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-11900 | Beta Function of x with y+1 by x+y over y | :$\map \Beta {x, y} = \dfrac {x + y} y \map \Beta {x, y + 1}$ | By definition of Beta function:
:$\ds \map \Beta {x + 1, y} = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t$
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d t} \rd t = u v - \int v \frac {\d u} {\d t} \rd t$
let:
{{begin-eqn}}
{{eqn | l = u
| r = t^x
| c =
}}
{{eqn | ll= \lea... | :$\map \Beta {x, y} = \dfrac {x + y} y \map \Beta {x, y + 1}$ | By definition of [[Definition:Beta Function|Beta function]]:
:$\ds \map \Beta {x + 1, y} = \int_0^1 t^x \paren {1 - t}^{y - 1} \rd t$
With a view to expressing the [[Definition:Primitive (Calculus)|primitive]] in the form:
:$\ds \int u \frac {\d v} {\d t} \rd t = u v - \int v \frac {\d u} {\d t} \rd t$
let:
{{begin-... | Beta Function of x with y+1 by x+y over y | https://proofwiki.org/wiki/Beta_Function_of_x_with_y+1_by_x+y_over_y | https://proofwiki.org/wiki/Beta_Function_of_x_with_y+1_by_x+y_over_y | [
"Beta Function"
] | [] | [
"Definition:Beta Function",
"Definition:Primitive (Calculus)",
"Power Rule for Derivatives",
"Primitive of Power",
"Integration by Parts",
"Beta Function of x+1 with y plus Beta Function of x with y+1"
] |
proofwiki-11901 | Partial Gamma Function expressed as Integral | Let $m \in \Z_{\ge 1}$.
Let $\map {\Gamma_m} x$ denote the partial Gamma function, defined as:
:$\map {\Gamma_m} x := \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} }$
Then:
:$\ds \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
for $x > 0$. | === Lemma ===
First we establish:
{{:Partial Gamma Function expressed as Integral/Lemma}}
The proof continues by induction on $m$.
For all $m \in \Z_{\ge 1}$, let $\map P m$ be the proposition:
:$\ds \map {\Gamma_m} x = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$ | Let $m \in \Z_{\ge 1}$.
Let $\map {\Gamma_m} x$ denote the [[Definition:Partial Gamma Function|partial Gamma function]], defined as:
:$\map {\Gamma_m} x := \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} }$
Then:
:$\ds \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
for ... | === [[Partial Gamma Function expressed as Integral/Lemma|Lemma]] ===
First we establish:
{{:Partial Gamma Function expressed as Integral/Lemma}}
The proof continues by [[Principle of Mathematical Induction|induction]] on $m$.
For all $m \in \Z_{\ge 1}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:... | Partial Gamma Function expressed as Integral | https://proofwiki.org/wiki/Partial_Gamma_Function_expressed_as_Integral | https://proofwiki.org/wiki/Partial_Gamma_Function_expressed_as_Integral | [
"Gamma Function"
] | [
"Definition:Gamma Function/Partial"
] | [
"Partial Gamma Function expressed as Integral/Lemma",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Partial Gamma Function expressed as Integral",
"Principle of Mathematical Induction"
] |
proofwiki-11902 | Partial Gamma Function expressed as Integral/Lemma | :$(1): \quad \ds \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
for $x > 0$. | Let:
{{begin-eqn}}
{{eqn | l = z
| r = \frac t m
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\d t} {\d z}
| r = m
| c =
}}
{{end-eqn}}
Recalculating the limits:
{{begin-eqn}}
{{eqn | l = t
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = z
| r = 0
| c =
}}
{{eqn... | :$(1): \quad \ds \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
for $x > 0$. | Let:
{{begin-eqn}}
{{eqn | l = z
| r = \frac t m
| c =
}}
{{eqn | ll= \leadsto
| l = \frac {\d t} {\d z}
| r = m
| c =
}}
{{end-eqn}}
Recalculating the limits:
{{begin-eqn}}
{{eqn | l = t
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = z
| r = 0
| c =
}}
{{e... | Partial Gamma Function expressed as Integral/Lemma | https://proofwiki.org/wiki/Partial_Gamma_Function_expressed_as_Integral/Lemma | https://proofwiki.org/wiki/Partial_Gamma_Function_expressed_as_Integral/Lemma | [
"Gamma Function"
] | [] | [
"Integration by Substitution"
] |
proofwiki-11903 | Derivative of Composite Function/Second Derivative | :${D_x}^2 w = {D_u}^2 w \paren { {D_x}^1 u}^2 + {D_u}^1 w {D_x}^2 u$ | For ease of understanding, let Leibniz's notation be used:
:$\dfrac {\d^k u} {\d x^k} := {D_x}^k u$
Then we have:
{{begin-eqn}}
{{eqn | l = {D_x}^2 w
| r = \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d x} }
| c =
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d u} \dfrac {\d u} {\d x} }
| c =... | :${D_x}^2 w = {D_u}^2 w \paren { {D_x}^1 u}^2 + {D_u}^1 w {D_x}^2 u$ | For ease of understanding, let [[Definition:Leibniz's Notation for Derivatives|Leibniz's notation]] be used:
:$\dfrac {\d^k u} {\d x^k} := {D_x}^k u$
Then we have:
{{begin-eqn}}
{{eqn | l = {D_x}^2 w
| r = \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d x} }
| c =
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\d... | Derivative of Composite Function/Second Derivative | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Second_Derivative | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Second_Derivative | [
"Derivative of Composite Function"
] | [] | [
"Definition:Derivative/Notation/Leibniz Notation",
"Derivative of Composite Function",
"Product Rule for Derivatives",
"Derivative of Composite Function"
] |
proofwiki-11904 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | In the summation:
:$\ds \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
the only element appearing is for $j ... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
the only ele... | Faà di Bruno's Formula/Example/0/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/0/Proof | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Definition:Summation",
"Definition:Continued Product",
"Definition:Continued Product/Vacuous Product"
] |
proofwiki-11905 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | In the summation:
:$\ds \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to consider $j = 0$ and $j = ... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to c... | Faà di Bruno's Formula/Example/1/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/1/Proof | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Definition:Summation",
"Definition:Summation",
"Definition:Summation/Vacuous Summation",
"Definition:Derivative/Higher Derivatives/Zeroth Derivative",
"Definition:Factorial",
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-11906 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | In the summation:
:$\ds \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to consider $j = 0, j =... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \paren {m!}^{k_m} }$
we nee... | Faà di Bruno's Formula/Example/2/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/2/Proof | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Definition:Summation",
"Definition:Derivative/Higher Derivatives/Zeroth Derivative",
"Definition:Factorial",
"Definition:Summation",
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-11907 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop =... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\... | Faà di Bruno's Formula/Proof 1 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Proof_1 | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Faà di Bruno's Formula/Example/0",
"Definition:Derivative/Higher Derivatives/Zeroth Derivative",
"Faà di Bruno's Formula/Example/1",
"Derivative of Composite Function",
"Faà di Bruno's Formula/Example/2",
"Derivative of Composite Functi... |
proofwiki-11908 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | For convenience, let:
:$w_j := D_u^j w$
:$u_k := D_x^k u$
Then:
{{begin-eqn}}
{{eqn | l = \map {D_x} {w_j}
| r = w_{j + 1} u_1
| c = Derivative of Composite Function
}}
{{eqn | l = \map {D_x} {u_k}
| r = u_{k + 1}
| c = {{Defof|Higher Derivative}}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = D... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | For convenience, let:
:$w_j := D_u^j w$
:$u_k := D_x^k u$
Then:
{{begin-eqn}}
{{eqn | l = \map {D_x} {w_j}
| r = w_{j + 1} u_1
| c = [[Derivative of Composite Function]]
}}
{{eqn | l = \map {D_x} {u_k}
| r = u_{k + 1}
| c = {{Defof|Higher Derivative}}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn... | Faà di Bruno's Formula/Proof 2 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Proof_2 | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Derivative of Composite Function",
"Derivative of Composite Function",
"Derivative of Composite Function/Second Derivative",
"Derivative of Composite Function/Third Derivative",
"Definition:Set Partition",
"Definition:Set Partition",
"Definition:Set",
"Definition:Set Partition",
"Definition:Bijecti... |
proofwiki-11909 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | We have that:
:$\dfrac {D_x^k u} {k!}$ is the coefficient of $z^k$ in $\map u {x + z}$
:$\dfrac {D_u^j w} {j!}$ is the coefficient of $y^j$ in $\map w {u + y}$.
Hence the coefficient of $z^n$ in $\map w {\map u {x + z} }$ is:
:$\dfrac {D_x^n w} {n!} = \ds \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | We have that:
:$\dfrac {D_x^k u} {k!}$ is the coefficient of $z^k$ in $\map u {x + z}$
:$\dfrac {D_u^j w} {j!}$ is the coefficient of $y^j$ in $\map w {u + y}$.
Hence the coefficient of $z^n$ in $\map w {\map u {x + z} }$ is:
:$\dfrac {D_x^n w} {n!} = \ds \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \su... | Faà di Bruno's Formula/Proof 3 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Proof_3 | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [] |
proofwiki-11910 | Faà di Bruno's Formula | Let $D_x^k u$ denote the $k$th derivative of a function $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \d... | $D_x^n$ can be expressed as a determinant:
:$D_x^n = \begin{vmatrix}
\dbinom {n - 1} 0 u_1 & \dbinom {n - 1} 1 u_2 & \dbinom {n - 1} 2 u_3 & \cdots & \dbinom {n - 1} {n - 2} u_{n - 1} & \dbinom {n - 1} {n - 1} u_n \\
-1 & \dbinom {n - 2} 0 u_1 & \dbinom {n - 2} 1 u_2 & \cdots & \dbinom {n - 2} ... | Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
Then:
:$\ds D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \math... | $D_x^n$ can be expressed as a [[Definition:Determinant of Matrix|determinant]]:
:$D_x^n = \begin{vmatrix}
\dbinom {n - 1} 0 u_1 & \dbinom {n - 1} 1 u_2 & \dbinom {n - 1} 2 u_3 & \cdots & \dbinom {n - 1} {n - 2} u_{n - 1} & \dbinom {n - 1} {n - 1} u_n \\
-1 & \dbinom {n - 2} 0 u_1 & \dbinom {n ... | Faà di Bruno's Formula/Proof 4 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Proof_4 | [
"Differential Calculus",
"Faà di Bruno's Formula"
] | [
"Definition:Derivative/Higher Derivatives/Higher Order",
"Definition:Real Function"
] | [
"Definition:Determinant/Matrix"
] |
proofwiki-11911 | Cardinality of Reduced Residue System | Let $n \ge 2$.
Let $\Z'_n$ be the reduced residue system modulo $n$.
Then:
:$\card {\Z'_n} = \map \phi n$
where $\map \phi n$ is the Euler phi function. | Recall the definition of $\Z'_n$:
:$\Z'_n = \set {\eqclass k n \in \Z_n: k \perp n}$
and the definition of $\map \phi n$:
:$\map \phi n = \card {\set {k: 1 \le k \le n, k \perp n} }$
The result follows from Integer is Congruent to Integer less than Modulus.
{{qed}} | Let $n \ge 2$.
Let $\Z'_n$ be the [[Definition:Reduced Residue System|reduced residue system modulo $n$]].
Then:
:$\card {\Z'_n} = \map \phi n$
where $\map \phi n$ is the [[Definition:Euler Phi Function|Euler phi function]]. | Recall the [[Definition:Reduced Residue System|definition]] of $\Z'_n$:
:$\Z'_n = \set {\eqclass k n \in \Z_n: k \perp n}$
and the [[Definition:Euler Phi Function|definition]] of $\map \phi n$:
:$\map \phi n = \card {\set {k: 1 \le k \le n, k \perp n} }$
The result follows from [[Integer is Congruent to Integer les... | Cardinality of Reduced Residue System | https://proofwiki.org/wiki/Cardinality_of_Reduced_Residue_System | https://proofwiki.org/wiki/Cardinality_of_Reduced_Residue_System | [
"Reduced Residue Systems",
"Euler Phi Function"
] | [
"Definition:Reduced Residue System",
"Definition:Euler Phi Function"
] | [
"Definition:Reduced Residue System",
"Definition:Euler Phi Function",
"Integer is Congruent to Integer less than Modulus"
] |
proofwiki-11912 | Equivalence of Definitions of Norm of Linear Functional/Corollary | For all $v \in V$, the following inequality holds:
:$\size {L v} \le \norm L \norm v$
{{explain|Define $\norm L$ and $\norm v$}} | If $v = \mathbf 0$ there is nothing to prove.
Let $v \ne \mathbf 0$.
By the definition of the supremum:
:$\dfrac {\size {L v} } {\norm v} \le \norm L_3 = \norm L$
{{explain| What is this $\norm L_3$}}
whence:
:$\size {L v} \le \norm L \norm v$
{{qed}} | For all $v \in V$, the following inequality holds:
:$\size {L v} \le \norm L \norm v$
{{explain|Define $\norm L$ and $\norm v$}} | If $v = \mathbf 0$ there is nothing to prove.
Let $v \ne \mathbf 0$.
By the definition of the [[Definition:Supremum of Set|supremum]]:
:$\dfrac {\size {L v} } {\norm v} \le \norm L_3 = \norm L$
{{explain| What is this $\norm L_3$}}
whence:
:$\size {L v} \le \norm L \norm v$
{{qed}} | Equivalence of Definitions of Norm of Linear Functional/Corollary | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Functional/Corollary | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Functional/Corollary | [
"Bounded Linear Functionals"
] | [] | [
"Definition:Supremum of Set"
] |
proofwiki-11913 | Faà di Bruno's Formula/Example/0/Proof | Consider Faà di Bruno's Formula:
{{:Faà di Bruno's Formula}}
When $n = 0$ we have:
{{:Faà di Bruno's Formula/Example/0}} | In the summation:
:$\ds \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
the only element appearing is for $j ... | Consider [[Faà di Bruno's Formula]]:
{{:Faà di Bruno's Formula}}
When $n = 0$ we have:
{{:Faà di Bruno's Formula/Example/0}} | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
the only ele... | Faà di Bruno's Formula/Example/0/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/0/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/0/Proof | [
"Faà di Bruno's Formula"
] | [
"Faà di Bruno's Formula"
] | [
"Definition:Summation",
"Definition:Continued Product",
"Definition:Continued Product/Vacuous Product"
] |
proofwiki-11914 | Faà di Bruno's Formula/Example/1/Proof | Consider Faà di Bruno's Formula:
{{:Faà di Bruno's Formula}}
When $n = 1$ we have:
{{:Faà di Bruno's Formula/Example/1}} | In the summation:
:$\ds \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to consider $j = 0$ and $j = ... | Consider [[Faà di Bruno's Formula]]:
{{:Faà di Bruno's Formula}}
When $n = 1$ we have:
{{:Faà di Bruno's Formula/Example/1}} | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to c... | Faà di Bruno's Formula/Example/1/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/1/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/1/Proof | [
"Faà di Bruno's Formula"
] | [
"Faà di Bruno's Formula"
] | [
"Definition:Summation",
"Definition:Summation",
"Definition:Summation/Vacuous Summation",
"Definition:Derivative/Higher Derivatives/Zeroth Derivative",
"Definition:Factorial",
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-11915 | Faà di Bruno's Formula/Example/2/Proof | Consider Faà di Bruno's Formula:
{{:Faà di Bruno's Formula}}
When $n = 2$ we have:
{{:Faà di Bruno's Formula/Example/2}} | In the summation:
:$\ds \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \paren {m!}^{k_m} }$
we need to consider $j = 0, j =... | Consider [[Faà di Bruno's Formula]]:
{{:Faà di Bruno's Formula}}
When $n = 2$ we have:
{{:Faà di Bruno's Formula/Example/2}} | In the [[Definition:Summation|summation]]:
:$\ds \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \paren {m!}^{k_m} }$
we nee... | Faà di Bruno's Formula/Example/2/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/2/Proof | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Example/2/Proof | [
"Faà di Bruno's Formula"
] | [
"Faà di Bruno's Formula"
] | [
"Definition:Summation",
"Definition:Derivative/Higher Derivatives/Zeroth Derivative",
"Definition:Factorial",
"Definition:Summation",
"Definition:Summation",
"Definition:Summation/Vacuous Summation"
] |
proofwiki-11916 | Faà di Bruno's Formula/Lemma 1 | :$\map {D_x} {\paren {D_x^m u}^{k_m} } = k_m \paren {D_x^m u}^{k_m - 1} D_x^{m + 1} u$ | {{begin-eqn}}
{{eqn | l = \map {D_x} {\paren {D_x^m u}^{k_m} }
| r = \paren {k_m \paren {D_x^m u}^{k_m - 1} } \map {D_x} {D_x^m u}
| c = Derivative of Power of Function
}}
{{eqn | r = k_m \paren {\paren {D_x^m u}^{k_m - 1} } D_x^{m + 1} u
| c = {{Defof|Higher Derivative}}
}}
{{end-eqn}}
{{qed}}
Catego... | :$\map {D_x} {\paren {D_x^m u}^{k_m} } = k_m \paren {D_x^m u}^{k_m - 1} D_x^{m + 1} u$ | {{begin-eqn}}
{{eqn | l = \map {D_x} {\paren {D_x^m u}^{k_m} }
| r = \paren {k_m \paren {D_x^m u}^{k_m - 1} } \map {D_x} {D_x^m u}
| c = [[Derivative of Power of Function]]
}}
{{eqn | r = k_m \paren {\paren {D_x^m u}^{k_m - 1} } D_x^{m + 1} u
| c = {{Defof|Higher Derivative}}
}}
{{end-eqn}}
{{qed}}
[... | Faà di Bruno's Formula/Lemma 1 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Lemma_1 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Lemma_1 | [
"Faà di Bruno's Formula"
] | [] | [
"Derivative of Power of Function",
"Category:Faà di Bruno's Formula"
] |
proofwiki-11917 | Faà di Bruno's Formula/Lemma 2 | :$\ds \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } } = \prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$ | {{begin-eqn}}
{{eqn | r = \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } }
| o =
| c =
}}
{{eqn | r = \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \map {D_x} {\prod_{m \mathop = 1}^r \paren {\paren {D_x^m u}^{k_m} } }
... | :$\ds \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } } = \prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$ | {{begin-eqn}}
{{eqn | r = \map {D_x} {\prod_{m \mathop = 1}^r \paren {\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } } }
| o =
| c =
}}
{{eqn | r = \prod_{m \mathop = 1}^r \paren {\dfrac 1 {k_m! \paren {m!}^{k_m} } } \map {D_x} {\prod_{m \mathop = 1}^r \paren {\paren {D_x^m u}^{k_m} } }
... | Faà di Bruno's Formula/Lemma 2 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Lemma_2 | https://proofwiki.org/wiki/Faà_di_Bruno's_Formula/Lemma_2 | [
"Faà di Bruno's Formula"
] | [] | [
"Product Rule for Derivatives/General Result",
"Faà di Bruno's Formula/Lemma 1",
"Category:Faà di Bruno's Formula"
] |
proofwiki-11918 | Supremum of Ideals is Upper Adjoint | Let $L = \struct {S, \vee, \preceq}$ be a bounded below continuous join semilattice.
Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids... | Define $d: S \to \map {\mathit {Ids} } L$
:$\forall t \in S: \map d t = \map \inf {f^{-1} \sqbrk {t^\succeq} }$
where
:$t^\succeq$ denotes the upper closure of $t$,
:$f^{-1} \sqbrk {t^\succeq} $ denotes the image of $t^\succeq$ over $f^{-1}$.
We will prove that
:$\forall t \in S: \map d t = \map \min {f^{-1} \sqbrk {t^... | Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Join Semilattice|join semilattice]].
Let $\map {\mathit {Ids} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
Let ... | Define $d: S \to \map {\mathit {Ids} } L$
:$\forall t \in S: \map d t = \map \inf {f^{-1} \sqbrk {t^\succeq} }$
where
:$t^\succeq$ denotes the [[Definition:Upper Closure of Element|upper closure]] of $t$,
:$f^{-1} \sqbrk {t^\succeq} $ denotes the [[Definition:Image of Subset under Relation|image]] of $t^\succeq$ over $... | Supremum of Ideals is Upper Adjoint | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Upper_Adjoint | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Upper_Adjoint | [
"Continuous Lattices",
"Galois Connections"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Join Semilattice",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Galois Connection"
] | [
"Definition:Upper Closure/Element",
"Definition:Image (Set Theory)/Relation/Subset",
"Continuous iff For Every Element There Exists Ideal Element Precedes Supremum",
"Definition:Ideal in Ordered Set",
"Definition:Ideal in Ordered Set",
"Definition:Lower Bound of Set",
"Definition:Lower Bound of Set",
... |
proofwiki-11919 | Supremum of Ideals is Increasing | Let $L = \struct {S, \preceq}$ be an up-complete ordered set.
Let $\map {\operatorname{Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\operatorname{Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\map {\operatorname{Ids} } L \times \map {\operatorname{Ids} } ... | Let $I, J \in \map {\operatorname{Ids} } L$ such that
:$I \precsim J$
By definition of $\precsim$:
:$I \subseteq J$
By definition of up-complete:
:$I$ and $J$ admit suprema in $L$.
By Supremum of Subset:
:$\sup I \preceq \sup J$
Thus by definition of $f$:
:$\map f I \preceq \map f J$
Hence $f$ is an increasing mapping.... | Let $L = \struct {S, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Ordered Set|ordered set]].
Let $\map {\operatorname{Ids} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
Let $P = \struct {\map {\operatorname{Ids} } L, \precsim}$ be an [[Defi... | Let $I, J \in \map {\operatorname{Ids} } L$ such that
:$I \precsim J$
By definition of $\precsim$:
:$I \subseteq J$
By definition of [[Definition:Up-Complete|up-complete]]:
:$I$ and $J$ admit [[Definition:Supremum of Set|suprema]] in $L$.
By [[Supremum of Subset]]:
:$\sup I \preceq \sup J$
Thus by definition of $f$... | Supremum of Ideals is Increasing | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Increasing | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Increasing | [
"Increasing Mappings"
] | [
"Definition:Up-Complete",
"Definition:Ordered Set",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Increasing/Mapping"
] | [
"Definition:Up-Complete",
"Definition:Supremum of Set",
"Supremum of Subset",
"Definition:Increasing/Mapping"
] |
proofwiki-11920 | Derivative of Composite Function/Third Derivative | :${D_x}^3 w = {D_u}^3 w \paren { {D_x}^1 u}^3 + 3 {D_u}^2 w {D_x}^2 u {D_x}^1 u + {D_u}^1 w {D_x}^3 u$ | For ease of understanding, let Leibniz's notation be used:
:$\dfrac {\d^k u} {\d x^k} := {D_x}^k u$
Then we have:
{{begin-eqn}}
{{eqn | l = {D_x}^3 w
| r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d x^2} }
| c =
}}
{{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} \paren {\dfrac {\d u} {\d ... | :${D_x}^3 w = {D_u}^3 w \paren { {D_x}^1 u}^3 + 3 {D_u}^2 w {D_x}^2 u {D_x}^1 u + {D_u}^1 w {D_x}^3 u$ | For ease of understanding, let [[Definition:Leibniz's Notation for Derivatives|Leibniz's notation]] be used:
:$\dfrac {\d^k u} {\d x^k} := {D_x}^k u$
Then we have:
{{begin-eqn}}
{{eqn | l = {D_x}^3 w
| r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d x^2} }
| c =
}}
{{eqn | r = \map {\dfrac \d {\d x} }... | Derivative of Composite Function/Third Derivative | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Third_Derivative | https://proofwiki.org/wiki/Derivative_of_Composite_Function/Third_Derivative | [
"Derivative of Composite Function"
] | [] | [
"Definition:Derivative/Notation/Leibniz Notation",
"Derivative of Composite Function/Second Derivative",
"Product Rule for Derivatives",
"Derivative of Composite Function",
"Definition:Derivative/Notation/Leibniz Notation",
"Category:Derivative of Composite Function"
] |
proofwiki-11921 | Number of Set Partitions by Number of Components | Let $S$ be a (finite) set whose cardinality is $n$.
Let $\map f {n, k}$ denote the number of different ways $S$ can be partitioned into $k$ components.
Then:
:$\ds \map f {n, k} = {n \brace k}$
where $\ds {n \brace k}$ denotes a Stirling number of the second kind. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \map f {n, k} = {n \brace k}$
$\map P 0$ is the degenerate case:
:$\ds \map f {0, k} = \delta_{0 k} = {0 \brace k}$
That is: the empty set can be partitioned one and only one way: into $0$ subsets.
Thus $\map P ... | Let $S$ be a [[Definition:Finite Set|(finite) set]] whose [[Definition:Cardinality|cardinality]] is $n$.
Let $\map f {n, k}$ denote the number of different ways $S$ can be [[Definition:Set Partition|partitioned]] into $k$ [[Definition:Component of Partition|components]].
Then:
:$\ds \map f {n, k} = {n \brace k}$
whe... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \map f {n, k} = {n \brace k}$
$\map P 0$ is the [[Definition:Degenerate Case|degenerate case]]:
:$\ds \map f {0, k} = \delta_{0 k} = {0 \bra... | Number of Set Partitions by Number of Components | https://proofwiki.org/wiki/Number_of_Set_Partitions_by_Number_of_Components | https://proofwiki.org/wiki/Number_of_Set_Partitions_by_Number_of_Components | [
"Combinatorics",
"Stirling Numbers",
"Number of Set Partitions by Number of Components",
"Set Partitions"
] | [
"Definition:Finite Set",
"Definition:Cardinality",
"Definition:Set Partition",
"Definition:Set Partition/Component",
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Degenerate Case",
"Definition:Empty Set",
"Definition:Set Partition",
"Definition:Subset",
"Definition:Set",
"Definition:Cardinality",
"Definition:Set Partition",
"Definition:Set Partition",
"Definition:Subset",
"Defi... |
proofwiki-11922 | Summation of Summation over Divisors of Function of Two Variables | Let $c, d, n \in \Z$.
Then:
:$\ds \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
where $c \divides d$ denotes that $c$ is a divisor of $d$. | From Exchange of Order of Summation with Dependency on Both Indices:
{{:Exchange of Order of Summation with Dependency on Both Indices}}
We have that:
:$\map R d$ is the propositional function:
::$d \divides n$
:$\map S {d, c}$ is the propositional function:
::$c \divides d$
Thus $\map {R'} {d, c}$ is the propositional... | Let $c, d, n \in \Z$.
Then:
:$\ds \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
where $c \divides d$ denotes that $c$ is a [[Definition:Divisor of Integer|divisor]] of $d$. | From [[Exchange of Order of Summation with Dependency on Both Indices]]:
{{:Exchange of Order of Summation with Dependency on Both Indices}}
We have that:
:$\map R d$ is the [[Definition:Propositional Function|propositional function]]:
::$d \divides n$
:$\map S {d, c}$ is the [[Definition:Propositional Function|propo... | Summation of Summation over Divisors of Function of Two Variables | https://proofwiki.org/wiki/Summation_of_Summation_over_Divisors_of_Function_of_Two_Variables | https://proofwiki.org/wiki/Summation_of_Summation_over_Divisors_of_Function_of_Two_Variables | [
"Summations",
"Divisors"
] | [
"Definition:Divisor (Algebra)/Integer"
] | [
"Exchange of Order of Summation with Dependency on Both Indices",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function",
"Definition:Propositional Function"
] |
proofwiki-11923 | Floor of Half of n+m plus Floor of Half of n-m+1 | Let $n, m \in \Z$ be integers.
:$\floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2} = n$
where $\floor x$ denotes the floor of $x$. | Either $n + m$ or $n - m + 1$ is even.
Thus:
:$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$
and so:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2}
| r = \floor {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}
| c = Sum of Floors not greater than Flo... | Let $n, m \in \Z$ be [[Definition:Integer|integers]].
:$\floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2} = n$
where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$. | Either $n + m$ or $n - m + 1$ is [[Definition:Even Integer|even]].
Thus:
:$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$
and so:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2}
| r = \floor {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}
| c = [[Su... | Floor of Half of n+m plus Floor of Half of n-m+1 | https://proofwiki.org/wiki/Floor_of_Half_of_n+m_plus_Floor_of_Half_of_n-m+1 | https://proofwiki.org/wiki/Floor_of_Half_of_n+m_plus_Floor_of_Half_of_n-m+1 | [
"Floor Function"
] | [
"Definition:Integer",
"Definition:Floor Function"
] | [
"Definition:Even Integer",
"Sum of Floors not greater than Floor of Sum"
] |
proofwiki-11924 | Ceiling of Half of n+m plus Ceiling of Half of n-m+1 | Let $n, m \in \Z$ be integers.
:$\ceiling {\dfrac {n + m} 2} + \ceiling {\dfrac {n - m + 1} 2} = n + 1$
where $\ceiling x$ denotes the ceiling of $x$. | Either $n + m$ or $n - m + 1$ is even.
Thus either $\dfrac {n + m} 2$ or $\dfrac {n - m + 1} 2$ is an integer.
So:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac {n + m} 2} + \ceiling {\dfrac {n - m + 1} 2}
| r = \ceiling {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}
| c = Sum of Ceilings not less than Ceiling of S... | Let $n, m \in \Z$ be [[Definition:Integer|integers]].
:$\ceiling {\dfrac {n + m} 2} + \ceiling {\dfrac {n - m + 1} 2} = n + 1$
where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$. | Either $n + m$ or $n - m + 1$ is [[Definition:Even Integer|even]].
Thus either $\dfrac {n + m} 2$ or $\dfrac {n - m + 1} 2$ is an [[Definition:Integer|integer]].
So:
{{begin-eqn}}
{{eqn | l = \ceiling {\dfrac {n + m} 2} + \ceiling {\dfrac {n - m + 1} 2}
| r = \ceiling {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}
... | Ceiling of Half of n+m plus Ceiling of Half of n-m+1 | https://proofwiki.org/wiki/Ceiling_of_Half_of_n+m_plus_Ceiling_of_Half_of_n-m+1 | https://proofwiki.org/wiki/Ceiling_of_Half_of_n+m_plus_Ceiling_of_Half_of_n-m+1 | [
"Ceiling Function"
] | [
"Definition:Integer",
"Definition:Ceiling Function"
] | [
"Definition:Even Integer",
"Definition:Integer",
"Sum of Ceilings not less than Ceiling of Sum"
] |
proofwiki-11925 | Floor of Non-Integer | Let $x \in \R$ be a real number.
Let $x \notin \Z$.
Then:
:$\floor x < x$
where $\floor x$ denotes the floor of $x$. | From Floor is between Number and One Less:
:$\floor x \le x$
From Real Number is Integer iff equals Floor:
:$x = \floor x \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\floor x \ne x$
and so:
:$\floor x < x$
{{qed}} | Let $x \in \R$ be a [[Definition:Real Number|real number]].
Let $x \notin \Z$.
Then:
:$\floor x < x$
where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$. | From [[Floor is between Number and One Less]]:
:$\floor x \le x$
From [[Real Number is Integer iff equals Floor]]:
:$x = \floor x \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\floor x \ne x$
and so:
:$\floor x < x$
{{qed}} | Floor of Non-Integer | https://proofwiki.org/wiki/Floor_of_Non-Integer | https://proofwiki.org/wiki/Floor_of_Non-Integer | [
"Floor Function"
] | [
"Definition:Real Number",
"Definition:Floor Function"
] | [
"Floor is between Number and One Less",
"Real Number is Integer iff equals Floor"
] |
proofwiki-11926 | Logarithm of Base | Let $b \in \R_{>0}$ such that $b \ne 1$.
Then:
:$\log_b b = 1$
where $\log_b$ denotes the logarithm to base $b$. | Let $a \in \R_{>0}$ such that $a \ne 1$.
Then:
{{begin-eqn}}
{{eqn | l = \log_b b
| r = \dfrac {\log_a b} {\log_a b}
| c = Change of Base of Logarithm
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
{{qed}}
Category:Logarithms
8p8v7a9h3w1b8jvsqgyq3scehh3d4y0 | Let $b \in \R_{>0}$ such that $b \ne 1$.
Then:
:$\log_b b = 1$
where $\log_b$ denotes the [[Definition:General Logarithm|logarithm]] to [[Definition:Base of Logarithm|base $b$]]. | Let $a \in \R_{>0}$ such that $a \ne 1$.
Then:
{{begin-eqn}}
{{eqn | l = \log_b b
| r = \dfrac {\log_a b} {\log_a b}
| c = [[Change of Base of Logarithm]]
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
{{qed}}
[[Category:Logarithms]]
8p8v7a9h3w1b8jvsqgyq3scehh3d4y0 | Logarithm of Base | https://proofwiki.org/wiki/Logarithm_of_Base | https://proofwiki.org/wiki/Logarithm_of_Base | [
"Logarithms"
] | [
"Definition:General Logarithm",
"Definition:Logarithm/Base"
] | [
"Change of Base of Logarithm",
"Category:Logarithms"
] |
proofwiki-11927 | Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x | Let $b \in \R$ be a real number.
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$
where $\floor x$ denotes the floor of $x$. | === Necessary Condition ===
Let:
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x}$
Let $x = b$.
Then:
:$\floor {\log_b b} = \floor {\log_b \floor b}$
{{begin-eqn}}
{{eqn | l = \floor {\log_b b}
| r = \floor {\log_b \floor b}
| c =
}}
{{eqn | ll= \leadsto
| l = \floor 1
|... | Let $b \in \R$ be a [[Definition:Real Number|real number]].
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$
where $\floor x$ denotes the [[Definition:Floor Function|floor of $x$]]. | === Necessary Condition ===
Let:
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x}$
Let $x = b$.
Then:
:$\floor {\log_b b} = \floor {\log_b \floor b}$
{{begin-eqn}}
{{eqn | l = \floor {\log_b b}
| r = \floor {\log_b \floor b}
| c =
}}
{{eqn | ll= \leadsto
| l = \floor 1
... | Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x/Proof 1 | https://proofwiki.org/wiki/Conditions_for_Floor_of_Log_base_b_of_x_to_equal_Floor_of_Log_base_b_of_Floor_of_x | https://proofwiki.org/wiki/Conditions_for_Floor_of_Log_base_b_of_x_to_equal_Floor_of_Log_base_b_of_Floor_of_x/Proof_1 | [
"Logarithms",
"Floor Function",
"Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x"
] | [
"Definition:Real Number",
"Definition:Floor Function"
] | [
"Logarithm of Base",
"Real Number is Integer iff equals Floor",
"Floor of Non-Integer",
"Definition:Contradiction",
"Proof by Contradiction",
"Integer equals Floor iff Number between Integer and One More",
"Power Function is Strictly Increasing over Positive Reals/Natural Exponent",
"Number not less t... |
proofwiki-11928 | Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x | Let $b \in \R$ be a real number.
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$
where $\floor x$ denotes the floor of $x$. | We have that:
:Logarithm is Strictly Increasing
and:
:Real Natural Logarithm Function is Continuous
Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say.
Then:
:$x = b^n$
It follows that:
:$x \in \Z \iff b \in \Z$
Thus by McEliece's Theorem (Integer Functions):
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor ... | Let $b \in \R$ be a [[Definition:Real Number|real number]].
:$\forall x \in \R_{\ge 1}: \floor {\log_b x} = \floor {\log_b \floor x} \iff b \in \Z_{> 1}$
where $\floor x$ denotes the [[Definition:Floor Function|floor of $x$]]. | We have that:
:[[Logarithm is Strictly Increasing]]
and:
:[[Real Natural Logarithm Function is Continuous]]
Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say.
