hcju/proofwikiqa huggingface dataset:
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_0       | r = 0       | c =  }} {{eqn | r = \dfrac {\left[{1 \atop 2}\right]} {0!}       | c = Unsigned Stirling Number of the First Kind of Number with Greater }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = H_1       | r = 1       | c =  }} {{eqn | r = \dfrac {\left[{2 \atop 2}\right]} {1!}       | c = Unsigned Stirling Number of the First Kind of Number with Self }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$H_k = \dfrac {\left[{ {k + 1} \atop 2}\right]} {k!}$ from which it is to be shown that: :$H_{k + 1} = \dfrac {\left[{ {k + 2} \atop 2}\right]} {\left({k + 1}\right)!}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \dfrac {\left[{ {k + 2} \atop 2}\right]} {\left({k + 1}\right)!}       | r = \dfrac {\left({k + 1}\right) \left[{ {k + 1} \atop 2}\right] + \left[{ {k + 1} \atop 1}\right]} {\left({k + 1}\right)!}       | c = {{Defof|Unsigned Stirling Numbers of the First Kind}} }} {{eqn | r = \dfrac {\left({k + 1}\right) \left[{ {k + 1} \atop 2}\right] + k!} {\left({k + 1}\right)!}       | c = Unsigned Stirling Number of the First Kind of n+1 with 1 }} {{eqn | r = \dfrac {\left[{ {k + 1} \atop 2}\right]} {k!} + \dfrac 1 {k + 1}       | c = simplifying }} {{eqn | r = H_k + \dfrac 1 {k + 1}       | c = Induction Hypothesis }} {{eqn | r = H_{k + 1}       | c = {{Defof|Harmonic Number}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: H_n = \dfrac {\left[{ {n + 1} \atop 2}\right]} {n!}$ {{qed}}
{{begin-eqn}} {{eqn | l = T \left({m + 1, n}\right) - T \left({m, n}\right)       | r = \left({H_{m + 1} + H_n - H_{\left({m + 1}\right) n} }\right) - \left({H_m + H_n - H_{m n} }\right)        | c =  }} {{eqn | r = \dfrac 1 {m + 1} + \left({H_{\left({m + 1}\right) n} - H_{m n} }\right)       | c =  }} {{eqn | r = \dfrac 1 {m + 1} - \left({\frac 1 {m n + 1} + \frac 1 {m n + 2} + \cdots + \frac 1 {m n + n} }\right)       | c =  }} {{eqn | o = \le       | r = \dfrac 1 {m + 1} - \left({\frac 1 {m n + n} + \frac 1 {m n + n} + \cdots + \frac 1 {m n + n} }\right)       | c =  }} {{eqn | r = \dfrac 1 {m + 1} - \frac n {n \left({m + 1}\right)}       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} Similarly: :$T \left({m, n + 1}\right) - T \left({m, n}\right) \le 0$ From Approximate Size of Sum of Harmonic Series, the limiting value of $T \left({m, n}\right)$ is the Euler-Mascheroni constant: :$\displaystyle \lim_{m, n \mathop \to \infty} H_m + H_n - H_{m n} = \gamma$ {{qed}}
{{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}}
From Summation over k to n of Natural Logarithm of k: :$\displaystyle \sum_{k \mathop = 1}^n \ln k = \map \ln {n!}$ 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 \ln n + n \paren {\ln n - \ln e}       | c =  }} {{eqn | r = \map \ln {\sqrt {2 \pi} } + \paren {n + \frac 1 2} \ln n - n       | c =  }} {{end-eqn}} Then we have that: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n H_k       | r = \paren {n + 1} H_n - n       | c = Sum of Sequence of Harmonic Numbers }} {{eqn | o = \approx       | r = \paren {n + 1} \paren {\ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } - n       | c = Approximate Size of Sum of Harmonic Series }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | o =        | r = \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^n \map \ln {n!}       | c =  }} {{eqn | o = \approx       | r = \paren {\paren {n + 1} \paren {\ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} } - n} - \paren {\ln \paren {\sqrt {2 \pi} } + \paren {n + \frac 1 2} \ln n - n}       | c =  }} {{eqn | r = n \ln n + n \gamma + \dfrac 1 2 - \dfrac 1 {12 n} + \dfrac 1 {120 n^3} + \ln n + \gamma + \dfrac 1 {2 n} - \dfrac 1 {12 n^2} + \dfrac 1 {120 n^4} - n - \map \ln {\sqrt {2 \pi} } - n \ln n - \frac 1 2 \ln n + n       | c =  }} {{eqn | r = \gamma n + \frac 1 2 \ln n + \dfrac 1 2 + \gamma - \map \ln {\sqrt {2 \pi} } + \dfrac 5 {12 n} - \dfrac 1 {12 n^2}  + \dfrac 1 {120 n^3} + \dfrac 1 {120 n^4}       | c =  }} {{end-eqn}} The result follows by ignoring the lower order terms in $n$. {{qed}}
