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-14500 | Wosets are Isomorphic to Each Other or Initial Segments | Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be well-ordered sets.
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is order isomorphic to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is order isomorphic to an initial segment in $\struct {T, \preceq_T}$
or:
:$\struct {T, \prece... | If the sets $S$ and $T$ considered are empty or singletons, the theorem holds vacuously or trivially.
Thus assume $S$ and $T$ each contain at least two elements.
Let $U = \struct {S, \preceq_S} \cup \struct {T, \preceq_T}$.
Define the following relation $\preceq$ on $U$:
:$\forall x, y \in U: x \preceq y$
{{iff}}:
:$x,... | Let $\struct {S, \preceq_S}$ and $\struct {T, \preceq_T}$ be [[Definition:Well-Ordered Set|well-ordered sets]].
Then precisely one of the following hold:
:$\struct {S, \preceq_S}$ is [[Definition:Order Isomorphic Well-Orderings|order isomorphic]] to $\struct {T, \preceq_T}$
or:
:$\struct {S, \preceq_S}$ is [[Definit... | If the [[Definition:Set|sets]] $S$ and $T$ considered are [[Definition:Empty Set|empty]] or [[Definition:Singleton|singletons]], the theorem holds [[Definition:Vacuous|vacuously]] or trivially.
Thus assume $S$ and $T$ each contain at least two elements.
Let $U = \struct {S, \preceq_S} \cup \struct {T, \preceq_T}$.
... | Wosets are Isomorphic to Each Other or Initial Segments/Proof Without Using Choice | https://proofwiki.org/wiki/Wosets_are_Isomorphic_to_Each_Other_or_Initial_Segments | https://proofwiki.org/wiki/Wosets_are_Isomorphic_to_Each_Other_or_Initial_Segments/Proof_Without_Using_Choice | [
"Wosets are Isomorphic to Each Other or Initial Segments",
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment"
] | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Singleton",
"Definition:Vacuous",
"Definition:Well-Ordering",
"Definition:Total Ordering",
"Definition:Reflexive Relation",
"Definition:Transitive Relation",
"Definition:Antisymmetric Relation",
"Definition:Total Ordering",
"Definition:Well-O... |
proofwiki-14501 | Minimal Uncountable Well-Ordered Set Unique up to Isomorphism | Let $\Omega, \Omega'$ be minimal uncountable well-ordered sets.
Then $\Omega$ is order isomorphic to $\Omega'$.
That is, the minimal uncountable well-ordered set is unique up to order isomorphism. | From Wosets are Isomorphic to Each Other or Initial Segments, precisely one of the following holds:
:$\Omega$ is order isomorphic to $\Omega'$
or:
:$\Omega$ is order isomorphic to an initial segment in $\Omega'$
or:
:$\Omega'$ is order isomorphic to an initial segment in $\Omega$.
By the definition of minimal uncountab... | Let $\Omega, \Omega'$ be [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered sets]].
Then $\Omega$ is [[Definition:Order Isomorphism|order isomorphic]] to $\Omega'$.
That is, the [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered set]] is [[Definition:Un... | From [[Wosets are Isomorphic to Each Other or Initial Segments]], precisely one of the following holds:
:$\Omega$ is [[Definition:Order Isomorphic Well-Orderings|order isomorphic]] to $\Omega'$
or:
:$\Omega$ is [[Definition:Order Isomorphic Well-Orderings|order isomorphic]] to an [[Definition:Initial Segment|initial... | Minimal Uncountable Well-Ordered Set Unique up to Isomorphism | https://proofwiki.org/wiki/Minimal_Uncountable_Well-Ordered_Set_Unique_up_to_Isomorphism | https://proofwiki.org/wiki/Minimal_Uncountable_Well-Ordered_Set_Unique_up_to_Isomorphism | [
"Well-Orderings",
"Uncountable Sets"
] | [
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Order Isomorphism",
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Unique",
"Definition:Order Isomorphism"
] | [
"Wosets are Isomorphic to Each Other or Initial Segments",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment",
"Definition:Order Isomorphism/Well-Orderings",
"Definition:Initial Segment",
"Definition:Minimal Uncountable Well-Ordered... |
proofwiki-14502 | Addition Rule for Gaussian Binomial Coefficients/Formulation 1 | :$\dbinom n m_q = \dbinom {n - 1} m_q + \dbinom {n - 1} {m - 1}_q q^{n - m}$ | By definition of Gaussian binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom n m_q
| o = :=
| r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
| c =
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \pa... | :$\dbinom n m_q = \dbinom {n - 1} m_q + \dbinom {n - 1} {m - 1}_q q^{n - m}$ | By definition of [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
{{begin-eqn}}
{{eqn | l = \binom n m_q
| o = :=
| r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
| c =
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \pare... | Addition Rule for Gaussian Binomial Coefficients/Formulation 1 | https://proofwiki.org/wiki/Addition_Rule_for_Gaussian_Binomial_Coefficients/Formulation_1 | https://proofwiki.org/wiki/Addition_Rule_for_Gaussian_Binomial_Coefficients/Formulation_1 | [
"Addition Rule for Gaussian Binomial Coefficients"
] | [] | [
"Definition:Gaussian Binomial Coefficient",
"Definition:Fraction/Denominator"
] |
proofwiki-14503 | Addition Rule for Gaussian Binomial Coefficients/Formulation 2 | :$\dbinom n m_q = \dbinom {n - 1} m_q q^m + \dbinom {n - 1} {m - 1}_q$ | By definition of Gaussian binomial coefficient:
{{begin-eqn}}
{{eqn | l = \binom n m_q
| o = :=
| r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
| c =
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \paren {1 - q^{n - m + 1} } } {\paren {1 - q^m} \pa... | :$\dbinom n m_q = \dbinom {n - 1} m_q q^m + \dbinom {n - 1} {m - 1}_q$ | By definition of [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
{{begin-eqn}}
{{eqn | l = \binom n m_q
| o = :=
| r = \prod_{j \mathop = 0}^{m - 1} \dfrac {1 - q^{n - j} } {1 - q^{j + 1} }
| c =
}}
{{eqn | r = \dfrac {\paren {1 - q^n} \paren {1 - q^{n - 1} } \dotsm \pare... | Addition Rule for Gaussian Binomial Coefficients/Formulation 2 | https://proofwiki.org/wiki/Addition_Rule_for_Gaussian_Binomial_Coefficients/Formulation_2 | https://proofwiki.org/wiki/Addition_Rule_for_Gaussian_Binomial_Coefficients/Formulation_2 | [
"Addition Rule for Gaussian Binomial Coefficients"
] | [] | [
"Definition:Gaussian Binomial Coefficient",
"Definition:Fraction/Denominator"
] |
proofwiki-14504 | Union of Initial Segments is Initial Segment or All of Woset | Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a union of initial segments defined by the elements of $A$.
Then either:
:$J = X$
or:
:$J$ is an initial segment of $X$. | Suppose the hypotheses of the theorem hold.
If $J = X$ then the theorem is satisfied.
Assume $J \ne X$.
Then $X \setminus J$ is non-empty.
By Subset of Well-Ordered Set is Well-Ordered, $X \setminus J$ is itself well-ordered.
Thus $X \setminus J$ has a smallest element; call it $b$.
We claim that $b \preccurlyeq y$ for... | Let $\struct {X, \preccurlyeq}$ be a [[Definition:Well-Ordered Set|well-ordered]] [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a [[Definition:Union of Family|union]] of [[Definition:Initial Segment|initial segments]] defined by ... | Suppose the hypotheses of the theorem hold.
If $J = X$ then the theorem is satisfied.
Assume $J \ne X$.
Then $X \setminus J$ is [[Definition:Non-Empty Set|non-empty]].
By [[Subset of Well-Ordered Set is Well-Ordered]], $X \setminus J$ is itself [[Definition:Well-Ordered Set|well-ordered]].
Thus $X \setminus J$ has... | Union of Initial Segments is Initial Segment or All of Woset/Proof 1 | https://proofwiki.org/wiki/Union_of_Initial_Segments_is_Initial_Segment_or_All_of_Woset | https://proofwiki.org/wiki/Union_of_Initial_Segments_is_Initial_Segment_or_All_of_Woset/Proof_1 | [
"Well-Orderings",
"Union of Initial Segments is Initial Segment or All of Woset"
] | [
"Definition:Well-Ordered Set",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Set Union/Family of Sets",
"Definition:Initial Segment",
"Definition:Element",
"Definition:Initial Segment"
] | [
"Definition:Non-Empty Set",
"Subset of Well-Ordered Set is Well-Ordered",
"Definition:Well-Ordered Set",
"Definition:Smallest Element",
"Definition:Initial Segment",
"Definition:Contradiction",
"Definition:Strictly Precede",
"Definition:Strictly Succeed",
"Definition:Initial Segment"
] |
proofwiki-14505 | Union of Initial Segments is Initial Segment or All of Woset | Let $\struct {X, \preccurlyeq}$ be a well-ordered non-empty set.
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a union of initial segments defined by the elements of $A$.
Then either:
:$J = X$
or:
:$J$ is an initial segment of $X$. | Suppose the hypotheses of the theorem hold.
For every $x \in A$, we have $S_x \subseteq X$ by the definition of initial segment.
By Union of Family of Subsets is Subset, we have $J \subseteq X$.
Consider $X \setminus J$.
If $X \setminus J$ is empty, then $X \subseteq J$ by Set Difference with Superset is Empty Set, and... | Let $\struct {X, \preccurlyeq}$ be a [[Definition:Well-Ordered Set|well-ordered]] [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $A \subseteq X$.
Let:
:$\ds J = \bigcup_{x \mathop \in A} S_x$
be a [[Definition:Union of Family|union]] of [[Definition:Initial Segment|initial segments]] defined by ... | Suppose the hypotheses of the theorem hold.
For every $x \in A$, we have $S_x \subseteq X$ by the [[Definition:Initial Segment|definition of initial segment]].
By [[Union of Family of Subsets is Subset]], we have $J \subseteq X$.
Consider $X \setminus J$.
If $X \setminus J$ is [[Definition:Empty Set|empty]], then $... | Union of Initial Segments is Initial Segment or All of Woset/Proof 2 | https://proofwiki.org/wiki/Union_of_Initial_Segments_is_Initial_Segment_or_All_of_Woset | https://proofwiki.org/wiki/Union_of_Initial_Segments_is_Initial_Segment_or_All_of_Woset/Proof_2 | [
"Well-Orderings",
"Union of Initial Segments is Initial Segment or All of Woset"
] | [
"Definition:Well-Ordered Set",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Set Union/Family of Sets",
"Definition:Initial Segment",
"Definition:Element",
"Definition:Initial Segment"
] | [
"Definition:Initial Segment",
"Union of Subsets is Subset/Family of Sets",
"Definition:Empty Set",
"Set Difference with Superset is Empty Set",
"Subset Relation is Antisymmetric",
"Definition:Non-Empty Set",
"Subset of Well-Ordered Set is Well-Ordered",
"Definition:Well-Ordered Set",
"Definition:Sma... |
proofwiki-14506 | Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound | Let $\Omega$ denote the minimal uncountable well-ordered set.
Let $\omega$ be a countable subset of $\Omega$.
Then $\omega$ has an upper bound in $\Omega$. | By the minimal uncountable well-ordered set, $\Omega$ is uncountable
Then $\omega \ne \Omega$, for the former is countable and the latter is not.
Consider the union:
:$\ds \bigcup_{x \mathop \in \omega} S_x$
of initial segments $S_x$ in $\omega$.
By the definition of $\Omega$, any $S_x$ is countable.
Thus $\ds \bigcup_... | Let $\Omega$ denote the [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered set]].
Let $\omega$ be a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $\Omega$.
Then $\omega$ has an [[Definition:Upper Bound of Set|upper bound]] in $\Omega$. | By the [[Definition:Minimal Uncountable Well-Ordered Set|minimal uncountable well-ordered set]], $\Omega$ is [[Definition:Uncountable|uncountable]]
Then $\omega \ne \Omega$, for the former is [[Definition:Countable Set|countable]] and the latter is not.
Consider the [[Definition:Union of Family|union]]:
:$\ds \bigcu... | Countable Subset of Minimal Uncountable Well-Ordered Set Has Upper Bound | https://proofwiki.org/wiki/Countable_Subset_of_Minimal_Uncountable_Well-Ordered_Set_Has_Upper_Bound | https://proofwiki.org/wiki/Countable_Subset_of_Minimal_Uncountable_Well-Ordered_Set_Has_Upper_Bound | [
"Order Theory"
] | [
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Countable Set",
"Definition:Subset",
"Definition:Upper Bound of Set"
] | [
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Uncountable",
"Definition:Countable Set",
"Definition:Set Union/Family of Sets",
"Definition:Initial Segment",
"Definition:Countable Set",
"Definition:Set Union/Countable Union",
"Definition:Countable Set",
"Countable Union of Countable ... |
proofwiki-14507 | Gaussian Binomial Coefficient of 1 | :$\dbinom 1 m_q = \delta_{0 m} + \delta_{1 m}$
That is:
:$\dbinom 1 m_q = \begin{cases} 1 & : m = 0 \text { or } m = 1 \\0 & : \text{otherwise} \end{cases}$
where $\dbinom 1 m_q$ denotes a Gaussian binomial coefficient. | By definition of Gaussian binomial coefficient:
:$\dbinom 1 m_q = \ds \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{k + 1} }$
When $m = 0$ the continued product on the {{RHS}} is vacuous, and so:
:$\dbinom 1 0_q = 1$
Let $m > 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dbinom 1 m_q
| r = \prod_{k \mathop... | :$\dbinom 1 m_q = \delta_{0 m} + \delta_{1 m}$
That is:
:$\dbinom 1 m_q = \begin{cases} 1 & : m = 0 \text { or } m = 1 \\0 & : \text{otherwise} \end{cases}$
where $\dbinom 1 m_q$ denotes a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]. | By definition of [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
:$\dbinom 1 m_q = \ds \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{1 - k} } {1 - q^{k + 1} }$
When $m = 0$ the [[Definition:Continued Product|continued product]] on the {{RHS}} is [[Definition:Vacuous Product|vacuous]], and ... | Gaussian Binomial Coefficient of 1 | https://proofwiki.org/wiki/Gaussian_Binomial_Coefficient_of_1 | https://proofwiki.org/wiki/Gaussian_Binomial_Coefficient_of_1 | [
"Gaussian Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient"
] | [
"Definition:Gaussian Binomial Coefficient",
"Definition:Continued Product",
"Definition:Continued Product/Vacuous Product",
"Definition:Fraction/Numerator",
"Category:Gaussian Binomial Coefficients"
] |
proofwiki-14508 | Negated Upper Index of Gaussian Binomial Coefficient | :$\dbinom r k_q = \paren {-1}^k \dbinom {k - r - 1} k_q q^{k r - k \paren {k - 1} / 2}$ | First note that:
{{begin-eqn}}
{{eqn | l = 1 - q^t
| r = q^{-t} \dfrac {1 - q^t} {q^{-t} }
| c =
}}
{{eqn | r = \dfrac {q^{-t} - 1} {q^{-t} }
| c =
}}
{{eqn | r = q^t \paren {q^{-t} - 1}
| c =
}}
{{eqn | n = 1
| r = -q^t \paren {1 - q^{-t} }
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn... | :$\dbinom r k_q = \paren {-1}^k \dbinom {k - r - 1} k_q q^{k r - k \paren {k - 1} / 2}$ | First note that:
{{begin-eqn}}
{{eqn | l = 1 - q^t
| r = q^{-t} \dfrac {1 - q^t} {q^{-t} }
| c =
}}
{{eqn | r = \dfrac {q^{-t} - 1} {q^{-t} }
| c =
}}
{{eqn | r = q^t \paren {q^{-t} - 1}
| c =
}}
{{eqn | n = 1
| r = -q^t \paren {1 - q^{-t} }
| c =
}}
{{end-eqn}}
Then:
{{begin... | Negated Upper Index of Gaussian Binomial Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Gaussian_Binomial_Coefficient | https://proofwiki.org/wiki/Negated_Upper_Index_of_Gaussian_Binomial_Coefficient | [
"Gaussian Binomial Coefficients",
"Negated Upper Index of Binomial Coefficient"
] | [] | [
"Permutation of Indices of Product",
"Closed Form for Triangular Numbers"
] |
proofwiki-14509 | Chu-Vandermonde Identity for Gaussian Binomial Coefficients | {{begin-eqn}}
{{eqn | l = \binom {r + s} n_q
| r = \sum_k \binom r k_q \binom s {n - k}_q q^{\paren {r - k} \paren {n - k} }
| c =
}}
{{eqn | r = \sum_k \binom r k_q \binom s {n - k}_q q^{\paren {s - n + k} k}
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop \in \Z} \dbinom {r + s} n_q q^{n \paren {n - 1} / 2} x^n
| r = \prod_{k \mathop = 1}^{r + s} \paren {1 + q^{k - 1} x}
| c = Gaussian Binomial Theorem
}}
{{eqn | r = \prod_{k \mathop = 1}^r \paren {1 + q^{k - 1} x} \prod_{k \mathop = r + 1}^s \paren {1 + q^{k - 1} x}... | {{begin-eqn}}
{{eqn | l = \binom {r + s} n_q
| r = \sum_k \binom r k_q \binom s {n - k}_q q^{\paren {r - k} \paren {n - k} }
| c =
}}
{{eqn | r = \sum_k \binom r k_q \binom s {n - k}_q q^{\paren {s - n + k} k}
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop \in \Z} \dbinom {r + s} n_q q^{n \paren {n - 1} / 2} x^n
| r = \prod_{k \mathop = 1}^{r + s} \paren {1 + q^{k - 1} x}
| c = [[Gaussian Binomial Theorem]]
}}
{{eqn | r = \prod_{k \mathop = 1}^r \paren {1 + q^{k - 1} x} \prod_{k \mathop = r + 1}^s \paren {1 + q^{k - 1... | Chu-Vandermonde Identity for Gaussian Binomial Coefficients | https://proofwiki.org/wiki/Chu-Vandermonde_Identity_for_Gaussian_Binomial_Coefficients | https://proofwiki.org/wiki/Chu-Vandermonde_Identity_for_Gaussian_Binomial_Coefficients | [
"Chu-Vandermonde Identity",
"Gaussian Binomial Coefficients"
] | [] | [
"Gaussian Binomial Theorem",
"Translation of Index Variable of Product"
] |
proofwiki-14510 | Symmetry Rule for Gaussian Binomial Coefficients | Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = \dbinom n {n - k}_q$
where $\dbinom n k_q$ is a Gaussian binomial coefficient. | If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k_q = \dbinom n {n - k}_q = 0$
{{explain|While we do indeed know that $\dbinom n k {{=}} 0$ for $k < 0$ and $k > n$, we haven't demonstrated anywhere that this still holds for $\dbinom n k_q$.}}
Let $0 \le k \le n$.
Conside... | Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = \dbinom n {n - k}_q$
where $\dbinom n k_q$ is a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]. | If $k < 0$ then $n - k > n$.
Similarly, if $k > n$, then $n - k < 0$.
In both cases:
:$\dbinom n k_q = \dbinom n {n - k}_q = 0$
{{explain|While we do indeed know that $\dbinom n k {{=}} 0$ for $k < 0$ and $k > n$, we haven't demonstrated anywhere that this still holds for $\dbinom n k_q$.}}
Let $0 \le k \le n$.
... | Symmetry Rule for Gaussian Binomial Coefficients | https://proofwiki.org/wiki/Symmetry_Rule_for_Gaussian_Binomial_Coefficients | https://proofwiki.org/wiki/Symmetry_Rule_for_Gaussian_Binomial_Coefficients | [
"Symmetry Rule for Binomial Coefficients",
"Gaussian Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient"
] | [] |
proofwiki-14511 | Reflection Rule for Gaussian Binomial Coefficients | Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = q^{k \paren {n - k} } \dbinom n k_{q^{-1} }$
where $\dbinom n k_q$ is a Gaussian binomial coefficient. | {{begin-eqn}}
{{eqn | l = \dbinom n k_{q^{-1} }
| r = \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - \paren {q^{-1} }^{n - j} } {1 - \paren {q^{-1} }^{j + 1} }
| c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \dfrac {\frac {q^{n - j} - 1} {q^{n - j} } } {\frac {q^{j + 1}... | Let $q \in \R_{\ne 1}, n \in \Z_{>0}, k \in \Z$.
Then:
:$\dbinom n k_q = q^{k \paren {n - k} } \dbinom n k_{q^{-1} }$
where $\dbinom n k_q$ is a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]. | {{begin-eqn}}
{{eqn | l = \dbinom n k_{q^{-1} }
| r = \prod_{j \mathop = 0}^{k - 1} \dfrac {1 - \paren {q^{-1} }^{n - j} } {1 - \paren {q^{-1} }^{j + 1} }
| c = {{Defof|Gaussian Binomial Coefficient}}
}}
{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \dfrac {\frac {q^{n - j} - 1} {q^{n - j} } } {\frac {q^{j + 1}... | Reflection Rule for Gaussian Binomial Coefficients | https://proofwiki.org/wiki/Reflection_Rule_for_Gaussian_Binomial_Coefficients | https://proofwiki.org/wiki/Reflection_Rule_for_Gaussian_Binomial_Coefficients | [
"Gaussian Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient"
] | [
"Closed Form for Triangular Numbers"
] |
proofwiki-14512 | Gaussian Binomial Theorem/Real Numbers | Let $r \in \R$ be a real number.