Then:
:$x = b^n$
It follows that:
:$x \in \Z \iff b \in \Z$
Thus by [[McEliece's Theorem (Integer Functions)]]:
:$\forall x \in \R_{\ge 1}: \floor {\l... | Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x/Proof 2 | https://proofwiki.org/wiki/Conditions_for_Floor_of_Log_base_b_of_x_to_equal_Floor_of_Log_base_b_of_Floor_of_x | https://proofwiki.org/wiki/Conditions_for_Floor_of_Log_base_b_of_x_to_equal_Floor_of_Log_base_b_of_Floor_of_x/Proof_2 | [
"Logarithms",
"Floor Function",
"Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x"
] | [
"Definition:Real Number",
"Definition:Floor Function"
] | [
"Logarithm is Strictly Increasing",
"Real Natural Logarithm Function is Continuous",
"McEliece's Theorem (Integer Functions)"
] |
proofwiki-11929 | Floor of x+m over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $\floor x$ denotes the floor of $x$. | Let:
:$y = x - \floor x$
:$M = \floor x + m$
We now have:
:$(1): \quad 0 \le y < 1$
and thus:
:$\floor y = 0$
By Division Theorem, we can write:
:$(2): \quad M = k n + r$
with $k \in \Z$ and $0 \le r \le n - 1$.
By $(1)$ and $(2)$:
:$(3): \quad 0 \le y + r < 1 + n - 1 = n$
We have:
{{begin-eqn}}
{{eqn | l = \floor {\fr... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $\floor x$ denotes the [[Definition:Floor Function|floor of $x$]]. | Let:
:$y = x - \floor x$
:$M = \floor x + m$
We now have:
:$(1): \quad 0 \le y < 1$
and thus:
:$\floor y = 0$
By [[Division Theorem]], we can write:
:$(2): \quad M = k n + r$
with $k \in \Z$ and $0 \le r \le n - 1$.
By $(1)$ and $(2)$:
:$(3): \quad 0 \le y + r < 1 + n - 1 = n$
We have:
{{begin-eqn}}
{{eqn | l = \f... | Floor of x+m over n/Proof 1 | https://proofwiki.org/wiki/Floor_of_x+m_over_n | https://proofwiki.org/wiki/Floor_of_x+m_over_n/Proof_1 | [
"Floor Function",
"Floor of x+m over n"
] | [
"Definition:Floor Function"
] | [
"Division Theorem",
"Floor of Number plus Integer"
] |
proofwiki-11930 | Floor of x+m over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $\floor x$ denotes the floor of $x$. | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$.
Let $\dfrac {x + m} n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | q = \exists s \in \Z
| l = \dfrac {x + m} n
| r = s
| c ... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
where $\floor x$ denotes the [[Definition:Floor Function|floor of $x$]]. | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both [[Definition:Strictly Increasing Real Function|strictly increasing]] and [[Definition:Continuous Real Function|continuous]] on the whole of $\R$.
Let $\dfrac {... | Floor of x+m over n/Proof 2 | https://proofwiki.org/wiki/Floor_of_x+m_over_n | https://proofwiki.org/wiki/Floor_of_x+m_over_n/Proof_2 | [
"Floor Function",
"Floor of x+m over n"
] | [
"Definition:Floor Function"
] | [
"Definition:Real Function",
"Definition:Strictly Increasing/Real Function",
"Definition:Continuous Real Function",
"McEliece's Theorem (Integer Functions)"
] |
proofwiki-11931 | McEliece's Theorem (Integer Functions) | Let $f: \R \to \R$ be a continuous, strictly increasing real function defined on an interval $A$.
Let:
:$\forall x \in A: \floor x \in A \text { and } \ceiling x \in A$
where:
:$\floor x$ denotes the floor of $x$
:$\ceiling x$ denotes the ceiling of $x$
Then:
:$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z... | Let $x \in A$.
Hence {{hypothesis}} we have that both $\floor x \in A$ and $\ceiling x \in A$. | Let $f: \R \to \R$ be a [[Definition:Continuous Real Function|continuous]], [[Definition:Strictly Increasing Real Function|strictly increasing real function]] defined on an [[Definition:Real Interval|interval]] $A$.
Let:
:$\forall x \in A: \floor x \in A \text { and } \ceiling x \in A$
where:
:$\floor x$ denotes the [... | Let $x \in A$.
Hence {{hypothesis}} we have that both $\floor x \in A$ and $\ceiling x \in A$. | McEliece's Theorem (Integer Functions) | https://proofwiki.org/wiki/McEliece's_Theorem_(Integer_Functions) | https://proofwiki.org/wiki/McEliece's_Theorem_(Integer_Functions) | [
"Floor Function",
"Ceiling Function"
] | [
"Definition:Continuous Real Function",
"Definition:Strictly Increasing/Real Function",
"Definition:Real Interval",
"Definition:Floor Function",
"Definition:Ceiling Function"
] | [] |
proofwiki-11932 | Ceiling of Non-Integer | Let $x \in \R$ be a real number.
Let $x \notin \Z$.
Then:
:$\ceiling x > x$
where $\ceiling x$ denotes the ceiling of $x$. | From Ceiling is between Number and One More:
:$\ceiling x \ge x$
From Real Number is Integer iff equals Ceiling:
:$x = \ceiling x \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\ceiling x \ne x$
and so:
:$\ceiling x > x$
{{qed}} | Let $x \in \R$ be a [[Definition:Real Number|real number]].
Let $x \notin \Z$.
Then:
:$\ceiling x > x$
where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling ]] of $x$. | From [[Ceiling is between Number and One More]]:
:$\ceiling x \ge x$
From [[Real Number is Integer iff equals Ceiling]]:
:$x = \ceiling x \iff x \in \Z$
But we have $x \notin \Z$.
So:
:$\ceiling x \ne x$
and so:
:$\ceiling x > x$
{{qed}} | Ceiling of Non-Integer | https://proofwiki.org/wiki/Ceiling_of_Non-Integer | https://proofwiki.org/wiki/Ceiling_of_Non-Integer | [
"Ceiling Function"
] | [
"Definition:Real Number",
"Definition:Ceiling Function"
] | [
"Ceiling is between Number and One More",
"Real Number is Integer iff equals Ceiling"
] |
proofwiki-11933 | Ceiling of x+m over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the ceiling of $x$. | {{begin-eqn}}
{{eqn | l = \ceiling {\dfrac {x + m} n}
| r = -\floor {-\dfrac {x + m} n}
| c = Floor of Negative equals Negative of Ceiling
}}
{{eqn | r = -\floor {\dfrac {-x - m} n}
| c =
}}
{{eqn | r = -\floor {\dfrac {\floor {-x} - m} n}
| c = Floor of $\dfrac {x + m} n$
}}
{{eqn | r = -\floo... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling of $x$]]. | {{begin-eqn}}
{{eqn | l = \ceiling {\dfrac {x + m} n}
| r = -\floor {-\dfrac {x + m} n}
| c = [[Floor of Negative equals Negative of Ceiling]]
}}
{{eqn | r = -\floor {\dfrac {-x - m} n}
| c =
}}
{{eqn | r = -\floor {\dfrac {\floor {-x} - m} n}
| c = [[Floor of x+m over n|Floor of $\dfrac {x + m... | Ceiling of x+m over n/Proof 1 | https://proofwiki.org/wiki/Ceiling_of_x+m_over_n | https://proofwiki.org/wiki/Ceiling_of_x+m_over_n/Proof_1 | [
"Ceiling Function",
"Ceiling of x+m over n"
] | [
"Definition:Ceiling Function"
] | [
"Floor of Negative equals Negative of Ceiling",
"Floor of x+m over n",
"Floor of Negative equals Negative of Ceiling",
"Floor of Negative equals Negative of Ceiling"
] |
proofwiki-11934 | Ceiling of x+m over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the ceiling of $x$. | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both strictly increasing and continuous on the whole of $\R$.
Let $\dfrac {x + m} n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | q = \exists s \in \Z
| l = \dfrac {x + m} n
| r = s
| c ... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling of $x$]]. | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \dfrac {x + m} n$
It is clear that $f$ is both [[Definition:Strictly Increasing Real Function|strictly increasing]] and [[Definition:Continuous Real Function|continuous]] on the whole of $\R$.
Let $\dfrac {... | Ceiling of x+m over n/Proof 2 | https://proofwiki.org/wiki/Ceiling_of_x+m_over_n | https://proofwiki.org/wiki/Ceiling_of_x+m_over_n/Proof_2 | [
"Ceiling Function",
"Ceiling of x+m over n"
] | [
"Definition:Ceiling Function"
] | [
"Definition:Real Function",
"Definition:Strictly Increasing/Real Function",
"Definition:Continuous Real Function",
"McEliece's Theorem (Integer Functions)"
] |
proofwiki-11935 | Summation over k of Floor of k over 2 | :$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \floor {\dfrac {n^2} 4}$ | By Permutation of Indices of Summation:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$
First take the case... | :$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \floor {\dfrac {n^2} 4}$ | By [[Permutation of Indices of Summation]]:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$
First take t... | Summation over k of Floor of k over 2 | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_k_over_2 | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_k_over_2 | [
"Floor Function",
"Summations"
] | [] | [
"Permutation of Indices of Summation",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Integer",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] |
proofwiki-11936 | Summation over k of Ceiling of k over 2 | :$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \ceiling {\dfrac {n \paren {n + 2} } 4}$ | By Permutation of Indices of Summation:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \sum_{k \mathop = 1}^n \ceiling {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2} }$
First tak... | :$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \ceiling {\dfrac {n \paren {n + 2} } 4}$ | By [[Permutation of Indices of Summation]]:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \sum_{k \mathop = 1}^n \ceiling {\dfrac {n + 1 - k} 2}$
and so:
:$\ds \sum_{k \mathop = 1}^n \ceiling {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\ceiling {\dfrac k 2} + \ceiling {\dfrac {n + 1 - k} 2} }$
Fi... | Summation over k of Ceiling of k over 2 | https://proofwiki.org/wiki/Summation_over_k_of_Ceiling_of_k_over_2 | https://proofwiki.org/wiki/Summation_over_k_of_Ceiling_of_k_over_2 | [
"Ceiling Function",
"Summations"
] | [] | [
"Permutation of Indices of Summation",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Integer",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Even Integer",
"Definition:Odd Integer"
] |
proofwiki-11937 | Summation over k of Floor of mk+x over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \dfrac {\paren {m - 1} \paren {n - 1} } 2 + \dfrac {d - 1} 2 + d \floor {\dfrac x d}$
where:
:$\floor x$ denotes the floor of $x$
:$d$ is the greatest common divisor of $m$ and $n$. | By definition of modulo 1:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n - \sum_{0 \mathop \le k \mathop < n} \fractpart {\dfrac {m k + x} n}$
where $\fractpart y$ in this context denotes the fractional part of $y$.
First we have:
{{begin-eq... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \dfrac {\paren {m - 1} \paren {n - 1} } 2 + \dfrac {d - 1} 2 + d \floor {\dfrac x d}$
where:
:$\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$
:$d$ is the [[Definition:... | By definition of [[Definition:Modulo 1|modulo 1]]:
:$\ds \sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {m k + x} n} = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n - \sum_{0 \mathop \le k \mathop < n} \fractpart {\dfrac {m k + x} n}$
where $\fractpart y$ in this context denotes the [[Definition:Fractional P... | Summation over k of Floor of mk+x over n | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_mk+x_over_n | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_mk+x_over_n | [
"Floor Function",
"Summations"
] | [
"Definition:Floor Function",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Modulo Operation/Modulo One",
"Definition:Fractional Part",
"Closed Form for Triangular Numbers",
"Definition:Integer",
"Definition:Summation",
"Closed Form for Triangular Numbers"
] |
proofwiki-11938 | Summation over k of Floor of x plus k over y | Let $x, y \in \R$ such that $y > 0$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < y} \floor {x + \dfrac k y} = \floor {x y + \floor {x + 1} \paren {\ceiling y - y} }$ | When $x$ increases by $1$, both sides increase by $\ceiling y$.
So we can assume $0 \le x < 1$.
When $x = 0$, both sides are equal to $0$.
When $x$ increases past $1 - \dfrac k y$ for $0 \le k < y$, both sides increase by $1$.
Hence the result.
{{qed}} | Let $x, y \in \R$ such that $y > 0$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < y} \floor {x + \dfrac k y} = \floor {x y + \floor {x + 1} \paren {\ceiling y - y} }$ | When $x$ increases by $1$, both sides increase by $\ceiling y$.
So we can assume $0 \le x < 1$.
When $x = 0$, both sides are equal to $0$.
When $x$ increases past $1 - \dfrac k y$ for $0 \le k < y$, both sides increase by $1$.
Hence the result.
{{qed}} | Summation over k of Floor of x plus k over y | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_x_plus_k_over_y | https://proofwiki.org/wiki/Summation_over_k_of_Floor_of_x_plus_k_over_y | [
"Floor Function",
"Summations"
] | [] | [] |
proofwiki-11939 | Summation over k of Ceiling of mk+x over n | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < n} \ceiling {\dfrac {m k + x} n} = \dfrac {\paren {m + 1} \paren {n - 1} } 2 - \dfrac {d - 1} 2 + d \ceiling {\dfrac x d}$
where:
:$\ceiling x$ denotes the ceiling of $x$
:$d$ is the greatest common divisor of $m$ and $n$. | {{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \ceiling {\dfrac {m k + x} n}
| r = \sum_{0 \mathop \le k \mathop < n} \paren {-\floor {-\dfrac {m k + x} n} }
| c = Floor of Negative equals Negative of Ceiling
}}
{{eqn | r = -\sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {-m k - x} n}
... | Let $m, n \in \Z$ such that $n > 0$.
Let $x \in \R$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < n} \ceiling {\dfrac {m k + x} n} = \dfrac {\paren {m + 1} \paren {n - 1} } 2 - \dfrac {d - 1} 2 + d \ceiling {\dfrac x d}$
where:
:$\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$
:$d$ is the [[D... | {{begin-eqn}}
{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \ceiling {\dfrac {m k + x} n}
| r = \sum_{0 \mathop \le k \mathop < n} \paren {-\floor {-\dfrac {m k + x} n} }
| c = [[Floor of Negative equals Negative of Ceiling]]
}}
{{eqn | r = -\sum_{0 \mathop \le k \mathop < n} \floor {\dfrac {-m k - x} n}
... | Summation over k of Ceiling of mk+x over n | https://proofwiki.org/wiki/Summation_over_k_of_Ceiling_of_mk+x_over_n | https://proofwiki.org/wiki/Summation_over_k_of_Ceiling_of_mk+x_over_n | [
"Ceiling Function",
"Summations"
] | [
"Definition:Ceiling Function",
"Definition:Greatest Common Divisor/Integers"
] | [
"Floor of Negative equals Negative of Ceiling",
"Summation over k of Floor of mk+x over n",
"Floor of Negative equals Negative of Ceiling"
] |
proofwiki-11940 | Semantic Consequence preserved in Supersignature | Let $\LL, \LL'$ be signatures for the language of predicate logic.
Let $\LL'$ be a supersignature of $\LL$.
Let $\mathbf A$ be an $\LL$-sentence.
Let $\Sigma$ be a set of $\LL$-sentences.
Then the following are equivalent:
:$\AA \models_{\mathrm{PL} } \mathbf A$ for all $\LL$-structures $\AA$ for which $\AA \models_{\m... | {{finish|
* every $\AA$ arises as the reduct of some $\AA'$;
* every $\AA'$ has the same valuations as its reduct.
}} | Let $\LL, \LL'$ be [[Definition:Signature for Predicate Logic|signatures]] for the [[Definition:Language of Predicate Logic|language of predicate logic]].
Let $\LL'$ be a [[Definition:Supersignature|supersignature]] of $\LL$.
Let $\mathbf A$ be an [[Definition:Sentence|$\LL$-sentence]].
Let $\Sigma$ be a [[Definitio... | {{finish|
* every $\AA$ arises as the reduct of some $\AA'$;
* every $\AA'$ has the same valuations as its reduct.
}} | Semantic Consequence preserved in Supersignature | https://proofwiki.org/wiki/Semantic_Consequence_preserved_in_Supersignature | https://proofwiki.org/wiki/Semantic_Consequence_preserved_in_Supersignature | [
"Model Theory for Predicate Logic"
] | [
"Definition:Signature (Logic)/Predicate Logic",
"Definition:Language of Predicate Logic",
"Definition:Subsignature/Supersignature",
"Definition:Classes of WFFs/Sentence",
"Definition:Set",
"Definition:Classes of WFFs/Sentence",
"Definition:Logical Equivalence",
"Definition:Structure for Predicate Logi... | [] |
proofwiki-11941 | Satisfiability preserved in Supersignature | Let $\LL, \LL'$ be signatures for the language of predicate logic.
Let $\LL'$ be a supersignature of $\LL$.
Let $\Sigma$ be a set of $\LL$-sentences.
Then the following are equivalent:
:$\AA \models_{\mathrm{PL} } \Sigma$ for some $\LL$-structure $\AA$
:$\AA' \models_{\mathrm{PL} } \Sigma$ for some $\LL'$-structure $\A... | {{finish|* every $\AA$ arises as the reduct of some $\AA'$;
* every $\AA'$ has the same valuations as its reduct.
}} | Let $\LL, \LL'$ be [[Definition:Signature for Predicate Logic|signatures]] for the [[Definition:Language of Predicate Logic|language of predicate logic]].
Let $\LL'$ be a [[Definition:Supersignature|supersignature]] of $\LL$.
Let $\Sigma$ be a [[Definition:Set|set]] of [[Definition:Sentence|$\LL$-sentences]].
Then ... | {{finish|* every $\AA$ arises as the reduct of some $\AA'$;
* every $\AA'$ has the same valuations as its reduct.
