semantic equivalence|context = Predicate Logic|contextview = predicate logic}} | Let $\AA$ be a structure for predicate logic.
Let $\sigma$ be an assignment for $\mathbf A \iff \mathbf B$ in $\AA$.
Then the value of $\mathbf A \iff \mathbf B$ under $\sigma$ is given by:
:$\map {f^\leftrightarrow} {\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma, \map {\operatorname{val}_\AA} {\mathbf B} \sq... | Let $\mathbf A, \mathbf B$ be [[Definition:WFF of Predicate Logic|WFFs of predicate logic]].
{{TFAE|def = Semantic Equivalence for Predicate Logic|view = semantic equivalence|context = Predicate Logic|contextview = predicate logic}} | Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
Let $\sigma$ be an [[Definition:Assignment for Formula|assignment]] for $\mathbf A \iff \mathbf B$ in $\AA$.
Then the [[Definition:Value of Formula under Assignment|value of $\mathbf A \iff \mathbf B$ under $\sigma$]] is given... | Equivalence of Definitions of Semantic Equivalence for Predicate Logic | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Semantic_Equivalence_for_Predicate_Logic | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Semantic_Equivalence_for_Predicate_Logic | [
"Definitions/Semantic Equivalence for Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Assignment for Structure/Formula",
"Definition:Value of Formula under Assignment",
"Definition:Biconditional/Truth Function",
"Definition:Structure for Predicate Logic/Formal Semantics/Well-Formed Formula",
"Definition:Tautology/Formal Semantics/Pred... |
proofwiki-11775 | Universal Closures are Semantically Equivalent | Let $\mathbf A$ be a WFF of predicate logic.
Let $\mathbf B, \mathbf B'$ be universal closures of $\mathbf A$.
Then $\mathbf B$ and $\mathbf B'$ are semantically equivalent. | Let $\AA$ be a structure for predicate logic.
Let $\mathbf B$ be any universal closure of $\mathbf A$.
Then $\mathbf B$ is a sentence of the form:
:$\forall x_1: \cdots \forall x_n: \mathbf A$
By definition of the models relation:
:$\AA \models_{\mathrm{PL}} \mathbf B$ {{iff}} $\map {\operatorname{val}_\AA} {\mathbf B}... | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\mathbf B, \mathbf B'$ be [[Definition:Universal Closure of Well-Formed Formula|universal closures]] of $\mathbf A$.
Then $\mathbf B$ and $\mathbf B'$ are [[Definition:Semantic Equivalence (Predicate Logic)|semantically equivalen... | Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
Let $\mathbf B$ be any [[Definition:Universal Closure of Well-Formed Formula|universal closure]] of $\mathbf A$.
Then $\mathbf B$ is a [[Definition:Sentence|sentence]] of the form:
:$\forall x_1: \cdots \forall x_n: \mathbf A... | Universal Closures are Semantically Equivalent | https://proofwiki.org/wiki/Universal_Closures_are_Semantically_Equivalent | https://proofwiki.org/wiki/Universal_Closures_are_Semantically_Equivalent | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Universal Closure of Well-Formed Formula",
"Definition:Semantic Equivalence/Predicate Logic"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Universal Closure of Well-Formed Formula",
"Definition:Classes of WFFs/Sentence",
"Definition:Structure for Predicate Logic/Formal Semantics/Sentence",
"Definition:Value of Formula under Assignment",
"Definition:Extension of Assignment",
"Value of ... |
proofwiki-11776 | Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset | Let $L = \struct {S, \vee, \wedge, \preceq}$ be an up-complete lattice.
Then
:$L$ is meet-continuous
{{iff}}:
:$\forall x \in S$, the directed subset $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$ | === Sufficient Condition ===
Let $L$ be meet-continuous.
Let $x$ be an element of $S$, $D$ be a directed subset of $S$ such that
:$x \preceq \sup D$
Thus
{{begin-eqn}}
{{eqn | l = x
| r = x \wedge \sup D
| c = Preceding iff Meet equals Less Operand
}}
{{eqn | r = \sup \set {x \wedge d: d \in D}
| c = ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice (Order Theory)|lattice]].