:$\ds \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k = \prod_{k \mathop \ge 0} \dfrac {1 + q^k x} {1 + q^{r + k} x}$
where:
:$\dbinom r k_q$ denotes a Gaussian binomial coefficient
:$x \in \R: \size x < 1$
:$q \in \R: \size q < 1$. | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss}} | Let $r \in \R$ be a [[Definition:Real Number|real number]].
:$\ds \sum_{k \mathop \in \Z} \dbinom r k_q q^{k \paren {k - 1} / 2} x^k = \prod_{k \mathop \ge 0} \dfrac {1 + q^k x} {1 + q^{r + k} x}$
where:
:$\dbinom r k_q$ denotes a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]
:$x \in \R: ... | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss}} | Gaussian Binomial Theorem/Real Numbers | https://proofwiki.org/wiki/Gaussian_Binomial_Theorem/Real_Numbers | https://proofwiki.org/wiki/Gaussian_Binomial_Theorem/Real_Numbers | [
"Gaussian Binomial Theorem"
] | [
"Definition:Real Number",
"Definition:Gaussian Binomial Coefficient"
] | [] |
proofwiki-14513 | Gaussian Binomial Theorem/Negation of Upper Index | Let $r \in \R$ be a real number.
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop \ge 0} \dfrac {1 + q^{k + r + 1} x} {1 + q^k x}
| r = \sum_{k \mathop \in \Z} \dbinom {-r - 1} k_q q^{k \left({k - 1}\right) / 2} \left({-q^{r + 1} x}\right)^k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \dbinom {k + r} x^k
... | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss}} | Let $r \in \R$ be a [[Definition:Real Number|real number]].
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop \ge 0} \dfrac {1 + q^{k + r + 1} x} {1 + q^k x}
| r = \sum_{k \mathop \in \Z} \dbinom {-r - 1} k_q q^{k \left({k - 1}\right) / 2} \left({-q^{r + 1} x}\right)^k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \... | {{ProofWanted}}
{{Namedfor|Carl Friedrich Gauss}} | Gaussian Binomial Theorem/Negation of Upper Index | https://proofwiki.org/wiki/Gaussian_Binomial_Theorem/Negation_of_Upper_Index | https://proofwiki.org/wiki/Gaussian_Binomial_Theorem/Negation_of_Upper_Index | [
"Gaussian Binomial Theorem"
] | [
"Definition:Real Number",
"Definition:Gaussian Binomial Coefficient"
] | [] |
proofwiki-14514 | Binomial Coefficient n Choose k by n Plus 1 Minus n Choose k + 1 | Let $\sequence {A_{n k} }$ be a sequence defined on $n, k \in \Z_{\ge 0}$ as:
:$A_{n k} = \begin{cases} 1 & : k = 0 \\
0 & : k \ne 0, n = 0 \\
A_{\paren {n - 1} k} + A_{\paren {n - 1} \paren {k - 1} } + \dbinom n k & : \text{otherwise}
\end{cases}$
Then the closed form for $A_{n k}$ is given as:
:$A_{n k} = \paren {n +... | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$A_{n k} = \paren {n + 1} \dbinom n k - \dbinom n {k + 1} + \dbinom n k$
=== Basis for the Induction ===
$\map P 0$ is the case $A_{0 k}$:
Let $k = 0$.
Then:
{{begin-eqn}}
{{eqn | l = A_{0 k}
| r = 1
| c ... | Let $\sequence {A_{n k} }$ be a [[Definition:Sequence|sequence]] defined on $n, k \in \Z_{\ge 0}$ as:
:$A_{n k} = \begin{cases} 1 & : k = 0 \\
0 & : k \ne 0, n = 0 \\
A_{\paren {n - 1} k} + A_{\paren {n - 1} \paren {k - 1} } + \dbinom n k & : \text{otherwise}
\end{cases}$
Then the closed form for $A_{n k}$ is given ... | 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]]:
:$A_{n k} = \paren {n + 1} \dbinom n k - \dbinom n {k + 1} + \dbinom n k$
=== Basis for the Induction ===
$\map P 0$ is the case $A_{0 k}$:
Let $... | Binomial Coefficient n Choose k by n Plus 1 Minus n Choose k + 1 | https://proofwiki.org/wiki/Binomial_Coefficient_n_Choose_k_by_n_Plus_1_Minus_n_Choose_k_+_1 | https://proofwiki.org/wiki/Binomial_Coefficient_n_Choose_k_by_n_Plus_1_Minus_n_Choose_k_+_1 | [
"Binomial Coefficients"
] | [
"Definition:Sequence"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Zero Choose n",
"Zero Choose n",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Binomial Coefficient n Choose k by n Plus 1 Minus n Choose k + 1",
"Pascal's Rule",
"Principle o... |
proofwiki-14515 | Cardinality of Set of Combinations with Repetition | Let $S$ be a finite set with $n$ elements
The number of $k$-combinations of $S$ with repetition is given by:
:$N = \dbinom {n + k - 1} k$ | Let the elements of $S$ be (totally) ordered in some way, by assigning an index to each element.
Thus let $S = \left\{ {a_1, a_2, a_3, \ldots, a_n}\right\}$.
Thus each $k$-combination of $S$ with repetition can be expressed as:
:$\left\{ {\left\{ {a_{r_1}, a_{r_1}, \ldots, a_{r_k} }\right\} }\right\}$
where:
:$r_1, r_2... | Let $S$ be a [[Definition:Finite Set|finite set]] with $n$ [[Definition:Element|elements]]
The number of [[Definition:Combination with Repetition|$k$-combinations of $S$ with repetition]] is given by:
:$N = \dbinom {n + k - 1} k$ | Let the [[Definition:Element|elements]] of $S$ be [[Definition:Total Ordering|(totally) ordered]] in some way, by assigning an [[Definition:Index (Indexing Set)|index]] to each [[Definition:Element|element]].
Thus let $S = \left\{ {a_1, a_2, a_3, \ldots, a_n}\right\}$.
Thus each [[Definition:Combination with Repetiti... | Cardinality of Set of Combinations with Repetition | https://proofwiki.org/wiki/Cardinality_of_Set_of_Combinations_with_Repetition | https://proofwiki.org/wiki/Cardinality_of_Set_of_Combinations_with_Repetition | [
"Combinations with Repetition"
] | [
"Definition:Finite Set",
"Definition:Element",
"Definition:Combination with Repetition"
] | [
"Definition:Element",
"Definition:Total Ordering",
"Definition:Indexing Set/Index",
"Definition:Element",
"Definition:Combination with Repetition",
"Definition:Element",
"Definition:Integer",
"Definition:Element",
"Definition:Subset",
"Definition:Set",
"Definition:Element",
"Cardinality of Set... |
proofwiki-14516 | Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
where:
:$\sqbrk {n \ge m}$ is Iverson's convention
:$\ds {n + 1 \brack k + 1}$ denotes an unsigned Stirling number of the first kind
:$\ds {k \brace m}$ denotes a Stirling number of the sec... | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$ | Let $m, n \in \Z_{\ge 0}$.
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$
where:
:$\sqbrk {n \ge m}$ is [[Definition:Iverson's Convention|Iverson's convention]]
:$\ds {n + 1 \brack k + 1}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsig... | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_k {n + 1 \brack k + 1} {k \brace m} \paren {-1}^{k - m} = \sqbrk {n \ge m} \dfrac {n!} {m!}$ | Sum over k of Unsigned Stirling Number of the First Kind of n+1 with k+1 by Stirling Number of the Second Kind of k with m by -1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_k+1_by_Stirling_Number_of_the_Second_Kind_of_k_with_m_by_-1^k-m | https://proofwiki.org/wiki/Sum_over_k_of_Unsigned_Stirling_Number_of_the_First_Kind_of_n+1_with_k+1_by_Stirling_Number_of_the_Second_Kind_of_k_with_m_by_-1^k-m | [
"Stirling Numbers"
] | [
"Definition:Iverson's Convention",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14517 | Dixon's Identity/General Case | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$ | From Sum over $k$ of $\dbinom {m - r + s} k$ by $\dbinom {n + r - s} {n - k}$ by $\dbinom {r + k} {m + n}$:
:$\ds \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n} = \binom r m \binom s n$
Setting $\tuple {m, n, r, s, k} \gets \tuple {m + k, l - k, m + n, n + l, j}$ into the... | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$ | From [[Sum over k of m+r+s Choose k by n+r-s Choose n-k by r+k Choose m+n|Sum over $k$ of $\dbinom {m - r + s} k$ by $\dbinom {n + r - s} {n - k}$ by $\dbinom {r + k} {m + n}$]]:
:$\ds \sum_{k \mathop \in \Z} \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m + n} = \binom r m \binom s n$
Setting $\tu... | Dixon's Identity/General Case | https://proofwiki.org/wiki/Dixon's_Identity/General_Case | https://proofwiki.org/wiki/Dixon's_Identity/General_Case | [
"Binomial Coefficients"
] | [] | [
"Sum over k of m+r+s Choose k by n+r-s Choose n-k by r+k Choose m+n",
"Symmetry Rule for Binomial Coefficients"
] |
proofwiki-14518 | Dixon's Identity | For $n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \binom {2 n} {n + k}^3 = \dfrac {\paren {3 n}!} {\paren {n!}^3}$ | Follows directly from Dixon's Identity/General Case:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$
setting $l = m = n$.
{{qed}}
{{Namedfor|Alfred Cardew Dixon|cat = Dixon, A.C.}} | For $n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \binom {2 n} {n + k}^3 = \dfrac {\paren {3 n}!} {\paren {n!}^3}$ | Follows directly from [[Dixon's Identity/General Case]]:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k} \dbinom {m + n} {m + k} \dbinom {n + l} {n + k} = \dfrac {\paren {l + m + n}!} {l! \, m! \, n!}$
setting $l = m = n$.
{{qed}}
{{Namedfor|Alfred Cardew Dixon|cat = Dixon, A.C.}} | Dixon's Identity | https://proofwiki.org/wiki/Dixon's_Identity | https://proofwiki.org/wiki/Dixon's_Identity | [
"Binomial Coefficients"
] | [] | [
"Dixon's Identity/General Case"
] |
proofwiki-14519 | Dixon's Identity/Gaussian Binomial Form/Formulation 2 | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k}_q \dbinom {m + n} {m + k}_q \dbinom {n + l} {n + k}_q = \dfrac {\paren {l + m + n}!_q} {l!_q \, m!_q \, n!_q}$
where:
:$\dbinom {l + m} {l + k}_q$ denotes a Gaussian binomial coefficient
:$l!_q$ is defined as $\ds \prod_{k... | {{ProofWanted}}
{{Namedfor|Alfred Cardew Dixon}} | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {l + m} {l + k}_q \dbinom {m + n} {m + k}_q \dbinom {n + l} {n + k}_q = \dfrac {\paren {l + m + n}!_q} {l!_q \, m!_q \, n!_q}$
where:
:$\dbinom {l + m} {l + k}_q$ denotes a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial co... | {{ProofWanted}}
{{Namedfor|Alfred Cardew Dixon}} | Dixon's Identity/Gaussian Binomial Form/Formulation 2 | https://proofwiki.org/wiki/Dixon's_Identity/Gaussian_Binomial_Form/Formulation_2 | https://proofwiki.org/wiki/Dixon's_Identity/Gaussian_Binomial_Form/Formulation_2 | [
"Gaussian Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient"
] | [] |
proofwiki-14520 | Dixon's Identity/Gaussian Binomial Form/Formulation 1 | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {m - r - s} k_q \dbinom {n + r - s} {n - k}_q \dbinom {r + k} {m + n}_q = \dbinom r m_q \dbinom s n_q$
where $\dbinom r m_q$ denotes a Gaussian binomial coefficient | {{ProofWanted}}
{{Namedfor|Alfred Cardew Dixon}} | For $l, m, n \in \Z_{\ge 0}$:
:$\ds \sum_{k \mathop \in \Z} \paren {-1}^k \dbinom {m - r - s} k_q \dbinom {n + r - s} {n - k}_q \dbinom {r + k} {m + n}_q = \dbinom r m_q \dbinom s n_q$
where $\dbinom r m_q$ denotes a [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]] | {{ProofWanted}}
{{Namedfor|Alfred Cardew Dixon}} | Dixon's Identity/Gaussian Binomial Form/Formulation 1 | https://proofwiki.org/wiki/Dixon's_Identity/Gaussian_Binomial_Form/Formulation_1 | https://proofwiki.org/wiki/Dixon's_Identity/Gaussian_Binomial_Form/Formulation_1 | [
"Gaussian Binomial Coefficients"
] | [
"Definition:Gaussian Binomial Coefficient"
] | [] |
proofwiki-14521 | Analytic Continuation of Generating Function of Dirichlet Series | Let $\ds \map \lambda s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a Dirichlet series
Let $c \in \R$ be greater than the abscissa of absolute convergence of $\lambda$ and greater than $0$.
Let $\ds \map g z = \sum_{k \mathop = 1}^\infty \map \lambda {c k} z^k $ be the generating function of $\map \lambda {c k... | We will first show that the series is meromorphic functions on $\C$ with simple poles at $n^c$.
For $\cmod z < R$, pick $M$ such that:
:$M^c > 2 R$
This is always possible, as:
:$c > 0$
Then for $n > M$:
{{begin-eqn}}
{{eqn | l = \cmod {1 - \frac z {n^c} }
| o = >
| r = 1 - \cmod {\frac z {n^c} }
| c... | Let $\ds \map \lambda s = \sum_{n \mathop = 1}^\infty \frac {a_n} {n^s}$ be a [[Definition:Dirichlet Series|Dirichlet series]]
Let $c \in \R$ be greater than the [[Existence of Abscissa of Absolute Convergence|abscissa of absolute convergence]] of $\lambda$ and greater than $0$.
Let $\ds \map g z = \sum_{k \mathop =... | We will first show that the [[Definition:Series|series]] is [[Definition:Meromorphic Function|meromorphic functions]] on $\C$ with [[Definition:Simple Pole|simple poles]] at $n^c$.
For $\cmod z < R$, pick $M$ such that:
:$M^c > 2 R$
This is always possible, as:
:$c > 0$
Then for $n > M$:
{{begin-eqn}}
{{eqn | l = ... | Analytic Continuation of Generating Function of Dirichlet Series | https://proofwiki.org/wiki/Analytic_Continuation_of_Generating_Function_of_Dirichlet_Series | https://proofwiki.org/wiki/Analytic_Continuation_of_Generating_Function_of_Dirichlet_Series | [
"Dirichlet Series"
] | [
"Definition:Dirichlet Series",
"Existence of Abscissa of Absolute Convergence",
"Definition:Generating Function",
"Definition:Analytic Continuation"
] | [
"Definition:Series",
"Definition:Meromorphic Function",
"Definition:Order of Pole/Simple Pole",
"Triangle Inequality",
"Existence of Abscissa of Absolute Convergence",
"Definition:Series",
"Definition:Convergent Series/Number Field",
"Definition:Addition/Summand",
"Definition:Meromorphic Function",
... |
proofwiki-14522 | Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Then there exists a unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$. | We have:
{{begin-eqn}}
{{eqn | l = \dbinom y {n + 1}
| o = \le
| r = \dbinom x {n + 1}
| c = Ordering of Binomial Coefficients
}}
{{eqn | o = \le
| r = \dbinom {y + 1} {n + 1}
| c =
}}
{{eqn | r = \dbinom y {n + 1} + \dbinom y n
| c = Pascal's Rule
}}
{{end-eqn}}
{{finish}} | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $x, y \in \R$ be [[Definition:Real Number|real numbers]] which satisfy:
:$n \le y \le x \le y + 1$
Then there exists a [[Definition:Unique|unique]] [[Definition:Real Number|real number]] $z$ such that:
:$\dbinom x {n + 1} = \dbinom y ... | We have:
{{begin-eqn}}
{{eqn | l = \dbinom y {n + 1}
| o = \le
| r = \dbinom x {n + 1}
| c = [[Ordering of Binomial Coefficients]]
}}
{{eqn | o = \le
| r = \dbinom {y + 1} {n + 1}
| c =
}}
{{eqn | r = \dbinom y {n + 1} + \dbinom y n
| c = [[Pascal's Rule]]
}}
{{end-eqn}}
{{finish}} | Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n | https://proofwiki.org/wiki/Uniqueness_of_Real_z_such_that_x_Choose_n+1_Equals_y_Choose_n+1_Plus_z_Choose_n | https://proofwiki.org/wiki/Uniqueness_of_Real_z_such_that_x_Choose_n+1_Equals_y_Choose_n+1_Plus_z_Choose_n | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Real Number",
"Definition:Unique",
"Definition:Real Number"
] | [
"Ordering of Binomial Coefficients",
"Pascal's Rule"
] |
proofwiki-14523 | X Choose n leq y Choose n + z Choose n-1 where n leq y leq x leq y+1 and n-1 leq z leq y | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Let $z$ be the unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
Its uniqueness is proved at Uniqueness of Real $z$ such that $\dbin... | If $z \ge n$, then from Ordering of Binomial Coefficients:
:$\dbinom z {n + 1} \le \dbinom y {n + 1}$
Otherwise $n - 1 \le z \le n$, and:
:$\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$
In either case:
:$(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$
Therefore:
{{begin-eqn}}
{{eqn | l = \dbinom {z + 1} {n + 1}
... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $x, y \in \R$ be [[Definition:Real Number|real numbers]] which satisfy:
:$n \le y \le x \le y + 1$
Let $z$ be the [[Definition:Unique|unique]] [[Definition:Real Number|real number]] $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + ... | If $z \ge n$, then from [[Ordering of Binomial Coefficients]]:
:$\dbinom z {n + 1} \le \dbinom y {n + 1}$
Otherwise $n - 1 \le z \le n$, and:
:$\dbinom z {n + 1} \le 0 \le \dbinom y {n + 1}$
In either case:
:$(1): \quad \dbinom z {n + 1} \le \dbinom y {n + 1}$
Therefore:
{{begin-eqn}}
{{eqn | l = \dbinom {z + 1... | X Choose n leq y Choose n + z Choose n-1 where n leq y leq x leq y+1 and n-1 leq z leq y | https://proofwiki.org/wiki/X_Choose_n_leq_y_Choose_n_+_z_Choose_n-1_where_n_leq_y_leq_x_leq_y+1_and_n-1_leq_z_leq_y | https://proofwiki.org/wiki/X_Choose_n_leq_y_Choose_n_+_z_Choose_n-1_where_n_leq_y_leq_x_leq_y+1_and_n-1_leq_z_leq_y | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Real Number",
"Definition:Unique",
"Definition:Real Number",
"Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n"
] | [
"Ordering of Binomial Coefficients",
"Pascal's Rule",
"Definition:By Hypothesis",
"Definition:Summation",
"Definition:Negative/Integer",
"Definition:Binomial Coefficient",
"Definition:Positive/Integer",
"Definition:Binomial Coefficient",
"Definition:Positive/Integer",
"Definition:Kronecker Delta",... |
proofwiki-14524 | Upper Bound for Binomial Coefficient | Let $n, k \in \Z$ such that $n \ge k \ge 0$.
Then:
:$\dbinom n k \le \left({\dfrac {n e} k}\right)^k$
where $\dbinom n k$ denotes a binomial coefficient. | From Lower and Upper Bound of Factorial, we have that:
:$\dfrac {k^k} {e^{k - 1} } \le k!$
so that:
:$(1): \quad \dfrac 1 {k!} \le \dfrac {e^{k - 1} } {k^k}$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n k
| r = \dfrac {n^\underline k} {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | o = \le
| r = \df... | Let $n, k \in \Z$ such that $n \ge k \ge 0$.
Then:
:$\dbinom n k \le \left({\dfrac {n e} k}\right)^k$
where $\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Lower and Upper Bound of Factorial]], we have that:
:$\dfrac {k^k} {e^{k - 1} } \le k!$
so that:
:$(1): \quad \dfrac 1 {k!} \le \dfrac {e^{k - 1} } {k^k}$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n k
| r = \dfrac {n^\underline k} {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | o = \le
... | Upper Bound for Binomial Coefficient | https://proofwiki.org/wiki/Upper_Bound_for_Binomial_Coefficient | https://proofwiki.org/wiki/Upper_Bound_for_Binomial_Coefficient | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Lower and Upper Bound of Factorial"
] |
proofwiki-14525 | Lower Bound for Binomial Coefficient | Let $n, k \in \Z$ such that $n \ge k \ge 0$.
Then:
:$\dbinom n k \ge \paren {\dfrac {\paren {n - k} e} k}^k \dfrac 1 {e k}$
where $\dbinom n k$ denotes a binomial coefficient. | From Lower and Upper Bound of Factorial, we have that:
:$k! \le \dfrac {k^{k + 1} } {e^{k - 1} }$
so that:
:$(1): \quad \dfrac 1 {k!} \ge \dfrac {e^{k - 1} } {k^{k + 1} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n k
| r = \dfrac {n^\underline k} {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | o = \ge
... | Let $n, k \in \Z$ such that $n \ge k \ge 0$.