}} | Satisfiability preserved in Supersignature | https://proofwiki.org/wiki/Satisfiability_preserved_in_Supersignature | https://proofwiki.org/wiki/Satisfiability_preserved_in_Supersignature | [
"Model Theory for Predicate Logic"
] | [
"Definition:Signature (Logic)/Predicate Logic",
"Definition:Language of Predicate Logic",
"Definition:Subsignature/Supersignature",
"Definition:Set",
"Definition:Classes of WFFs/Sentence",
"Definition:Logical Equivalence",
"Definition:Structure for Predicate Logic",
"Definition:Structure for Predicate... | [] |
proofwiki-11942 | Floor Function is Replicative | The floor function is a replicative function in the sense that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 0}^{n - 1} \floor {x + \frac k n} = \floor {n x}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \floor {x + \dfrac k n}
| r = \floor {x n + \floor {x + 1} \paren {\ceiling n - n} }
| c = Summation over $k$ of $\floor {x + \dfrac k y}$, setting $y = n$
}}
{{eqn | r = \floor {x n + \floor {x + 1} \paren {n - n} }
| c = Real Number is Integer ... | The [[Definition:Floor Function|floor function]] is a [[Definition:Replicative Function|replicative function]] in the sense that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 0}^{n - 1} \floor {x + \frac k n} = \floor {n x}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \floor {x + \dfrac k n}
| r = \floor {x n + \floor {x + 1} \paren {\ceiling n - n} }
| c = [[Summation over k of Floor of x plus k over y|Summation over $k$ of $\floor {x + \dfrac k y}$]], setting $y = n$
}}
{{eqn | r = \floor {x n + \floor {x + 1} \pa... | Floor Function is Replicative | https://proofwiki.org/wiki/Floor_Function_is_Replicative | https://proofwiki.org/wiki/Floor_Function_is_Replicative | [
"Floor Function",
"Replicative Functions"
] | [
"Definition:Floor Function",
"Definition:Replicative Function"
] | [
"Summation over k of Floor of x plus k over y",
"Real Number is Integer iff equals Ceiling"
] |
proofwiki-11943 | Subtract Half is Replicative Function | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x - \dfrac 1 2$
Then $f$ is a replicative function. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {x - \frac 1 2 + \frac k n}
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \frac {n \paren {n - 1}... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = x - \dfrac 1 2$
Then $f$ is a [[Definition:Replicative Function|replicative function]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {x - \frac 1 2 + \frac k n}
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k
| c =
}}
{{eqn | r = n x - \frac n 2 + \frac 1 n \frac {n \paren {n - 1}... | Subtract Half is Replicative Function | https://proofwiki.org/wiki/Subtract_Half_is_Replicative_Function | https://proofwiki.org/wiki/Subtract_Half_is_Replicative_Function | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Replicative Function"
] | [
"Closed Form for Triangular Numbers",
"Definition:Replicative Function"
] |
proofwiki-11944 | Membership of Set of Integers is Replicative Function | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z}$
where $\sqbrk {\cdots}$ is Iverson's convention.
Then $f$ is a replicative function. | First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$.
Thus there can be no more than one $k$ such that $0 \le k < n$ such that:
:$x + \dfrac k n \in \Z$
Hence:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$
First let:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x ... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z}$
where $\sqbrk {\cdots}$ is [[Definition:Iverson's Convention|Iverson's convention]].
Then $f$ is a [[Definition:Replicative Function|replicative function]]. | First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$.
Thus there can be no more than one $k$ such that $0 \le k < n$ such that:
:$x + \dfrac k n \in \Z$
Hence:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$
First let:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk... | Membership of Set of Integers is Replicative Function | https://proofwiki.org/wiki/Membership_of_Set_of_Integers_is_Replicative_Function | https://proofwiki.org/wiki/Membership_of_Set_of_Integers_is_Replicative_Function | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Iverson's Convention",
"Definition:Replicative Function"
] | [
"Definition:Multiple/Integer",
"Definition:Contradiction",
"Definition:Replicative Function"
] |
proofwiki-11945 | Membership of Set of Strictly Positive Integers is Replicative Function | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z_{> 0} }$
where $\sqbrk \cdots$ is Iverson's convention.
Then $f$ is a replicative function. | Let $x \in \R$ such that $x > 0$.
Then for all $k \in \Z$ such that $0 \le k < n$:
:$x + \dfrac k n \in \Z_{> 0}$
and so from Membership of Set of Integers is Replicative Function:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = \sqbrk {n x \in \Z_{> 0} }$
Let $x \le 0$.
Then for all $k \in \Z... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \sqbrk {x \in \Z_{> 0} }$
where $\sqbrk \cdots$ is [[Definition:Iverson's Convention|Iverson's convention]].
Then $f$ is a [[Definition:Replicative Function|replicative function]]. | Let $x \in \R$ such that $x > 0$.
Then for all $k \in \Z$ such that $0 \le k < n$:
:$x + \dfrac k n \in \Z_{> 0}$
and so from [[Membership of Set of Integers is Replicative Function]]:
:$\ds \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = \sqbrk {n x \in \Z_{> 0} }$
Let $x \le 0$.
Then for all ... | Membership of Set of Strictly Positive Integers is Replicative Function | https://proofwiki.org/wiki/Membership_of_Set_of_Strictly_Positive_Integers_is_Replicative_Function | https://proofwiki.org/wiki/Membership_of_Set_of_Strictly_Positive_Integers_is_Replicative_Function | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Iverson's Convention",
"Definition:Replicative Function"
] | [
"Membership of Set of Integers is Replicative Function"
] |
proofwiki-11946 | Membership of Equivalence Class of m mod pi is Replicative Function | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \sqbrk {\exists r \in \Q, \exists m \in \Z: x = r \pi + m}$
where $\sqbrk {\cdots}$ is Iverson's convention.
Then $f$ is a replicative function. | Let $\map f x = 1$.
Then:
:$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$
and:
{{begin-eqn}}
{{eqn | l = x
| r = r \pi + m
| c =
}}
{{eqn | ll= \leadsto
| l = n x
| r = \paren {r n} \pi + m n
| c =
}}
{{end-eqn}}
But $r n \in \Q$ and $m n \in \Z$.
So:
:$\map f x = 1 \implies \map f ... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \sqbrk {\exists r \in \Q, \exists m \in \Z: x = r \pi + m}$
where $\sqbrk {\cdots}$ is [[Definition:Iverson's Convention|Iverson's convention]].
Then $f$ is a [[Definition:Replicative Function|replicative ... | Let $\map f x = 1$.
Then:
:$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$
and:
{{begin-eqn}}
{{eqn | l = x
| r = r \pi + m
| c =
}}
{{eqn | ll= \leadsto
| l = n x
| r = \paren {r n} \pi + m n
| c =
}}
{{end-eqn}}
But $r n \in \Q$ and $m n \in \Z$.
So:
:$\map f x = 1 \implies \ma... | Membership of Equivalence Class of m mod pi is Replicative Function | https://proofwiki.org/wiki/Membership_of_Equivalence_Class_of_m_mod_pi_is_Replicative_Function | https://proofwiki.org/wiki/Membership_of_Equivalence_Class_of_m_mod_pi_is_Replicative_Function | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Iverson's Convention",
"Definition:Replicative Function"
] | [
"Definition:Coefficient",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:Integer",
"Definition:Multiple/Integer",
"Definition:Replicative Function"
] |
proofwiki-11947 | Logarithm of Absolute Value of 2 times Sine of pi x is Replicative Function | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \log \, \size {2 \sin \pi x}$
Then $f$ is a replicative function. | We have that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \log \, \size {2 \sin \pi \paren {x + \frac k n} }
| c =
}}
{{eqn | r = \log \prod_{k \mathop = 0}^{n - 1} \size {2 \sin \pi \paren {x + \frac k n} }
| c = Sum of Logarithms
... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \log \, \size {2 \sin \pi x}$
Then $f$ is a [[Definition:Replicative Function|replicative function]]. | We have that:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \log \, \size {2 \sin \pi \paren {x + \frac k n} }
| c =
}}
{{eqn | r = \log \prod_{k \mathop = 0}^{n - 1} \size {2 \sin \pi \paren {x + \frac k n} }
| c = [[Sum of Logarith... | Logarithm of Absolute Value of 2 times Sine of pi x is Replicative Function | https://proofwiki.org/wiki/Logarithm_of_Absolute_Value_of_2_times_Sine_of_pi_x_is_Replicative_Function | https://proofwiki.org/wiki/Logarithm_of_Absolute_Value_of_2_times_Sine_of_pi_x_is_Replicative_Function | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Replicative Function"
] | [
"Sum of Logarithms",
"Definition:Replicative Function",
"Product Formula for Sine"
] |
proofwiki-11948 | Supremum of Ideals is Upper Adjoint implies Lattice is Continuous | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below up-complete lattice.
Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.
Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids... | We will prove that
:$\forall x \in S: \exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
Let $x \in S$.
Define $I := \map \inf {f^{-1} \sqbrk {x^\succeq} }$.
By definition of $P$:
:$I \in \map {\mathit {Ids} } L$
We will prove t... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Up-Complete|up-complete]] [[Definition:Lattice (Order Theory)|lattice]].
Let $\map {\mathit {Ids} } L$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
Let $P = ... | We will prove that
:$\forall x \in S: \exists I \in \map {\mathit {Ids} } L: x \preceq \sup I \land \forall J \in \map {\mathit {Ids} } L: x \preceq \sup J \implies I \subseteq J$
Let $x \in S$.
Define $I := \map \inf {f^{-1} \sqbrk {x^\succeq} }$.
By definition of $P$:
:$I \in \map {\mathit {Ids} } L$
We will pro... | Supremum of Ideals is Upper Adjoint implies Lattice is Continuous | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Upper_Adjoint_implies_Lattice_is_Continuous | https://proofwiki.org/wiki/Supremum_of_Ideals_is_Upper_Adjoint_implies_Lattice_is_Continuous | [
"Continuous Lattices",
"Galois Connections"
] | [
"Definition:Bounded Below Set",
"Definition:Up-Complete",
"Definition:Lattice (Order Theory)",
"Definition:Set of Sets",
"Definition:Ideal in Ordered Set",
"Definition:Ordered Set",
"Definition:Mapping",
"Definition:Galois Connection",
"Definition:Continuous Ordered Set"
] | [
"Definition:Upper Closure/Element",
"Definition:Image (Set Theory)/Relation/Subset",
"Definition:Infimum of Set",
"Definition:Galois Connection",
"Definition:Mapping",
"Definition:Galois Connection",
"Galois Connection is Expressed by Minimum",
"Definition:Smallest Element",
"Definition:Image (Set T... |
proofwiki-11949 | Sum of Replicative Functions is Replicative | Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ both be replicative functions.
Then the pointwise sum of $f$ and $g$ is also a replicative function. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map {\paren {f + g} } {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {\map f {x + \frac k n} + \map g {x + \frac k n} }
| c = {{Defof|Pointwise Sum}}
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} + \sum_{k \mathop = 0}^{... | Let $f: \R \to \R$ and $g: \R \to \R$ be [[Definition:Real Function|real functions]].
Let $f$ and $g$ both be [[Definition:Replicative Function|replicative functions]].
Then the [[Definition:Pointwise Sum|pointwise sum]] of $f$ and $g$ is also a [[Definition:Replicative Function|replicative function]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map {\paren {f + g} } {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} \paren {\map f {x + \frac k n} + \map g {x + \frac k n} }
| c = {{Defof|Pointwise Sum}}
}}
{{eqn | r = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} + \sum_{k \mathop = 0}^{... | Sum of Replicative Functions is Replicative | https://proofwiki.org/wiki/Sum_of_Replicative_Functions_is_Replicative | https://proofwiki.org/wiki/Sum_of_Replicative_Functions_is_Replicative | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Replicative Function",
"Definition:Pointwise Addition",
"Definition:Replicative Function"
] | [
"Definition:Replicative Function"
] |
proofwiki-11950 | Constant Multiple of Replicative Function is Replicative | Let $f: \R \to \R$ be a real function.
Let $f$ be a replicative function.
Let $c \in \R$ be a constant.
Let $g: \R \to \R$ be the real function defined as:
:$\map g x = c \map f x$
Then $g$ is also a replicative function. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} c \map f {x + \frac k n}
| c =
}}
{{eqn | r = c \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| c =
}}
{{eqn | r = c \map f {n x}
| c = {{Defof|Replicative Function}}
}}
{{eqn... | Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
Let $f$ be a [[Definition:Replicative Function|replicative function]].
Let $c \in \R$ be a [[Definition:Constant|constant]].
Let $g: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\map g x = c \map f x$
Then $g$ is also... | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n}
| r = \sum_{k \mathop = 0}^{n - 1} c \map f {x + \frac k n}
| c =
}}
{{eqn | r = c \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
| c =
}}
{{eqn | r = c \map f {n x}
| c = {{Defof|Replicative Function}}
}}
{{eqn... | Constant Multiple of Replicative Function is Replicative | https://proofwiki.org/wiki/Constant_Multiple_of_Replicative_Function_is_Replicative | https://proofwiki.org/wiki/Constant_Multiple_of_Replicative_Function_is_Replicative | [
"Replicative Functions"
] | [
"Definition:Real Function",
"Definition:Replicative Function",
"Definition:Constant",
"Definition:Real Function",
"Definition:Replicative Function"
] | [
"Definition:Replicative Function"
] |
proofwiki-11951 | Replicative Function of x minus Floor of x is Replicative | Let $f: \R \to \R$ be a real function.
Let $f$ be a replicative function.
Let $g: \R \to \R$ be the real function defined as:
:$\map g x = \map f {x - \floor x}$
Then $g$ is also a replicative function. | === Lemma ===
{{:Replicative Function of x minus Floor of x is Replicative/Lemma}}{{qed|lemma}}
First observe:
:$\ds \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n} = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }$
{{begin-eqn}}
{{eqn | l = \map g {n x}
| r = \map f {n x - \floo... | Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
Let $f$ be a [[Definition:Replicative Function|replicative function]].
Let $g: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\map g x = \map f {x - \floor x}$
Then $g$ is also a [[Definition:Replicative Function|replic... | === [[Replicative Function of x minus Floor of x is Replicative/Lemma|Lemma]] ===
{{:Replicative Function of x minus Floor of x is Replicative/Lemma}}{{qed|lemma}}
First observe:
:$\ds \sum_{k \mathop = 0}^{n - 1} \map g {x + \frac k n} = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n - \floor {x + \frac k n} }$
... | Replicative Function of x minus Floor of x is Replicative | https://proofwiki.org/wiki/Replicative_Function_of_x_minus_Floor_of_x_is_Replicative | https://proofwiki.org/wiki/Replicative_Function_of_x_minus_Floor_of_x_is_Replicative | [
"Replicative Functions",
"Replicative Function of x minus Floor of x is Replicative"
] | [
"Definition:Real Function",
"Definition:Replicative Function",
"Definition:Real Function",
"Definition:Replicative Function"
] | [
"Replicative Function of x minus Floor of x is Replicative/Lemma",
"Definition:Addition/Integers",
"Definition:Integer",
"Real Number minus Floor",
"Number not less than Integer iff Floor not less than Integer",
"Replicative Function of x minus Floor of x is Replicative/Lemma",
"Replicative Function of ... |
proofwiki-11952 | Closed Form for Sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... | Let $a_1, a_2, a_3, \ldots$ be the integer sequence:
:$\sequence {a_n} = 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots$
Then:
:$a_n = \ceiling {\dfrac {\sqrt {1 + 8 n} - 1} 2}$ | From Closed Form for Triangular Numbers, for $n = 1, 3, 6, 10, \ldots$:
:$n = \dfrac {a_n \paren {a_n + 1} } 2$
Thus by the Quadratic Formula:
$a_n = \dfrac {-1 \pm \sqrt {1 + 8 n} } 2$
In this context it is the positive root that is required.
The result follows by definition of ceiling function.
{{qed}} | Let $a_1, a_2, a_3, \ldots$ be the [[Definition:Integer Sequence|integer sequence]]:
:$\sequence {a_n} = 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots$
Then:
:$a_n = \ceiling {\dfrac {\sqrt {1 + 8 n} - 1} 2}$ | From [[Closed Form for Triangular Numbers]], for $n = 1, 3, 6, 10, \ldots$:
:$n = \dfrac {a_n \paren {a_n + 1} } 2$
Thus by the [[Quadratic Formula]]:
$a_n = \dfrac {-1 \pm \sqrt {1 + 8 n} } 2$
In this context it is the [[Definition:Positive Real Number|positive]] root that is required.
The result follows by definit... | Closed Form for Sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... | https://proofwiki.org/wiki/Closed_Form_for_Sequence_1,_2,_2,_3,_3,_3,_4,_4,_4,_4,_... | https://proofwiki.org/wiki/Closed_Form_for_Sequence_1,_2,_2,_3,_3,_3,_4,_4,_4,_4,_... | [
"Sequences",
"Ceiling Function",
"Closed Forms"
] | [
"Definition:Integer Sequence"
] | [
"Closed Form for Triangular Numbers",
"Solution to Quadratic Equation",
"Definition:Positive/Real Number",
"Definition:Ceiling Function"
] |
proofwiki-11953 | Sum of Sequence as Summation of Difference of Adjacent Terms | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n a_k
| r = \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{... | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$ | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n a_k
| r = \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{... | Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 1 | https://proofwiki.org/wiki/Sum_of_Sequence_as_Summation_of_Difference_of_Adjacent_Terms | https://proofwiki.org/wiki/Sum_of_Sequence_as_Summation_of_Difference_of_Adjacent_Terms/Proof_1 | [
"Summations",
"Sum of Sequence as Summation of Difference of Adjacent Terms"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Translation of Index Variable of Summation"
] |
proofwiki-11954 | Sum of Sequence as Summation of Difference of Adjacent Terms | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$ | From Abel's Lemma: Formulation 2, after renaming and reassigning variables:
:$\ds \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} \map {A_k} {a_k - a_{k + 1} } + A_n a_n$
where:
:$\sequence a$ and $\sequence b$ are sequences in $\R$
:$\ds A_n = \sum_{k \mathop = 1}^n {b_k}$ be the partial sum of $\sequenc... | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$ | From [[Abel's Lemma: Formulation 2]], after renaming and reassigning variables:
:$\ds \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} \map {A_k} {a_k - a_{k + 1} } + A_n a_n$
where:
:$\sequence a$ and $\sequence b$ are [[Definition:Real Sequence|sequences in $\R$]]
:$\ds A_n = \sum_{k \mathop = 1}^n {b_... | Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 2 | https://proofwiki.org/wiki/Sum_of_Sequence_as_Summation_of_Difference_of_Adjacent_Terms | https://proofwiki.org/wiki/Sum_of_Sequence_as_Summation_of_Difference_of_Adjacent_Terms/Proof_2 | [
"Summations",
"Sum of Sequence as Summation of Difference of Adjacent Terms"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Abel's Lemma/Formulation 2",
"Definition:Real Sequence",
"Definition:Series/Sequence of Partial Sums"
] |
proofwiki-11955 | Continuous Lattice and Way Below implies Preceding implies Preceding | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a continuous complete lattice.