Then
:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]
{{iff}}:
:$\forall x \in S$, the [[Definition:Directed Subset|directed subset]] $D$ of $S: x \preceq \sup D \impl... | === Sufficient Condition ===
Let $L$ be [[Definition:Meet-Continuous Lattice|meet-continuous]].
Let $x$ be an [[Definition:Element|element]] of $S$, $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
:$x \preceq \sup D$
Thus
{{begin-eqn}}
{{eqn | l = x
| r = x \wedge \sup D
| c = [[... | Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset | https://proofwiki.org/wiki/Meet-Continuous_iff_if_Element_Precedes_Supremum_of_Directed_Subset_then_Element_equals_Supremum_of_Meet_of_Element_by_Directed_Subset | https://proofwiki.org/wiki/Meet-Continuous_iff_if_Element_Precedes_Supremum_of_Directed_Subset_then_Element_equals_Supremum_of_Meet_of_Element_by_Directed_Subset | [
"Meet-Continuous Lattices"
] | [
"Definition:Up-Complete",
"Definition:Lattice (Order Theory)",
"Definition:Meet-Continuous Lattice",
"Definition:Directed Subset"
] | [
"Definition:Meet-Continuous Lattice",
"Definition:Element",
"Definition:Directed Subset",
"Preceding iff Meet equals Less Operand",
"Definition:Directed Subset",
"Definition:Directed Subset",
"Definition:Meet-Continuous Lattice"
] |
proofwiki-11777 | Continuous Lattice is Meet-Continuous | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Then $L$ is meet-continuous. | Let $x \in S$, $D$ be a directed subset of $S$ such that
:$x \preceq \sup D$
By Way Below Closure is Directed in Bounded Below Join Semilattice:
:$x^\ll$ is directed.
By definition of continuous:
:$L$ is up-complete and satisfies the axiom of approximation.
By definition of up-complete:
:$x^\ll$ admits a supremum.
By L... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Then $L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]. | Let $x \in S$, $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
:$x \preceq \sup D$
By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
:$x^\ll$ is [[Definition:Directed Subset|directed]].
By definition of [[Definition:Continuous Ordered Set|continuous]]:
:$L$ is [[Definiti... | Continuous Lattice is Meet-Continuous | https://proofwiki.org/wiki/Continuous_Lattice_is_Meet-Continuous | https://proofwiki.org/wiki/Continuous_Lattice_is_Meet-Continuous | [
"Meet-Continuous Lattices",
"Continuous Lattices"
] | [
"Definition:Bounded Below",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)",
"Definition:Meet-Continuous Lattice"
] | [
"Definition:Directed Subset",
"Way Below Closure is Directed in Bounded Below Join Semilattice",
"Definition:Directed Subset",
"Definition:Continuous Ordered Set",
"Definition:Up-Complete",
"Axiom:Axiom of Approximation",
"Definition:Up-Complete",
"Definition:Supremum of Set",
"Lower Closure of Elem... |
proofwiki-11778 | Number to Power of Zero Rising is One | :$x^{\overline 0} = 1$ | {{begin-eqn}}
{{eqn | l = x^{\overline 0}
| r = \prod_{j \mathop = 0}^{-1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = 1
| c = Product is Vacuous
}}
{{end-eqn}}
{{qed}}
Category:Rising Factorials
1tfwsij49sdapn8zmyovlleahasydr4 | :$x^{\overline 0} = 1$ | {{begin-eqn}}
{{eqn | l = x^{\overline 0}
| r = \prod_{j \mathop = 0}^{-1} \paren {x + j}
| c = {{Defof|Rising Factorial}}
}}
{{eqn | r = 1
| c = Product is [[Definition:Vacuous Product|Vacuous]]
}}
{{end-eqn}}
{{qed}}
[[Category:Rising Factorials]]
1tfwsij49sdapn8zmyovlleahasydr4 | Number to Power of Zero Rising is One | https://proofwiki.org/wiki/Number_to_Power_of_Zero_Rising_is_One | https://proofwiki.org/wiki/Number_to_Power_of_Zero_Rising_is_One | [
"Rising Factorials"
] | [] | [
"Definition:Continued Product/Vacuous Product",
"Category:Rising Factorials"
] |
proofwiki-11779 | Number to Power of Zero Falling is One | :$x^{\underline 0} = 1$ | {{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = \prod_{j \mathop = 0}^{-1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = 1
| c = Product is Vacuous
}}
{{end-eqn}}
{{qed}}
Category:Falling Factorials
0thb0bl3pugwn0hxx3fa89xkjtu0dbp | :$x^{\underline 0} = 1$ | {{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = \prod_{j \mathop = 0}^{-1} \paren {x - j}
| c = {{Defof|Falling Factorial}}
}}
{{eqn | r = 1
| c = Product is [[Definition:Vacuous Product|Vacuous]]
}}
{{end-eqn}}
{{qed}}
[[Category:Falling Factorials]]
0thb0bl3pugwn0hxx3fa89xkjtu0dbp | Number to Power of Zero Falling is One | https://proofwiki.org/wiki/Number_to_Power_of_Zero_Falling_is_One | https://proofwiki.org/wiki/Number_to_Power_of_Zero_Falling_is_One | [
"Falling Factorials"
] | [] | [
"Definition:Continued Product/Vacuous Product",
"Category:Falling Factorials"
] |
proofwiki-11780 | Product of Number by its Falling Factorial | Let $x^{\underline n}$ denote the $n$th falling factorial power of $x$.