When $x = -1$ we have that $1 + \dfrac 1 x = 0$, so $\ln \left({1 + \dfrac 1 x}\right)$ is undefined. Let $S_n = \displaystyle \sum_{k \mathop \in \Z} \binom n k x^k H_k$ 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} \left({\binom n k + \binom n {k - 1} }\right) x^k H_k       | c = Pascal's Rule }} {{eqn | r = S_n + x \sum_{k \mathop \ge 1} \left({\binom n {k - 1} }\right) x^{k - 1} \left({H_{k - 1} + \frac 1 k}\right)       | c =  }} {{eqn | r = S_n + x S_n + \sum_{k \mathop \ge 1} \binom n {k - 1} x^k \frac 1 k       | c =  }} {{eqn | r = \left({1 + x}\right) S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k x^k       | c = (look up result for this) }} {{end-eqn}} Setting $x = -1$: {{begin-eqn}} {{eqn | l = S_{n + 1}       | r = \left({1 + -1}\right) S_n + \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \left({-1}\right)^k       | c =  }} {{eqn | r = \frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n + 1} k \left({-1}\right)^k       | c =  }} {{eqn | ll= \leadsto       | l = S_n       | r = \frac 1 n \sum_{k \mathop \ge 1} \binom n k \left({-1}\right)^k       | c =  }} {{eqn | r = \frac 1 n \left({\sum_{k \mathop \ge 0} \binom n k \left({-1}\right)^k - \binom n 0}\right)       | c =  }} {{eqn | r = \frac 1 n \left({0 - \binom n 0}\right)       | c = Alternating Sum and Difference of Binomial Coefficients for Given n }} {{eqn | r = \frac {-1} n       | c = Binomial Coefficient with Zero }} {{end-eqn}} {{qed}}
{{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}       | c =  }} {{eqn | r = -\paren {\dfrac {H_{n + 1} } n - \dfrac {H_2} 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1} \paren {H_{k + 2} - H_{k + 1} } }        | c = Abel's Lemma: Formulation 1 }} {{eqn | r = H_2 - \dfrac {H_{n + 1} } n + \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1} \dfrac 1 {k + 2}       | c = simplification }} {{eqn | r = \frac 3 2 - \dfrac {H_{n + 1} } n + \sum_{k \mathop = 1}^{n - 1} \paren {\dfrac 1 {k + 1} - \dfrac 1 {k + 2} }       | c = separating the fractions, and Harmonic Number $H_2$ }} {{eqn | r = \frac 3 2 - \dfrac {H_{n + 1} } n + \frac 1 2 - \dfrac 1 {n + 1}       | c = Telescoping Series }} {{eqn | r = 2 - \frac 1 n \paren {H_n + \dfrac 1 {n + 1} } - \dfrac 1 {n + 1}       | c =  }} {{eqn | r = 2 - \frac {H_n} n - \dfrac 1 {n \paren {n + 1} } - \dfrac 1 {n + 1}       | c =  }} {{eqn | r = 2 - \frac {H_n} n - \paren {\dfrac 1 n - \dfrac 1 {n + 1} } - \dfrac 1 {n + 1}       | c =  }} {{eqn | r = 2 - \frac {H_n} n - \dfrac 1 n       | c = after simplification }} {{end-eqn}} {{qed}}
By definition of the general harmonic numbers: :$\displaystyle \zeta \left({r}\right) = \lim_{n \mathop \to \infty} H_n^{\left({r}\right)} = \sum_{k \mathop \ge 1} \frac 1 {k^r}$ From Sequence of General Harmonic Numbers Converges for Index Greater than 1: {{begin-eqn}} {{eqn | l = \zeta \left({1000}\right)       | o = \le       | r = \dfrac {2^{1000} } {2^{1000} - 1}       | c =  }} {{eqn | r = \dfrac 1 {1 - 2^{-1000} }       | c =  }} {{end-eqn}} which is $1 \cdot 000 \ldots$ to a good few hundred decimal places. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} k       | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j k}       | 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 {\sum_{k \mathop = 1}^n \dfrac 1 k}^2 + \paren {\sum_{k \mathop = 1}^n \dfrac 1 {k^2} } }       | c = Summation of Products of n Numbers taken m at a time with Repetitions }} {{eqn | r = \dfrac { {H_n}^2 + H_n^{\paren 2} } 2       | c = {{Defof|Harmonic Number}} and {{Defof|General Harmonic Numbers}} }} {{end-eqn}} Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {k + 1}       | r = \sum_{k \mathop = 1}^n \paren{\dfrac {H_{k + 1} } {k + 1} - \dfrac 1 {\left({k + 1}\right) \left({k + 1}\right)} }       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \dfrac {H_{k + 1} } {k + 1} - \sum_{k \mathop = 1}^n \dfrac 1 {\left({k + 1}\right)^2}       | c =  }} {{eqn | r = \sum_{k \mathop = 2}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 2}^{n + 1} \dfrac 1 {k^2}       | c = Translation of Index Variable of Summation }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \dfrac {H_1} 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - \dfrac 1 {1^2} }       | c =  }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - 1 - \paren{\sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2} - 1}       | c = Harmonic Number $H_1$ }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - \sum_{k \mathop = 1}^{n + 1} \dfrac 1 {k^2}       | c = simplifying }} {{eqn | r = \sum_{k \mathop = 1}^{n + 1} \dfrac {H_k} k - H_{n + 1}^{\left({2}\right)}       | c = {{Defof|General Harmonic Numbers}} }} {{eqn | r = \dfrac { {H_{n + 1} }^2 + H_{n + 1}^{\left({2}\right)} } 2 -  H_{n + 1}^{\left({2}\right)}       | c = Summation to n of kth Harmonic Number over k }} {{eqn | r = \dfrac { {H_{n + 1} }^2 + H_{n + 1}^{\left({2}\right)} - 2 H_{n + 1}^{\left({2}\right)} } 2       | c =  }} {{eqn | r = \dfrac { {H_{n + 1} }^2 - H_{n + 1}^{\left({2}\right)} } 2       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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 Equality, $\triangle ABD = \triangle ACD$. Thus $AD = DC$. The result follows by definition of bisection. {{qed}}
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 Equality, $\triangle ABD = \triangle ACD$. Thus $\angle ADB = \angle ADC$. By Two Angles on Straight Line make Two Right Angles, $\angle ADB + \angle ADC$ equals $2$ right angles. Thus each of $\angle ADB$ and $\angle ADC$ are right angles. The result follows by definition of perpendicular. {{qed}}
By definition of equilateral triangle, any two of its sides 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 any two internal angles of $\triangle ABC$ are equal. Hence all $3$ internal angles of $\triangle ABC$ are equal. {{qed}}
By definition of equilateral triangle, any two of its internal angles 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 $\triangle ABC$ are equal. Hence all $3$ sides of $\triangle ABC$ are equal. Hence the result by definition of equilateral triangle. {{qed}}
{{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 $S_\beta$. By hypothesis $S_\alpha$ and $S_\beta$ are order isomorphic. Thus there is an order isomorphism between $S_\beta$ and an initial segment of $S_\beta$. This contradicts No Isomorphism from Woset to Initial Segment. Thus $S_\alpha = S_\beta$. {{qed}}
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}}
$\triangle CDE$ 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 Equality, $\triangle CDE$ and $\triangle CDF$ are congruent. Hence the result. {{qed}}
$\angle EFH, \angle FEG, \angle EGH, \angle FHG$ 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}}
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, $AC$ bisects $BD$. From Bisector of Apex of Isosceles Triangle is Perpendicular to Base, $AC$ bisects $BD$ at right angles. Hence the result. {{qed}}