Let $a, b \in S$.
Let:
:$\forall c \in S: c \ll a \implies c \preceq b$
Then $a \preceq b$ | By definition of way below closure:
:$\forall c \in a^\ll: c \preceq b$
By definition:
:$b$ is an upper bound for $a^\ll$
By definition of supremum:
:$\map \sup {a^\ll} \preceq b$
By definition of continuous:
:$L$ satisfies the axiom of approximation.
Thus by the axiom of approximation:
:$a = \map \sup {a^\ll} \preceq ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Continuous Ordered Set|continuous]] [[Definition:Complete Lattice|complete lattice]].
Let $a, b \in S$.
Let:
:$\forall c \in S: c \ll a \implies c \preceq b$
Then $a \preceq b$ | By definition of [[Definition:Way Below Closure|way below closure]]:
:$\forall c \in a^\ll: c \preceq b$
By definition:
:$b$ is an [[Definition:Upper Bound of Set|upper bound]] for $a^\ll$
By definition of [[Definition:Supremum of Set|supremum]]:
:$\map \sup {a^\ll} \preceq b$
By definition of [[Definition:Continuou... | Continuous Lattice and Way Below implies Preceding implies Preceding | https://proofwiki.org/wiki/Continuous_Lattice_and_Way_Below_implies_Preceding_implies_Preceding | https://proofwiki.org/wiki/Continuous_Lattice_and_Way_Below_implies_Preceding_implies_Preceding | [
"Continuous Lattices"
] | [
"Definition:Continuous Ordered Set",
"Definition:Complete Lattice"
] | [
"Definition:Way Below Closure",
"Definition:Upper Bound of Set",
"Definition:Supremum of Set",
"Definition:Continuous Ordered Set",
"Axiom:Axiom of Approximation",
"Axiom:Axiom of Approximation"
] |
proofwiki-11956 | Sum over k of Floor of Log base b of k | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\log_b k} = \paren {n + 1} \floor {\log_b n} - \dfrac {b^{\floor {\log_b n} + 1} - b} {b - 1}$ | From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \floor {\log_b k} = n \floor {\log_b n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\flo... | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\log_b k} = \paren {n + 1} \floor {\log_b n} - \dfrac {b^{\floor {\log_b n} + 1} - b} {b - 1}$ | From [[Sum of Sequence as Summation of Difference of Adjacent Terms]]:
:$(1): \quad \ds \sum_{k \mathop = 1}^n \floor {\log_b k} = n \floor {\log_b n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\map {\log_b} {k + 1} } - \floor {\log_b k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren... | Sum over k of Floor of Log base b of k | https://proofwiki.org/wiki/Sum_over_k_of_Floor_of_Log_base_b_of_k | https://proofwiki.org/wiki/Sum_over_k_of_Floor_of_Log_base_b_of_k | [
"Summations",
"Floor Function",
"Logarithms"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Sum of Sequence as Summation of Difference of Adjacent Terms",
"Definition:Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Iverson's Convention",
"Sum of Geometric Sequence"
] |
proofwiki-11957 | Sum over k of Floor of Root k | Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = \floor {\sqrt n} \paren {n - \dfrac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6}$ | From Sum of Sequence as Summation of Difference of Adjacent Terms:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = n \floor {\sqrt n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - ... | Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
Let $b \in \Z$ such that $b \ge 2$.
Then:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = \floor {\sqrt n} \paren {n - \dfrac {\paren {2 \floor {\sqrt n} + 5} \paren {\floor {\sqrt n} - 1} } 6}$ | From [[Sum of Sequence as Summation of Difference of Adjacent Terms]]:
:$\ds \sum_{k \mathop = 1}^n \floor {\sqrt k} = n \floor {\sqrt n} - \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1} } - \floor {\sqrt k} }$
Let $S$ be defined as:
:$\ds S := \sum_{k \mathop = 1}^{n - 1} k \paren {\floor {\sqrt {k + 1... | Sum over k of Floor of Root k | https://proofwiki.org/wiki/Sum_over_k_of_Floor_of_Root_k | https://proofwiki.org/wiki/Sum_over_k_of_Floor_of_Root_k | [
"Summations",
"Floor Function"
] | [
"Definition:Strictly Positive/Integer"
] | [
"Sum of Sequence as Summation of Difference of Adjacent Terms",
"Definition:Square Number",
"Definition:Summation",
"Sum of Sequence of Squares"
] |
proofwiki-11958 | Top is Meet Irreducible | Let $\left({S, \wedge, \preceq}\right)$ be a bounded above meet semilattice.
Then $\top$ is meet irreducible
where $\top$ denotes the greatest element in $S$. | Let $x, y \in S$ such that
:$\top = x \wedge y$
By Meet Precedes Operands
:$\top \preceq x$ and $\top \preceq y$
By definition of greatest element:
:$x \preceq \top$
Thus by definition of antisymmetry:
:$\top = x$
{{qed}} | Let $\left({S, \wedge, \preceq}\right)$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Meet Semilattice|meet semilattice]].
Then $\top$ is [[Definition:Meet Irreducible Element|meet irreducible]]
where $\top$ denotes the [[Definition:Greatest Element|greatest element]] in $S$. | Let $x, y \in S$ such that
:$\top = x \wedge y$
By [[Meet Precedes Operands]]
:$\top \preceq x$ and $\top \preceq y$
By definition of [[Definition:Greatest Element|greatest element]]:
:$x \preceq \top$
Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
:$\top = x$
{{qed}} | Top is Meet Irreducible | https://proofwiki.org/wiki/Top_is_Meet_Irreducible | https://proofwiki.org/wiki/Top_is_Meet_Irreducible | [
"Order Theory",
"Meet Irreducible Elements"
] | [
"Definition:Bounded Above Set",
"Definition:Meet Semilattice",
"Definition:Meet Irreducible Element",
"Definition:Greatest Element"
] | [
"Meet Precedes Operands",
"Definition:Greatest Element",
"Definition:Antisymmetric Relation"
] |
proofwiki-11959 | Meet Irreducible iff Finite Infimum equals Element | Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.
Let $x \in S$.
Then
:$x$ is meet irreducible
{{iff}}
:for every non-empty finite subset $A$ of $S$: $x = \inf A \implies x \in A$ | === Sufficient Condition ===
Let $x$ be meet irreducible.
We will prove the result by induction on cardinality of $A$. | Let $L = \left({S, \wedge, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
Let $x \in S$.
Then
:$x$ is [[Definition:Meet Irreducible Element|meet irreducible]]
{{iff}}
:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S$: $x... | === Sufficient Condition ===
Let $x$ be [[Definition:Meet Irreducible Element|meet irreducible]].
We will prove the result by [[Principle of Mathematical Induction|induction]] on [[Definition:Cardinality|cardinality]] of $A$. | Meet Irreducible iff Finite Infimum equals Element | https://proofwiki.org/wiki/Meet_Irreducible_iff_Finite_Infimum_equals_Element | https://proofwiki.org/wiki/Meet_Irreducible_iff_Finite_Infimum_equals_Element | [
"Order Theory",
"Meet Irreducible Elements"
] | [
"Definition:Meet Semilattice",
"Definition:Meet Irreducible Element",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset"
] | [
"Definition:Meet Irreducible Element",
"Principle of Mathematical Induction",
"Definition:Cardinality",
"Definition:Meet Irreducible Element"
] |
proofwiki-11960 | Sum over k of Sum over j of Floor of n + jb^k over b^k+1 | Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.
Then:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = n$
where $\floor {\, \cdot \,}$ denotes the floor function. | We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }
| r = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b}
| c =
}}
{{eq... | Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.
Then:
:$\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = n$
where $\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]]. | We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:
{{begin-eqn}}
{{eqn | l = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }
| r = \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b}
| c =
}}
{{e... | Sum over k of Sum over j of Floor of n + jb^k over b^k+1 | https://proofwiki.org/wiki/Sum_over_k_of_Sum_over_j_of_Floor_of_n_+_jb^k_over_b^k+1 | https://proofwiki.org/wiki/Sum_over_k_of_Sum_over_j_of_Floor_of_n_+_jb^k_over_b^k+1 | [
"Summations",
"Floor Function"
] | [
"Definition:Floor Function"
] | [
"Summation over k of Floor of mk+x over n"
] |
proofwiki-11961 | Sum over j of Function of Floor of mj over n | Let $f$ be a real function.
Then:
:$\ds \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + n \map f {m - 1}$ | {{begin-eqn}}
{{eqn | l = r
| m = \floor {\dfrac {m j} n}
| c =
}}
{{eqn | ll= \leadsto
| l = r
| o = \le
| m = \dfrac {m j} n
| mo= <
| r = r + 1
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {r n} m
| o = \le
| m = j
| mo= <
| r = \dfrac {\p... | Let $f$ be a [[Definition:Real Function|real function]].
Then:
:$\ds \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + n \map f {m - 1}$ | {{begin-eqn}}
{{eqn | l = r
| m = \floor {\dfrac {m j} n}
| c =
}}
{{eqn | ll= \leadsto
| l = r
| o = \le
| m = \dfrac {m j} n
| mo= <
| r = r + 1
| c =
}}
{{eqn | ll= \leadsto
| l = \dfrac {r n} m
| o = \le
| m = j
| mo= <
| r = \dfrac {\p... | Sum over j of Function of Floor of mj over n | https://proofwiki.org/wiki/Sum_over_j_of_Function_of_Floor_of_mj_over_n | https://proofwiki.org/wiki/Sum_over_j_of_Function_of_Floor_of_mj_over_n | [
"Summations",
"Floor Function"
] | [
"Definition:Real Function"
] | [
"Definition:Integer"
] |
proofwiki-11962 | Equivalence of Definitions of Legendre Symbol | Let $p$ be an odd prime.
Let $a \in \Z$.
{{TFAE|def = Legendre Symbol}} | {{ProofWanted}}
Category:Legendre Symbol
0mcyc6lpkl926gdog8ux9xww3otz0x4 | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Let $a \in \Z$.
{{TFAE|def = Legendre Symbol}} | {{ProofWanted}}
[[Category:Legendre Symbol]]
0mcyc6lpkl926gdog8ux9xww3otz0x4 | Equivalence of Definitions of Legendre Symbol | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Legendre_Symbol | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Legendre_Symbol | [
"Legendre Symbol"
] | [
"Definition:Odd Prime"
] | [
"Category:Legendre Symbol"
] |
proofwiki-11963 | Upper Closure of Element without Element is Filter implies Element is Meet Irreducible | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $x \in S$.
Let
:$x^\succeq \setminus \set x$ be a filter in $L$.
Then $x$ is meet irreducible. | Let $a, b \in S$.
{{AimForCont}}:
:$x = a \wedge b$ and $x \ne a$ and $x \ne b$
By Meet Precedes Operands:
:$x \preceq b$ and $x \preceq a$
By definition of upper closure of element:
:$b, a \in x^\succeq$
By definitions of singleton and difference:
:$b, a \in x^\succeq \setminus \set x$
By definition of filtered:
:$\ex... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $x \in S$.
Let
:$x^\succeq \setminus \set x$ be a [[Definition:Filter in Ordered Set|filter]] in $L$.
Then $x$ is [[Definition:Meet Irreducible Element|meet irreducible]]. | Let $a, b \in S$.
{{AimForCont}}:
:$x = a \wedge b$ and $x \ne a$ and $x \ne b$
By [[Meet Precedes Operands]]:
:$x \preceq b$ and $x \preceq a$
By definition of [[Definition:Upper Closure of Element|upper closure of element]]:
:$b, a \in x^\succeq$
By definitions of [[Definition:Singleton|singleton]] and [[Definiti... | Upper Closure of Element without Element is Filter implies Element is Meet Irreducible | https://proofwiki.org/wiki/Upper_Closure_of_Element_without_Element_is_Filter_implies_Element_is_Meet_Irreducible | https://proofwiki.org/wiki/Upper_Closure_of_Element_without_Element_is_Filter_implies_Element_is_Meet_Irreducible | [
"Order Theory",
"Meet Irreducible Elements"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Filter in Ordered Set",
"Definition:Meet Irreducible Element"
] | [
"Meet Precedes Operands",
"Definition:Upper Closure/Element",
"Definition:Singleton",
"Definition:Set Difference",
"Definition:Filtered Subset",
"Definition:Infimum of Set",
"Definition:Upper Section",
"Definition:Singleton"
] |
proofwiki-11964 | Maximal Element of Complement of Filter is Meet Irreducible | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a lattice.
Let $F$ be a filter in $L$.
Let $p \in S$.
Let $p = \map \max {\relcomp S F}$.
Then $p$ is meet irreducible. | Let $x, y \in S$.
{{AimForCont}}
:$p = x \wedge y$ and $p \ne x$ and $p \ne y$
By Meet Precedes Operands:
:$p \preceq x$ and $p \preceq y$
By definition of $\prec$:
:$p \prec x$ and $p \prec y$
We will prove that
:$x \notin F$ or $y \notin F$
{{AimForCont}}
:$x \in F \land y \in F$
By definition of filtered:
:$\exists ... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice (Order Theory)|lattice]].
Let $F$ be a [[Definition:Filter in Ordered Set|filter]] in $L$.
Let $p \in S$.
Let $p = \map \max {\relcomp S F}$.
Then $p$ is [[Definition:Meet Irreducible Element|meet irreducible]]. | Let $x, y \in S$.
{{AimForCont}}
:$p = x \wedge y$ and $p \ne x$ and $p \ne y$
By [[Meet Precedes Operands]]:
:$p \preceq x$ and $p \preceq y$
By definition of $\prec$:
:$p \prec x$ and $p \prec y$
We will prove that
:$x \notin F$ or $y \notin F$
{{AimForCont}}
:$x \in F \land y \in F$
By definition of [[Definiti... | Maximal Element of Complement of Filter is Meet Irreducible | https://proofwiki.org/wiki/Maximal_Element_of_Complement_of_Filter_is_Meet_Irreducible | https://proofwiki.org/wiki/Maximal_Element_of_Complement_of_Filter_is_Meet_Irreducible | [
"Order Theory",
"Meet Irreducible Elements"
] | [
"Definition:Lattice (Order Theory)",
"Definition:Filter in Ordered Set",
"Definition:Meet Irreducible Element"
] | [
"Meet Precedes Operands",
"Definition:Filtered Subset",
"Definition:Infimum of Set",
"Definition:Upper Section",
"Definition:Greatest Element",
"Definition:Relative Complement",
"Definition:Greatest Element"
] |
proofwiki-11965 | De Polignac's Formula/Technique | When calculating $\mu$, the easiest way to calculate the next term is simply to divide the previous term by $p$ and discard the remainder:
:$\floor {\dfrac n {p^{k + 1} } } = \floor {\floor {\dfrac n {p^k} } / p}$ | From Floor of $\dfrac {x + m} n$: Corollary:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
which is valid for all integers $m, n$ such that $n > 0$.
In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$.
{{qed}} | When calculating $\mu$, the easiest way to calculate the next term is simply to divide the previous term by $p$ and discard the remainder:
:$\floor {\dfrac n {p^{k + 1} } } = \floor {\floor {\dfrac n {p^k} } / p}$ | From [[Floor of x+m over n/Corollary|Floor of $\dfrac {x + m} n$: Corollary]]:
:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
which is valid for all [[Definition:Integer|integers]] $m, n$ such that $n > 0$.
In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$.
{{qed}} | De Polignac's Formula/Technique | https://proofwiki.org/wiki/De_Polignac's_Formula/Technique | https://proofwiki.org/wiki/De_Polignac's_Formula/Technique | [
"De Polignac's Formula"
] | [] | [
"Floor of x+m over n/Corollary",
"Definition:Integer"
] |
proofwiki-11966 | Not Preceding implies There Exists Meet Irreducible Element Not Preceding | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Let $x, y \in S$ such that
:$y \npreceq x$
Then
:$\exists p \in S: p$ is meet irreducible and $x \preceq p$ and $y \npreceq p$ | By definition of continuous:
:$L$ satisfies the axiom of approximation
and
:$\forall x \in S: x^\ll$ is directed.
By Axiom of Approximation in Up-Complete Semilattice
:$\exists u \in S: u \ll y \land u \npreceq x$
By Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure:
:there exists way bel... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Let $x, y \in S$ such that
:$y \npreceq x$
Then
:$\exists p \in S: p$ is [[Definition:Meet Irreducible Element|meet irreduci... | By definition of [[Definition:Continuous Ordered Set|continuous]]:
:$L$ satisfies the [[Axiom:Axiom of Approximation|axiom of approximation]]
and
:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]].