Then:
:$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x x^{\underline n}
| r = \paren {x - n + n} x^{\underline n}
| c =
}}
{{eqn | r = \paren {x - n} x^{\underline n} + n x^{\underline n}
| c =
}}
{{eqn | r = \paren {x - n} \prod_{j \mathop = 0}^{n - 1} \paren {x - j} + n x^{\underline n}
| c = {{Defof|Falling Factorial... | Let $x^{\underline n}$ denote the [[Definition:Falling Factorial|$n$th falling factorial power]] of $x$.
Then:
:$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$ | {{begin-eqn}}
{{eqn | l = x x^{\underline n}
| r = \paren {x - n + n} x^{\underline n}
| c =
}}
{{eqn | r = \paren {x - n} x^{\underline n} + n x^{\underline n}
| c =
}}
{{eqn | r = \paren {x - n} \prod_{j \mathop = 0}^{n - 1} \paren {x - j} + n x^{\underline n}
| c = {{Defof|Falling Factorial... | Product of Number by its Falling Factorial | https://proofwiki.org/wiki/Product_of_Number_by_its_Falling_Factorial | https://proofwiki.org/wiki/Product_of_Number_by_its_Falling_Factorial | [
"Falling Factorials"
] | [
"Definition:Falling Factorial"
] | [
"Category:Falling Factorials"
] |
proofwiki-11781 | Product of Number by its Rising Factorial | Let $x^{\overline n}$ denote the $n$th rising factorial power of $x$.
Then:
:$x x^{\overline n} = x^{\overline {n + 1} } - n x^{\overline n}$ | {{begin-eqn}}
{{eqn | l = x x^{\overline n}
| r = \paren {x + n - n} x^{\overline n}
| c =
}}
{{eqn | r = \paren {x + n} x^{\overline n} - n x^{\overline n}
| c =
}}
{{eqn | r = \paren {x + n} \prod_{j \mathop = 0}^{n - 1} \paren {x + j} - n x^{\overline n}
| c = {{Defof|Rising Factorial}}
}}
... | Let $x^{\overline n}$ denote the [[Definition:Rising Factorial|$n$th rising factorial power]] of $x$.
Then:
:$x x^{\overline n} = x^{\overline {n + 1} } - n x^{\overline n}$ | {{begin-eqn}}
{{eqn | l = x x^{\overline n}
| r = \paren {x + n - n} x^{\overline n}
| c =
}}
{{eqn | r = \paren {x + n} x^{\overline n} - n x^{\overline n}
| c =
}}
{{eqn | r = \paren {x + n} \prod_{j \mathop = 0}^{n - 1} \paren {x + j} - n x^{\overline n}
| c = {{Defof|Rising Factorial}}
}}
... | Product of Number by its Rising Factorial | https://proofwiki.org/wiki/Product_of_Number_by_its_Rising_Factorial | https://proofwiki.org/wiki/Product_of_Number_by_its_Rising_Factorial | [
"Rising Factorials"
] | [
"Definition:Rising Factorial"
] | [
"Category:Rising Factorials"
] |
proofwiki-11782 | Value of Formula under Assignment Determined by Free Variables | Let $\mathbf A$ be a WFF of predicate logic.