=== 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}} === Necessary Condition === Let $AB = CD$ and $AB \parallel CD$. Then from Lines Joining Equal and Parallel Straight Lines are Parallel, $AD \parallel BC$. Thus we have $AB \parallel CD$ and $AD \parallel BC$, and so by definition $ABCD$ is a parallelogram. {{qed}}
=== 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 $AB \parallel CD$ and $AD \parallel BC$. {{qed|lemma}} === Necessary Condition === Let $ABCD$ be such that: :either $AB = CD$ and $AD = BC$ :or $AB \parallel CD$ and $AD \parallel BC$. First suppose that $AB = CD$ and $AD = BC$. Then from Opposite Sides Equal implies Parallelogram, $ABCD$ is a parallelogram. Now suppose that $AB \parallel CD$ and $AD \parallel BC$. Then $ABCD$ is a parallelogram by definition. Hence the result by Proof by Cases. {{qed}}
=== 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}} === Necessary Condition === Let $ABCD$ be such that: :$\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$. {{ProofWanted|Requires a diagram with a construction that allows the use of Equal Alternate Angles implies Parallel Lines, for example.}}
=== 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}} === Necessary Condition === Let $ABCD$ be such that: : $AD$ is a bisector of $BC$ and : $BC$ is a bisector of $AD$. Then from Quadrilateral with Bisecting Diagonals is Parallelogram, $ABCD$ is a parallelogram. {{qed}}
{{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 Equality we have that $\triangle ABC = \triangle EFG$. From Quadrilateral is Parallelogram iff Both Pairs of Opposite Angles are Equal: :$\angle ABC = \angle ADC$ :$\angle EFG = \angle EHG$ From Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel, $AB = CD$ and $EF = GF$. Thus again by Triangle Side-Angle-Side Equality we have that $\triangle ADC = \triangle EHG$. So: : $\triangle ABC + \triangle ADC = \triangle EFG + \triangle EHG$ and the result follows. {{qed}}
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}}
:400px {{ProofWanted|easy enough but boring </br> Construct line parallel to $BC$ passing through $D$. </br> Use Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side and Opposite Sides of Parallelogram are Equal}}
This is a direct application of Parallel Line in Triangle Cuts Sides Proportionally. {{qed}}
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 parallelogram. By definition of a parallelogram, $AB \parallel CF$. From Opposite Sides and Angles of Parallelogram are Equal, $AD = CF$. But $AD = DB$ as $D$ is the midpoint of $AB$. So $DB = CF$ and $DB \parallel CF$. From Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel, $BCFD$ is a parallelogram. Thus also by Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel $DF = BC$ and $DF \parallel BC$. As $DE = EF$, $DE$ is the midpoint of $DF$ and so $DE = \dfrac 1 2 DF$. Thus $DE = \dfrac 1 2 BC$ and $DE \parallel BC$. Hence the result. {{qed}}
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 follows from Diameters of Parallelogram Bisect each other. {{qed}}
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 $\angle BAC$ is a right angle it follows that $\angle BEC$ is also a right angle. Thus by definition $ABEC$ is a rectangle. From Diagonals of Rectangle are Equal: :$AE = BC$ From above, $D$ is the midpoint of both $AE$ and $BC$. Thus $AD = BD$ and hence the result. {{qed}}
: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: :$AB = DC$ :$\angle ADC = \angle DAB$ :$AD$ is common to both $\triangle ADC$ and $\triangle DAB$ and so by Triangle Side-Angle-Side Equality it follows that $\triangle ADC = \triangle DAB$. Thus $AC = BD$. {{qed}} Category:Rectangles qf0a94ak8cop2n19e3772780ji7ov3i