Then:
:$\dbinom n k \ge \paren {\dfrac {\paren {n - k} e} k}^k \dfrac 1 {e k}$
where $\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | From [[Lower and Upper Bound of Factorial]], we have that:
:$k! \le \dfrac {k^{k + 1} } {e^{k - 1} }$
so that:
:$(1): \quad \dfrac 1 {k!} \ge \dfrac {e^{k - 1} } {k^{k + 1} }$
Then:
{{begin-eqn}}
{{eqn | l = \dbinom n k
| r = \dfrac {n^\underline k} {k!}
| c = {{Defof|Binomial Coefficient}}
}}
{{eqn | ... | Lower Bound for Binomial Coefficient | https://proofwiki.org/wiki/Lower_Bound_for_Binomial_Coefficient | https://proofwiki.org/wiki/Lower_Bound_for_Binomial_Coefficient | [
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient"
] | [
"Lower and Upper Bound of Factorial"
] |
proofwiki-14526 | Sum over k of n Choose k by p^k by (1-p)^n-k by Absolute Value of k-np | Let $n \in \Z_{\ge 0}$ be a non-negative integer.
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom n k p^k \paren {1 - p}^{n - k} \size {k - n p} = 2 \ceiling {n p} \dbinom n {\ceiling {n p} } p^{\ceiling {n p} } \paren {1 - p}^{n - 1 - \ceiling {n p} }$ | Let $t_k = k \dbinom n k p^k \paren {1 - p}^{n + 1 - k}$.
Then:
:$t_k - t_{k + 1} = \dbinom n k p^k \paren {1 - p}^{n - k} \paren {k - n p}$
Thus the stated summation is:
:$\ds \sum_{k \mathop < \ceiling {n p} } \paren {t_{k + 1} - t_k} + \sum_{k \mathop \ge \ceiling {n p} } \paren {t_k - t_{k + 1} } = 2 t_{\ceiling {n... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom n k p^k \paren {1 - p}^{n - k} \size {k - n p} = 2 \ceiling {n p} \dbinom n {\ceiling {n p} } p^{\ceiling {n p} } \paren {1 - p}^{n - 1 - \ceiling {n p} }$ | Let $t_k = k \dbinom n k p^k \paren {1 - p}^{n + 1 - k}$.
Then:
:$t_k - t_{k + 1} = \dbinom n k p^k \paren {1 - p}^{n - k} \paren {k - n p}$
Thus the stated summation is:
:$\ds \sum_{k \mathop < \ceiling {n p} } \paren {t_{k + 1} - t_k} + \sum_{k \mathop \ge \ceiling {n p} } \paren {t_k - t_{k + 1} } = 2 t_{\ceiling... | Sum over k of n Choose k by p^k by (1-p)^n-k by Absolute Value of k-np | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_p^k_by_(1-p)^n-k_by_Absolute_Value_of_k-np | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_p^k_by_(1-p)^n-k_by_Absolute_Value_of_k-np | [
"Binomial Coefficients"
] | [
"Definition:Positive/Integer"
] | [] |
proofwiki-14527 | Sequence of General Harmonic Numbers Converges for Index Greater than 1 | Let $\map {H^{\paren r} } n$ denote the general harmonic number:
:$\ds \map {H^{\paren r} } n = \sum_{k \mathop = 1}^n \frac 1 {k^r}$
for $r \in \R_{>0}$.
Let $r > 1$.
Then as $n \to \infty$, $\map {H^{\paren r} } n$ is convergent with an upper bound of $\dfrac {2^{r - 1} } {2^{r - 1} - 1}$. | Let $m \in \N$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \harm r {2^{m - 1} }
| r = \harm r {2^{m - 1} - 1} + \dfrac 1 {\paren {2^{m - 1} }^r} + \dfrac 1 {\paren {2^{m - 1} + 1}^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} + \paren {2^{m - 1} - 1} }^r}
| c =
}}
{{eqn | o = <
| r = \harm r {2^{m - 1}... | Let $\map {H^{\paren r} } n$ denote the [[Definition:General Harmonic Numbers|general harmonic number]]:
:$\ds \map {H^{\paren r} } n = \sum_{k \mathop = 1}^n \frac 1 {k^r}$
for $r \in \R_{>0}$.
Let $r > 1$.
Then as $n \to \infty$, $\map {H^{\paren r} } n$ is [[Definition:Convergent Series of Numbers|convergent]] w... | Let $m \in \N$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \harm r {2^{m - 1} }
| r = \harm r {2^{m - 1} - 1} + \dfrac 1 {\paren {2^{m - 1} }^r} + \dfrac 1 {\paren {2^{m - 1} + 1}^r} + \cdots + \dfrac 1 {\paren {2^{m - 1} + \paren {2^{m - 1} - 1} }^r}
| c =
}}
{{eqn | o = <
| r = \harm r {2^{m - ... | Sequence of General Harmonic Numbers Converges for Index Greater than 1/Proof | https://proofwiki.org/wiki/Sequence_of_General_Harmonic_Numbers_Converges_for_Index_Greater_than_1 | https://proofwiki.org/wiki/Sequence_of_General_Harmonic_Numbers_Converges_for_Index_Greater_than_1/Proof | [
"Sequence of General Harmonic Numbers Converges for Index Greater than 1",
"General Harmonic Numbers"
] | [
"Definition:Harmonic Numbers/General Definition",
"Definition:Convergent Series/Number Field",
"Definition:Upper Bound"
] | [
"Ordering of Reciprocals",
"Sum of Geometric Sequence",
"Definition:Series/Sequence of Partial Sums",
"Monotone Convergence Theorem (Real Analysis)",
"Definition:Convergent Series/Number Field",
"Definition:Upper Bound"
] |
proofwiki-14528 | Inductive Construction of Sigma-Algebra Generated by Collection of Subsets | Let $\EE$ be a set of sets which are subsets of some set $X$.
Let $\map \sigma \EE$ be the $\sigma$-algebra generated by $\EE$.
Then $\map \sigma \EE$ can be constructed inductively.
The construction is as follows:
Let $\Omega$ denote the minimal uncountable well-ordered set.
Let $\alpha$ be an arbitrary initial segme... | === Step 1 ===
We will show that:
:$\map \sigma \EE \subseteq \EE_\Omega$.
Define:
:$\OO := \set {o \in \Omega: \EE_o \in \map \sigma \EE}$
By the definition of a $\sigma$-algebra:
:$\EE_1 \subseteq \map \sigma \EE$.
From $\sigma$-Algebra of Countable Sets, if $\beta$ immediately precedes $\alpha$, then $\EE_\alpha$ i... | Let $\EE$ be a [[Definition:Set of Sets|set of sets]] which are [[Definition:Subset|subsets]] of some [[Definition:Set|set]] $X$.
Let $\map \sigma \EE$ be the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\EE$]].
Then $\map \sigma \EE$ can be constructed [[Induction on... | === Step 1 ===
We will show that:
:$\map \sigma \EE \subseteq \EE_\Omega$.
Define:
:$\OO := \set {o \in \Omega: \EE_o \in \map \sigma \EE}$
By the [[Definition:Sigma-Algebra|definition of a $\sigma$-algebra]]:
:$\EE_1 \subseteq \map \sigma \EE$.
From [[Sigma-Algebra of Countable Sets|$\sigma$-Algebra of Countabl... | Inductive Construction of Sigma-Algebra Generated by Collection of Subsets | https://proofwiki.org/wiki/Inductive_Construction_of_Sigma-Algebra_Generated_by_Collection_of_Subsets | https://proofwiki.org/wiki/Inductive_Construction_of_Sigma-Algebra_Generated_by_Collection_of_Subsets | [
"Sigma-Algebras"
] | [
"Definition:Set of Sets",
"Definition:Subset",
"Definition:Set",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Principle of Mathematical Induction/Well-Ordered Set",
"Definition:Minimal Uncountable Well-Ordered Set",
"Definition:Ordinal/Definition 3",
"Definition:Immediate Predecesso... | [
"Definition:Sigma-Algebra",
"Sigma-Algebra of Countable Sets",
"Definition:Immediate Predecessor Element",
"Definition:Sigma-Algebra",
"Definition:Strictly Precede",
"Definition:Set Union/Countable Union",
"Definition:Measurable Set",
"Definition:Sigma-Algebra",
"Definition:Set Union/Family of Sets"... |
proofwiki-14529 | Generating Function for Bernoulli Polynomials | Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial.
Then the generating function for $B_n$ is:
:$\ds \frac {t e^{t x} } {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k$ | By definition of the generating function for Bernoulli numbers:
:$\ds \frac t {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k$
By Power Series Expansion for Exponential Function:
:$\ds e^{t x} = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k$
Thus:
:$\ds \frac {t e^{t x} } {e^t - 1} = \paren {\sum_{k \ma... | Let $\map {B_n} x$ denote the $n$th [[Definition:Bernoulli Polynomial|Bernoulli polynomial]].
Then the [[Definition:Generating Function|generating function]] for $B_n$ is:
:$\ds \frac {t e^{t x} } {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {\map {B_k} x} {k!} t^k$ | By definition of the [[Definition:Bernoulli Numbers/Generating Function|generating function for Bernoulli numbers]]:
:$\ds \frac t {e^t - 1} = \sum_{k \mathop = 0}^\infty \frac {B_k} {k!} t^k$
By [[Power Series Expansion for Exponential Function]]:
:$\ds e^{t x} = \sum_{k \mathop = 0}^\infty \frac {x^k} {k!} t^k$
... | Generating Function for Bernoulli Polynomials | https://proofwiki.org/wiki/Generating_Function_for_Bernoulli_Polynomials | https://proofwiki.org/wiki/Generating_Function_for_Bernoulli_Polynomials | [
"Bernoulli Polynomials",
"Examples of Generating Functions"
] | [
"Definition:Bernoulli Polynomial",
"Definition:Generating Function"
] | [
"Definition:Bernoulli Numbers/Generating Function",
"Power Series Expansion for Exponential Function",
"Category:Bernoulli Polynomials",
"Category:Examples of Generating Functions"
] |
proofwiki-14530 | Symmetry of Bernoulli Polynomial | Let $\map {B_n} x$ denote the nth Bernoulli polynomial.
Then:
:$\map {B_n} {1 - x} = \paren {-1}^n \map {B_n} x$ | Let $\map G {t, x}$ denote the Generating Function for Bernoulli Polynomials:
:$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map G {t, 1 - x}
| r = \frac {t e^{t \paren {1 - x} } } {e^t - 1}
}}
{{eqn | r = \frac {t e^{t - t x} } {e^t - 1}
}}
{{eqn | r = \frac {t e^{-t x} } {1 - ... | Let $\map {B_n} x$ denote the nth [[Definition:Bernoulli Polynomial|Bernoulli polynomial]].
Then:
:$\map {B_n} {1 - x} = \paren {-1}^n \map {B_n} x$ | Let $\map G {t, x}$ denote the [[Generating Function for Bernoulli Polynomials]]:
:$\map G {t, x} = \dfrac {t e^{t x} } {e^t - 1}$
Then:
{{begin-eqn}}
{{eqn | l = \map G {t, 1 - x}
| r = \frac {t e^{t \paren {1 - x} } } {e^t - 1}
}}
{{eqn | r = \frac {t e^{t - t x} } {e^t - 1}
}}
{{eqn | r = \frac {t e^{-t x} ... | Symmetry of Bernoulli Polynomial | https://proofwiki.org/wiki/Symmetry_of_Bernoulli_Polynomial | https://proofwiki.org/wiki/Symmetry_of_Bernoulli_Polynomial | [
"Bernoulli Polynomials"
] | [
"Definition:Bernoulli Polynomial"
] | [
"Generating Function for Bernoulli Polynomials",
"Category:Bernoulli Polynomials"
] |
proofwiki-14531 | Value of Odd Bernoulli Polynomial at One Half | Let $\map {B_n} x$ denote the $n$th Bernoulli polynomial.
Then:
:$\map {B_{2 n + 1} } {\dfrac 1 2} = 0$ | {{begin-eqn}}
{{eqn | l = \map {B_{2 n + 1} } {1 - x}
| r = \paren {-1}^{2 n + 1} \map {B_{2 n + 1} } x
| c = Symmetry of Bernoulli Polynomial
}}
{{eqn | r = \paren {-1} \map {B_{2 n + 1} } x
}}
{{eqn | ll= \leadsto
| l = \map {B_{2 n + 1} } {\frac 1 2}
| r = \paren {-1} \map {B_{2 n + 1} } {\fr... | Let $\map {B_n} x$ denote the $n$th [[Definition:Bernoulli Polynomial|Bernoulli polynomial]].
Then:
:$\map {B_{2 n + 1} } {\dfrac 1 2} = 0$ | {{begin-eqn}}
{{eqn | l = \map {B_{2 n + 1} } {1 - x}
| r = \paren {-1}^{2 n + 1} \map {B_{2 n + 1} } x
| c = [[Symmetry of Bernoulli Polynomial]]
}}
{{eqn | r = \paren {-1} \map {B_{2 n + 1} } x
}}
{{eqn | ll= \leadsto
| l = \map {B_{2 n + 1} } {\frac 1 2}
| r = \paren {-1} \map {B_{2 n + 1} } ... | Value of Odd Bernoulli Polynomial at One Half | https://proofwiki.org/wiki/Value_of_Odd_Bernoulli_Polynomial_at_One_Half | https://proofwiki.org/wiki/Value_of_Odd_Bernoulli_Polynomial_at_One_Half | [
"Bernoulli Polynomials"
] | [
"Definition:Bernoulli Polynomial"
] | [
"Symmetry of Bernoulli Polynomial",
"Category:Bernoulli Polynomials"
] |
proofwiki-14532 | Harmonic Number is Greater than Logarithm plus Gamma | :$H_n > \ln n + \gamma$
where:
:$H_n$ denotes the $n$th harmonic number
:$\gamma$ denotes the Euler-Mascheroni constant. | From Approximate Size of Sum of Harmonic Series:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
where $0 < \epsilon < \dfrac 1 {252 n^6}$.
We have that:
:$\dfrac 1 {2 n} + \dfrac 1 {120 n^4} > \dfrac 1 {12 n^2} + \dfrac 1 {252 n^6}$
and the result follows.
{{qed}} | :$H_n > \ln n + \gamma$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$\gamma$ denotes the [[Definition:Euler-Mascheroni Constant|Euler-Mascheroni constant]]. | From [[Approximate Size of Sum of Harmonic Series]]:
:$H_n \approx \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - \epsilon$
where $0 < \epsilon < \dfrac 1 {252 n^6}$.
We have that:
:$\dfrac 1 {2 n} + \dfrac 1 {120 n^4} > \dfrac 1 {12 n^2} + \dfrac 1 {252 n^6}$
and the result follows.
{{... | Harmonic Number is Greater than Logarithm plus Gamma | https://proofwiki.org/wiki/Harmonic_Number_is_Greater_than_Logarithm_plus_Gamma | https://proofwiki.org/wiki/Harmonic_Number_is_Greater_than_Logarithm_plus_Gamma | [
"Harmonic Numbers"
] | [
"Definition:Harmonic Numbers",
"Definition:Euler-Mascheroni Constant"
] | [
"Approximate Size of Sum of Harmonic Series"
] |
proofwiki-14533 | Summation to n of Power of k over k | :$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th harmonic number
:$\dbinom n k$ denotes a binomial coefficient. | The proof proceeds by induction over $n$.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \ds \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \sum_{k \math... | :$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] over $n$.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \ds \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
=== Basis for the Inducti... | Summation to n of Power of k over k/Proof 1 | https://proofwiki.org/wiki/Summation_to_n_of_Power_of_k_over_k | https://proofwiki.org/wiki/Summation_to_n_of_Power_of_k_over_k/Proof_1 | [
"Harmonic Numbers",
"Binomial Coefficients",
"Summation to n of Power of k over k"
] | [
"Definition:Harmonic Numbers",
"Definition:Binomial Coefficient"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Binomial Coefficient with Self",
"Harmonic Numbers/Examples/H1",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Summation to n of Power of k over k/Proof 1",
"Pascal's Rule",
"... |
proofwiki-14534 | Summation to n of Power of k over k | :$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th harmonic number
:$\dbinom n k$ denotes a binomial coefficient. | Differentiating {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d x} \paren {\sum_{k \mathop = 1}^n \dfrac {x^k} k}
| r = \sum_{k \mathop = 1}^n\dfrac \d {\d x}\paren {\dfrac {x^k} k}
| c = Applications of Linear Combination of Derivatives
}}
{{eqn | r = \sum_{k \mathop = 1}^n k \dfrac {x... | :$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]. | [[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$:
{{begin-eqn}}
{{eqn | l = \dfrac \d {\d x} \paren {\sum_{k \mathop = 1}^n \dfrac {x^k} k}
| r = \sum_{k \mathop = 1}^n\dfrac \d {\d x}\paren {\dfrac {x^k} k}
| c = Applications of [[Linear Combination of Derivatives]]
}}
{{eqn | r ... | Summation to n of Power of k over k/Proof 2 | https://proofwiki.org/wiki/Summation_to_n_of_Power_of_k_over_k | https://proofwiki.org/wiki/Summation_to_n_of_Power_of_k_over_k/Proof_2 | [
"Harmonic Numbers",
"Binomial Coefficients",
"Summation to n of Power of k over k"
] | [
"Definition:Harmonic Numbers",
"Definition:Binomial Coefficient"
] | [
"Definition:Differentiation",
"Linear Combination of Derivatives",
"Power Rule for Derivatives",
"Translation of Index Variable of Summation",
"Sum of Geometric Sequence",
"Integration by Substitution",
"Binomial Theorem",
"Binomial Coefficient with Zero",
"Linear Combination of Integrals/Indefinite... |
proofwiki-14535 | Harmonic Number as Unsigned Stirling Number of First Kind over Factorial | :$H_n = \dfrac { {n + 1} \brack 2} {n!}$
where:
:$H_n$ denotes the $n$th harmonic number
:$n!$ denotes the $n$th factorial
:$\ds { {n + 1} \brack 2}$ denotes an unsigned Stirling number of the first kind. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$H_n = \dfrac { {n + 1} \brack 2} {n!}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = H_0
| r = 0
| c =
}}
{{eqn | r = \dfrac {1 \brack 2} {0!}
| c = Unsigned Stirling Number of the First Kind of Numb... | :$H_n = \dfrac { {n + 1} \brack 2} {n!}$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$n!$ denotes the [[Definition:Factorial|$n$th factorial]]
:$\ds { {n + 1} \brack 2}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$H_n = \dfrac { {n + 1} \brack 2} {n!}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = H_0
| r = 0
| c =
}}
{{eqn | r = \dfrac {1 \brack... | Harmonic Number as Unsigned Stirling Number of First Kind over Factorial | https://proofwiki.org/wiki/Harmonic_Number_as_Unsigned_Stirling_Number_of_First_Kind_over_Factorial | https://proofwiki.org/wiki/Harmonic_Number_as_Unsigned_Stirling_Number_of_First_Kind_over_Factorial | [
"Harmonic Numbers",
"Stirling Numbers",
"Factorials"
] | [
"Definition:Harmonic Numbers",
"Definition:Factorial",
"Definition:Stirling Numbers of the First Kind/Unsigned"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Unsigned Stirling Number of the First Kind of Number with Greater",
"Principle of Mathematical Induction"
] |
proofwiki-14536 | Sum of Harmonic Numbers approaches Harmonic Number of Product of Indices | Let $m, n \in \Z_{>0}$ be (strictly) positive integers.
Let:
:$\map T {m, n} := H_m + H_n - H_{m n}$
where $H_n$ denotes the $n$th harmonic number.
Then as $m$ or $n$ increases, $\map T {m, n}$ never increases, and reaches its minimum when $m$ and $n$ approach infinity. | {{begin-eqn}}
{{eqn | l = \map T {m + 1, n} - \map T {m, n}
| r = \left({H_{m + 1} + H_n - H_{\paren {m + 1} n} }\right) - \paren {H_m + H_n - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} + \paren {H_{\paren {m + 1} n} - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} - \paren {\frac 1 {m n + 1} ... | Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
Let:
:$\map T {m, n} := H_m + H_n - H_{m n}$
where $H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]].