Let $\AA$ be a structure for predicate logic.
Let $\sigma, \sigma'$ be assignments for $\mathbf A$ in $\AA$ such that:
:For each free variable $x$ of $\mathbf A$, $\map \sigma x = \map {\sigma'} x$
Then:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {\oper... | Proceed by the Principle of Structural Induction applied to the bottom-up specification of predicate logic.
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {p_\AA} {\map {\operatorname{val}_\AA} {\tau_1} \sqbrk \sigma, \ldots, \map {\operatorname{... | Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
Let $\sigma, \sigma'$ be [[Definition:Assignment for Formula|assignments for $\mathbf A$ in $\AA$]] such that:
:For each [[Definition:Free Vari... | Proceed by the [[Principle of Structural Induction]] applied to the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]].
If $\mathbf A = \map p {\tau_1, \ldots, \tau_n}$, then:
:$\map {\operatorname{val}_\AA} {\mathbf A} \sqbrk \sigma = \map {p_\AA} {\map {\operatorna... | Value of Formula under Assignment Determined by Free Variables | https://proofwiki.org/wiki/Value_of_Formula_under_Assignment_Determined_by_Free_Variables | https://proofwiki.org/wiki/Value_of_Formula_under_Assignment_Determined_by_Free_Variables | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Structure for Predicate Logic",
"Definition:Assignment for Structure/Formula",
"Definition:Free Variable",
"Definition:Value of Formula under Assignment"
] | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Quantifier",
"Definition:Free Variable",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Assignment for Structure/Formula",
"Definition:Assignment for Structure/Term",
"Value of Term under A... |
proofwiki-11783 | First Inversion Formula for Stirling Numbers | For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brack k}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the second kind
:$\delta_{m n}$ denotes the Kronecker delta. | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$ | For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
:$\ds {k \brace m}$ denotes a [[Definition:Stirling Numbers of the Second... | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brack k} {k \brace m} \paren {-1}^{n - k} = \delta_{m n}$ | First Inversion Formula for Stirling Numbers | https://proofwiki.org/wiki/First_Inversion_Formula_for_Stirling_Numbers | https://proofwiki.org/wiki/First_Inversion_Formula_for_Stirling_Numbers | [
"Stirling Numbers"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Stirling Numbers of the Second Kind",
"Definition:Kronecker Delta"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11784 | Unsigned Stirling Number of the First Kind of 1 | :$\ds {1 \brack n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = {1 \brack n}
| r = 0 \times {0 \brack n} + {0 \brack n - 1}
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | r = {0 \brack n - 1}
| c =
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn ... | :$\ds {1 \brack n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = {1 \brack n}
| r = 0 \times {0 \brack n} + {0 \brack n - 1}
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn | r = {0 \brack n - 1}
| c =
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
| c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
}}
{{eqn ... | Unsigned Stirling Number of the First Kind of 1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [] |
proofwiki-11785 | Signed Stirling Number of the First Kind of 1 | :$\map s {1, n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = \map s {1, n}
| r = \map s {0, n - 1} - 0 \times \map s {0, n}
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \map s {0, n - 1}
| c =
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn |... | :$\map s {1, n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = \map s {1, n}
| r = \map s {0, n - 1} - 0 \times \map s {0, n}
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn | r = \map s {0, n - 1}
| c =
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
| c = {{Defof|Signed Stirling Numbers of the First Kind}}
}}
{{eqn |... | Signed Stirling Number of the First Kind of 1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_1 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_1 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [] |
proofwiki-11786 | Stirling Number of the Second Kind of 1 | :$\ds {1 \brace n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = {1 \brace n}
| r = n \times {0 \brace n} + {0 \brace n - 1}
| c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = n \times \delta_{0 n} + \delta_{0 \paren {n - 1} }
| c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
... | :$\ds {1 \brace n} = \delta_{1 n}$ | {{begin-eqn}}
{{eqn | l = {1 \brace n}