By [[Axiom of Approximation in Up-Complete Semilattice]]
:$\exists u \in S: u \ll y \land u \npreceq x$... | Not Preceding implies There Exists Meet Irreducible Element Not Preceding | https://proofwiki.org/wiki/Not_Preceding_implies_There_Exists_Meet_Irreducible_Element_Not_Preceding | https://proofwiki.org/wiki/Not_Preceding_implies_There_Exists_Meet_Irreducible_Element_Not_Preceding | [
"Order Theory",
"Meet Irreducible Elements"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)",
"Definition:Meet Irreducible Element"
] | [
"Definition:Continuous Ordered Set",
"Axiom:Axiom of Approximation",
"Definition:Directed Subset",
"Axiom of Approximation in Up-Complete Semilattice",
"Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure",
"Definition:Way Below Open",
"Definition:Filter in Ordered Set",
"... |
proofwiki-11967 | Termial on Real Numbers is Extension of Integers | The termial function as defined on the real numbers is an extension of its definition on the integers $\Z$. | From the definition of the termial function on the integers:
:$\ds n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$
From Closed Form for Triangular Numbers, we have that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$
This agrees with the definition of the termial function on ... | The [[Definition:Termial|termial function]] as [[Definition:Termial/Real Numbers|defined]] on the [[Definition:Real Number|real numbers]] is an [[Definition:Extension of Mapping|extension]] of its definition on the [[Definition:Integer|integers]] $\Z$. | From the definition of the [[Definition:Termial|termial function]] on the [[Definition:Integer|integers]]:
:$\ds n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$
From [[Closed Form for Triangular Numbers]], we have that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$
This [[... | Termial on Real Numbers is Extension of Integers | https://proofwiki.org/wiki/Termial_on_Real_Numbers_is_Extension_of_Integers | https://proofwiki.org/wiki/Termial_on_Real_Numbers_is_Extension_of_Integers | [
"Termial Function"
] | [
"Definition:Termial",
"Definition:Termial/Real Numbers",
"Definition:Real Number",
"Definition:Extension of Mapping",
"Definition:Integer"
] | [
"Definition:Termial",
"Definition:Integer",
"Closed Form for Triangular Numbers",
"Definition:Agreement/Mappings",
"Definition:Termial/Real Numbers",
"Definition:Real Number",
"Definition:Extension of Mapping"
] |
proofwiki-11968 | Floor of m+n-1 over n | Let $m, n \in \Z$ such that $n > 0$.
Then:
:$\floor {\dfrac {m + n - 1} n} = \ceiling {\dfrac m n}$
The identity does not necessarily apply for $n < 0$. | First let $n > 0$ as stated.
Suppose $n \divides m$.
Then $m = k n$ for some $k \in \Z$.
It follows that:
:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 - \dfrac 1 n} = k$
and:
:$\ceiling {\dfrac m n} = k$
Now suppose $n \nmid m$.
Since $n > 0$, we have $m = k n + r$ for some $k \in\Z$ and $r \in \N$, $0 < r < n$.
Th... | Let $m, n \in \Z$ such that $n > 0$.
Then:
:$\floor {\dfrac {m + n - 1} n} = \ceiling {\dfrac m n}$
The identity does not necessarily apply for $n < 0$. | First let $n > 0$ as stated.
Suppose $n \divides m$.
Then $m = k n$ for some $k \in \Z$.
It follows that:
:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 - \dfrac 1 n} = k$
and:
:$\ceiling {\dfrac m n} = k$
Now suppose $n \nmid m$.
Since $n > 0$, we have $m = k n + r$ for some $k \in\Z$ and $r \in \N$, $0 < r... | Floor of m+n-1 over n | https://proofwiki.org/wiki/Floor_of_m+n-1_over_n | https://proofwiki.org/wiki/Floor_of_m+n-1_over_n | [
"Floor Function"
] | [] | [] |
proofwiki-11969 | Characteristics of Floor and Ceiling Function | Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then either:
:$\forall x \in \Q: \map f x = \floor x$
or:
:$\forall x \in \Q: \map f x = \ceiling x$ | From $(1)$, by induction we have:
:$\forall n \in \N: \map f {x + n} = \map f x + n$
and
:$\forall n \in\N: \map f {x - n} = \map f x - n$
and therefore, in particular:
:$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$
From $(2)$, we get
{{begin-eqn}}
{{eqn | l = \map f 0
| r = \map f {\map f 0}
}}
{{eqn | ... | Let $f: \R \to \Z$ be an [[Definition:Integer-Valued Function|integer-valued function]] which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then either:
:$\forall x \in \Q: \map f x = \floor x$
or:
:$\for... | From $(1)$, by [[Principle of Mathematical Induction|induction]] we have:
:$\forall n \in \N: \map f {x + n} = \map f x + n$
and
:$\forall n \in\N: \map f {x - n} = \map f x - n$
and therefore, in particular:
:$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$
From $(2)$, we get
{{begin-eqn}}
{{eqn | l = \map f... | Characteristics of Floor and Ceiling Function | https://proofwiki.org/wiki/Characteristics_of_Floor_and_Ceiling_Function | https://proofwiki.org/wiki/Characteristics_of_Floor_and_Ceiling_Function | [
"Floor Function",
"Ceiling Function"
] | [
"Definition:Integer-Valued Function"
] | [
"Principle of Mathematical Induction",
"Definition:Integer-Valued Function",
"Principle of Mathematical Induction",
"Definition:Rational Number",
"Definition:Integer-Valued Function",
"Floor of Negative equals Negative of Ceiling",
"Definition:Rational Number"
] |
proofwiki-11970 | Characteristics of Floor and Ceiling Function/Real Domain | Let $f: \R \to \Z$ be an integer-valued function which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then it is not necessarily the case that either:
:$\forall x \in \R: \map f x = \floor x$
or:
:$\forall x... | Let $h: \R \to \R$ be a real function such that for all $x, y \in \R$:
{{begin-eqn}}
{{eqn | n = 3
| l = \map h 1
| r = 1
| c =
}}
{{eqn | n = 4
| l = \map h x + \map h y
| r = \map h {x + y}
| c =
}}
{{end-eqn}}
Consider the integer-valued function $f: \R \to \Z$ defined as:
:$\ma... | Let $f: \R \to \Z$ be an [[Definition:Integer-Valued Function|integer-valued function]] which satisfies both of the following:
:$(1): \quad \map f {x + 1} = \map f x + 1$
:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
Then it is not necessarily the case that either:
:$\forall x \in... | Let $h: \R \to \R$ be a [[Definition:Real Function|real function]] such that for all $x, y \in \R$:
{{begin-eqn}}
{{eqn | n = 3
| l = \map h 1
| r = 1
| c =
}}
{{eqn | n = 4
| l = \map h x + \map h y
| r = \map h {x + y}
| c =
}}
{{end-eqn}}
Consider the [[Definition:Integer-Val... | Characteristics of Floor and Ceiling Function/Real Domain | https://proofwiki.org/wiki/Characteristics_of_Floor_and_Ceiling_Function/Real_Domain | https://proofwiki.org/wiki/Characteristics_of_Floor_and_Ceiling_Function/Real_Domain | [
"Floor Function",
"Ceiling Function"
] | [
"Definition:Integer-Valued Function"
] | [
"Definition:Real Function",
"Definition:Integer-Valued Function",
"Definition:Additive Function",
"Additive Function is Linear for Rational Factors",
"Additive Function is Linear for Rational Factors",
"Rational Numbers form Subfield of Real Numbers",
"Vector Space on Field Extension is Vector Space",
... |
proofwiki-11971 | Continuous Replicative Function | Let $f: \R \to \R$ be a real function.
Let $f$ be continuous on $\R$.
Let $f$ also be a replicative function.
Then $f$ is of the form:
:$\map f x = \paren {x - \dfrac 1 2} a$
where $a \in \R$. | Let $f$ be a replicative function.
Then:
:$\forall n > 0: \map f {n x + 1} - \map f {n x} = \map f {x + 1} - \map f x$
If $f$ is then also continuous:
:$\forall x \in \R: \map f {x + 1} - \map f x$
and so:
:$\map g x = \map f x - c \floor x$
is both replicative and periodic.
We have:
:$\ds \int_0^1 e^{2 \pi i n x} \map... | Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
Let $f$ be [[Definition:Continuous Real Function|continuous]] on $\R$.
Let $f$ also be a [[Definition:Replicative Function|replicative function]].
Then $f$ is of the form:
:$\map f x = \paren {x - \dfrac 1 2} a$
where $a \in \R$. | Let $f$ be a [[Definition:Replicative Function|replicative function]].
Then:
:$\forall n > 0: \map f {n x + 1} - \map f {n x} = \map f {x + 1} - \map f x$
If $f$ is then also [[Definition:Continuous Real Function|continuous]]:
:$\forall x \in \R: \map f {x + 1} - \map f x$
and so:
:$\map g x = \map f x - c \floor x$... | Continuous Replicative Function | https://proofwiki.org/wiki/Continuous_Replicative_Function | https://proofwiki.org/wiki/Continuous_Replicative_Function | [
"Replicative Functions",
"Continuous Functions"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Replicative Function"
] | [
"Definition:Replicative Function",
"Definition:Continuous Real Function",
"Definition:Replicative Function",
"Definition:Periodic Function/Real",
"Definition:Fourier Series"
] |
proofwiki-11972 | Sum over k to p over 2 of Floor of 2kq over p | Let $p \in \Z$ be an odd prime.
Let $q \in \Z$ be an odd integer and $p \nmid q$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {2 k q} p} \equiv \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {k q} p} \pmod 2$ | When $k < \dfrac p 4$ we have:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac {\paren {p - 1 - 2 k} q} p}
| r = \floor {q - \dfrac {\paren {2 k + 1} q} p}
}}
{{eqn | r = q + \floor {-\dfrac {\paren {2 k + 1} q} p}
| c = Floor of Number plus Integer
}}
{{eqn | r = q - \ceiling {\dfrac {\paren {2 k + 1} q} p}
... | Let $p \in \Z$ be an [[Definition:Odd Prime|odd prime]].
Let $q \in \Z$ be an [[Definition:Odd Integer|odd integer]] and $p \nmid q$.
Then:
:$\ds \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {2 k q} p} \equiv \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {k q} p} \pmod 2$ | When $k < \dfrac p 4$ we have:
{{begin-eqn}}
{{eqn | l = \floor {\dfrac {\paren {p - 1 - 2 k} q} p}
| r = \floor {q - \dfrac {\paren {2 k + 1} q} p}
}}
{{eqn | r = q + \floor {-\dfrac {\paren {2 k + 1} q} p}
| c = [[Floor of Number plus Integer]]
}}
{{eqn | r = q - \ceiling {\dfrac {\paren {2 k + 1} q} p}
... | Sum over k to p over 2 of Floor of 2kq over p | https://proofwiki.org/wiki/Sum_over_k_to_p_over_2_of_Floor_of_2kq_over_p | https://proofwiki.org/wiki/Sum_over_k_to_p_over_2_of_Floor_of_2kq_over_p | [
"Prime Numbers"
] | [
"Definition:Odd Prime",
"Definition:Odd Integer"
] | [
"Floor of Number plus Integer",
"Floor of Negative equals Negative of Ceiling",
"Floor equals Ceiling iff Integer",
"Definition:Integer"
] |
proofwiki-11973 | Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous lattice.
Let $x, y \in S$ such that
:$x \ll y$
where $\ll$ denotes the way below relation.
Then there exists a way below open filter in $L$: $y \in F \land F \subseteq x^\gg$
where $x^\gg$ denotes the way above closure of $x$. | We will prove that
:$x^\gg$ is way below open.
Let $z \in x^\gg$.
By definition of way above closure:
:$x \ll z$
By Way Below has Interpolation Property:
:$\exists x' \in S: x \ll x' \land x' \ll z$
Thus by definition of way above closure:
:$x' \in x^\gg$
Thus
:$x' \ll z$
{{qed|lemma}}
Then:
:$\forall z \in x^\gg: \exi... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice (Order Theory)|lattice]].
Let $x, y \in S$ such that
:$x \ll y$
where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
Then t... | We will prove that
:$x^\gg$ is [[Definition:Way Below Open|way below open]].
Let $z \in x^\gg$.
By definition of [[Definition:Way Above Closure|way above closure]]:
:$x \ll z$
By [[Way Below has Interpolation Property]]:
:$\exists x' \in S: x \ll x' \land x' \ll z$
Thus by definition of [[Definition:Way Above Closu... | Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure | https://proofwiki.org/wiki/Way_Below_implies_There_Exists_Way_Below_Open_Filter_Subset_of_Way_Above_Closure | https://proofwiki.org/wiki/Way_Below_implies_There_Exists_Way_Below_Open_Filter_Subset_of_Way_Above_Closure | [
"Continuous Lattices"
] | [
"Definition:Bounded Below Set",
"Definition:Continuous Ordered Set",
"Definition:Lattice (Order Theory)",
"Definition:Element is Way Below",
"Definition:Way Below Open",
"Definition:Filter in Ordered Set",
"Definition:Way Above Closure"
] | [
"Definition:Way Below Open",
"Definition:Way Above Closure",
"Way Below has Interpolation Property",
"Definition:Way Above Closure",
"Axiom:Axiom of Choice",
"Definition:Mapping",
"Definition:Way Above Closure",
"Definition:Reflexivity",
"Way Below implies Preceding",
"Definition:Transitive",
"P... |
proofwiki-11974 | Powers of Group Elements/Sum of Indices/Additive Notation | :$\forall m, n \in \Z: \paren {m \cdot g} + \paren {n \cdot g} = \paren {m + n} \cdot g$ | All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
where in this context the group operation is $+$ and $n$th power of $g$ is denoted $n g$.
{{qed}} | :$\forall m, n \in \Z: \paren {m \cdot g} + \paren {n \cdot g} = \paren {m + n} \cdot g$ | All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Sum of Indices|Index Laws for Monoids: Sum of Indices]]:
:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
where in this context the [[Definition:Group Operati... | Powers of Group Elements/Sum of Indices/Additive Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Sum_of_Indices/Additive_Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Sum_of_Indices/Additive_Notation | [
"Group Theory",
"Index Laws"
] | [] | [
"Definition:Group",
"Definition:Invertible Element",
"Index Laws for Monoids/Sum of Indices",
"Definition:Group Product/Group Law",
"Definition:Power of Element/Group"
] |
proofwiki-11975 | Powers of Group Elements/Negative Index/Additive Notation | :$\forall n \in \Z: -\paren {n g} = \paren {-n} g = n \paren {-g}$ | All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Sum of Indices:
:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
where in this context:
:the group operation is $+$
:the $n$th power of $g$ is denoted $n g$
:the inverse of $g$ is $-g$.
{{qed}} | :$\forall n \in \Z: -\paren {n g} = \paren {-n} g = n \paren {-g}$ | All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Sum of Indices|Index Laws for Monoids: Sum of Indices]]:
:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
where in this context:
:the [[Def... | Powers of Group Elements/Negative Index/Additive Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Negative_Index/Additive_Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Negative_Index/Additive_Notation | [
"Group Theory",
"Index Laws"
] | [] | [
"Definition:Group",
"Definition:Invertible Element",
"Index Laws for Monoids/Sum of Indices",
"Definition:Group Product/Group Law",
"Definition:Power of Element/Group",
"Definition:Inverse (Abstract Algebra)/Inverse"
] |
proofwiki-11976 | Powers of Group Elements/Product of Indices/Additive Notation | :$\forall m, n \in \Z: n \cdot \paren {m \cdot g} = \paren {m \times n} \cdot g = m \cdot \paren {n \cdot g}$ | All elements of a group are invertible, so we can directly use the result from Index Laws for Monoids: Product of Indices:
:$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$
where in this context:
:the group operation is $+$
:the $n$th power of $g$ is denoted $n \cdot g$
{{qed}} | :$\forall m, n \in \Z: n \cdot \paren {m \cdot g} = \paren {m \times n} \cdot g = m \cdot \paren {n \cdot g}$ | All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Product of Indices|Index Laws for Monoids: Product of Indices]]:
:$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$
where in this context:
:the ... | Powers of Group Elements/Product of Indices/Additive Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Product_of_Indices/Additive_Notation | https://proofwiki.org/wiki/Powers_of_Group_Elements/Product_of_Indices/Additive_Notation | [
"Group Theory",
"Index Laws"
] | [] | [
"Definition:Group",
"Definition:Invertible Element",
"Index Laws for Monoids/Product of Indices",
"Definition:Group Product/Group Law",
"Definition:Power of Element/Group"
] |
proofwiki-11977 | Power of Product in Abelian Group/Additive Notation | :$\forall x, y \in G: \forall k \in \Z: k \cdot \paren {x + y} = \paren {k \cdot x} + \paren {k \cdot y}$ | By definition of abelian group, $x$ and $y$ commute.
That is:
:$x + y = y + x$
The result follows from Power of Product of Commutative Elements in Group.
{{qed}} | :$\forall x, y \in G: \forall k \in \Z: k \cdot \paren {x + y} = \paren {k \cdot x} + \paren {k \cdot y}$ | By definition of [[Definition:Abelian Group|abelian group]], $x$ and $y$ [[Definition:Commute|commute]].
That is:
:$x + y = y + x$
The result follows from [[Power of Product of Commutative Elements in Group]].
{{qed}} | Power of Product in Abelian Group/Additive Notation | https://proofwiki.org/wiki/Power_of_Product_in_Abelian_Group/Additive_Notation | https://proofwiki.org/wiki/Power_of_Product_in_Abelian_Group/Additive_Notation | [
"Abelian Groups"
] | [] | [
"Definition:Abelian Group",
"Definition:Commutative/Elements",
"Power of Product of Commutative Elements in Group"
] |
proofwiki-11978 | Power of Idempotent Element | Let $\struct {S, \circ}$ be an algebraic structure.
Let $s \in S$ be an idempotent element with respect to $\circ$.
Then:
:$\forall n \in \Z_{> 0}: s^n = s$
where $s^n$ is defined as:
:$s^n = \begin{cases} s & : n = 1 \\
s^{n - 1} \circ s & : n > 1 \end{cases}$ | The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$s^n = s$
$\map P 1$ is the case:
:$s^1 = s$
which holds by definition.
Thus $\map P 1$ is seen to hold. | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $s \in S$ be an [[Definition:Idempotent Element|idempotent element]] with respect to $\circ$.
Then:
:$\forall n \in \Z_{> 0}: s^n = s$
where $s^n$ is defined as:
:$s^n = \begin{cases} s & : n = 1 \\
s^{n - ... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$s^n = s$
$\map P 1$ is the case:
:$s^1 = s$
which holds by definition.
Thus $\map P 1$ is seen to hold. | Power of Idempotent Element | https://proofwiki.org/wiki/Power_of_Idempotent_Element | https://proofwiki.org/wiki/Power_of_Idempotent_Element | [
"Idempotence"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Idempotence/Element"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-11979 | Powers of Group Element Commute | Let $\struct {G, \circ}$ be a group.
Let $g \in G$.
Let $m, n \in \N_{>0}$.
Then:
:$\forall m, n \in \N_{>0}: g^n \circ g^m = g^m \circ g^n$ | By definition, a group is also a semigroup.