Then as $m$ or $n$ increases, $\map T {m, n}$ never increases, and reaches its minimum when $m$ a... | {{begin-eqn}}
{{eqn | l = \map T {m + 1, n} - \map T {m, n}
| r = \left({H_{m + 1} + H_n - H_{\paren {m + 1} n} }\right) - \paren {H_m + H_n - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} + \paren {H_{\paren {m + 1} n} - H_{m n} }
| c =
}}
{{eqn | r = \dfrac 1 {m + 1} - \paren {\frac 1 {m n + 1} ... | Sum of Harmonic Numbers approaches Harmonic Number of Product of Indices | https://proofwiki.org/wiki/Sum_of_Harmonic_Numbers_approaches_Harmonic_Number_of_Product_of_Indices | https://proofwiki.org/wiki/Sum_of_Harmonic_Numbers_approaches_Harmonic_Number_of_Product_of_Indices | [
"Harmonic Numbers"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Harmonic Numbers",
"Definition:Infinity"
] | [
"Approximate Size of Sum of Harmonic Series",
"Definition:Euler-Mascheroni Constant"
] |
proofwiki-14537 | Summation over k to n of Natural Logarithm of k | :$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$
where $n!$ denotes the $n$th factorial. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \ln k
| r = \ln \prod_{k \mathop = 1}^n k
| c = Summation of General Logarithms
}}
{{eqn | r = \map \ln {n!}
| c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}} | :$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$
where $n!$ denotes the $n$th [[Definition:Factorial|factorial]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \ln k
| r = \ln \prod_{k \mathop = 1}^n k
| c = [[Summation of General Logarithms]]
}}
{{eqn | r = \map \ln {n!}
| c = {{Defof|Factorial}}
}}
{{end-eqn}}
{{qed}} | Summation over k to n of Natural Logarithm of k | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Natural_Logarithm_of_k | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Natural_Logarithm_of_k | [
"Factorials",
"Natural Logarithms"
] | [
"Definition:Factorial"
] | [
"Summation of General Logarithms"
] |
proofwiki-14538 | Difference between Summation of Natural Logarithms and Summation of Harmonic Numbers | :$\ds \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!} \approx \gamma n + \dfrac {\ln n} 2 + 0 \cdotp 158$
where:
:$H_k$ denotes the $k$th harmonic number
:$n!$ denotes the $n$th factorial
:$\gamma$ denotes the Euler-Mascheroni constant. | From Summation over k to n of Natural Logarithm of k:
:$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$
Then:
{{begin-eqn}}
{{eqn | l = \map \ln {n!}
| o = \approx
| r = \map \ln {\sqrt {2 \pi n} \paren {\dfrac n e}^n}
| c = Stirling's Formula
}}
{{eqn | r = \frac 1 2 \map \ln {2 \pi} + \frac 1 2 \l... | :$\ds \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!} \approx \gamma n + \dfrac {\ln n} 2 + 0 \cdotp 158$
where:
:$H_k$ denotes the $k$th [[Definition:Harmonic Number|harmonic number]]
:$n!$ denotes the $n$th [[Definition:Factorial|factorial]]
:$\gamma$ denotes the [[Definition:Euler-Mascheroni Const... | From [[Summation over k to n of Natural Logarithm of k]]:
:$\ds \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$
Then:
{{begin-eqn}}
{{eqn | l = \map \ln {n!}
| o = \approx
| r = \map \ln {\sqrt {2 \pi n} \paren {\dfrac n e}^n}
| c = [[Stirling's Formula]]
}}
{{eqn | r = \frac 1 2 \map \ln {2 \pi} + \f... | Difference between Summation of Natural Logarithms and Summation of Harmonic Numbers | https://proofwiki.org/wiki/Difference_between_Summation_of_Natural_Logarithms_and_Summation_of_Harmonic_Numbers | https://proofwiki.org/wiki/Difference_between_Summation_of_Natural_Logarithms_and_Summation_of_Harmonic_Numbers | [
"Harmonic Numbers",
"Natural Logarithms"
] | [
"Definition:Harmonic Numbers",
"Definition:Factorial",
"Definition:Euler-Mascheroni Constant"
] | [
"Summation over k to n of Natural Logarithm of k",
"Stirling's Formula",
"Sum of Sequence of Harmonic Numbers",
"Approximate Size of Sum of Harmonic Series"
] |
proofwiki-14539 | Sum over k of n Choose k by x to the k by kth Harmonic Number/x = -1 | While for $x \in \R_{>0}$ be a real number:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$
when $x = -1$ we have:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \dfrac {-1} n$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$H_k$ denote... | When $x = -1$ we have that $1 + \dfrac 1 x = 0$, so $\map \ln {1 + \dfrac 1 x}$ is undefined.
Let $S_n = \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$
Then:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
| r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \paren {\bin... | While for $x \in \R_{>0}$ be a [[Definition:Real Number|real number]]:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \paren {x + 1}^n \paren {H_n - \map \ln {1 + \frac 1 x} } + \epsilon$
when $x = -1$ we have:
:$\ds \sum_{k \mathop \in \Z} \binom n k x^k H_k = \dfrac {-1} n$
where:
:$\dbinom n k$ denotes a [[D... | When $x = -1$ we have that $1 + \dfrac 1 x = 0$, so $\map \ln {1 + \dfrac 1 x}$ is undefined.
Let $S_n = \ds \sum_{k \mathop \in \Z} \binom n k x^k H_k$
Then:
{{begin-eqn}}
{{eqn | l = S_{n + 1}
| r = \sum_{k \mathop \in \Z} \binom {n + 1} k x^k H_k
| c =
}}
{{eqn | r = \sum_{k \mathop \in \Z} \paren {... | Sum over k of n Choose k by x to the k by kth Harmonic Number/x = -1 | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_x_to_the_k_by_kth_Harmonic_Number/x_=_-1 | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_x_to_the_k_by_kth_Harmonic_Number/x_=_-1 | [
"Harmonic Numbers",
"Binomial Coefficients"
] | [
"Definition:Real Number",
"Definition:Binomial Coefficient",
"Definition:Harmonic Numbers"
] | [
"Pascal's Rule",
"Alternating Sum and Difference of Binomial Coefficients for Given n",
"Binomial Coefficient with Zero"
] |
proofwiki-14540 | Summation to n of Reciprocal of k by k-1 of Harmonic Number | :$\ds \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k = 2 - \dfrac {H_n} n - \dfrac 1 n$
where $H_n$ denotes the $n$th harmonic number. | {{begin-eqn}}
{{eqn | l = \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k
| r = \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {\paren {k + 1} k} H_{k + 1}
| c = Translation of Index Variable of Summation
}}
{{eqn | r = -\sum_{k \mathop = 1}^{n - 1} \paren {\dfrac 1 {k + 1} - \dfrac 1 k} H_{k + 1... | :$\ds \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k = 2 - \dfrac {H_n} n - \dfrac 1 n$
where $H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]. | {{begin-eqn}}
{{eqn | l = \sum_{1 \mathop < k \mathop \le n} \dfrac 1 {k \paren {k - 1} } H_k
| r = \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {\paren {k + 1} k} H_{k + 1}
| c = [[Translation of Index Variable of Summation]]
}}
{{eqn | r = -\sum_{k \mathop = 1}^{n - 1} \paren {\dfrac 1 {k + 1} - \dfrac 1 k} H_{k... | Summation to n of Reciprocal of k by k-1 of Harmonic Number | https://proofwiki.org/wiki/Summation_to_n_of_Reciprocal_of_k_by_k-1_of_Harmonic_Number | https://proofwiki.org/wiki/Summation_to_n_of_Reciprocal_of_k_by_k-1_of_Harmonic_Number | [
"Harmonic Numbers"
] | [
"Definition:Harmonic Numbers"
] | [
"Translation of Index Variable of Summation",
"Abel's Lemma/Formulation 1",
"Harmonic Numbers/Examples/H2",
"Definition:Telescoping Series"
] |
proofwiki-14541 | Riemann Zeta Function of 1000 | To at least $100$ decimal places:
:$\map \zeta {1000} \approx 1$
where $\zeta$ denotes the Riemann zeta function. | By definition of the general harmonic numbers:
:$\ds \map \zeta r = \lim_{n \mathop \to \infty} H_n^{\paren r} = \sum_{k \mathop \ge 1} \frac 1 {k^r}$
From Sequence of General Harmonic Numbers Converges for Index Greater than 1:
{{begin-eqn}}
{{eqn | l = \map \zeta {1000}
| o = \le
| r = \dfrac {2^{1000} } ... | To at least $100$ [[Definition:Decimal Place|decimal places]]:
:$\map \zeta {1000} \approx 1$
where $\zeta$ denotes the [[Definition:Riemann Zeta Function|Riemann zeta function]]. | By definition of the [[Definition:General Harmonic Numbers|general harmonic numbers]]:
:$\ds \map \zeta r = \lim_{n \mathop \to \infty} H_n^{\paren r} = \sum_{k \mathop \ge 1} \frac 1 {k^r}$
From [[Sequence of General Harmonic Numbers Converges for Index Greater than 1]]:
{{begin-eqn}}
{{eqn | l = \map \zeta {1000}
... | Riemann Zeta Function of 1000 | https://proofwiki.org/wiki/Riemann_Zeta_Function_of_1000 | https://proofwiki.org/wiki/Riemann_Zeta_Function_of_1000 | [
"Riemann Zeta Function"
] | [
"Definition:Decimal Expansion/Decimal Place",
"Definition:Riemann Zeta Function"
] | [
"Definition:Harmonic Numbers/General Definition",
"Sequence of General Harmonic Numbers Converges for Index Greater than 1",
"Definition:Decimal Expansion/Decimal Place"
] |
proofwiki-14542 | Summation to n of kth Harmonic Number over k | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} k = \dfrac { {H_n}^2 + \harm 2 n } 2$
where:
:$H_n$ denotes the $n$th harmonic number
:$\harm 2 n$ denotes the general harmonic number of order $2$ evaluated at $n$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} k
| r = \sum_{k \mathop = 1}^n \dfrac 1 k \sum_{j \mathop = 1}^k \dfrac 1 j
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 j \dfrac 1 k
| c =
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {... | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} k = \dfrac { {H_n}^2 + \harm 2 n } 2$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$\harm 2 n$ denotes the [[Definition:General Harmonic Numbers|general harmonic number of order $2$]] evaluated at $n$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} k
| r = \sum_{k \mathop = 1}^n \dfrac 1 k \sum_{j \mathop = 1}^k \dfrac 1 j
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 j \dfrac 1 k
| c =
}}
{{eqn | r = \dfrac 1 2 \paren {\paren {... | Summation to n of kth Harmonic Number over k | https://proofwiki.org/wiki/Summation_to_n_of_kth_Harmonic_Number_over_k | https://proofwiki.org/wiki/Summation_to_n_of_kth_Harmonic_Number_over_k | [
"Harmonic Numbers",
"General Harmonic Numbers"
] | [
"Definition:Harmonic Numbers",
"Definition:Harmonic Numbers/General Definition"
] | [
"Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2"
] |
proofwiki-14543 | Summation to n of kth Harmonic Number over k+1 | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1} = \dfrac { {H_{n + 1} }^2 - \harm 2 {n + 1} } 2$
where:
:$H_n$ denotes the $n$th harmonic number
:$\harm 2 n$ denotes the general harmonic number of order $2$ evaluated at $n$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1}
| r = \sum_{k \mathop = 1}^n \dfrac 1 {k + 1} \paren {H_{k + 1} - \dfrac 1 {\paren {k + 1} } }
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\dfrac {H_{k + 1} } {k + 1} - \dfrac 1 {\paren {k + 1} \paren {k ... | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1} = \dfrac { {H_{n + 1} }^2 - \harm 2 {n + 1} } 2$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$\harm 2 n$ denotes the [[Definition:General Harmonic Numbers|general harmonic number of order $2$]] evaluated at $n$. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1}
| r = \sum_{k \mathop = 1}^n \dfrac 1 {k + 1} \paren {H_{k + 1} - \dfrac 1 {\paren {k + 1} } }
| c = {{Defof|Harmonic Number}}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {\dfrac {H_{k + 1} } {k + 1} - \dfrac 1 {\paren {k + 1} \paren {k ... | Summation to n of kth Harmonic Number over k+1 | https://proofwiki.org/wiki/Summation_to_n_of_kth_Harmonic_Number_over_k+1 | https://proofwiki.org/wiki/Summation_to_n_of_kth_Harmonic_Number_over_k+1 | [
"Harmonic Numbers",
"General Harmonic Numbers"
] | [
"Definition:Harmonic Numbers",
"Definition:Harmonic Numbers/General Definition"
] | [
"Translation of Index Variable of Summation",
"Harmonic Numbers/Examples/H1",
"Summation to n of kth Harmonic Number over k"
] |
proofwiki-14544 | Sum of External Angles of Polygon equals Four Right Angles | Let the external angles of a polygon be generated in the same direction going around the polygon.
Then the sum of all these external angles equals $4$ right angles, that is, $360 \degrees$. | Let $V$ be an arbitrary vertex of a polygon $P$.
Let $T$ be the sum of all the external angles of $P$
Let $A$ be the size of the internal angle of $V$.
Let $E$ be the size of the external angle of $V$.
By definition, $E = 180 \degrees - A$.
Let $S$ denote the sum of all internal angles of $P$.
Hence we have:
:$T = n \t... | Let the [[Definition:External Angle of Polygon|external angles]] of a [[Definition:Polygon|polygon]] be generated in the same direction going around the [[Definition:Polygon|polygon]].
Then the [[Definition:Real Addition|sum]] of all these [[Definition:External Angle|external angles]] equals $4$ [[Definition:Right Ang... | Let $V$ be an arbitrary [[Definition:Vertex of Polygon|vertex]] of a [[Definition:Polygon|polygon]] $P$.
Let $T$ be the sum of all the [[Definition:External Angle|external angles]] of $P$
Let $A$ be the [[Definition:Measure of Angle|size]] of the [[Definition:Internal Angle|internal angle]] of $V$.
Let $E$ be the [... | Sum of External Angles of Polygon equals Four Right Angles | https://proofwiki.org/wiki/Sum_of_External_Angles_of_Polygon_equals_Four_Right_Angles | https://proofwiki.org/wiki/Sum_of_External_Angles_of_Polygon_equals_Four_Right_Angles | [
"External Angles",
"Polygons"
] | [
"Definition:External Angle of Polygon",
"Definition:Polygon",
"Definition:Polygon",
"Definition:Addition/Real Numbers",
"Definition:Polygon/External Angle",
"Definition:Right Angle"
] | [
"Definition:Polygon/Vertex",
"Definition:Polygon",
"Definition:Polygon/External Angle",
"Definition:Angular Measure",
"Definition:Polygon/Internal Angle",
"Definition:Angular Measure",
"Definition:Polygon/External Angle",
"Definition:Polygon/Internal Angle",
"Sum of Internal Angles of Polygon"
] |
proofwiki-14545 | Bisector of Apex of Isosceles Triangle also Bisects Base | Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.
Then $AD$ bisects $BC$. | By definition of isosceles triangle, $AB = AC$.
By definition of bisector, $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle ACD$.
Thus $AD = DC$.
The result follows by definition of bisection.
{{qed}} | Let $\triangle ABC$ be an [[Definition:Isosceles Triangle|isosceles triangle]] whose [[Definition:Apex of Isosceles Triangle|apex]] is $A$.
Let $AD$ be the [[Definition:Bisection|bisector]] of $\angle BAC$ such that $AD$ [[Definition:Intersection (Geometry)|intersects]] $BC$ at $D$.
Then $AD$ [[Definition:Bisection|... | By definition of [[Definition:Isosceles Triangle|isosceles triangle]], $AB = AC$.
By definition of [[Definition:Bisection|bisector]], $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by [[Triangle Side-Angle-Side Congruence]], $\triangle ABD = \triangle ACD$.
Thus $AD = DC$.
The result follows by d... | Bisector of Apex of Isosceles Triangle also Bisects Base | https://proofwiki.org/wiki/Bisector_of_Apex_of_Isosceles_Triangle_also_Bisects_Base | https://proofwiki.org/wiki/Bisector_of_Apex_of_Isosceles_Triangle_also_Bisects_Base | [
"Isosceles Triangles"
] | [
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Triangle (Geometry)/Isosceles/Apex",
"Definition:Bisection",
"Definition:Intersection (Geometry)",
"Definition:Bisection"
] | [
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Bisection",
"Triangle Side-Angle-Side Congruence",
"Definition:Bisection"
] |
proofwiki-14546 | Bisector of Apex of Isosceles Triangle is Perpendicular to Base | Let $\triangle ABC$ be an isosceles triangle whose apex is $A$.
Let $AD$ be the bisector of $\angle BAC$ such that $AD$ intersects $BC$ at $D$.
Then $AD$ is perpendicular to $BC$. | By definition of isosceles triangle, $AB = AC$.
By definition of bisector, $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by Triangle Side-Angle-Side Congruence, $\triangle ABD = \triangle ACD$.
Thus $\angle ADB = \angle ADC$.
By Two Angles on Straight Line make Two Right Angles, $\angle ADB + \angle ... | Let $\triangle ABC$ be an [[Definition:Isosceles Triangle|isosceles triangle]] whose [[Definition:Apex of Isosceles Triangle|apex]] is $A$.
Let $AD$ be the [[Definition:Bisection|bisector]] of $\angle BAC$ such that $AD$ [[Definition:Intersection (Geometry)|intersects]] $BC$ at $D$.
Then $AD$ is [[Definition:Perpend... | By definition of [[Definition:Isosceles Triangle|isosceles triangle]], $AB = AC$.
By definition of [[Definition:Bisection|bisector]], $\angle BAD = \angle CAD$.
By construction, $AD$ is common.
Thus by [[Triangle Side-Angle-Side Congruence]], $\triangle ABD = \triangle ACD$.
Thus $\angle ADB = \angle ADC$.
By [[Tw... | Bisector of Apex of Isosceles Triangle is Perpendicular to Base | https://proofwiki.org/wiki/Bisector_of_Apex_of_Isosceles_Triangle_is_Perpendicular_to_Base | https://proofwiki.org/wiki/Bisector_of_Apex_of_Isosceles_Triangle_is_Perpendicular_to_Base | [
"Isosceles Triangles"
] | [
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Triangle (Geometry)/Isosceles/Apex",
"Definition:Bisection",
"Definition:Intersection (Geometry)",
"Definition:Right Angle/Perpendicular"
] | [
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Bisection",
"Triangle Side-Angle-Side Congruence",
"Two Angles on Straight Line make Two Right Angles",
"Definition:Right Angle",
"Definition:Right Angle",
"Definition:Right Angle/Perpendicular"
] |
proofwiki-14547 | Equilateral Triangle is Equiangular | Let $\triangle ABC$ be an equilateral triangle.
Then $\triangle ABC$ is also equiangular, with each internal angle equal to $60 \degrees$. | Let $\triangle ABC$ be an equilateral triangle.
By definition of equilateral triangle, any two of the sides of $\triangle ABC$ are equal.
{{WLOG}}, let $AB = AC$.
Then by Isosceles Triangle has Two Equal Angles:
:$\angle ABC = \angle ACB$
As the choice of equal sides was arbitrary, it follows that every two of the inte... | Let $\triangle ABC$ be an [[Definition:Equilateral Triangle|equilateral triangle]].
Then $\triangle ABC$ is also [[Definition:Equiangular Polygon|equiangular]], with each [[Definition:Internal Angle|internal angle]] equal to $60 \degrees$. | Let $\triangle ABC$ be an [[Definition:Equilateral Triangle|equilateral triangle]].
By definition of [[Definition:Equilateral Triangle|equilateral triangle]], any two of the [[Definition:Side of Polygon|sides]] of $\triangle ABC$ are equal.
{{WLOG}}, let $AB = AC$.
Then by [[Isosceles Triangle has Two Equal Angles]]... | Equilateral Triangle is Equiangular | https://proofwiki.org/wiki/Equilateral_Triangle_is_Equiangular | https://proofwiki.org/wiki/Equilateral_Triangle_is_Equiangular | [
"Equilateral Triangles"
] | [
"Definition:Triangle (Geometry)/Equilateral",
"Definition:Polygon/Equiangular",
"Definition:Polygon/Internal Angle"
] | [
"Definition:Triangle (Geometry)/Equilateral",
"Definition:Triangle (Geometry)/Equilateral",
"Definition:Polygon/Side",
"Isosceles Triangle has Two Equal Angles",
"Definition:Polygon/Side",
"Definition:Polygon/Internal Angle",
"Definition:Polygon/Internal Angle",
"Sum of Angles of Triangle equals Two R... |
proofwiki-14548 | Equiangular Triangle is Equilateral | Let $\triangle ABC$ be equiangular.
Then $\triangle ABC$ is an equilateral triangle. | Let $\triangle ABC$ be equiangular.
By definition of equiangular polygon, any two of the internal angles of $\triangle ABC$ are equal.
{{WLOG}}, let $\angle ABC = \angle ACB$.
Then by Triangle with Two Equal Angles is Isosceles, $AB = AC$.
As the choice of equal angles was arbitrary, it follows that any two sides of $\... | Let $\triangle ABC$ be [[Definition:Equiangular Polygon|equiangular]].
Then $\triangle ABC$ is an [[Definition:Equilateral Triangle|equilateral triangle]]. | Let $\triangle ABC$ be [[Definition:Equiangular Polygon|equiangular]].
By definition of [[Definition:Equiangular Polygon|equiangular polygon]], any two of the [[Definition:Internal Angle|internal angles]] of $\triangle ABC$ are equal.
{{WLOG}}, let $\angle ABC = \angle ACB$.
Then by [[Triangle with Two Equal Angles ... | Equiangular Triangle is Equilateral | https://proofwiki.org/wiki/Equiangular_Triangle_is_Equilateral | https://proofwiki.org/wiki/Equiangular_Triangle_is_Equilateral | [
"Equilateral Triangles"
] | [
"Definition:Polygon/Equiangular",
"Definition:Triangle (Geometry)/Equilateral"
] | [
"Definition:Polygon/Equiangular",
"Definition:Polygon/Equiangular",
"Definition:Polygon/Internal Angle",
"Triangle with Two Equal Angles is Isosceles",
"Definition:Polygon/Internal Angle",
"Definition:Polygon/Side",
"Definition:Polygon/Side",
"Definition:Triangle (Geometry)/Equilateral"
] |
proofwiki-14549 | Characterization of Strictly Increasing Mapping on Woset | Let $J$ and $E$ be well-ordered sets.
Let $h: J \to E$ be a mapping.
Let $S_\alpha$ denote an initial segment determined by $\alpha$.
{{TFAE}}
:$(1): \quad h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
:$(2): \quad \forall \alpha \in J: \map h \alpha = \map \min {E \setminus ... | === $(1)$ implies $(2)$ ===
Suppose $h$ satisfies:
:$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$
Then for any $x,y \in J$:
{{begin-eqn}}
{{eqn | l = x
| o = \prec
| r = y
}}
{{eqn | ll= \leadsto
| l = \map h x
| o = \prec
| r = \map h y
|... | Let $J$ and $E$ be [[Definition:Well-Ordered Set|well-ordered sets]].
Let $h: J \to E$ be a [[Definition:Mapping|mapping]].
Let $S_\alpha$ denote an [[Definition:Initial Segment|initial segment determined by $\alpha$]].