| r = n \times {0 \brace n} + {0 \brace n - 1}
| c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = n \times \delta_{0 n} + \delta_{0 \paren {n - 1} }
| c = {{Defof|Stirling Numbers of the Second Kind}}
}}
{{eqn | r = \delta_{0 \paren {n - 1} }
... | Stirling Number of the Second Kind of 1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_1 | [
"Stirling Numbers"
] | [] | [] |
proofwiki-11787 | Second Inversion Formula for Stirling Numbers | For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brace k}$ denotes a Stirling number of the second kind
:$\ds {k \brack m}$ denotes an unsigned Stirling number of the first kind
:$\delta_{m n}$ denotes the Kronecker delta. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$ | For all $m, n \in \Z_{\ge 0}$:
:$\ds \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$
where:
:$\ds {n \brace k}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
:$\ds {k \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|un... | The proof proceeds by [[Second Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \forall m \in \Z_{\ge 0}: \sum_k {n \brace k} {k \brack m} \paren {-1}^{n - k} = \delta_{m n}$ | Second Inversion Formula for Stirling Numbers | https://proofwiki.org/wiki/Second_Inversion_Formula_for_Stirling_Numbers | https://proofwiki.org/wiki/Second_Inversion_Formula_for_Stirling_Numbers | [
"Stirling Numbers"
] | [
"Definition:Stirling Numbers of the Second Kind",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Kronecker Delta"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11788 | Zero Choose n | :$\dbinom 0 n = \delta_{0 n}$ | :$\dbinom m n = \begin{cases}\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\&\\0 & : \text { otherwise } \end{cases}$ | :$\dbinom 0 n = \delta_{0 n}$ | :$\dbinom m n = \begin{cases}\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\&\\0 & : \text { otherwise } \end{cases}$ | Zero Choose n | https://proofwiki.org/wiki/Zero_Choose_n | https://proofwiki.org/wiki/Zero_Choose_n | [
"Examples of Binomial Coefficients"
] | [] | [] |
proofwiki-11789 | Unsigned Stirling Number of the First Kind of 0 | :$\ds {0 \brack n} = \delta_{0 n}$ | By definition of unsigned Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = 1
| c = Number to Power of Zero Falling is One
}}
{{eqn | r = x^0
| c = {{Defof|Integer Power}}
}}
{{end-eqn}... | :$\ds {0 \brack n} = \delta_{0 n}$ | By definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
$\ds x^{\underline 0} = \sum_k \paren {-1}^{0 - k} {0 \brack k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = 1
| c = [[Number to Power of Zero Falling is One]]
}}
... | Unsigned Stirling Number of the First Kind of 0 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_0 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_0 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Number to Power of Zero Falling is One"
] |
proofwiki-11790 | Signed Stirling Number of the First Kind of 0 | :$\map s {0, n} = \delta_{0 n}$ | By definition of signed Stirling number of the first kind:
$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = 1
| c = Number to Power of Zero Falling is One
}}
{{eqn | r = x^0
| c = {{Defof|Integer Power}}
}}
{{end-eqn}}
Thus, in the expres... | :$\map s {0, n} = \delta_{0 n}$ | By definition of [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]]:
$\ds x^{\underline 0} = \sum_k \map s {0, k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^{\underline 0}
| r = 1
| c = [[Number to Power of Zero Falling is One]]
}}
{{eqn | r = x^0
|... | Signed Stirling Number of the First Kind of 0 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_0 | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_0 | [
"Examples of Stirling Numbers of the First Kind"
] | [] | [
"Definition:Stirling Numbers of the First Kind/Signed",
"Number to Power of Zero Falling is One"
] |
proofwiki-11791 | Stirling Number of the Second Kind of 0 | :$\ds {0 \brace n} = \delta_{0 n}$ | By definition of Stirling numbers of the second kind:
$\ds x^{\underline 0} = \sum_k {0 \brace k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^0
| r = 1
| c = {{Defof|Integer Power}}
}}
{{eqn | r = x^{\underline 0}
| c = Number to Power of Zero Falling is One
}}
{{end-eqn}}
Thus, in the expression:
... | :$\ds {0 \brace n} = \delta_{0 n}$ | By definition of [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]]:
$\ds x^{\underline 0} = \sum_k {0 \brace k} x^k$
Thus we have:
{{begin-eqn}}
{{eqn | l = x^0
| r = 1
| c = {{Defof|Integer Power}}
}}
{{eqn | r = x^{\underline 0}
| c = [[Number to Power of Zer... | Stirling Number of the Second Kind of 0 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_0 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_0 | [
"Examples of Stirling Numbers of the Second Kind"