The result follows as a special case of Powers of Semigroup Element Commute
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $g \in G$.
Let $m, n \in \N_{>0}$.
Then:
:$\forall m, n \in \N_{>0}: g^n \circ g^m = g^m \circ g^n$ | By definition, a [[Definition:Group|group]] is also a [[Definition:Semigroup|semigroup]].
The result follows as a special case of [[Powers of Semigroup Element Commute]]
{{Qed}} | Powers of Group Element Commute | https://proofwiki.org/wiki/Powers_of_Group_Element_Commute | https://proofwiki.org/wiki/Powers_of_Group_Element_Commute | [
"Group Theory",
"Powers (Abstract Algebra)",
"Commutativity"
] | [
"Definition:Group"
] | [
"Definition:Group",
"Definition:Semigroup",
"Powers of Semigroup Element Commute"
] |
proofwiki-11980 | Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers | Let $\R_{>0}$ be the set of strictly positive real numbers, that is:
:$\R_{>0} = \set {x \in \R: x > 0}$
The structure $\struct {\R_{>0}, \times}$ forms a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero, that is:
:$\R_{\ne 0} = \R \setminus \set 0$ | From Non-Zero Real Numbers under Multiplication form Abelian Group we have that $\struct {\R_{\ne 0}, \times}$ is a group.
We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
Now, verify that the conditions for Two-Step Subgroup Test are satisfied: | Let $\R_{>0}$ be the set of [[Definition:Strictly Positive Real Number|strictly positive real numbers]], that is:
:$\R_{>0} = \set {x \in \R: x > 0}$
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\R_{>0}, \times}$ forms a [[Definition:Subgroup|subgroup]] of $\struct {\R_{\ne 0}, \times}... | From [[Non-Zero Real Numbers under Multiplication form Abelian Group]] we have that $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Group|group]].
We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
Now, verify that the conditions for [[Two-Step Subgroup Test]] are satisfied: | Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers | https://proofwiki.org/wiki/Strictly_Positive_Real_Numbers_under_Multiplication_form_Subgroup_of_Non-Zero_Real_Numbers | https://proofwiki.org/wiki/Strictly_Positive_Real_Numbers_under_Multiplication_form_Subgroup_of_Non-Zero_Real_Numbers | [
"Real Multiplication",
"Subgroups"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Subgroup",
"Definition:Set",
"Definition:Real Number",
"Definition:Zero (Number)"
] | [
"Non-Zero Real Numbers under Multiplication form Abelian Group",
"Definition:Group",
"Two-Step Subgroup Test",
"Two-Step Subgroup Test"
] |
proofwiki-11981 | Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers | Let $\Q_{> 0}$ be the set of strictly positive rational numbers, that is $\Q_{> 0} = \set { x \in \Q: x > 0}$.
The structure $\struct {\Q_{> 0}, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of rational numbers without zero: $\Q_{\ne 0} = \Q \setminus \set 0$. | From Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group we have that $\struct {\Q_{\ne 0}, \times}$ is a group.
We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$.
Let $a, b \in \Q_{> 0}$.
Then:
:$a b \in \Q_{\ne 0}$ and $ab > 0$
Hence:
:$a b \in \Q_{> 0}$
Let $a \in \Q_{> 0}$.
Th... | Let $\Q_{> 0}$ be the set of [[Definition:Strictly Positive Rational Number|strictly positive rational numbers]], that is $\Q_{> 0} = \set { x \in \Q: x > 0}$.
The [[Definition:Algebraic Structure with One Operation|structure]] $\struct {\Q_{> 0}, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{\ne 0},... | From [[Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group]] we have that $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Group|group]].
We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$.
Let $a, b \in \Q_{> 0}$.
Then:
:$a b \in \Q_{\ne 0}$ and $ab > 0$
Hence:
:$a b \in \Q_{>... | Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_under_Multiplication_form_Subgroup_of_Non-Zero_Rational_Numbers | https://proofwiki.org/wiki/Strictly_Positive_Rational_Numbers_under_Multiplication_form_Subgroup_of_Non-Zero_Rational_Numbers | [
"Rational Multiplication",
"Subgroups"
] | [
"Definition:Strictly Positive/Rational Number",
"Definition:Algebraic Structure/One Operation",
"Definition:Subgroup",
"Definition:Rational Number",
"Definition:Zero (Number)"
] | [
"Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group",
"Definition:Group",
"Two-Step Subgroup Test",
"Definition:Subgroup",
"Category:Rational Multiplication",
"Category:Subgroups"
] |
proofwiki-11982 | Union of Upper Sections is Upper | Let $\struct {S, \preceq}$ be a preordered set.
Let $A$ be a set of subsets of $S$ such that:
:$\forall X \in A: X$ is an upper section.
Then:
:$\bigcup A$ is also an upper section. | Let $x \in \bigcup A, y \in S$ such that:
:$x \preceq y$
By definition of union:
:$\exists Y \in A: x \in Y$
By assumption:
:$Y$ is an upper section.
By definition of upper section:
:$y \in Y$
Thus by definition of union:
:$y \in \bigcup A$
Hence
:$\bigcup A$ is an upper section.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Preordered Set|preordered set]].
Let $A$ be a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$ such that:
:$\forall X \in A: X$ is an [[Definition:Upper Section|upper section]].
Then:
:$\bigcup A$ is also an [[Definition:Upper Section|upper section]]... | Let $x \in \bigcup A, y \in S$ such that:
:$x \preceq y$
By definition of [[Definition:Union of Set of Sets|union]]:
:$\exists Y \in A: x \in Y$
By assumption:
:$Y$ is an [[Definition:Upper Section|upper section]].
By definition of [[Definition:Upper Section|upper section]]:
:$y \in Y$
Thus by definition of [[Defin... | Union of Upper Sections is Upper | https://proofwiki.org/wiki/Union_of_Upper_Sections_is_Upper | https://proofwiki.org/wiki/Union_of_Upper_Sections_is_Upper | [
"Upper Sections",
"Set Union"
] | [
"Definition:Preordering/Preordered Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Upper Section",
"Definition:Upper Section"
] | [
"Definition:Set Union/Set of Sets",
"Definition:Upper Section",
"Definition:Upper Section",
"Definition:Set Union/Set of Sets",
"Definition:Upper Section"
] |
proofwiki-11983 | Set of Rotations in Space about Fixed Point forms Infinite Group | Let $\SS$ be a rigid body in space.
Let $O$ be a fixed point in space.
The set of all rotations of $\SS$ through some straight line through $O$ forms an infinite group. | {{ProofWanted|Needs the mathematical definition of a rotation in space to be defined before we start.}} | Let $\SS$ be a [[Definition:Rigid Body|rigid body]] in [[Definition:Ordinary Space|space]].
Let $O$ be a fixed [[Definition:Point|point]] in [[Definition:Ordinary Space|space]].
The [[Definition:Set|set]] of all [[Definition:Space Rotation|rotations]] of $\SS$ through some [[Definition:Straight Line|straight line]] t... | {{ProofWanted|Needs the mathematical definition of a rotation in space to be defined before we start.}} | Set of Rotations in Space about Fixed Point forms Infinite Group | https://proofwiki.org/wiki/Set_of_Rotations_in_Space_about_Fixed_Point_forms_Infinite_Group | https://proofwiki.org/wiki/Set_of_Rotations_in_Space_about_Fixed_Point_forms_Infinite_Group | [
"Examples of Groups"
] | [
"Definition:Body/Rigid",
"Definition:Ordinary Space",
"Definition:Point",
"Definition:Ordinary Space",
"Definition:Set",
"Definition:Rotation (Geometry)/Space",
"Definition:Line/Straight Line",
"Definition:Infinite Group"
] | [] |
proofwiki-11984 | Group of Rotations about Fixed Point is not Abelian | Let $\SS$ be a rigid body in space.
Let $O$ be a fixed point in space.
Let $\GG$ be the group of all rotations of $\SS$ around $O$.
Then $\GG$ is not an abelian group. | Let $\SS$ be a square lamina.
Let $O$ be the center of $\SS$.
Recall the definition of the symmetry group of the square $D_4$:
{{:Definition:Symmetry Group of Square}}
We have that:
:Reflection $t_x$ can be achieved by a rotation of $\SS$ of $\pi$ radians about $x$.
:Reflection $t_y$ can be achieved by a rotation of $\... | Let $\SS$ be a [[Definition:Rigid Body|rigid body]] in [[Definition:Ordinary Space|space]].
Let $O$ be a fixed [[Definition:Point|point]] in [[Definition:Ordinary Space|space]].
Let $\GG$ be the [[Definition:Group|group]] of all [[Definition:Space Rotation|rotations]] of $\SS$ around $O$.
Then $\GG$ is not an [[Def... | Let $\SS$ be a [[Definition:Square (Geometry)|square]] [[Definition:Lamina|lamina]].
Let $O$ be the center of $\SS$.
Recall the definition of the [[Definition:Symmetry Group of Square|symmetry group of the square]] $D_4$:
{{:Definition:Symmetry Group of Square}}
We have that:
:Reflection $t_x$ can be achieved by a... | Group of Rotations about Fixed Point is not Abelian | https://proofwiki.org/wiki/Group_of_Rotations_about_Fixed_Point_is_not_Abelian | https://proofwiki.org/wiki/Group_of_Rotations_about_Fixed_Point_is_not_Abelian | [
"Examples of Groups"
] | [
"Definition:Body/Rigid",
"Definition:Ordinary Space",
"Definition:Point",
"Definition:Ordinary Space",
"Definition:Group",
"Definition:Rotation (Geometry)/Space",
"Definition:Abelian Group"
] | [
"Definition:Quadrilateral/Square",
"Definition:Lamina",
"Definition:Symmetry Group of Square",
"Definition:Rotation (Geometry)/Space",
"Definition:Angular Measure/Radian",
"Definition:Rotation (Geometry)/Space",
"Definition:Angular Measure/Radian",
"Definition:Subgroup",
"Symmetry Group of Square is... |
proofwiki-11985 | Union of Filtered Sets is Filtered | Let $\struct {S, \preceq}$ be a preordered set.
Let $A$ be a set of subsets of $S$ such that
:$\forall X \in A: X$ is filtered
and
:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
Then:
:$\bigcup A$ is also filtered. | Let $x, y \in \bigcup A$.
By definition of union:
:$\exists X \in A: x \in X$
and
:$\exists Y \in A: y \in Y$
By assumption:
:$\exists Z \in A: X \cup Y \subseteq Z$
By definition of union:
:$x, y \in X \cup Y$
By definition of subset:
:$x, y \in Z$
By assumption:
:$Z$ is filtered.
By definition of filtered:
:$\exists ... | Let $\struct {S, \preceq}$ be a [[Definition:Preordered Set|preordered set]].
Let $A$ be a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$ such that
:$\forall X \in A: X$ is [[Definition:Filtered Subset|filtered]]
and
:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
Then:
:$\bigcup... | Let $x, y \in \bigcup A$.
By definition of [[Definition:Union of Set of Sets|union]]:
:$\exists X \in A: x \in X$
and
:$\exists Y \in A: y \in Y$
By assumption:
:$\exists Z \in A: X \cup Y \subseteq Z$
By definition of [[Definition:Set Union|union]]:
:$x, y \in X \cup Y$
By definition of [[Definition:Subset|subset]... | Union of Filtered Sets is Filtered | https://proofwiki.org/wiki/Union_of_Filtered_Sets_is_Filtered | https://proofwiki.org/wiki/Union_of_Filtered_Sets_is_Filtered | [
"Preorder Theory"
] | [
"Definition:Preordering/Preordered Set",
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Filtered Subset",
"Definition:Filtered Subset"
] | [
"Definition:Set Union/Set of Sets",
"Definition:Set Union",
"Definition:Subset",
"Definition:Filtered Subset",
"Definition:Filtered Subset",
"Definition:Set Union/Set of Sets",
"Definition:Filtered Subset"
] |
proofwiki-11986 | Special Orthogonal Group is Group | Let $k$ be a field.
The $n$th orthogonal group on $k$ is a group. | A direct corollary of Special Orthogonal Group is Subgroup of Orthogonal Group.
{{qed}} | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
The $n$th [[Definition:Special Orthogonal Group|orthogonal group]] on $k$ is a [[Definition:Group|group]]. | A direct corollary of [[Special Orthogonal Group is Subgroup of Orthogonal Group]].
{{qed}} | Special Orthogonal Group is Group | https://proofwiki.org/wiki/Special_Orthogonal_Group_is_Group | https://proofwiki.org/wiki/Special_Orthogonal_Group_is_Group | [
"Orthogonal Groups"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Special Orthogonal Group",
"Definition:Group"
] | [
"Special Orthogonal Group is Subgroup of Orthogonal Group"
] |
proofwiki-11987 | Unit Matrix is its own Inverse | The inverse of the unit matrix $\mathbf I_n$ of order $n$ is $\mathbf I_n$.
That is, a unit matrix it its own inverse. | By definition, a unit matrix is a diagonal matrix.
From Inverse of Diagonal Matrix, the inverse of a diagonal matrix:
:<nowiki>$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$</nowiki>
is the diagonal matrix... | The [[Definition:Inverse Matrix|inverse]] of the [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is $\mathbf I_n$.
That is, a [[Definition:Unit Matrix|unit matrix]] it its own [[Definition:Inverse Matrix|inverse]]. | By definition, a [[Definition:Unit Matrix|unit matrix]] is a [[Definition:Diagonal Matrix|diagonal matrix]].
From [[Inverse of Diagonal Matrix]], the [[Definition:Inverse Matrix|inverse]] of a [[Definition:Diagonal Matrix|diagonal matrix]]:
:<nowiki>$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} &... | Unit Matrix is its own Inverse | https://proofwiki.org/wiki/Unit_Matrix_is_its_own_Inverse | https://proofwiki.org/wiki/Unit_Matrix_is_its_own_Inverse | [
"Unit Matrices",
"Inverse Matrices"
] | [
"Definition:Inverse Matrix",
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Unit Matrix",
"Definition:Inverse Matrix"
] | [
"Definition:Unit Matrix",
"Definition:Diagonal Matrix",
"Inverse of Diagonal Matrix",
"Definition:Inverse Matrix",
"Definition:Diagonal Matrix",
"Definition:Diagonal Matrix",
"Definition:Unit Matrix",
"Category:Unit Matrices",
"Category:Inverse Matrices"
] |
proofwiki-11988 | Way Above Closure is Subset of Upper Closure of Element | Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then:
:$x^\gg \subseteq x^\succeq$
where
:$x^\gg$ denotes the way above closure of $x$
:$x^\succeq$ denotes the upper closure of $x$. | Let $y \in x^\gg$.
By definition of way above closure:
:$x \ll y$
where $\ll$ denotes the way below relation.
By Way Below implies Preceding:
:$x \preceq y$
Thus by definition of upper closure of element:
:$y \in x^\succeq$
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $x \in S$.
Then:
:$x^\gg \subseteq x^\succeq$
where
:$x^\gg$ denotes the [[Definition:Way Above Closure|way above closure]] of $x$
:$x^\succeq$ denotes the [[Definition:Upper Closure of Element|upper closure]] of $x$. | Let $y \in x^\gg$.
By definition of [[Definition:Way Above Closure|way above closure]]:
:$x \ll y$
where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
By [[Way Below implies Preceding]]:
:$x \preceq y$
Thus by definition of [[Definition:Upper Closure of Element|upper closure of element]]:... | Way Above Closure is Subset of Upper Closure of Element | https://proofwiki.org/wiki/Way_Above_Closure_is_Subset_of_Upper_Closure_of_Element | https://proofwiki.org/wiki/Way_Above_Closure_is_Subset_of_Upper_Closure_of_Element | [
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Way Above Closure",
"Definition:Upper Closure/Element"
] | [
"Definition:Way Above Closure",
"Definition:Element is Way Below",
"Way Below implies Preceding",
"Definition:Upper Closure/Element"
] |
proofwiki-11989 | Diagonal Matrix is Symmetric | Let $D$ be a diagonal matrix.
Then $D$ is symmetric. | By definition of diagonal matrix:
:$\forall j, k: j \ne k \implies a_{jk} = 0 = a_{kj}$
So by definition of transpose of $D$:
:$D = D^\intercal$
where $D^\intercal$ denotes the transpose.
Hence the result, by definition of symmetric matrix.
{{qed}}
Category:Symmetric Matrices
Category:Diagonal Matrices
dv7n9kzvzryg6tju... | Let $D$ be a [[Definition:Diagonal Matrix|diagonal matrix]].
Then $D$ is [[Definition:Symmetric Matrix|symmetric]]. | By definition of [[Definition:Diagonal Matrix|diagonal matrix]]:
:$\forall j, k: j \ne k \implies a_{jk} = 0 = a_{kj}$
So by definition of [[Definition:Transpose of Matrix|transpose]] of $D$:
:$D = D^\intercal$
where $D^\intercal$ denotes the [[Definition:Transpose of Matrix|transpose]].
Hence the result, by definiti... | Diagonal Matrix is Symmetric | https://proofwiki.org/wiki/Diagonal_Matrix_is_Symmetric | https://proofwiki.org/wiki/Diagonal_Matrix_is_Symmetric | [
"Symmetric Matrices",
"Diagonal Matrices"
] | [
"Definition:Diagonal Matrix",
"Definition:Symmetric Matrix"
] | [
"Definition:Diagonal Matrix",
"Definition:Transpose of Matrix",
"Definition:Transpose of Matrix",
"Definition:Symmetric Matrix",
"Category:Symmetric Matrices",
"Category:Diagonal Matrices"
] |
proofwiki-11990 | Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.
Let $X$ be upper way below open subset of $S$.
Let $x \in S$ such that
:$x \in \relcomp S X$
Then
:$\exists m \in S: x \preceq m \land m = \max \relcomp S X$ | Define $A := \set {C \in \map {\mathit {Chains} } L: C \subseteq \relcomp S X \land x \in C}$
where $\map {\mathit {Chains} } L$ denotes the set of all chains of $L$.