{{TFAE}}
:$(1): \quad h$ is [[Definition:Strictly Increasing Mapping|strictly increasing]] and ... | === $(1)$ implies $(2)$ ===
Suppose $h$ satisfies:
:$h$ is [[Definition:Strictly Increasing Mapping|strictly increasing]] and its [[Definition:Image Set of Mapping|image]] is either all of $E$ or an [[Definition:Initial Segment|initial segment]] of $E$
Then for any $x,y \in J$:
{{begin-eqn}}
{{eqn | l = x
| o... | Characterization of Strictly Increasing Mapping on Woset | https://proofwiki.org/wiki/Characterization_of_Strictly_Increasing_Mapping_on_Woset | https://proofwiki.org/wiki/Characterization_of_Strictly_Increasing_Mapping_on_Woset | [
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Mapping",
"Definition:Initial Segment",
"Definition:Strictly Increasing/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Initial Segment",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Smallest Element"
] | [
"Definition:Strictly Increasing/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Initial Segment",
"Definition:Strictly Increasing/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Initial Segment",
"Definition:Initial Segment"
] |
proofwiki-14550 | No Order Isomophism Between Distinct Initial Segments of Woset | Let $E$ be a well-ordered set.
Let $S_\alpha, S_\beta$ be initial segments of $E$ that are order isomorphic.
Then $S_\alpha = S_\beta$. | {{AimForCont}} $S_\alpha \ne S_\beta$.
Then $\alpha \ne \beta$.
By the trichotomy law, $\alpha \prec \beta$ or $\beta \prec \alpha$.
{{WLOG}} assume $\alpha \prec \beta$.
Then $S_\alpha \subsetneqq S_\beta$.
That is, $S_\alpha$ is an initial segment of $S_\beta$.
We have {{hypothesis}} that $S_\alpha$ and $S_\beta$ are... | Let $E$ be a [[Definition:Well-Ordered Set|well-ordered set]].
Let $S_\alpha, S_\beta$ be [[Definition:Initial Segment|initial segments]] of $E$ that are [[Definition:Order Isomorphism|order isomorphic]].
Then $S_\alpha = S_\beta$. | {{AimForCont}} $S_\alpha \ne S_\beta$.
Then $\alpha \ne \beta$.
By the [[Trichotomy Law (Ordering)|trichotomy law]], $\alpha \prec \beta$ or $\beta \prec \alpha$.
{{WLOG}} assume $\alpha \prec \beta$.
Then $S_\alpha \subsetneqq S_\beta$.
That is, $S_\alpha$ is an [[Definition:Initial Segment|initial segment]] of $... | No Order Isomophism Between Distinct Initial Segments of Woset | https://proofwiki.org/wiki/No_Order_Isomophism_Between_Distinct_Initial_Segments_of_Woset | https://proofwiki.org/wiki/No_Order_Isomophism_Between_Distinct_Initial_Segments_of_Woset | [
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Initial Segment",
"Definition:Order Isomorphism"
] | [
"Trichotomy Law (Ordering)",
"Definition:Initial Segment",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism",
"Definition:Initial Segment",
"Definition:Contradiction",
"Well-Ordered Class is not Isomorphic to Initial Segment"
] |
proofwiki-14551 | Perpendicular is Shortest Straight Line from Point to Straight Line | Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Then the length of $CD$ is less than the length of all other line segments that can be drawn from $C$ to $AB$. | Let $E$ on $AB$ such that $E$ is different from $D$.
Then $CDE$ forms a right triangle where $CE$ is the hypotenuse.
By Pythagoras's Theorem:
:$CD^2 + DE^2 = CE^2$
and so $CD < CE$.
{{qed}} | Let $AB$ be a [[Definition:Straight Line|straight line]].
Let $C$ be a [[Definition:Point|point]] which is not on $AB$.
Let $D$ be a [[Definition:Point|point]] on $AB$ such that $CD$ is [[Definition:Perpendicular|perpendicular]] to $AB$.
Then the [[Definition:Length of Line|length]] of $CD$ is less than the [[Defin... | Let $E$ on $AB$ such that $E$ is different from $D$.
Then $CDE$ forms a [[Definition:Right Triangle|right triangle]] where $CE$ is the [[Definition:Hypotenuse|hypotenuse]].
By [[Pythagoras's Theorem]]:
:$CD^2 + DE^2 = CE^2$
and so $CD < CE$.
{{qed}} | Perpendicular is Shortest Straight Line from Point to Straight Line | https://proofwiki.org/wiki/Perpendicular_is_Shortest_Straight_Line_from_Point_to_Straight_Line | https://proofwiki.org/wiki/Perpendicular_is_Shortest_Straight_Line_from_Point_to_Straight_Line | [
"Perpendicular Distance from Straight Line in Plane to Point"
] | [
"Definition:Line/Straight Line",
"Definition:Point",
"Definition:Point",
"Definition:Right Angle/Perpendicular",
"Definition:Linear Measure/Length",
"Definition:Linear Measure/Length",
"Definition:Line/Segment"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse",
"Pythagoras's Theorem"
] |
proofwiki-14552 | Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal | Let $AB$ be a straight line.
Let $C$ be a point which is not on $AB$.
Let $D$ be a point on $AB$ such that $CD$ is perpendicular to $AB$.
Let $E, F$ be points on $AB$ such that $\angle DCE = \angle DCF$.
Then $CE = CF$. | $\triangle CDE$ and $\triangle CDF$ are right triangle where $CE$ and $CF$ are the hypotenuses.
We have:
:$\angle CDE = \angle CDF$ as both are right angles.
:$\angle DCE = \angle DCF$ by hypothesis.
:$CD$ is common.
Thus by Triangle Angle-Side-Angle Congruence, $\triangle CDE$ and $\triangle CDF$ are congruent.
Hence ... | Let $AB$ be a [[Definition:Straight Line|straight line]].
Let $C$ be a [[Definition:Point|point]] which is not on $AB$.
Let $D$ be a [[Definition:Point|point]] on $AB$ such that $CD$ is [[Definition:Perpendicular|perpendicular]] to $AB$.
Let $E, F$ be [[Definition:Point|points]] on $AB$ such that $\angle DCE = \angl... | $\triangle CDE$ and $\triangle CDF$ are [[Definition:Right Triangle|right triangle]] where $CE$ and $CF$ are the [[Definition:Hypotenuse|hypotenuses]].
We have:
:$\angle CDE = \angle CDF$ as both are [[Definition:Right Angle|right angles]].
:$\angle DCE = \angle DCF$ [[Definition:By Hypothesis|by hypothesis]].
:$CD$... | Straight Lines which make Equal Angles with Perpendicular to Straight Line are Equal | https://proofwiki.org/wiki/Straight_Lines_which_make_Equal_Angles_with_Perpendicular_to_Straight_Line_are_Equal | https://proofwiki.org/wiki/Straight_Lines_which_make_Equal_Angles_with_Perpendicular_to_Straight_Line_are_Equal | [
"Lines",
"Angles"
] | [
"Definition:Line/Straight Line",
"Definition:Point",
"Definition:Point",
"Definition:Right Angle/Perpendicular",
"Definition:Point"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse",
"Definition:Right Angle",
"Definition:By Hypothesis",
"Triangle Angle-Side-Angle Congruence",
"Definition:Congruence (Geometry)"
] |
proofwiki-14553 | Distance between Two Parallel Straight Lines is Everywhere the Same | Let $AB$ and $CD$ be parallel straight lines.
Let perpendiculars $EF$ and $GH$ be drawn from $AB$ to $CD$, where $E, G$ are on $AB$ and $F, H$ are on $CD$.
Then $EF = GH$.
That is, the distance between $AB$ and $CD$ is the same everywhere along their length. | $\angle EFH, \angle FEG, \angle EGH, \angle FHG$ are all right angles.
Then $EF$ and $GH$ are parallel.
Thus $\Box EFHG$ is by definition a parallelogram.
From Opposite Sides and Angles of Parallelogram are Equal it follows that $EF = GH$.
{{qed}} | Let $AB$ and $CD$ be [[Definition:Parallel Lines|parallel]] [[Definition:Straight Line|straight lines]].
Let [[Definition:Perpendicular|perpendiculars]] $EF$ and $GH$ be drawn from $AB$ to $CD$, where $E, G$ are on $AB$ and $F, H$ are on $CD$.
Then $EF = GH$.
That is, the [[Definition:Distance between Parallel Line... | $\angle EFH, \angle FEG, \angle EGH, \angle FHG$ are all [[Definition:Right Angle|right angles]].
Then $EF$ and $GH$ are [[Definition:Parallel Lines|parallel]].
Thus $\Box EFHG$ is by definition a [[Definition:Parallelogram|parallelogram]].
From [[Opposite Sides and Angles of Parallelogram are Equal]] it follows tha... | Distance between Two Parallel Straight Lines is Everywhere the Same | https://proofwiki.org/wiki/Distance_between_Two_Parallel_Straight_Lines_is_Everywhere_the_Same | https://proofwiki.org/wiki/Distance_between_Two_Parallel_Straight_Lines_is_Everywhere_the_Same | [
"Parallel Lines"
] | [
"Definition:Parallel (Geometry)/Lines",
"Definition:Line/Straight Line",
"Definition:Right Angle/Perpendicular",
"Definition:Distance between Parallel Lines",
"Definition:Linear Measure/Length"
] | [
"Definition:Right Angle",
"Definition:Parallel (Geometry)/Lines",
"Definition:Quadrilateral/Parallelogram",
"Opposite Sides and Angles of Parallelogram are Equal"
] |
proofwiki-14554 | Diagonals of Rhombus Bisect Each Other at Right Angles | Let $ABCD$ be a rhombus.
The diagonals $AC$ and $BD$ of $ABCD$ bisect each other at right angles. | By the definition of a rhombus, $AB = AD = BC = DC$.
{{WLOG}}, consider the diagonal $BD$.
Thus:
: $\triangle ABD$ is an isosceles triangle whose apex is $A$ and whose base is $BD$.
By Diagonals of Rhombus Bisect Angles, $AC$ bisects $\angle BAD$.
From Bisector of Apex of Isosceles Triangle also Bisects Base, $AC$ bise... | Let $ABCD$ be a [[Definition:Rhombus|rhombus]].
The [[Definition:Diagonal of Parallelogram|diagonals]] $AC$ and $BD$ of $ABCD$ [[Definition:Bisection|bisect]] each other at [[Definition:Right Angle|right angles]]. | By the definition of a [[Definition:Rhombus|rhombus]], $AB = AD = BC = DC$.
{{WLOG}}, consider the [[Definition:Diagonal of Parallelogram|diagonal]] $BD$.
Thus:
: $\triangle ABD$ is an [[Definition:Isosceles Triangle|isosceles triangle]] whose [[Definition:Apex of Isosceles Triangle|apex]] is $A$ and whose [[Definiti... | Diagonals of Rhombus Bisect Each Other at Right Angles | https://proofwiki.org/wiki/Diagonals_of_Rhombus_Bisect_Each_Other_at_Right_Angles | https://proofwiki.org/wiki/Diagonals_of_Rhombus_Bisect_Each_Other_at_Right_Angles | [
"Parallelograms"
] | [
"Definition:Quadrilateral/Rhombus",
"Definition:Diameter of Parallelogram",
"Definition:Bisection",
"Definition:Right Angle"
] | [
"Definition:Quadrilateral/Rhombus",
"Definition:Diameter of Parallelogram",
"Definition:Triangle (Geometry)/Isosceles",
"Definition:Triangle (Geometry)/Isosceles/Apex",
"Definition:Triangle (Geometry)/Isosceles/Base",
"Diagonals of Rhombus Bisect Angles",
"Definition:Bisection",
"Bisector of Apex of I... |
proofwiki-14555 | Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel | Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:$AB = CD$ and $AB \parallel CD$
where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$. | === Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then $AB \parallel CD$ by definition.
From Opposite Sides and Angles of Parallelogram are Equal it follows that $AB = CD$.
{{qed|lemma}} | Let $ABCD$ be a [[Definition:Quadrilateral|quadrilateral]].
Then:
:$ABCD$ is a [[Definition:Parallelogram|parallelogram]]
{{iff}}:
:$AB = CD$ and $AB \parallel CD$
where $AB \parallel CD$ denotes that $AB$ is [[Definition:Parallel Lines|parallel]] to $CD$. | === Sufficient Condition ===
Let $ABCD$ be a [[Definition:Parallelogram|parallelogram]].
Then $AB \parallel CD$ by definition.
From [[Opposite Sides and Angles of Parallelogram are Equal]] it follows that $AB = CD$.
{{qed|lemma}} | Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_One_Pair_of_Opposite_Sides_is_Equal_and_Parallel | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_One_Pair_of_Opposite_Sides_is_Equal_and_Parallel | [
"Parallelograms"
] | [
"Definition:Quadrilateral",
"Definition:Quadrilateral/Parallelogram",
"Definition:Parallel (Geometry)/Lines"
] | [
"Definition:Quadrilateral/Parallelogram",
"Opposite Sides and Angles of Parallelogram are Equal",
"Definition:Quadrilateral/Parallelogram"
] |
proofwiki-14556 | Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel | Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:either $AB = CD$ and $AD = BC$
:or $AB \parallel CD$ and $AD \parallel BC$
where $AB \parallel CD$ denotes that $AB$ is parallel to $CD$. | === Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by definition:
:$AB \parallel CD$ and $AD \parallel BC$
and from Opposite Sides and Angles of Parallelogram are Equal:
:$AB = CD$ and $AD = BC$
From Conjunction implies Disjunction, it follows that:
:either $AB = CD$ and $AD = BC$
:or $AB \parallel CD$ an... | Let $ABCD$ be a [[Definition:Quadrilateral|quadrilateral]].
Then:
:$ABCD$ is a [[Definition:Parallelogram|parallelogram]]
{{iff}}:
:either $AB = CD$ and $AD = BC$
:or $AB \parallel CD$ and $AD \parallel BC$
where $AB \parallel CD$ denotes that $AB$ is [[Definition:Parallel Lines|parallel]] to $CD$. | === Sufficient Condition ===
Let $ABCD$ be a [[Definition:Parallelogram|parallelogram]].
Then by definition:
:$AB \parallel CD$ and $AD \parallel BC$
and from [[Opposite Sides and Angles of Parallelogram are Equal]]:
:$AB = CD$ and $AD = BC$
From [[Conjunction implies Disjunction]], it follows that:
:either $AB = C... | Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Both_Pairs_of_Opposite_Sides_are_Equal_or_Parallel | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Both_Pairs_of_Opposite_Sides_are_Equal_or_Parallel | [
"Parallelograms"
] | [
"Definition:Quadrilateral",
"Definition:Quadrilateral/Parallelogram",
"Definition:Parallel (Geometry)/Lines"
] | [
"Definition:Quadrilateral/Parallelogram",
"Opposite Sides and Angles of Parallelogram are Equal",
"Conjunction implies Disjunction",
"Definition:Quadrilateral/Parallelogram",
"Definition:Quadrilateral/Parallelogram"
] |
proofwiki-14557 | Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal | Let $ABCD$ be a quadrilateral.
Then:
:$ABCD$ is a parallelogram
{{iff}}:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$. | === Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by Opposite Sides and Angles of Parallelogram are Equal:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.
{{qed|lemma}} | Let $ABCD$ be a [[Definition:Quadrilateral|quadrilateral]].
Then:
:$ABCD$ is a [[Definition:Parallelogram|parallelogram]]
{{iff}}:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$. | === Sufficient Condition ===
Let $ABCD$ be a [[Definition:Parallelogram|parallelogram]].
Then by [[Opposite Sides and Angles of Parallelogram are Equal]]:
:$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$.
{{qed|lemma}} | Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Both_Pairs_of_Opposite_Angles_are_Equal | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Both_Pairs_of_Opposite_Angles_are_Equal | [
"Parallelograms"
] | [
"Definition:Quadrilateral",
"Definition:Quadrilateral/Parallelogram"
] | [
"Definition:Quadrilateral/Parallelogram",
"Opposite Sides and Angles of Parallelogram are Equal"
] |
proofwiki-14558 | Quadrilateral is Parallelogram iff Diagonals Bisect each other | Let $ABCD$ be a quadrilateral.
Then:
: $ABCD$ is a parallelogram
{{iff}}:
:both:
:: $AD$ is a bisector of $BC$
:and:
:: $BC$ is a bisector of $AD$. | === Sufficient Condition ===
Let $ABCD$ be a parallelogram.
Then by Diameters of Parallelogram Bisect each other:
: $AD$ is a bisector of $BC$
and
: $BC$ is a bisector of $AD$.
{{qed|lemma}} | Let $ABCD$ be a [[Definition:Quadrilateral|quadrilateral]].
Then:
: $ABCD$ is a [[Definition:Parallelogram|parallelogram]]
{{iff}}:
:both:
:: $AD$ is a [[Definition:Bisection|bisector]] of $BC$
:and:
:: $BC$ is a [[Definition:Bisection|bisector]] of $AD$. | === Sufficient Condition ===
Let $ABCD$ be a [[Definition:Parallelogram|parallelogram]].
Then by [[Diameters of Parallelogram Bisect each other]]:
: $AD$ is a [[Definition:Bisection|bisector]] of $BC$
and
: $BC$ is a [[Definition:Bisection|bisector]] of $AD$.
{{qed|lemma}} | Quadrilateral is Parallelogram iff Diagonals Bisect each other | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Diagonals_Bisect_each_other | https://proofwiki.org/wiki/Quadrilateral_is_Parallelogram_iff_Diagonals_Bisect_each_other | [
"Parallelograms"
] | [
"Definition:Quadrilateral",
"Definition:Quadrilateral/Parallelogram",
"Definition:Bisection",
"Definition:Bisection"
] | [
"Definition:Quadrilateral/Parallelogram",
"Diameters of Parallelogram Bisect each other",
"Definition:Bisection",
"Definition:Bisection",
"Definition:Bisection",
"Definition:Bisection",
"Definition:Quadrilateral/Parallelogram"
] |
proofwiki-14559 | Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal | Let $ABCD$ and $EFGH$ be parallelograms.
Then $ABCD$ and $EFGH$ are congruent if:
: $2$ adjacent sides of $ABCD$ are equal to $2$ corresponding adjacent sides of $EFGH$
: the angle between those $2$ adjacent sides on both $ABCD$ and $EFGH$ are equal. | {{WLOG}} let the $2$ adjacent sides of $ABCD$ be $AB$ and $BC$.
Let the $2$ corresponding adjacent sides of $EFGH$ be $EF$ and $FG$ such that $AB = EF$ and $BC = FG$.
Hence let $\angle ABC = \angle EFG$.
From Triangle Side-Angle-Side Congruence:
:$\triangle ABC = \triangle EFG$
From Quadrilateral is Parallelogram iff B... | Let $ABCD$ and $EFGH$ be [[Definition:Parallelogram|parallelograms]].
Then $ABCD$ and $EFGH$ are [[Definition:Congruence (Geometry)|congruent]] if:
: $2$ [[Definition:Adjacent Sides|adjacent sides]] of $ABCD$ are equal to $2$ corresponding [[Definition:Adjacent Sides|adjacent sides]] of $EFGH$
: the [[Definition:Plane... | {{WLOG}} let the $2$ [[Definition:Adjacent Sides|adjacent sides]] of $ABCD$ be $AB$ and $BC$.
Let the $2$ corresponding [[Definition:Adjacent Sides|adjacent sides]] of $EFGH$ be $EF$ and $FG$ such that $AB = EF$ and $BC = FG$.
Hence let $\angle ABC = \angle EFG$.
From [[Triangle Side-Angle-Side Congruence]]:
:$\tri... | Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal | https://proofwiki.org/wiki/Parallelograms_are_Congruent_if_Two_Adjacent_Sides_and_Included_Angle_are_respectively_Equal | https://proofwiki.org/wiki/Parallelograms_are_Congruent_if_Two_Adjacent_Sides_and_Included_Angle_are_respectively_Equal | [
"Parallelograms"
] | [
"Definition:Quadrilateral/Parallelogram",
"Definition:Congruence (Geometry)",
"Definition:Polygon/Adjacent/Sides",
"Definition:Polygon/Adjacent/Sides",
"Definition:Angle",
"Definition:Polygon/Adjacent/Sides"
] | [
"Definition:Polygon/Adjacent/Sides",
"Definition:Polygon/Adjacent/Sides",
"Triangle Side-Angle-Side Congruence",
"Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal",
"Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel",
"Triangle Side-Angle-Side ... |
proofwiki-14560 | Rectangles with Equal Bases and Equal Altitudes are Congruent | Let $ABCD$ and $EFGH$ be rectangles.
Then $ABCD$ and $EFGH$ are congruent if:
: the base of $ABCD$ equals the base of $EFGH$
: the altitude of $ABCD$ equals the altitude of $EFGH$. | A rectangle is a parallelogram whose vertices are right angles.
Thus the altitudes of $ABCD$ and of $EFGH$ coincide with the sides of $ABCD$ and $EFGH$ which are adjacent to the bases.
The result then follows from Parallelograms are Congruent if Two Adjacent Sides and Included Angle are respectively Equal.
{{qed}} | Let $ABCD$ and $EFGH$ be [[Definition:Rectangle|rectangles]].
Then $ABCD$ and $EFGH$ are [[Definition:Congruence (Geometry)|congruent]] if:
: the [[Definition:Base of Parallelogram|base]] of $ABCD$ equals the [[Definition:Base of Parallelogram|base]] of $EFGH$
: the [[Definition:Altitude of Parallelogram|altitude]] of... | A [[Definition:Rectangle|rectangle]] is a [[Definition:Parallelogram|parallelogram]] whose [[Definition:Vertex of Polygon|vertices]] are [[Definition:Right Angle|right angles]].