] | [] | [
"Definition:Stirling Numbers of the Second Kind",
"Number to Power of Zero Falling is One"
] |
proofwiki-11792 | Unsigned Stirling Number of the First Kind of Number with Self | :$\ds {n \brack n} = 1$ | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brack n} = 1$ | :$\ds {n \brack n} = 1$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds {n \brack n} = 1$ | Unsigned Stirling Number of the First Kind of Number with Self | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Self | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Self | [
"Stirling Numbers"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11793 | Unsigned Stirling Number of the First Kind of Number with Greater | Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
:$\ds {n \brack k} = 0$ | By definition, unsigned Stirling number of the first kind are defined as the polynomial coefficients $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the {{... | Let $\ds {n \brack k}$ denote an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
Then:
:$\ds {n \brack k} = 0$ | By definition, [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\ds {n \brack k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \paren {-1}^{n - k} {n \brack k} x^k$
wher... | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_1 | [
"Stirling Numbers",
"Unsigned Stirling Number of the First Kind of Number with Greater"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Coefficient of Polynomial",
"Definition:Falling Factorial",
"Definition:Expression",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Coefficient of Polynomial"
] |
proofwiki-11794 | Unsigned Stirling Number of the First Kind of Number with Greater | Let $\ds {n \brack k}$ denote an unsigned Stirling number of the first kind.
Then:
:$\ds {n \brack k} = 0$ | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P v$ be the proposition:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from Unsigned Stirling Number of the First Kind of 0.
So by definition of Kronecker delta:
:$\fo... | Let $\ds {n \brack k}$ denote an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
Then:
:$\ds {n \brack k} = 0$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P v$ be the [[Definition:Proposition|proposition]]:
:$\ds k > n \implies {n \brack k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brack k} = \delta_{0 k}$
from [[Unsigned Stirli... | Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2 | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater | https://proofwiki.org/wiki/Unsigned_Stirling_Number_of_the_First_Kind_of_Number_with_Greater/Proof_2 | [
"Stirling Numbers",
"Unsigned Stirling Number of the First Kind of Number with Greater"
] | [
"Definition:Stirling Numbers of the First Kind/Unsigned"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Unsigned Stirling Number of the First Kind of 0",
"Definition:Kronecker Delta",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Unsigned Stirling Number of the First Kind of Number ... |
proofwiki-11795 | Signed Stirling Number of the First Kind of Number with Greater | Let $\map s {n, k}$ denote a signed Stirling number of the first kind.
Then:
:$\map s {n, k} = 0$ | By definition, the signed Stirling numbers of the first kind are defined as the polynomial coefficients $\map s {n, k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \map s {n, k} x^k$
where $x^{\underline n}$ denotes the $n$th falling factorial of $x$.
Both of the expressions on the {{LHS}} and {{RHS}} a... | Let $\map s {n, k}$ denote a [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]].
Then:
:$\map s {n, k} = 0$ | By definition, the [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling numbers of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\map s {n, k}$ which satisfy the equation:
:$\ds x^{\underline n} = \sum_k \map s {n, k} x^k$
where $x^{\underline n}$ ... | Signed Stirling Number of the First Kind of Number with Greater | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_Number_with_Greater | https://proofwiki.org/wiki/Signed_Stirling_Number_of_the_First_Kind_of_Number_with_Greater | [
"Stirling Numbers"
] | [
"Definition:Stirling Numbers of the First Kind/Signed"
] | [
"Definition:Stirling Numbers of the First Kind/Signed",
"Definition:Coefficient of Polynomial",
"Definition:Falling Factorial",
"Definition:Expression",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Coefficient of Polynomial"
] |
proofwiki-11796 | Stirling Number of the Second Kind of Number with Greater | Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$ | By definition, the Stirling numbers of the second kind are defined as the coefficients $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the $k$th falling factorial of $x$.