We will prove that
:$\forall Z: Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X} \implies \bigcup Z \in A... | Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
Let $X$ be [[Definition:Upper Section|upper]] [[Definition:Way Below Open|way below open]] [[Definition:Subset|subset]] of $S$.
Let $x \in S$ such that
:$x \in \relcomp S X$
Then
:$\exists m \in S: x \preceq m \land ... | Define $A := \set {C \in \map {\mathit {Chains} } L: C \subseteq \relcomp S X \land x \in C}$
where $\map {\mathit {Chains} } L$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Chain (Order Theory)|chains]] of $L$.
We will prove that
:$\forall Z: Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y ... | Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement | https://proofwiki.org/wiki/Upper_Way_Below_Open_Subset_Complement_is_Non_Empty_implies_There_Exists_Maximal_Element_of_Complement | https://proofwiki.org/wiki/Upper_Way_Below_Open_Subset_Complement_is_Non_Empty_implies_There_Exists_Maximal_Element_of_Complement | [
"Complete Lattices"
] | [
"Definition:Complete Lattice",
"Definition:Upper Section",
"Definition:Way Below Open",
"Definition:Subset"
] | [
"Definition:Set of Sets",
"Definition:Chain (Order Theory)",
"Definition:Chain (Order Theory)",
"Definition:Set Union/Set of Sets",
"Definition:Subset",
"Definition:Connected Relation",
"Definition:Non-Empty Set",
"Definition:Set Union/Set of Sets",
"Union of Subsets is Subset/Set of Sets",
"Singl... |
proofwiki-11991 | Unit Matrix is Orthogonal | The unit matrix $\mathbf I_n$ of order $n$ is orthogonal. | By Unit Matrix is its own Inverse the inverse $I_n^{-1}$ of $I_n$ is $I_n$.
By definition a unit matrix is a diagonal matrix.
Hence by Diagonal Matrix is Symmetric:
:$I_n = I_n^\intercal$
where $I_n^\intercal$ is the transpose of $I_n$.
Thus:
:$I_n^{-1} = I_n^\intercal$
and the result follows by definition of orthogona... | The [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is [[Definition:Orthogonal Matrix|orthogonal]]. | By [[Unit Matrix is its own Inverse]] the [[Definition:Inverse Matrix|inverse]] $I_n^{-1}$ of $I_n$ is $I_n$.
By definition a [[Definition:Unit Matrix|unit matrix]] is a [[Definition:Diagonal Matrix|diagonal matrix]].
Hence by [[Diagonal Matrix is Symmetric]]:
:$I_n = I_n^\intercal$
where $I_n^\intercal$ is the [[Def... | Unit Matrix is Orthogonal | https://proofwiki.org/wiki/Unit_Matrix_is_Orthogonal | https://proofwiki.org/wiki/Unit_Matrix_is_Orthogonal | [
"Unit Matrices",
"Orthogonal Matrices"
] | [
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Orthogonal Matrix"
] | [
"Unit Matrix is its own Inverse",
"Definition:Inverse Matrix",
"Definition:Unit Matrix",
"Definition:Diagonal Matrix",
"Diagonal Matrix is Symmetric",
"Definition:Transpose of Matrix",
"Definition:Orthogonal Matrix",
"Category:Unit Matrices",
"Category:Orthogonal Matrices"
] |
proofwiki-11992 | Unit Matrix is Proper Orthogonal | The unit matrix $\mathbf I_n$ of order $n$ is proper orthogonal. | By Unit Matrix is Orthogonal, $\mathbf I_n$ is an orthogonal matrix.
By Determinant of Unit Matrix, the determinant of $\mathbf I_n$ is equal to $1$.
The result follows by definition of proper orthogonal.
{{qed}}
Category:Unit Matrices
Category:Proper Orthogonal Matrices
3xdpyj4v5c1djkul1buy2fq721ardd4 | The [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is [[Definition:Proper Orthogonal Matrix|proper orthogonal]]. | By [[Unit Matrix is Orthogonal]], $\mathbf I_n$ is an [[Definition:Orthogonal Matrix|orthogonal matrix]].
By [[Determinant of Unit Matrix]], the [[Definition:Determinant of Matrix|determinant]] of $\mathbf I_n$ is equal to $1$.
The result follows by definition of [[Definition:Proper Orthogonal Matrix|proper orthogona... | Unit Matrix is Proper Orthogonal | https://proofwiki.org/wiki/Unit_Matrix_is_Proper_Orthogonal | https://proofwiki.org/wiki/Unit_Matrix_is_Proper_Orthogonal | [
"Unit Matrices",
"Proper Orthogonal Matrices"
] | [
"Definition:Unit Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Proper Orthogonal Matrix"
] | [
"Unit Matrix is Orthogonal",
"Definition:Orthogonal Matrix",
"Determinant of Unit Matrix",
"Definition:Determinant/Matrix",
"Definition:Proper Orthogonal Matrix",
"Category:Unit Matrices",
"Category:Proper Orthogonal Matrices"
] |
proofwiki-11993 | Inverse of Orthogonal Matrix is Orthogonal | Let $\mathbf A$ be an orthogonal matrix.
Then its inverse $\mathbf A^{-1}$ is also orthogonal. | By definition of orthogonal matrix:
:$\mathbf A^\intercal = \mathbf A^{-1}$
where $\mathbf A^\intercal$ is the transpose of $\mathbf A$.
By Inverse of Inverse of Matrix:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
By Transpose of Transpose of Matrix:
:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
Thus we have:
... | Let $\mathbf A$ be an [[Definition:Orthogonal Matrix|orthogonal matrix]].
Then its [[Definition:Inverse Matrix|inverse]] $\mathbf A^{-1}$ is also [[Definition:Orthogonal Matrix|orthogonal]]. | By definition of [[Definition:Orthogonal Matrix|orthogonal matrix]]:
:$\mathbf A^\intercal = \mathbf A^{-1}$
where $\mathbf A^\intercal$ is the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
By [[Inverse of Inverse of Matrix]]:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
By [[Transpose of Transpose of ... | Inverse of Orthogonal Matrix is Orthogonal | https://proofwiki.org/wiki/Inverse_of_Orthogonal_Matrix_is_Orthogonal | https://proofwiki.org/wiki/Inverse_of_Orthogonal_Matrix_is_Orthogonal | [
"Orthogonal Matrices",
"Inverse Matrices"
] | [
"Definition:Orthogonal Matrix",
"Definition:Inverse Matrix",
"Definition:Orthogonal Matrix"
] | [
"Definition:Orthogonal Matrix",
"Definition:Transpose of Matrix",
"Inverse of Inverse of Matrix",
"Transpose of Transpose of Matrix",
"Definition:Orthogonal Matrix",
"Category:Orthogonal Matrices",
"Category:Inverse Matrices"
] |
proofwiki-11994 | Inverse of Inverse of Matrix | Let $\mathbf A$ be a nonsingular matrix.
Then:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
That is, a nonsingular matrix equals the inverse of its inverse. | By definition of inverse matrix:
:$\mathbf A^{-1} \mathbf A = \mathbf I$
where $\mathbf I$ is the unit matrix.
Thus the inverse of $\mathbf A^{-1}$ is $\mathbf A$.
Hence the result.
{{qed}}
Category:Inverse Matrices
9l2e2amuqajjgd2tbsewqgpmsc67u4p | Let $\mathbf A$ be a [[Definition:Nonsingular Matrix|nonsingular matrix]].
Then:
:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
That is, a [[Definition:Nonsingular Matrix|nonsingular matrix]] equals the [[Definition:Inverse Matrix|inverse]] of its [[Definition:Inverse Matrix|inverse]]. | By definition of [[Definition:Inverse Matrix|inverse matrix]]:
:$\mathbf A^{-1} \mathbf A = \mathbf I$
where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
Thus the [[Definition:Inverse Matrix|inverse]] of $\mathbf A^{-1}$ is $\mathbf A$.
Hence the result.
{{qed}}
[[Category:Inverse Matrices]]
9l2e2amuqa... | Inverse of Inverse of Matrix | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Matrix | https://proofwiki.org/wiki/Inverse_of_Inverse_of_Matrix | [
"Inverse Matrices"
] | [
"Definition:Nonsingular Matrix",
"Definition:Nonsingular Matrix",
"Definition:Inverse Matrix",
"Definition:Inverse Matrix"
] | [
"Definition:Inverse Matrix",
"Definition:Unit Matrix",
"Definition:Inverse Matrix",
"Category:Inverse Matrices"
] |
proofwiki-11995 | Inverse of Proper Orthogonal Matrix is Proper Orthogonal | Let $\mathbf A$ be a proper orthogonal matrix.
Then its inverse $\mathbf A^{-1}$ is also proper orthogonal. | By definition of proper orthogonal matrix:
:$\mathbf A$ is orthogonal
:the determinant of $\mathbf A$ is equal to $1$.
By Inverse of Orthogonal Matrix is Orthogonal, $\mathbf A^{-1}$ is orthogonal.
By Determinant of Inverse Matrix:
:$\det \mathbf A^{-1} = \dfrac 1 {\det \mathbf A} = 1$
Hence, by definition, $\mathbf A^... | Let $\mathbf A$ be a [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]].
Then its [[Definition:Inverse Matrix|inverse]] $\mathbf A^{-1}$ is also [[Definition:Proper Orthogonal Matrix|proper orthogonal]]. | By definition of [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]]:
:$\mathbf A$ is [[Definition:Orthogonal Matrix|orthogonal]]
:the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$ is equal to $1$.
By [[Inverse of Orthogonal Matrix is Orthogonal]], $\mathbf A^{-1}$ is [[Definition:Orthog... | Inverse of Proper Orthogonal Matrix is Proper Orthogonal | https://proofwiki.org/wiki/Inverse_of_Proper_Orthogonal_Matrix_is_Proper_Orthogonal | https://proofwiki.org/wiki/Inverse_of_Proper_Orthogonal_Matrix_is_Proper_Orthogonal | [
"Proper Orthogonal Matrices",
"Inverse Matrices"
] | [
"Definition:Proper Orthogonal Matrix",
"Definition:Inverse Matrix",
"Definition:Proper Orthogonal Matrix"
] | [
"Definition:Proper Orthogonal Matrix",
"Definition:Orthogonal Matrix",
"Definition:Determinant/Matrix",
"Inverse of Orthogonal Matrix is Orthogonal",
"Definition:Orthogonal Matrix",
"Determinant of Inverse Matrix",
"Definition:Proper Orthogonal Matrix",
"Category:Proper Orthogonal Matrices",
"Catego... |
proofwiki-11996 | Special Orthogonal Group is Subgroup of Orthogonal Group | Let $k$ be a field.
Let $\map {\operatorname O} {n, k}$ be the $n$th orthogonal group on $k$.
Let $\map {\operatorname {SO} } {n, k}$ be the $n$th special orthogonal group on $k$.
Then $\map {\operatorname {SO} } {n, k}$ is a subgroup of $\map {\operatorname O} {n, k}$. | We have that Unit Matrix is Proper Orthogonal, so $\map {\operatorname {SO} } {n, k}$ is not empty.
Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are proper orthogonal.
Then by Inverse of Proper Orthogonal Matrix is Proper Orthogonal:
:$\mathbf B^{-1}... | Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $\map {\operatorname O} {n, k}$ be the $n$th [[Definition:Orthogonal Group|orthogonal group]] on $k$.
Let $\map {\operatorname {SO} } {n, k}$ be the $n$th [[Definition:Special Orthogonal Group|special orthogonal group]] on $k$.
Then $\map {\operatornam... | We have that [[Unit Matrix is Proper Orthogonal]], so $\map {\operatorname {SO} } {n, k}$ is not [[Definition:Empty Set|empty]].
Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$.
Then, by definition, $\mathbf A$ and $\mathbf B$ are [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
Then by [[... | Special Orthogonal Group is Subgroup of Orthogonal Group | https://proofwiki.org/wiki/Special_Orthogonal_Group_is_Subgroup_of_Orthogonal_Group | https://proofwiki.org/wiki/Special_Orthogonal_Group_is_Subgroup_of_Orthogonal_Group | [
"Orthogonal Groups"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Orthogonal Group",
"Definition:Special Orthogonal Group",
"Definition:Subgroup"
] | [
"Unit Matrix is Proper Orthogonal",
"Definition:Empty Set",
"Definition:Proper Orthogonal Matrix",
"Inverse of Proper Orthogonal Matrix is Proper Orthogonal",
"Definition:Proper Orthogonal Matrix",
"Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix",
"Definition:Proper Orthogonal Matrix"... |
proofwiki-11997 | Negative Matrix is Inverse for Hadamard Product | Let $\struct {G, \cdot}$ be a group whose identity is $e$.
Let $\map {\MM_G} {m, n}$ be a $m \times n$ matrix space over $\struct {G, \cdot}$.
Let $\mathbf A$ be an element of $\map {\MM_G} {m, n}$.
Let $-\mathbf A$ be the negative of $\mathbf A$.
Then $-\mathbf A$ is the inverse for the operation $\circ$, where $\circ... | Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_G} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \paren {-\mathbf A}
| r = \sqbrk a_{m n} \circ \paren {-\sqbrk a_{m n} }
| c = Definition of $\mathbf A$
}}
{{eqn | r = \sqbrk a_{m n} \circ \sqbrk {a^{-1} }_{m n}
| c = {{Defof|Negative Matrix|su... | Let $\struct {G, \cdot}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\map {\MM_G} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {G, \cdot}$.
Let $\mathbf A$ be an [[Definition:Element|element]] of $\map {\MM_G} {m, n}$.
Let $-\mathbf ... | Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_G} {m, n}$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \circ \paren {-\mathbf A}
| r = \sqbrk a_{m n} \circ \paren {-\sqbrk a_{m n} }
| c = Definition of $\mathbf A$
}}
{{eqn | r = \sqbrk a_{m n} \circ \sqbrk {a^{-1} }_{m n}
| c = {{Defof|Negative Matrix|... | Negative Matrix is Inverse for Hadamard Product | https://proofwiki.org/wiki/Negative_Matrix_is_Inverse_for_Hadamard_Product | https://proofwiki.org/wiki/Negative_Matrix_is_Inverse_for_Hadamard_Product | [
"Negative Matrices",
"Hadamard Product"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Matrix Space",
"Definition:Element",
"Definition:Negative Matrix/General Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Hadamard Product"
] | [
"Zero Matrix is Identity for Hadamard Product"
] |
proofwiki-11998 | Cancellable Infinite Semigroup is not necessarily Group | Let $\struct {S, \circ}$ be a semigroup whose underlying set is infinite.
Let $\struct {S, \circ}$ be such that all elements of $S$ are cancellable.
Then it is not necessarily the case that $\struct {S, \circ}$ is a group. | Consider the semigroup $\struct {\N, +}$.
From Natural Numbers under Addition form Commutative Semigroup, $\struct {\N, +}$ forms a semigroup.
From Natural Numbers are Infinite, the underlying set of $\struct {\N, +}$ is infinite.
From Natural Number Addition is Cancellable, all elements of $\struct {\N, +}$ are cancel... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] whose [[Definition:Underlying Set of Structure|underlying set]] is [[Definition:Infinite Set|infinite]].
Let $\struct {S, \circ}$ be such that all [[Definition:Element|elements]] of $S$ are [[Definition:Cancellable Element|cancellable]].
Then it is not... | Consider the [[Definition:Semigroup|semigroup]] $\struct {\N, +}$.
From [[Natural Numbers under Addition form Commutative Semigroup]], $\struct {\N, +}$ forms a [[Definition:Semigroup|semigroup]].
From [[Natural Numbers are Infinite]], the [[Definition:Underlying Set of Structure|underlying set]] of $\struct {\N, +}$... | Cancellable Infinite Semigroup is not necessarily Group | https://proofwiki.org/wiki/Cancellable_Infinite_Semigroup_is_not_necessarily_Group | https://proofwiki.org/wiki/Cancellable_Infinite_Semigroup_is_not_necessarily_Group | [
"Semigroups",
"Cancellability"
] | [
"Definition:Semigroup",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Infinite Set",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Group"
] | [
"Definition:Semigroup",
"Natural Numbers under Addition form Commutative Semigroup",
"Definition:Semigroup",
"Natural Numbers are Infinite",
"Definition:Underlying Set/Abstract Algebra",
"Definition:Infinite Set",
"Natural Number Addition is Cancellable",
"Definition:Cancellable Element",
"Natural N... |
proofwiki-11999 | Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection | Let $\struct {S, \circ}$ be a finite semigroup.
Let $a \in S$ be left cancellable.
Then the left regular representation $\lambda_a$ of $\struct {S, \circ}$ with respect to $a$ is a bijection. | By Left Cancellable iff Left Regular Representation Injective, $\lambda_a$ is an injection.
By hypothesis, $S$ is finite.
From Injection from Finite Set to Itself is Surjection, $\lambda_a$ is a surjection.
Thus $\lambda_a$ is injective and surjective, and therefore a bijection.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
Let $a \in S$ be [[Definition:Left Cancellable Element|left cancellable]].
Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ of $\struct {S, \circ}$ with respect to $a$ is a [[Defi... | By [[Left Cancellable iff Left Regular Representation Injective]], $\lambda_a$ is an [[Definition:Injection|injection]].
[[Definition:By Hypothesis|By hypothesis]], $S$ is [[Definition:Finite Set|finite]].
From [[Injection from Finite Set to Itself is Surjection]], $\lambda_a$ is a [[Definition:Surjection|surjection]... | Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection | https://proofwiki.org/wiki/Left_Regular_Representation_wrt_Left_Cancellable_Element_on_Finite_Semigroup_is_Bijection | https://proofwiki.org/wiki/Left_Regular_Representation_wrt_Left_Cancellable_Element_on_Finite_Semigroup_is_Bijection | [
"Semigroups",
"Regular Representations",
"Cancellability"
] | [
"Definition:Finite",
"Definition:Semigroup",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Bijection"
] | [
"Left Cancellable iff Left Regular Representation Injective",
"Definition:Injection",
"Definition:By Hypothesis",
"Definition:Finite Set",
"Injection from Finite Set to Itself is Surjection",
"Definition:Surjection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.