Thus the [[Definition:Altitude of Parallelogram|altitudes]] of $ABCD$ and of $EFGH$ coincide with the [[Definition:Side of Polygon|sides]] of... | Rectangles with Equal Bases and Equal Altitudes are Congruent | https://proofwiki.org/wiki/Rectangles_with_Equal_Bases_and_Equal_Altitudes_are_Congruent | https://proofwiki.org/wiki/Rectangles_with_Equal_Bases_and_Equal_Altitudes_are_Congruent | [
"Rectangles"
] | [
"Definition:Quadrilateral/Rectangle",
"Definition:Congruence (Geometry)",
"Definition:Parallelogram/Base",
"Definition:Parallelogram/Base",
"Definition:Parallelogram/Altitude",
"Definition:Parallelogram/Altitude"
] | [
"Definition:Quadrilateral/Rectangle",
"Definition:Quadrilateral/Parallelogram",
"Definition:Polygon/Vertex",
"Definition:Right Angle",
"Definition:Parallelogram/Altitude",
"Definition:Polygon/Side",
"Definition:Polygon/Adjacent/Sides",
"Definition:Parallelogram/Base",
"Parallelograms are Congruent i... |
proofwiki-14561 | Parallel Lines which intercept Equal Segments on Transversals | Let $3$ or more parallel lines intersect equal line segments on one transversal.
Then those same $3$ or more parallel lines intersect equal line segments on every transversal. | {{ProofWanted|Someone who understands what this is really trying to say?}} | Let $3$ or more [[Definition:Parallel Lines|parallel lines]] intersect equal [[Definition:Line Segment|line segments]] on one [[Definition:Transversal|transversal]].
Then those same $3$ or more [[Definition:Parallel Lines|parallel lines]] intersect equal [[Definition:Line Segment|line segments]] on every [[Definition:... | {{ProofWanted|Someone who understands what this is really trying to say?}} | Parallel Lines which intercept Equal Segments on Transversals | https://proofwiki.org/wiki/Parallel_Lines_which_intercept_Equal_Segments_on_Transversals | https://proofwiki.org/wiki/Parallel_Lines_which_intercept_Equal_Segments_on_Transversals | [
"Parallel Lines"
] | [
"Definition:Parallel (Geometry)/Lines",
"Definition:Line/Segment",
"Definition:Transversal",
"Definition:Parallel (Geometry)/Lines",
"Definition:Line/Segment",
"Definition:Transversal"
] | [] |
proofwiki-14562 | Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side | Let $ABC$ be a triangle.
Let $DE$ be a straight line parallel to $BC$.
Let $DE$ bisect $AB$.
Then $DE$ also bisects $AC$.
That is, $DE$ is a midline of $\triangle ABC$.
400px | This is a direct application of the Parallel Transversal Theorem.
{{qed}} | Let $ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
Let $DE$ be a [[Definition:Straight Line|straight line]] [[Definition:Parallel Lines|parallel]] to $BC$.
Let $DE$ [[Definition:Bisection|bisect]] $AB$.
Then $DE$ also [[Definition:Bisection|bisects]] $AC$.
That is, $DE$ is a [[Definition:Midline of Triang... | This is a direct application of the [[Parallel Transversal Theorem]].
{{qed}} | Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side | https://proofwiki.org/wiki/Line_Parallel_to_Side_of_Triangle_which_Bisects_One_Side_also_Bisects_Other_Side | https://proofwiki.org/wiki/Line_Parallel_to_Side_of_Triangle_which_Bisects_One_Side_also_Bisects_Other_Side | [
"Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Line/Straight Line",
"Definition:Parallel (Geometry)/Lines",
"Definition:Bisection",
"Definition:Bisection",
"Definition:Midline of Triangle",
"File:Midline of Triangle.png"
] | [
"Parallel Transversal Theorem"
] |
proofwiki-14563 | Midline Theorem | The midline of a triangle is parallel to the third side of that triangle and half its length. | 400px
Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ through $AB$ and $AC$.
Extend $DE$ to $DF$ so $DE = EF$.
As $E$ is the midpoint of $AC$, the diagonals of the quadrilateral $ADCF$ bisect each other.
From Quadrilateral with Bisecting Diagonals is Parallelogram, $ADCF$ is a parallelogra... | The [[Definition:Midline of Triangle|midline]] of a [[Definition:Triangle (Geometry)|triangle]] is [[Definition:Parallel Lines|parallel]] to the third [[Definition:Side of Polygon|side]] of that [[Definition:Triangle (Geometry)|triangle]] and half its [[Definition:Length of Line|length]]. | [[File:Midline Theorem.png|400px]]
Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
Let $DE$ be the [[Definition:Midline of Triangle|midline]] of $\triangle ABC$ through $AB$ and $AC$.
Extend $DE$ to $DF$ so $DE = EF$.
As $E$ is the [[Definition:Midpoint of Line|midpoint]] of $AC$, the [[Defini... | Midline Theorem | https://proofwiki.org/wiki/Midline_Theorem | https://proofwiki.org/wiki/Midline_Theorem | [
"Midline Theorem",
"Triangles",
"Named Theorems"
] | [
"Definition:Midline of Triangle",
"Definition:Triangle (Geometry)",
"Definition:Parallel (Geometry)/Lines",
"Definition:Polygon/Side",
"Definition:Triangle (Geometry)",
"Definition:Linear Measure/Length"
] | [
"File:Midline Theorem.png",
"Definition:Triangle (Geometry)",
"Definition:Midline of Triangle",
"Definition:Line/Midpoint",
"Definition:Diameter of Quadrilateral",
"Definition:Quadrilateral",
"Definition:Bisection",
"Quadrilateral with Bisecting Diagonals is Parallelogram",
"Definition:Quadrilateral... |
proofwiki-14564 | Midline and Median of Triangle Bisect Each Other | Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ which bisects $AB$ and $AC$.
Let $AF$ be the median of $ABC$ which bisects $BC$.
Then $AF$ and $DE$ bisect each other. | 400px
Construct the midlines $DF$ and $EF$.
Then by the Midline Theorem $DF \parallel AE$ and $EF \parallel AD$.
Thus by Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel, $\Box ADFE$ is a parallelogram.
By construction, $AF$ and $DE$ are the diagonals of $\Box ADFE$.
The result foll... | Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
Let $DE$ be the [[Definition:Midline of Triangle|midline]] of $\triangle ABC$ which [[Definition:Bisection|bisects]] $AB$ and $AC$.
Let $AF$ be the [[Definition:Median of Triangle|median]] of $ABC$ which [[Definition:Bisection|bisects]] $BC$.
The... | [[File:Midline and Median of Triangle Bisect Each Other.png|400px]]
Construct the [[Definition:Midline of Triangle|midlines]] $DF$ and $EF$.
Then by the [[Midline Theorem]] $DF \parallel AE$ and $EF \parallel AD$.
Thus by [[Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel]], $\Bo... | Midline and Median of Triangle Bisect Each Other | https://proofwiki.org/wiki/Midline_and_Median_of_Triangle_Bisect_Each_Other | https://proofwiki.org/wiki/Midline_and_Median_of_Triangle_Bisect_Each_Other | [
"Triangles",
"Medians of Triangles"
] | [
"Definition:Triangle (Geometry)",
"Definition:Midline of Triangle",
"Definition:Bisection",
"Definition:Median of Triangle",
"Definition:Bisection",
"Definition:Bisection"
] | [
"File:Midline and Median of Triangle Bisect Each Other.png",
"Definition:Midline of Triangle",
"Midline Theorem",
"Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel",
"Definition:Quadrilateral/Parallelogram",
"Definition:Diameter of Quadrilateral",
"Diameters of Paral... |
proofwiki-14565 | Median to Hypotenuse of Right Triangle equals Half Hypotenuse | Let $\triangle ABC$ be a right triangle such that $BC$ is the hypotenuse.
Let $AD$ be the median to $BC$.
Then $AD$ is half of $BC$. | 400px
Construct $BE$ and $CE$ parallel to $AC$ and $AB$ respectively.
Then by definition $ABEC$ is a parallelogram.
By construction, $BC$ is a diagonal of $ABEC$ such that $AD$ is a bisector of it.
Thus by Quadrilateral is Parallelogram iff Diagonals Bisect each other, $AE$ is also a bisector of $ABEC$.
As $\angle BAC$... | Let $\triangle ABC$ be a [[Definition:Right Triangle|right triangle]] such that $BC$ is the [[Definition:Hypotenuse|hypotenuse]].
Let $AD$ be the [[Definition:Median of Triangle|median]] to $BC$.
Then $AD$ is half of $BC$. | [[File:Median to Hypotenuse of Right Triangle equals Half Hypotenuse.png|400px]]
Construct $BE$ and $CE$ [[Definition:Parallel Lines|parallel]] to $AC$ and $AB$ respectively.
Then by definition $ABEC$ is a [[Definition:Parallelogram|parallelogram]].
By construction, $BC$ is a [[Definition:Diagonal of Parallelogram|d... | Median to Hypotenuse of Right Triangle equals Half Hypotenuse | https://proofwiki.org/wiki/Median_to_Hypotenuse_of_Right_Triangle_equals_Half_Hypotenuse | https://proofwiki.org/wiki/Median_to_Hypotenuse_of_Right_Triangle_equals_Half_Hypotenuse | [
"Right Triangles",
"Medians of Triangles"
] | [
"Definition:Triangle (Geometry)/Right-Angled",
"Definition:Triangle (Geometry)/Right-Angled/Hypotenuse",
"Definition:Median of Triangle"
] | [
"File:Median to Hypotenuse of Right Triangle equals Half Hypotenuse.png",
"Definition:Parallel (Geometry)/Lines",
"Definition:Quadrilateral/Parallelogram",
"Definition:Diameter of Parallelogram",
"Definition:Bisection",
"Quadrilateral is Parallelogram iff Diagonals Bisect each other",
"Definition:Bisect... |
proofwiki-14566 | Diagonals of Rectangle are Equal | The diagonals of a rectangle are equal. | :400px
Let $ABCD$ be a rectangle.
The diagonals of $ABCD$ are $AC$ and $BD$.
Then $\angle ADC = \angle DAB$ as both are right angles by definition of rectangle.
By Rectangle is Parallelogram, $ABCD$ is also a type of parallelogram.
Thus by Opposite Sides and Angles of Parallelogram are Equal $AB = DC$.
Thus we have:
:$... | The [[Definition:Diagonal of Quadrilateral|diagonals]] of a [[Definition:Rectangle|rectangle]] are equal. | :[[File:Diagonals of Rectangle are Equal.png|400px]]
Let $ABCD$ be a [[Definition:Rectangle|rectangle]].
The [[Definition:Diagonal of Quadrilateral|diagonals]] of $ABCD$ are $AC$ and $BD$.
Then $\angle ADC = \angle DAB$ as both are [[Definition:Right Angle|right angles]] by definition of [[Definition:Rectangle|recta... | Diagonals of Rectangle are Equal | https://proofwiki.org/wiki/Diagonals_of_Rectangle_are_Equal | https://proofwiki.org/wiki/Diagonals_of_Rectangle_are_Equal | [
"Rectangles"
] | [
"Definition:Diameter of Quadrilateral",
"Definition:Quadrilateral/Rectangle"
] | [
"File:Diagonals of Rectangle are Equal.png",
"Definition:Quadrilateral/Rectangle",
"Definition:Diameter of Quadrilateral",
"Definition:Right Angle",
"Definition:Quadrilateral/Rectangle",
"Rectangle is Parallelogram",
"Definition:Quadrilateral/Parallelogram",
"Opposite Sides and Angles of Parallelogram... |
proofwiki-14567 | Summation to n of Square of kth Harmonic Number | :$\ds \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$
where $H_k$ denotes the $k$th harmonic number. | Consider:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| r = \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} - 1} {k + 1}
| c = factorizing
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
| c = Sum of Summations equals Summat... | :$\ds \sum_{k \mathop = 1}^n {H_k}^2 = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n$
where $H_k$ denotes the $k$th [[Definition:Harmonic Number|harmonic number]]. | Consider:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}
| r = \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} - 1} {k + 1}
| c = factorizing
}}
{{eqn | r = \sum_{k \mathop = 1}^{n - 1} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}
| c = [[Sum of Summations equals Sum... | Summation to n of Square of kth Harmonic Number | https://proofwiki.org/wiki/Summation_to_n_of_Square_of_kth_Harmonic_Number | https://proofwiki.org/wiki/Summation_to_n_of_Square_of_kth_Harmonic_Number | [
"Harmonic Numbers"
] | [
"Definition:Harmonic Numbers"
] | [
"Sum of Summations equals Summation of Sum",
"Summation of Unity over Elements",
"Definition:Harmonic Numbers",
"Abel's Lemma",
"Sum of Sequence of Harmonic Numbers",
"Sum of Summations equals Summation of Sum",
"Sum of Sequence of Harmonic Numbers"
] |
proofwiki-14568 | Summation of Odd Reciprocals in terms of Harmonic Numbers | :$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$
where $H_n$ denotes the $n$th harmonic number. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}
| r = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} + \sum_{k \mathop = 1}^n \dfrac 1 {2 k} - \sum_{k \mathop = 1}^n \dfrac 1 {2 k}
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^{2 n} \dfrac 1 k - \dfrac 1 2 \sum_{k \mathop = 1}^n \dfrac 1 k
... | :$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = H_{2 n} - \dfrac {H_n} 2$
where $H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}
| r = \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} + \sum_{k \mathop = 1}^n \dfrac 1 {2 k} - \sum_{k \mathop = 1}^n \dfrac 1 {2 k}
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^{2 n} \dfrac 1 k - \dfrac 1 2 \sum_{k \mathop = 1}^n \dfrac 1 k
... | Summation of Odd Reciprocals in terms of Harmonic Numbers | https://proofwiki.org/wiki/Summation_of_Odd_Reciprocals_in_terms_of_Harmonic_Numbers | https://proofwiki.org/wiki/Summation_of_Odd_Reciprocals_in_terms_of_Harmonic_Numbers | [
"Harmonic Numbers"
] | [
"Definition:Harmonic Numbers"
] | [] |
proofwiki-14569 | Numerator of p-1th Harmonic Number is Divisible by Prime p | Let $p$ be an odd prime.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p$. | Add the terms of $H_{p - 1}$ using the definition of rational addition to obtain $\dfrac m n$.
Do not cancel common prime factors from $m$ and $n$.
It is seen that $n = \paren {p - 1}!$
Hence $p$ is not a divisor of $n$.
The numerator $m$ is seen to be:
:$m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \c... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Consider the [[Definition:Harmonic Number|harmonic number]] $H_{p - 1}$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
The [[Definition:Numerator|numerator]] of $H_{p - 1}$ is [[Definition:Divisor of Integer|divisible]] by $p$. | Add the terms of $H_{p - 1}$ using the definition of [[Definition:Rational Addition|rational addition]] to obtain $\dfrac m n$.
Do not cancel common [[Definition:Prime Factor|prime factors]] from $m$ and $n$.
It is seen that $n = \paren {p - 1}!$
Hence $p$ is not a [[Definition:Divisor of Integer|divisor]] of $n$.
... | Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 1 | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_Prime_p | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_Prime_p/Proof_1 | [
"Harmonic Numbers",
"Numerator of p-1th Harmonic Number is Divisible by Prime p"
] | [
"Definition:Odd Prime",
"Definition:Harmonic Numbers",
"Definition:Rational Number/Canonical Form",
"Definition:Fraction/Numerator",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Addition/Rational Numbers",
"Definition:Prime Factor",
"Definition:Divisor (Algebra)/Integer",
"Definition:Fraction/Numerator",
"Definition:Multiple/Integer",
"Wilson's Theorem",
"Definition:Abelian Group",
"Inverse in Group is Unique",
"Definition:Set",
"Definition:Set",
"Closed For... |
proofwiki-14570 | Numerator of p-1th Harmonic Number is Divisible by Prime p | Let $p$ be an odd prime.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p$. | Note that for any integer $x$:
{{begin-eqn}}
{{eqn | l = x^p - x
| o = \equiv
| r = 0
| rr= \pmod p
| c = Corollary $1$ to Fermat's Little Theorem
}}
{{eqn | o = \equiv
| r = x^{\overline p}
| rr= \pmod p
| c = Divisibility of Product of Consecutive Integers
}}
{{eqn | o = \equ... | Let $p$ be an [[Definition:Odd Prime|odd prime]].
Consider the [[Definition:Harmonic Number|harmonic number]] $H_{p - 1}$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
The [[Definition:Numerator|numerator]] of $H_{p - 1}$ is [[Definition:Divisor of Integer|divisible]] by $p$. | Note that for any [[Definition:Integer|integer]] $x$:
{{begin-eqn}}
{{eqn | l = x^p - x
| o = \equiv
| r = 0
| rr= \pmod p
| c = [[Fermat's Little Theorem/Corollary 1|Corollary $1$ to Fermat's Little Theorem]]
}}
{{eqn | o = \equiv
| r = x^{\overline p}
| rr= \pmod p
| c = [[Di... | Numerator of p-1th Harmonic Number is Divisible by Prime p/Proof 2 | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_Prime_p | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_Prime_p/Proof_2 | [
"Harmonic Numbers",
"Numerator of p-1th Harmonic Number is Divisible by Prime p"
] | [
"Definition:Odd Prime",
"Definition:Harmonic Numbers",
"Definition:Rational Number/Canonical Form",
"Definition:Fraction/Numerator",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Integer",
"Fermat's Little Theorem/Corollary 1",
"Divisibility of Product of Consecutive Integers",
"Sum over k of Unsigned Stirling Numbers of First Kind by x^k",
"Definition:Stirling Numbers of the First Kind/Unsigned",
"Definition:Kronecker Delta",
"Harmonic Number as Unsigned Stirling Nu... |
proofwiki-14571 | Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3 | Let $p$ be a prime number such that $p > 3$.
Consider the harmonic number $H_{p - 1}$ expressed in canonical form.
The numerator of $H_{p - 1}$ is divisible by $p^2$. | {{ProofWanted|research and invoke Wolstenholme's Theorem}} | Let $p$ be a [[Definition:Prime Number|prime number]] such that $p > 3$.
Consider the [[Definition:Harmonic Number|harmonic number]] $H_{p - 1}$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
The [[Definition:Numerator|numerator]] of $H_{p - 1}$ is [[Definition:Divisor of Integer|divis... | {{ProofWanted|research and invoke [[Wolstenholme's Theorem]]}} | Numerator of p-1th Harmonic Number is Divisible by p^2 for Prime Greater than 3 | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_p^2_for_Prime_Greater_than_3 | https://proofwiki.org/wiki/Numerator_of_p-1th_Harmonic_Number_is_Divisible_by_p^2_for_Prime_Greater_than_3 | [
"Harmonic Numbers"
] | [
"Definition:Prime Number",
"Definition:Harmonic Numbers",
"Definition:Rational Number/Canonical Form",
"Definition:Fraction/Numerator",
"Definition:Divisor (Algebra)/Integer"
] | [
"Wolstenholme's Theorem"
] |
proofwiki-14572 | Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals | Let:
: $S = \dfrac p q = \ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$
where $\dfrac p q$ is the canonical form of $S$.
Let $n = 2^k m$ where $m$ is odd.
Then the largest power of $2$ that divides $p$ is $2^{2 k}$. | We have that:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = \sum_{i \mathop = 0}^{M - 1} \paren {\dfrac 1 {i \times 2^{r + 1} + 1} + \dfrac 1 {i \times 2^{r + 1} + 3} + \cdots + \dfrac 1 {\paren {i + 1} \times 2^{r + 1} - 1} }$
where $k = 2^r M$ where $M$ is odd.
Collect the $r$ terms in the parenthesis on the {{RH... | Let:
: $S = \dfrac p q = \ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1}$
where $\dfrac p q$ is the [[Definition:Canonical Form of Rational Number|canonical form]] of $S$.
Let $n = 2^k m$ where $m$ is [[Definition:Odd Integer|odd]].
Then the largest [[Definition:Integer Power|power of $2$]] that [[Definition:Divisor o... | We have that:
:$\ds \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = \sum_{i \mathop = 0}^{M - 1} \paren {\dfrac 1 {i \times 2^{r + 1} + 1} + \dfrac 1 {i \times 2^{r + 1} + 3} + \cdots + \dfrac 1 {\paren {i + 1} \times 2^{r + 1} - 1} }$
where $k = 2^r M$ where $M$ is [[Definition:Odd Integer|odd]].
Collect the $r$ terms i... | Highest Power of 2 Dividing Numerator of Sum of Odd Reciprocals | https://proofwiki.org/wiki/Highest_Power_of_2_Dividing_Numerator_of_Sum_of_Odd_Reciprocals | https://proofwiki.org/wiki/Highest_Power_of_2_Dividing_Numerator_of_Sum_of_Odd_Reciprocals | [
"Harmonic Numbers"
] | [
"Definition:Rational Number/Canonical Form",
"Definition:Odd Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Odd Integer",
"Definition:Parenthesis",
"Definition:Fraction/Denominator",
"Definition:Fraction/Numerator",
"Definition:Odd Integer",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Odd Integer",
"Definition:Congruence (Number Theory)/Residue",
"Definition:Odd Integer",
"D... |
proofwiki-14573 | Summation of Power Series by Harmonic Sequence | Consider the power series:
:$\map f x = \ds \sum_{k \mathop \ge 0} a_k x^k$
Let $\map f x$ converge for $x = x_0$.