Both of the expressions on the {{LHS}} and {{RHS}} are polynomials ... | Let $\ds {n \brace k}$ denote a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
Then:
:$\ds {n \brace k} = 0$ | By definition, the [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]] are defined as the [[Definition:Coefficient|coefficients]] $\ds {n \brace k}$ which satisfy the equation:
:$\ds x^n = \sum_k {n \brace k} x^{\underline k}$
where $x^{\underline k}$ denotes the [[Definition:Falling... | Stirling Number of the Second Kind of Number with Greater/Proof 1 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_1 | [
"Stirling Numbers",
"Stirling Number of the Second Kind of Number with Greater"
] | [
"Definition:Stirling Numbers of the Second Kind"
] | [
"Definition:Stirling Numbers of the Second Kind",
"Definition:Coefficient",
"Definition:Falling Factorial",
"Definition:Expression",
"Definition:Polynomial/Real Numbers",
"Definition:Degree of Polynomial",
"Definition:Coefficient of Polynomial"
] |
proofwiki-11797 | Stirling Number of the Second Kind of Number with Greater | Let $\ds {n \brace k}$ denote a Stirling number of the second kind.
Then:
:$\ds {n \brace k} = 0$ | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from Stirling Number of the Second Kind of 0.
So by definition of Kronecker delta:
:$\forall k \... | Let $\ds {n \brace k}$ denote a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
Then:
:$\ds {n \brace k} = 0$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds k > n \implies {n \brace k} = 0$
=== Basis for the Induction ===
$\map P 0$ is the case:
:$\ds {0 \brace k} = \delta_{0 k}$
from [[Stirling Number... | Stirling Number of the Second Kind of Number with Greater/Proof 2 | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Greater/Proof_2 | [
"Stirling Numbers",
"Stirling Number of the Second Kind of Number with Greater"
] | [
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Stirling Number of the Second Kind of 0",
"Definition:Kronecker Delta",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Stirling Number of the Second Kind of Number with Greater/Pro... |
proofwiki-11798 | Stirling Number of the Second Kind of Number with Self | :$\ds {n \brace n} = 1$ | The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds {n \brace n} = 1$ | :$\ds {n \brace n} = 1$ | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds {n \brace n} = 1$ | Stirling Number of the Second Kind of Number with Self | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Self | https://proofwiki.org/wiki/Stirling_Number_of_the_Second_Kind_of_Number_with_Self | [
"Stirling Numbers"
] | [] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11799 | Value of Term under Assignment Determined by Variables | Let $\tau$ be a term of predicate logic.
Let $\AA$ be a structure for predicate logic.
Let $\sigma, \sigma'$ be assignments for $\tau$ in $\AA$ such that:
:For each variable $x$ occurring in $\tau$, $\map \sigma x = \map {\sigma'} x$
Then:
:$\map {\operatorname{val}_\AA} \tau \sqbrk \sigma = \map {\operatorname{val}_\A... | Proceed by the Principle of Structural Induction applied to the definition of a term.
If $\tau = x$, then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma
| r = \map \sigma x
| c = {{Defof|Value of Term under Assignment|value under $\sigma$}}
}}
{{eqn | r = \map {\sigma'} x
... | Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]].
Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
Let $\sigma, \sigma'$ be [[Definition:Assignment for Formula|assignments for $\tau$ in $\AA$]] such that:
:For each [[Definition:Variable (Logic)|v... | Proceed by the [[Principle of Structural Induction]] applied to the definition of a [[Definition:Term (Predicate Logic)|term]].
If $\tau = x$, then:
{{begin-eqn}}
{{eqn | l = \map {\operatorname{val}_\AA} \tau \sqbrk \sigma
| r = \map \sigma x
| c = {{Defof|Value of Term under Assignment|value under $\si... | Value of Term under Assignment Determined by Variables | https://proofwiki.org/wiki/Value_of_Term_under_Assignment_Determined_by_Variables | https://proofwiki.org/wiki/Value_of_Term_under_Assignment_Determined_by_Variables | [
"Predicate Logic"
] | [
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Definition:Structure for Predicate Logic",
"Definition:Assignment for Structure/Formula",
"Definition:Variable/Predicate Logic",
"Definition:Occurrence (Predicate Logic)",
"Definition:Value of Term under Assignment"
] | [
"Principle of Structural Induction",
"Definition:Language of Predicate Logic/Formal Grammar/Term",
"Principle of Structural Induction",
"Category:Predicate Logic"
] |
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