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 Summation of Sum }} {{eqn | r = n - 1 - \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k + 1}       | c = Summation of Unity over Elements }} {{eqn | r = n - 1 - \paren {H_n - 1}       | c = Harmonic Number and compensate for $\dfrac 1 1$ term }} {{eqn | n = 1       | r = n - H_n       | c = simplification }} {{end-eqn}} {{refactor|The above result is worth publishing on its own page as a result in its own right. I'm surprised we don't already have it.}} Using Summation by Parts:  {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n {H_k}^2       | r = H_n \sum_{k \mathop = 1}^n H_k - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {\sum_{i \mathop = 1}^k {H_i} } \paren {H_{k + 1} - H_k} }       | c = }} {{eqn | r = H_n \paren {\paren {n + 1} H_n - n} - \sum_{k \mathop = 1}^{n - 1} \paren {\paren {k + 1} H_k - k} \paren {H_{k + 1} - H_k}       | c = Sum of Sequence of Harmonic Numbers twice }} {{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} \dfrac {\paren {k + 1} H_k - k} {k + 1}       | c = $H_{k + 1} - H_k = \dfrac 1 {k+1}$ }} {{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \sum_{k \mathop = 1}^{n - 1} H_k + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}        | c = Sum of Summations equals Summation of Sum }} {{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren { \sum_{k \mathop = 1}^n H_k - H_n} + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}        | c = increase range of summation and compensate }} {{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {\paren {n + 1} H_n - n - H_n } + \sum_{k \mathop = 1}^{n - 1} \dfrac k {k + 1}        | c = Sum of Sequence of Harmonic Numbers }} {{eqn | r = \paren {n + 1} {H_n}^2 - n H_n - \paren {n + 1} H_n + n + H_n + n - H_n       | c = $(1)$ }} {{eqn | r = \paren {n + 1} {H_n}^2 - \paren {2 n + 1} H_n + 2 n       | c = simplification }} {{end-eqn}} {{qed}}
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: :$m = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1}$ Thus it is sufficient to show that $m$ is a multiple of $p$. Each term in this sum is an integer of the form $\dfrac {\paren {p - 1}!} k$. For each $k \in \set {1, 2, \ldots, p - 1}$, define $k'= - \dfrac {\paren {p - 1}!} k \bmod p$. By Wilson's Theorem :$k k' \equiv -\paren {p - 1}! \equiv 1 \pmod p$ Therefore :$k' \equiv k^{-1} \pmod p$ From the corollary to Reduced Residue System under Multiplication forms Abelian Group: :$\struct {\Z'_p, \times}$ is an abelian group. Since Inverse in Group is Unique, the set: :$\set {1', 2', \ldots, \paren {p - 1}'}$ is merely the set: :$\set {1, 2, \ldots, p - 1}$ in a different order. Thus  {{begin-eqn}} {{eqn | l = m       | r = \dfrac {\paren {p - 1}!} 1 + \dfrac {\paren {p - 1}!} 2 + \cdots + \dfrac {\paren {p - 1}!} {p - 1} }} {{eqn | o = \equiv       | r = 1 + 2 + \cdots + p - 1        | rr= \pmod p }} {{eqn | o = \equiv       | r = \frac {p \paren {p - 1} } 2       | rr= \pmod p       | c = Closed Form for Triangular Numbers }} {{eqn | o = \equiv       | r = 0       | rr= \pmod p       | c =  }} {{end-eqn}} {{qed}}
From Polynomial x^p - x is Congruent mod p to x to the p-1 Rising: :$x^{\overline p} \equiv x^p - x$ {{TheoremWanted|The above is Knuth 4.6.2-6}} Thus from Sum over k of Unsigned Stirling Numbers of First Kind by x^k: :$\displaystyle \left[{p \atop k}\right] \equiv \delta_{k p} - \delta _{k 1}$ where: :$\displaystyle \left[{p \atop k}\right]$ denotes an unsigned Stirling number of the first kind :$\delta$ is the Kronecker delta. The result follows from Harmonic Number as Unsigned Stirling Number of First Kind over Factorial. {{finish|Considerable work is needed to explain some of the above. For a start, the result in 4.6.2-6 appears to be for the falling, not rising, factorial, which looks on the surface like a mistake.}}