Then:
:$\ds \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$
where $H_n$ denotes the $n$th harmonic number. | {{begin-eqn}}
{{eqn | l = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y
| r = \int_0^1 \dfrac {\sum_{k \mathop \ge 0} a_k {x_0}^k - \sum_{k \mathop \ge 0} a_k {x_0}^k y^k} {1 - y} \rd y
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \dfrac {1 - y^k} {1 - y} \rd y
| ... | Consider the [[Definition:Power Series|power series]]:
:$\map f x = \ds \sum_{k \mathop \ge 0} a_k x^k$
Let $\map f x$ [[Definition:Convergent Series of Numbers|converge]] for $x = x_0$.
Then:
:$\ds \sum_{k \mathop \ge 0} a_k {x_0}^k H_k = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y$
where $H_n$... | {{begin-eqn}}
{{eqn | l = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y
| r = \int_0^1 \dfrac {\sum_{k \mathop \ge 0} a_k {x_0}^k - \sum_{k \mathop \ge 0} a_k {x_0}^k y^k} {1 - y} \rd y
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \dfrac {1 - y^k} {1 - y} \rd y
| ... | Summation of Power Series by Harmonic Sequence | https://proofwiki.org/wiki/Summation_of_Power_Series_by_Harmonic_Sequence | https://proofwiki.org/wiki/Summation_of_Power_Series_by_Harmonic_Sequence | [
"Harmonic Numbers"
] | [
"Definition:Power Series",
"Definition:Convergent Series/Number Field",
"Definition:Harmonic Numbers"
] | [
"Sum of Geometric Sequence",
"Primitive of Power"
] |
proofwiki-14574 | Summation over k to n of Harmonic Numbers over n+1-k | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k} = {H_{n + 1} }^2 - H_{n + 1}^{\paren 2}$
where:
:$H_n$ denotes the $n$th harmonic number
:$H_n^{\paren 2}$ denotes a general harmonic number. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k}
| r = \sum_{k \mathop = 1}^n \dfrac {H_{n + 1 - k} } k
| c = Permutation of Indices of Summation
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j \paren {n + 1 - k} }
| c = {{Defof|Harmonic Number}}
}}
{{... | :$\ds \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k} = {H_{n + 1} }^2 - H_{n + 1}^{\paren 2}$
where:
:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
:$H_n^{\paren 2}$ denotes a [[Definition:General Harmonic Numbers|general harmonic number]]. | {{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k}
| r = \sum_{k \mathop = 1}^n \dfrac {H_{n + 1 - k} } k
| c = [[Permutation of Indices of Summation]]
}}
{{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j \paren {n + 1 - k} }
| c = {{Defof|Harmonic Number}}
}... | Summation over k to n of Harmonic Numbers over n+1-k | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Harmonic_Numbers_over_n+1-k | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Harmonic_Numbers_over_n+1-k | [
"Harmonic Numbers",
"General Harmonic Numbers"
] | [
"Definition:Harmonic Numbers",
"Definition:Harmonic Numbers/General Definition"
] | [
"Permutation of Indices of Summation"
] |
proofwiki-14575 | Extension of Harmonic Number to Non-Integer Argument | Let $\map H x$ be the real function defined as:
:$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$
where:
:$\map \Gamma x$ denotes the Gamma function
:$\map {\Gamma'} x$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.
Then $H$ is an extension of ... | For $n \in \N$:
{{begin-eqn}}
{{eqn | l = \map H n
| r = \gamma + \frac {\map {\Gamma'} {n + 1} } {\map \Gamma {n + 1} }
}}
{{eqn | r = \gamma - \gamma + \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {n + m} }
| c = Reciprocal times Derivative of Gamma Function
}}
{{eqn | r = \lim_{k \mathop \to \... | Let $\map H x$ be the [[Definition:Real Function|real function]] defined as:
:$\map H x = \gamma + \dfrac {\map {\Gamma'} {x + 1} } {\map \Gamma {x + 1} }$
where:
:$\map \Gamma x$ denotes the [[Definition:Gamma Function|Gamma function]]
:$\map {\Gamma'} x$ denotes the [[Definition:Derivative|derivative]] of the [[Defin... | For $n \in \N$:
{{begin-eqn}}
{{eqn | l = \map H n
| r = \gamma + \frac {\map {\Gamma'} {n + 1} } {\map \Gamma {n + 1} }
}}
{{eqn | r = \gamma - \gamma + \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {n + m} }
| c = [[Reciprocal times Derivative of Gamma Function]]
}}
{{eqn | r = \lim_{k \mathop ... | Extension of Harmonic Number to Non-Integer Argument | https://proofwiki.org/wiki/Extension_of_Harmonic_Number_to_Non-Integer_Argument | https://proofwiki.org/wiki/Extension_of_Harmonic_Number_to_Non-Integer_Argument | [
"Gamma Function",
"Harmonic Numbers"
] | [
"Definition:Real Function",
"Definition:Gamma Function",
"Definition:Derivative",
"Definition:Gamma Function",
"Definition:Euler-Mascheroni Constant",
"Definition:Extension of Mapping",
"Definition:Mapping",
"Definition:Harmonic Numbers"
] | [
"Reciprocal times Derivative of Gamma Function",
"Linear Combination of Convergent Series"
] |
proofwiki-14576 | Initial Segment Determined by Smallest Element is Empty | Let $\struct {S, \preceq}$ be a well-ordered set, where $S$ is non-empty.
Let $s_0 = \min S$, the smallest element of $S$.
Then the initial segment determined by $s_0$, $S_{s_0}$, is empty. | By the definition of initial segment:
:$S_{s_0} := \set {b \in S: b \prec s_0}$
By the definition of smallest element:
:$\forall b \in S: s_0 \preceq b$
Thus $S_{s_0}$ is empty.
{{qed}}
Category:Initial Segments
Category:Well-Orderings
cxsawbph07ijlb2ilxdcb4u5ecu9848 | Let $\struct {S, \preceq}$ be a [[Definition:Well-Ordered Set|well-ordered set]], where $S$ is [[Definition:Non-Empty Set|non-empty]].
Let $s_0 = \min S$, the [[Definition:Smallest Element|smallest element]] of $S$.
Then the [[Definition:Initial Segment|initial segment determined by $s_0$]], $S_{s_0}$, is [[Definiti... | By the definition of [[Definition:Initial Segment|initial segment]]:
:$S_{s_0} := \set {b \in S: b \prec s_0}$
By the definition of [[Definition:Smallest Element|smallest element]]:
:$\forall b \in S: s_0 \preceq b$
Thus $S_{s_0}$ is [[Definition:Empty Set|empty]].
{{qed}}
[[Category:Initial Segments]]
[[Category:... | Initial Segment Determined by Smallest Element is Empty | https://proofwiki.org/wiki/Initial_Segment_Determined_by_Smallest_Element_is_Empty | https://proofwiki.org/wiki/Initial_Segment_Determined_by_Smallest_Element_is_Empty | [
"Initial Segments",
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Non-Empty Set",
"Definition:Smallest Element",
"Definition:Initial Segment",
"Definition:Empty Set"
] | [
"Definition:Initial Segment",
"Definition:Smallest Element",
"Definition:Empty Set",
"Category:Initial Segments",
"Category:Well-Orderings"
] |
proofwiki-14577 | Strictly Increasing Mapping Between Wosets Implies Order Isomorphism | Let $J$ and $E$ be well-ordered sets.
Let there exist a mapping $k: J \to E$ which is strictly increasing.
Then $J$ is order isomorphic to $E$ or an initial segment of $E$. | If the sets considered are empty or singletons, the theorem holds vacuously or trivially.
Suppose $J, E$ both have at least two elements.
Let $e_0 = \min E$, the smallest element of $E$.
Define the mapping:
:$h: J \to E$:
:$\map h \alpha = \begin {cases} \map \min {E \setminus h \sqbrk {S_\alpha} } & : h \sqbrk {S_\alp... | Let $J$ and $E$ be [[Definition:Well-Ordered Set|well-ordered sets]].
Let there exist a [[Definition:Mapping|mapping]] $k: J \to E$ which is [[Definition:Strictly Increasing Mapping|strictly increasing]].
Then $J$ is [[Definition:Order Isomorphism|order isomorphic]] to $E$ or an [[Definition:Initial Segment|initial ... | If the [[Definition:Set|sets]] considered are [[Definition:Empty Set|empty]] or [[Definition:Singleton|singletons]], the theorem holds [[Definition:Vacuous|vacuously]] or trivially.
Suppose $J, E$ both have at least two [[Definition:Element|elements]].
Let $e_0 = \min E$, the [[Definition:Smallest Element|smallest e... | Strictly Increasing Mapping Between Wosets Implies Order Isomorphism | https://proofwiki.org/wiki/Strictly_Increasing_Mapping_Between_Wosets_Implies_Order_Isomorphism | https://proofwiki.org/wiki/Strictly_Increasing_Mapping_Between_Wosets_Implies_Order_Isomorphism | [
"Well-Orderings"
] | [
"Definition:Well-Ordered Set",
"Definition:Mapping",
"Definition:Strictly Increasing/Mapping",
"Definition:Order Isomorphism",
"Definition:Initial Segment"
] | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Singleton",
"Definition:Vacuous",
"Definition:Element",
"Definition:Smallest Element",
"Definition:Mapping",
"Definition:Initial Segment",
"Definition:Image (Set Theory)/Mapping/Subset",
"Principle of Recursive Definition for Well-Ordered Sets"... |
proofwiki-14578 | Equality to Initial Segment Imposes Well-Ordering | Let $X$ be a set.
Let $\AA$ be the set of all ordered pairs $\struct {A, <}$ such that $A$ is a subset of $X$ and $<$ is a strict well-ordering of $A$.
Define $\prec$ as:
:$\struct {A, <} \prec \struct {A', <'}$
{{iff}}
:$\struct {A, <}$ equals an initial segment of $\struct {A', <'}$.
Let $\BB$ be a set of ordered pa... | If the set $X$ considered is empty or a singleton, the lemma holds vacuously or trivially.
Thus assume $X$ contains at least two elements.
We first prove that $\prec$ is a strict partial ordering on $\AA$.
From the definition of initial segment, no $\struct {A, <}$ can equal an initial segment of itself.
Thus $\prec$ i... | Let $X$ be a [[Definition:Set|set]].
Let $\AA$ be the [[Definition:Set|set]] of all [[Definition:Ordered Pair|ordered pairs]] $\struct {A, <}$ such that $A$ is a [[Definition:Subset|subset]] of $X$ and $<$ is a [[Definition:Strict Well-Ordering|strict well-ordering]] of $A$.
Define $\prec$ as:
:$\struct {A, <} \prec... | If the [[Definition:Set|set]] $X$ considered is [[Definition:Empty Set|empty]] or a [[Definition:Singleton|singleton]], the [[Definition:Lemma|lemma]] holds [[Definition:Vacuous|vacuously]] or trivially.
Thus assume $X$ contains at least two [[Definition:Element|elements]].
We first prove that $\prec$ is a [[Definit... | Equality to Initial Segment Imposes Well-Ordering | https://proofwiki.org/wiki/Equality_to_Initial_Segment_Imposes_Well-Ordering | https://proofwiki.org/wiki/Equality_to_Initial_Segment_Imposes_Well-Ordering | [
"Well-Orderings",
"Set Equality"
] | [
"Definition:Set",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Subset",
"Definition:Strict Well-Ordering",
"Definition:Equals",
"Definition:Initial Segment",
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Order",
"Definition:Union",
"Definition:Set Union",
"Definition:R... | [
"Definition:Set",
"Definition:Empty Set",
"Definition:Singleton",
"Definition:Lemma",
"Definition:Vacuous",
"Definition:Element",
"Definition:Strict Partial Ordering",
"Definition:Initial Segment",
"Definition:Initial Segment",
"Definition:Antireflexive Relation",
"Definition:Initial Segment",
... |
proofwiki-14579 | Empty Mapping is Injective | Let $\nu: \O \to T$ be an empty mapping.
Then $\nu$ is an injection. | There are no elements in the domain of $\nu$.
Thus $\nu$ is an injection vacuously.
{{qed}} | Let $\nu: \O \to T$ be an [[Definition:Empty Mapping|empty mapping]].
Then $\nu$ is an [[Definition:Injection|injection]]. | There are no [[Definition:Element|elements]] in the [[Definition:Domain of Mapping|domain]] of $\nu$.
Thus $\nu$ is an [[Definition:Injection|injection]] [[Definition:Vacuous Truth|vacuously]].
{{qed}} | Empty Mapping is Injective | https://proofwiki.org/wiki/Empty_Mapping_is_Injective | https://proofwiki.org/wiki/Empty_Mapping_is_Injective | [
"Injections",
"Empty Mapping"
] | [
"Definition:Empty Mapping",
"Definition:Injection"
] | [
"Definition:Element",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Injection",
"Definition:Vacuous Truth"
] |
proofwiki-14580 | Upper and Lower Bound of Fibonacci Number | For all $n \in \N_{> 0}$:
:$\phi^{n - 2} \le F_n \le \phi^{n - 1}$
where:
:$F_n$ is the $n$th Fibonacci number
:$\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$ | From Fibonacci Number greater than Golden Section to Power less Two:
:$F_n \ge \phi^{n - 2}$
From Fibonacci Number less than Golden Section to Power less One:
:$F_n \le \phi^{n - 1}$
{{qed}} | For all $n \in \N_{> 0}$:
:$\phi^{n - 2} \le F_n \le \phi^{n - 1}$
where:
:$F_n$ is the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]]
:$\phi$ is the [[Definition:Golden Section|golden section]]: $\phi = \dfrac {1 + \sqrt 5} 2$ | From [[Fibonacci Number greater than Golden Section to Power less Two]]:
:$F_n \ge \phi^{n - 2}$
From [[Fibonacci Number less than Golden Section to Power less One]]:
:$F_n \le \phi^{n - 1}$
{{qed}} | Upper and Lower Bound of Fibonacci Number | https://proofwiki.org/wiki/Upper_and_Lower_Bound_of_Fibonacci_Number | https://proofwiki.org/wiki/Upper_and_Lower_Bound_of_Fibonacci_Number | [
"Fibonacci Numbers",
"Golden Mean"
] | [
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Fibonacci Number greater than Golden Section to Power less Two",
"Fibonacci Number less than Golden Section to Power less One"
] |
proofwiki-14581 | Generating Function for Fibonacci Numbers | Let $\map G z$ be the function defined as:
:$\map G z = \dfrac z {1 - z - z^2}$
Then $\map G z$ is a generating function for the Fibonacci numbers. | Let the form of $\map G z$ be assumed as:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{k \mathop \ge 0} F_k z^k
| c =
}}
{{eqn | r = F_0 + F_1 z + F_2 z^2 + F_3 z^3 + F_4 z^4 + \cdots
| c =
}}
{{eqn | r = 0 + z + z^2 + 2 z^3 + 3 z^4 + \cdots
| c =
}}
{{end-eqn}}
where $F_n$ denotes the $n$t... | Let $\map G z$ be the [[Definition:Real Function|function]] defined as:
:$\map G z = \dfrac z {1 - z - z^2}$
Then $\map G z$ is a [[Definition:Generating Function|generating function]] for the [[Definition:Fibonacci Number|Fibonacci numbers]]. | Let the form of $\map G z$ be assumed as:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{k \mathop \ge 0} F_k z^k
| c =
}}
{{eqn | r = F_0 + F_1 z + F_2 z^2 + F_3 z^3 + F_4 z^4 + \cdots
| c =
}}
{{eqn | r = 0 + z + z^2 + 2 z^3 + 3 z^4 + \cdots
| c =
}}
{{end-eqn}}
where $F_n$ denotes the [[... | Generating Function for Fibonacci Numbers | https://proofwiki.org/wiki/Generating_Function_for_Fibonacci_Numbers | https://proofwiki.org/wiki/Generating_Function_for_Fibonacci_Numbers | [
"Fibonacci Numbers",
"Examples of Generating Functions"
] | [
"Definition:Real Function",
"Definition:Generating Function",
"Definition:Fibonacci Number"
] | [
"Definition:Fibonacci Number"
] |
proofwiki-14582 | Summation over k to n of Product of kth with n-kth Fibonacci Numbers | :$\ds \sum_{k \mathop = 0}^n F_k F_{n - k} = \dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5$
where $F_n$ denotes the $n$th Fibonacci number. | From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
:$\map G z = \dfrac z {1 - z - z^2}$
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \dfrac z {1 - z - z^2}
| c =
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }... | :$\ds \sum_{k \mathop = 0}^n F_k F_{n - k} = \dfrac {\paren {n - 1} F_n + 2n F_{n - 1} } 5$
where $F_n$ denotes the [[Definition:Fibonacci Number|$n$th Fibonacci number]]. | From [[Generating Function for Fibonacci Numbers]], a [[Definition:Generating Function|generating function]] for the [[Definition:Fibonacci Number|Fibonacci numbers]] is:
:$\map G z = \dfrac z {1 - z - z^2}$
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \dfrac z {1 - z - z^2}
| c =
}}
{{eqn | r = \dfra... | Summation over k to n of Product of kth with n-kth Fibonacci Numbers | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Product_of_kth_with_n-kth_Fibonacci_Numbers | https://proofwiki.org/wiki/Summation_over_k_to_n_of_Product_of_kth_with_n-kth_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Fibonacci Number"
] | [
"Generating Function for Fibonacci Numbers",
"Definition:Generating Function",
"Definition:Fibonacci Number",
"Definition:Partial Fractions Expansion",
"Power Series Expansion for Reciprocal of Square of 1 - z",
"Generating Function for Fibonacci Numbers",
"Translation of Index Variable of Summation",
... |
proofwiki-14583 | Fibonacci Numbers which equal their Index | The only Fibonacci numbers which equal their index are:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{eqn | l = F_5
| r = 5
}}
{{end-eqn}} | By definition of the Fibonacci numbers:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{end-eqn}}
Then it is observed that $F_5 = 5$.
After that, for $n > 5$, we have that $F_n > n$.
{{qed}} | The only [[Definition:Fibonacci Number|Fibonacci numbers]] which equal their [[Definition:Index (Indexing Set)|index]] are:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{eqn | l = F_5
| r = 5
}}
{{end-eqn}} | By definition of the [[Definition:Fibonacci Number|Fibonacci numbers]]:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{end-eqn}}
Then it is observed that $F_5 = 5$.
After that, for $n > 5$, we have that $F_n > n$.
{{qed}} | Fibonacci Numbers which equal their Index | https://proofwiki.org/wiki/Fibonacci_Numbers_which_equal_their_Index | https://proofwiki.org/wiki/Fibonacci_Numbers_which_equal_their_Index | [
"Fibonacci Numbers"
] | [
"Definition:Fibonacci Number",
"Definition:Indexing Set/Index"
] | [
"Definition:Fibonacci Number"
] |
proofwiki-14584 | Fibonacci Numbers which equal the Square of their Index | The only Fibonacci numbers which equal the square of their index are:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{eqn | l = F_{12}
| r = 12^2 = 144
}}
{{end-eqn}} | By definition of the Fibonacci numbers:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{end-eqn}}
Then it is observed that $F_{12} = 144$.
After that, for $n > 12$, we have that $F_n > n^2$.
{{qed}} | The only [[Definition:Fibonacci Number|Fibonacci numbers]] which equal the [[Definition:Square (Algebra)|square]] of their [[Definition:Index (Indexing Set)|index]] are:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{eqn | l = F_{12}
| r = 12^2 = 144
}}
{{end-eqn}} | By definition of the [[Definition:Fibonacci Number|Fibonacci numbers]]:
{{begin-eqn}}
{{eqn | l = F_0
| r = 0
}}
{{eqn | l = F_1
| r = 1
}}
{{end-eqn}}
Then it is observed that $F_{12} = 144$.
After that, for $n > 12$, we have that $F_n > n^2$.
{{qed}} | Fibonacci Numbers which equal the Square of their Index | https://proofwiki.org/wiki/Fibonacci_Numbers_which_equal_the_Square_of_their_Index | https://proofwiki.org/wiki/Fibonacci_Numbers_which_equal_the_Square_of_their_Index | [
"Fibonacci Numbers"
] | [
"Definition:Fibonacci Number",
"Definition:Square/Function",
"Definition:Indexing Set/Index"
] | [
"Definition:Fibonacci Number"
] |
proofwiki-14585 | Cassini's Identity/Negative Indices | Let $n \in \Z_{<0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then Cassini's Identity:
:$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$
continues to hold. | Let $n \in \Z_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = F_{-\paren {n + 1} } F_{-\paren {n - 1} } - {F_{-n} }^2
| r = \paren {-1}^{n + 2} F_{n + 1} \paren {-1}^n F_{n - 1} - \paren {\paren {-1}^{n + 1} F_n}^2
| c = Fibonacci Number with Negative Index
}}
{{eqn | r = \paren {-1}^{2 n + 2} F_{n + 1} F_{n - 1} -... | Let $n \in \Z_{<0}$ be a [[Definition:Negative Integer|negative integer]].
Let $F_n$ be the $n$th [[Definition:Fibonacci Number for Negative Index|Fibonacci number (as extended to negative integers)]].
Then [[Cassini's Identity]]:
:$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$
continues to hold. | Let $n \in \Z_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = F_{-\paren {n + 1} } F_{-\paren {n - 1} } - {F_{-n} }^2
| r = \paren {-1}^{n + 2} F_{n + 1} \paren {-1}^n F_{n - 1} - \paren {\paren {-1}^{n + 1} F_n}^2
| c = [[Fibonacci Number with Negative Index]]
}}
{{eqn | r = \paren {-1}^{2 n + 2} F_{n + 1} F_{n ... | Cassini's Identity/Negative Indices | https://proofwiki.org/wiki/Cassini's_Identity/Negative_Indices | https://proofwiki.org/wiki/Cassini's_Identity/Negative_Indices | [
"Cassini's Identity"
] | [
"Definition:Negative/Integer",
"Definition:Fibonacci Number/Negative",
"Cassini's Identity"
] | [
"Fibonacci Number with Negative Index",
"Definition:Power (Algebra)/Even Power"
] |
proofwiki-14586 | Honsberger's Identity/Negative Indices | Let $n \in \Z_{< 0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then Honsberger's Identity:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
continues to hold, whether $m$ or $n$ are positive or negative. | The proof proceeds by induction.
For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
This can equivalently be written:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$F_{m - n} = F_{m - 1} F_{-n} + F_m F_{-n + 1}$
$\map P 0$ is the case:
{{begin-e... | Let $n \in \Z_{< 0}$ be a [[Definition:Negative Integer|negative integer]].
Let $F_n$ be the $n$th [[Definition:Fibonacci Number for Negative Index|Fibonacci number (as extended to negative integers)]].
Then [[Honsberger's Identity]]:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
continues to hold, whether $m$ or $n... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\le 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
This can equivalently be written:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition... | Honsberger's Identity/Negative Indices | https://proofwiki.org/wiki/Honsberger's_Identity/Negative_Indices | https://proofwiki.org/wiki/Honsberger's_Identity/Negative_Indices | [
"Honsberger's Identity"
] | [
"Definition:Negative/Integer",
"Definition:Fibonacci Number/Negative",
"Honsberger's Identity",
"Definition:Positive/Integer",
"Definition:Negative/Integer"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14587 | Euler-Binet Formula/Negative Index | Let $n \in \Z_{< 0}$ be a negative integer.
Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).
Then the Euler-Binet Formula:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} = \dfrac {\phi^n - \hat \phi^n} {\phi - \hat \phi}$
continues to hold. | Let $n \in \Z_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\phi^{-n} - \hat \phi^{-n} } {\sqrt 5}
| r = \dfrac 1 {\sqrt 5} \paren {\phi^{-n} - \paren {-\dfrac 1 \phi}^{-n} }
| c = Definition of $\hat \phi$
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {\phi^n} - \paren {-1}^n \phi^n}
| c = Exp... | Let $n \in \Z_{< 0}$ be a [[Definition:Negative Integer|negative integer]].
Let $F_n$ be the $n$th [[Definition:Fibonacci Number for Negative Index|Fibonacci number (as extended to negative integers)]].
Then the [[Euler-Binet Formula]]:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} = \dfrac {\phi^n - \hat \phi^n}... | Let $n \in \Z_{> 0}$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac {\phi^{-n} - \hat \phi^{-n} } {\sqrt 5}
| r = \dfrac 1 {\sqrt 5} \paren {\phi^{-n} - \paren {-\dfrac 1 \phi}^{-n} }
| c = Definition of $\hat \phi$
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {\phi^n} - \paren {-1}^n \phi^n}
| c = [... | Euler-Binet Formula/Negative Index | https://proofwiki.org/wiki/Euler-Binet_Formula/Negative_Index | https://proofwiki.org/wiki/Euler-Binet_Formula/Negative_Index | [
"Euler-Binet Formula"
] | [
"Definition:Negative/Integer",
"Definition:Fibonacci Number/Negative",
"Euler-Binet Formula"
] | [
"Exponent Combination Laws/Negative Power",
"Euler-Binet Formula",
"Definition:Positive/Integer",
"Fibonacci Number with Negative Index"
] |
proofwiki-14588 | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less | Let $n \in \Z$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\phi$ denotes the golden mean. | === Positive Index ===
First the result is proved for positive integers.
{{:Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index}}{{qed|lemma}} | Let $n \in \Z$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the [[Definition:Fibonacci Number|$n$th Fibonacci number]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | === [[Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index|Positive Index]] ===
First the result is proved for [[Definition:Positive Integer|positive integers]].
{{:Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index}}{{qed|lemma}} | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less | [
"Fibonacci Numbers",
"Golden Mean",
"Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less"
] | [
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index",
"Definition:Positive/Integer",
"Definition:Positive/Integer"
] |
proofwiki-14589 | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index | Let $n \in \Z_{\ge 0}$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\phi$ denotes the golden mean. | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$\phi^n = F_n \phi + F_{n - 1}$
$P \left({0}\right)$ is the case:
{{begin-eqn}}
{{eqn | l = F_0 \times \phi + F_{-1}
| r = F_0 \times \phi + \left({-1}\right)^0 F_1
| c = Fibonacci Number with Negative... | Let $n \in \Z_{\ge 0}$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the [[Definition:Fibonacci Number|$n$th Fibonacci number]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:$\phi^n = F_n \phi + F_{n - 1}$
$P \left({0}\right)$ is the case:
{{begin-eqn}}
{{eqn | l = F_0 \times \phi + F_{-1}
| r = F_0 \times \p... | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less/Positive_Index | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less/Positive_Index | [
"Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less"
] | [
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Fibonacci Number with Negative Index",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Square of Golden Mean equals One plus Golden Mean",
"Fibonacci Number by Golden... |
proofwiki-14590 | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index | Let $n \in \Z_{\le 0}$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the $n$th Fibonacci number as extended to negative indices
:$\phi$ denotes the golden mean. | The proof proceeds by induction.
For all $n \in \Z_{\le 0}$, let $\map P n$ be the proposition:
:$\phi^n = F_n \phi + F_{n - 1}$
This can equivalently be expressed as:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\phi^{-n} = F_{-n} \phi + F_{-n - 1}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l =... | Let $n \in \Z_{\le 0}$.
Then:
:$\phi^n = F_n \phi + F_{n - 1}$
where:
:$F_n$ denotes the [[Definition:Fibonacci Number for Negative Index|$n$th Fibonacci number as extended to negative indices]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\le 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\phi^n = F_n \phi + F_{n - 1}$
This can equivalently be expressed as:
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposi... | Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less/Negative_Index | https://proofwiki.org/wiki/Fibonacci_Number_by_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less/Negative_Index | [
"Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less"
] | [
"Definition:Fibonacci Number/Negative",
"Definition:Golden Mean"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Proposition",
"Fibonacci Number with Negative Index",
"Fibonacci Number with Negative Index",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Fibonacci Number with Nega... |
proofwiki-14591 | Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less | Let $n \in \Z$.
Then:
:$\hat \phi^n = F_n \hat \phi + F_{n - 1}$
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\hat \phi$ denotes the $1$ minus the golden mean:
::$\hat \phi := 1 - \phi$ | {{begin-eqn}}
{{eqn | l = F_n \hat \phi + F_{n - 1}
| r = F_n \paren {-\dfrac 1 \phi} + F_{n - 1}
| c = Reciprocal Form of One Minus Golden Mean
}}
{{eqn | r = -\dfrac 1 \phi \paren {F_n - \phi F_{n - 1} }
| c =
}}
{{eqn | r = -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} - \phi \paren {-1}^n F_{... | Let $n \in \Z$.
Then:
:$\hat \phi^n = F_n \hat \phi + F_{n - 1}$
where:
:$F_n$ denotes the [[Definition:Fibonacci Number|$n$th Fibonacci number]]
:$\hat \phi$ denotes the [[Definition:One Minus Golden Mean|$1$ minus the golden mean]]:
::$\hat \phi := 1 - \phi$ | {{begin-eqn}}
{{eqn | l = F_n \hat \phi + F_{n - 1}
| r = F_n \paren {-\dfrac 1 \phi} + F_{n - 1}
| c = [[Reciprocal Form of One Minus Golden Mean]]
}}
{{eqn | r = -\dfrac 1 \phi \paren {F_n - \phi F_{n - 1} }
| c =
}}
{{eqn | r = -\dfrac 1 \phi \paren {\paren {-1}^{n + 1} F_{-n} - \phi \paren {-1}^n... | Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less | https://proofwiki.org/wiki/Fibonacci_Number_by_One_Minus_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less | https://proofwiki.org/wiki/Fibonacci_Number_by_One_Minus_Golden_Mean_plus_Fibonacci_Number_of_Index_One_Less | [
"Fibonacci Numbers",
"Golden Mean",
"Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less"
] | [
"Definition:Fibonacci Number",
"Definition:Golden Mean/One Minus Golden Mean"
] | [
"Reciprocal Form of One Minus Golden Mean",
"Fibonacci Number with Negative Index",
"Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less"
] |
proofwiki-14592 | Reciprocal Form of One Minus Golden Mean | :$\hat \phi = - \dfrac 1 \phi$
where:
:$\phi$ denotes the golden mean
:$\hat \phi$ denotes one minus the golden mean: $\hat \phi = 1 - \phi$. | {{begin-eqn}}
{{eqn | l = \hat \phi
| r = 1 - \phi
| c =
}}
{{eqn | r = -\left({\phi - 1}\right)
| c =
}}
{{eqn | r = -\frac 1 \phi
| c = {{Defof|Golden Mean|index = 3}}
}}
{{end-eqn}}
{{qed}}
Category:Golden Mean
k9cbb5cuwgltozh1uhujnnwmhw70wd4 | :$\hat \phi = - \dfrac 1 \phi$
where:
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]
:$\hat \phi$ denotes [[Definition:One Minus Golden Mean|one minus the golden mean]]: $\hat \phi = 1 - \phi$. | {{begin-eqn}}
{{eqn | l = \hat \phi
| r = 1 - \phi
| c =
}}
{{eqn | r = -\left({\phi - 1}\right)
| c =
}}
{{eqn | r = -\frac 1 \phi
| c = {{Defof|Golden Mean|index = 3}}
}}
{{end-eqn}}
{{qed}}
[[Category:Golden Mean]]
k9cbb5cuwgltozh1uhujnnwmhw70wd4 | Reciprocal Form of One Minus Golden Mean | https://proofwiki.org/wiki/Reciprocal_Form_of_One_Minus_Golden_Mean | https://proofwiki.org/wiki/Reciprocal_Form_of_One_Minus_Golden_Mean | [
"Golden Mean"
] | [
"Definition:Golden Mean",
"Definition:Golden Mean/One Minus Golden Mean"
] | [
"Category:Golden Mean"
] |
proofwiki-14593 | Closed Form of One Minus Golden Mean | :$\hat \phi = \dfrac {1 - \sqrt 5} 2$
where:
:$\hat \phi$ denotes one minus the golden mean: $\hat \phi = 1 - \phi$. | {{begin-eqn}}
{{eqn | l = \hat \phi
| r = 1 - \phi
| c =
}}
{{eqn | r = 1 - \dfrac {1 + \sqrt 5} 2
| c = {{Defof|Golden Mean|index = 2}}
}}
{{eqn | r = \dfrac {2 - \left({1 + \sqrt 5}\right)} 2
| c = common denominator
}}
{{eqn | r = \dfrac {1 - \sqrt 5} 2
| c =
}}
{{end-eqn}}
{{qed}}
Ca... | :$\hat \phi = \dfrac {1 - \sqrt 5} 2$
where:
:$\hat \phi$ denotes [[Definition:One Minus Golden Mean|one minus the golden mean]]: $\hat \phi = 1 - \phi$. | {{begin-eqn}}
{{eqn | l = \hat \phi
| r = 1 - \phi
| c =
}}
{{eqn | r = 1 - \dfrac {1 + \sqrt 5} 2
| c = {{Defof|Golden Mean|index = 2}}
}}
{{eqn | r = \dfrac {2 - \left({1 + \sqrt 5}\right)} 2
| c = common denominator
}}
{{eqn | r = \dfrac {1 - \sqrt 5} 2
| c =
}}
{{end-eqn}}
{{qed}}
[... | Closed Form of One Minus Golden Mean | https://proofwiki.org/wiki/Closed_Form_of_One_Minus_Golden_Mean | https://proofwiki.org/wiki/Closed_Form_of_One_Minus_Golden_Mean | [
"Golden Mean"
] | [
"Definition:Golden Mean/One Minus Golden Mean"
] | [
"Category:Golden Mean"
] |
proofwiki-14594 | Second Order Fibonacci Number in terms of Fibonacci Numbers | The second order Fibonacci number $\FF_n$ can be expressed in terms of Fibonacci numbers as:
:$\dfrac {3 n + 3} 5 F_n - \dfrac n 5 F_{n + 1}$ | Let $\map \GG z = \ds \sum_{n \mathop \ge 0} \mathop F_n z^n$ be a generating function for $\FF_n$.
Then we have:
{{begin-eqn}}
{{eqn | l = \paren {1 - z - z^2} \map \GG z
| r = \paren {\FF_0 + \FF_1 z + \FF_2 z^2 + \FF_3 z^3 + \FF_4 z^4 + \cdots}
| c =
}}
{{eqn | o =
| ro= -
| r = \paren {\FF... | The [[Definition:Second Order Fibonacci Number|second order Fibonacci number]] $\FF_n$ can be expressed in terms of [[Definition:Fibonacci Number|Fibonacci numbers]] as:
:$\dfrac {3 n + 3} 5 F_n - \dfrac n 5 F_{n + 1}$ | Let $\map \GG z = \ds \sum_{n \mathop \ge 0} \mathop F_n z^n$ be a [[Definition:Generating Function|generating function]] for $\FF_n$.
Then we have:
{{begin-eqn}}
{{eqn | l = \paren {1 - z - z^2} \map \GG z
| r = \paren {\FF_0 + \FF_1 z + \FF_2 z^2 + \FF_3 z^3 + \FF_4 z^4 + \cdots}
| c =
}}
{{eqn | o =
... | Second Order Fibonacci Number in terms of Fibonacci Numbers | https://proofwiki.org/wiki/Second_Order_Fibonacci_Number_in_terms_of_Fibonacci_Numbers | https://proofwiki.org/wiki/Second_Order_Fibonacci_Number_in_terms_of_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Second Order Fibonacci Number",
"Definition:Fibonacci Number"
] | [
"Definition:Generating Function",
"Definition:Generating Function",
"Definition:Fibonacci Number",
"Generating Function for Fibonacci Numbers",
"Summation over k to n of Product of kth with n-kth Fibonacci Numbers"
] |
proofwiki-14595 | General Fibonacci Number in terms of Fibonacci Numbers | Let $r$ and $s$ be numbers, usually integers but not necessarily so limited.
Let $\sequence {a_n}$ be the general Fibonacci sequence:
:<nowiki>$a_n = \begin{cases}
r & : n = 0 \\
s & : n = 1 \\
a_{n - 2} + a_{n - 1} & : n > 1
\end{cases}$</nowiki>
Then $a_n$ can be expressed in Fibonacci numbers as:
:$a_n = F_{n - 1} r... | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n = F_{n - 1} r + F_n s$ | Let $r$ and $s$ be [[Definition:Number|numbers]], usually [[Definition:Integer|integers]] but not necessarily so limited.
Let $\sequence {a_n}$ be the [[Definition:General Fibonacci Sequence|general Fibonacci sequence]]:
:<nowiki>$a_n = \begin{cases}
r & : n = 0 \\
s & : n = 1 \\
a_{n - 2} + a_{n - 1} & : n > 1
\end{c... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$a_n = F_{n - 1} r + F_n s$ | General Fibonacci Number in terms of Fibonacci Numbers | https://proofwiki.org/wiki/General_Fibonacci_Number_in_terms_of_Fibonacci_Numbers | https://proofwiki.org/wiki/General_Fibonacci_Number_in_terms_of_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Number",
"Definition:Integer",
"Definition:General Fibonacci Sequence",
"Definition:Fibonacci Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14596 | Fibonacci Number plus Constant in terms of Fibonacci Numbers | Let $c$ be a number.
Let $\sequence {b_n}$ be the sequence defined as:
:$b_n = \begin{cases}
0 & : n = 0 \\
1 & : n = 1 \\
b_{n - 2} + b_{n - 1} + c & : n > 1
\end{cases}$
Then $\sequence {b_n}$ can be expressed in Fibonacci numbers as:
:$b_n = c F_{n - 1} + \paren {c + 1} F_n - c$ | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$b_n = c F_{n - 1} + \paren {c + 1} F_n - c$ | Let $c$ be a [[Definition:Number|number]].
Let $\sequence {b_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$b_n = \begin{cases}
0 & : n = 0 \\
1 & : n = 1 \\
b_{n - 2} + b_{n - 1} + c & : n > 1
\end{cases}$
Then $\sequence {b_n}$ can be expressed in [[Definition:Fibonacci Number|Fibonacci numbers]] as:
:... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$b_n = c F_{n - 1} + \paren {c + 1} F_n - c$ | Fibonacci Number plus Constant in terms of Fibonacci Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Constant_in_terms_of_Fibonacci_Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Constant_in_terms_of_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Number",
"Definition:Sequence",
"Definition:Fibonacci Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14597 | Fibonacci Number plus Binomial Coefficient in terms of Fibonacci Numbers | Let $m \in \Z_{>0}$ be a positive integer.
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 2} + a_{n - 1} + \dbinom {n - 2} m & : n > 1 \end{cases}$
where $\dbinom {n - 2} m$ denotes a binonial coefficient.
Then $\sequence {a_n}$ can be expressed in Fibonacc... | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n = F_{m + 1} F_{n - 1} + \paren {F_{m + 2} + 1} F_n - \ds \sum_{k \mathop = 0}^m \binom {n + m - k} k$ | Let $m \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]].
Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 2} + a_{n - 1} + \dbinom {n - 2} m & : n > 1 \end{cases}$
where $\dbinom {n - 2} m$ denotes a [[Definition:Bin... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$a_n = F_{m + 1} F_{n - 1} + \paren {F_{m + 2} + 1} F_n - \ds \sum_{k \mathop = 0}^m \binom {n + m - k} k$ | Fibonacci Number plus Binomial Coefficient in terms of Fibonacci Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Binomial_Coefficient_in_terms_of_Fibonacci_Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Binomial_Coefficient_in_terms_of_Fibonacci_Numbers | [
"Fibonacci Numbers",
"Binomial Coefficients"
] | [
"Definition:Positive/Integer",
"Definition:Sequence",
"Definition:Binomial Coefficient",
"Definition:Fibonacci Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14598 | Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers | Let $\map f n$ and $\map g n$ be arbitrary arithmetic functions.
Let $\sequence {a_n}$ be the sequence defined as:
:$a_n = \begin{cases}
0 & : n = 0 \\
1 & : n = 1 \\
a_{n - 1} + a_{n - 2} + \map f {n - 2} & : n > 1
\end{cases}$
Let $\sequence {b_n}$ be the sequence defined as:
:$b_n = \begin{cases}
0 & : n = 0 \\
1 & ... | === Lemma ===
{{:Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma}}{{qed|lemma}}
Hence also:
:$b_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map g k$
Thus:
{{begin-eqn}}
{{eqn | l = c_n
| r = F_n + \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \paren {x \map f k + y \map g k... | Let $\map f n$ and $\map g n$ be arbitrary [[Definition:Arithmetic Function|arithmetic functions]].
Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$a_n = \begin{cases}
0 & : n = 0 \\
1 & : n = 1 \\
a_{n - 1} + a_{n - 2} + \map f {n - 2} & : n > 1
\end{cases}$
Let $\sequence {b_n}$ be the ... | === [[Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma|Lemma]] ===
{{:Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma}}{{qed|lemma}}
Hence also:
:$b_n = F_n + \ds \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \map g k$
Thus:
{{begin-eqn}}
{{eqn | l = c_n
... | Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Arbitrary_Function_in_terms_of_Fibonacci_Numbers | https://proofwiki.org/wiki/Fibonacci_Number_plus_Arbitrary_Function_in_terms_of_Fibonacci_Numbers | [
"Fibonacci Numbers",
"Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers"
] | [
"Definition:Arithmetic Function",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Sequence",
"Definition:Fibonacci Number"
] | [
"Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma"
] |
proofwiki-14599 | Vajda's Identity/Formulation 1 | :$F_{n + i} F_{n + j} - F_n F_{n + i + j} = \paren {-1}^n F_i F_j$ | From Honsberger's Identity:
{{begin-eqn}}
{{eqn | l = F_{n + i}
| r = F_n F_{i - 1} + F_{n + 1} F_i
}}
{{eqn | l = F_{n + j}
| r = F_n F_{j - 1} + F_{n + 1} F_j
}}
{{eqn | l = F_{n + i + j}
| r = F_{i - 1} F_{n + j} + F_i F_{n + j + 1}
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | o =
| r = F_... | :$F_{n + i} F_{n + j} - F_n F_{n + i + j} = \paren {-1}^n F_i F_j$ | From [[Honsberger's Identity]]:
{{begin-eqn}}
{{eqn | l = F_{n + i}
| r = F_n F_{i - 1} + F_{n + 1} F_i
}}
{{eqn | l = F_{n + j}
| r = F_n F_{j - 1} + F_{n + 1} F_j
}}
{{eqn | l = F_{n + i + j}
| r = F_{i - 1} F_{n + j} + F_i F_{n + j + 1}
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | o =
... | Vajda's Identity/Formulation 1 | https://proofwiki.org/wiki/Vajda's_Identity/Formulation_1 | https://proofwiki.org/wiki/Vajda's_Identity/Formulation_1 | [
"Vajda's Identity"
] | [] | [
"Honsberger's Identity",
"Fibonacci Number in terms of Larger Fibonacci Numbers",
"Fibonacci Number with Negative Index",
"Category:Vajda's Identity"
] |
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