hcju/proofwikiqa huggingface dataset:
From Sequence of Wythoff Pairs contains all Positive Integers exactly Once Each, every positive integer can be found in the sequence of Wythoff pairs. From Difference between Terms of Wythoff Pair, the first coordinate is the smaller of the coordinates of the Wythoff pair. So consider a given Wythoff pair. Let $p$ be the smallest positive integer which has not yet appeared in the sequence. It has to appear somewhere. The terms of both the lower and upper Wythoff sequences are in ascending order. $n$ must appear in the next Wythoff pair, otherwise a larger positive integer will appear before it when it eventually does appear. $n$ cannot appear as the second coordinate, or it will be less than whatever number appears as the first coordinate. Hence $n$ appears as the first coordinate of the next Wythoff pair. {{qed}}
By definition, the $n$th Wythoff pair is $\tuple {\floor {n \phi}, \floor {n \phi^2} }$. Thus the coordinates of the sequence of Wythoff pairs are the terms of the lower and upper Wythoff sequences. By definition: :the lower Wythoff sequence is the Beatty sequence $\BB_x$ on the golden section $\phi$. :the upper Wythoff sequence is the complementary Beatty sequence on the golden section $\phi$. Also by definition, the complementary Beatty sequence on $x$ is the integer sequence formed by the integers which are missing from $\BB_x$. Hence the result. {{qed}}
From Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \map \zeta z = \prod_p \frac 1 {1 - p^{-z} }$ where $p$ ranges over the prime numbers. Thus: {{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \prod_p \frac 1 {1 - p^{-2} }       | c =  }} {{eqn | r = \prod_p \frac {p^2} {p^2 - 1}       | c = multiplying top and bottom by $p^2$ }} {{eqn | r = \prod_p \frac {p^2} {\paren {p - 1} \paren {p + 1} }       | c = Difference of Two Squares }} {{eqn | r = \prod_p \paren {\frac p {p - 1} } \paren {\frac p {p + 1} }       | c =  }} {{end-eqn}} which is the result required. {{qed}}
By definition of filter in ordered set: :$F$ is non-empty and upper. By definition of non-empty set: :$\exists x: x \in F$ By definition of greatest element: :$x \preceq \top$ Thus by definition of upper set: :$\top \in F$ {{qed}}
By Finite Infima Set and Upper Closure is Smallest Filter: :$X \subseteq {\operatorname{fininfs}\left({X}\right)}^\succeq$ By definition of non-empty set: :${\operatorname{fininfs}\left({X}\right)}^\succeq$ is a non-empty set. We will prove that :$\operatorname{fininfs}\left({X}\right)$ is filtered. Let $x, y \in \operatorname{fininfs}\left({X}\right)$ By definition of finite infima set: :$\exists A \in \mathit{Fin}\left({X}\right): x = \inf A \land A$ admits an infimum and :$\exists B \in \mathit{Fin}\left({X}\right): y = \inf B \land B$ admits an infimum where $\mathit{Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$. Define $C = A \cup B$. By Union of Subsets is Subset: :$C \subseteq X$ By Union of Finite Sets is Finite: :$C$ is finite. Then :$C \in \mathit{Fin}\left({X}\right)$ By Existence of Non-Empty Finite Infima in Meet Semilattice: :$C \ne \varnothing \implies C$ admits an infimum. By Union is Empty iff Sets are Empty: :$C = \varnothing \implies A = \varnothing$ So :$C$ admits an infimum. By definition of finite infima set: :$\inf C \in \operatorname{fininfs}\left({X}\right)$ By Set is Subset of Union: :$A \subseteq C$ and $B \subseteq C$ Thus by Infimum of Subset: :$\inf C \preceq x$ and $\inf C \preceq y$ Hence $\operatorname{fininfs}\left({X}\right)$ is filtered. {{qed|lemma}} By Filtered iff Upper Closure Filtered: :${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filtered. By Upper Closure is Upper Set: :${\operatorname{fininfs}\left({X}\right)}^\succeq$ is upper. Hence ${\operatorname{fininfs}\left({X}\right)}^\succeq$ is filter in $P$. {{qed}}
Let $N$ be divisible by $2$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod 2 }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod 2 }} {{eqn | ll= \leadstoandfrom       | l = a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}       | o = \equiv       | r = 0 \pmod 2 }} {{eqn | ll= \leadstoandfrom       | l = a_0       | o = \equiv       | r = 0 \pmod 2       | c = as $10 \equiv 0 \pmod 2$ }} {{end-eqn}} {{qed}}
First note that: :$10^r = 2^r 5^r$ and so: :$2^r \divides 10^r$ where $\divides$ denotes divisibility. Thus: :$\forall s \in \Z: s \ge r: 2^r \divides 10^s$ but: :$\forall s \in \Z: s < r: 2^r \nmid 10^s$ Thus let $N$ be divisible by $2^r$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod {2^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod {2^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r + \sum_{k \mathop = r + 1}^n a_k 10^k       | o = \equiv       | r = 0 \pmod {2^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r + 10^r \sum_{k \mathop = r + 1}^n a_k 10^{k - r}       | o = \equiv       | r = 0 \pmod {2^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r       | o = \equiv       | r = 0 \pmod {2^r}       | c = as $10^r \equiv 0 \pmod {2^r}$ }} {{end-eqn}} Hence the result. {{qed}}
=== Necessary Condition === Let $a$ be the smallest of $m$ consecutive (strictly) positive integers, where $m \ge 2$. From Sum of Arithmetic Sequence, their sum is $\dfrac {m \paren {2 a + m - 1} } 2$.  {{AimForCont}} $\dfrac {m \paren {2 a + m - 1} } 2$ is a power of $2$. Then $m$ and $2 a + m - 1$ must also be powers of $2$. Since $m \ge 2$, $m$ must be even. Then $2 a + m - 1$ is odd. Together with $2 a + m - 1$ being a power of $2$, we have $2 a + m - 1 = 1$. However, $2 a + m - 1 > 2 \times 0 + 2 - 1 = 1$. This is a contradiction. Therefore any sum of $2$ or more consecutive integers cannot be a power of $2$. {{qed|lemma}} === Sufficient Condition === Let $n$ be an integer that is not a power of $2$. Then $n$ contains an odd factor greater than $1$. Let $d = 2 m + 1$ be such a factor. Then: :$\dfrac n d - m, \dfrac n d - m + 1, \dots, \dfrac n d + m$ is a sequence of $2 m + 1$ consecutive integers. From Sum of Arithmetic Sequence: {{begin-eqn}} {{eqn | l = \paren {\dfrac n d - m} + \paren {\dfrac n d - m + 1} + \cdots + \paren {\dfrac n d + m}       | r = \dfrac {\paren {2 m + 1} \paren {\dfrac n d - m + \dfrac n d + m} } 2       | c =  }} {{eqn | r = \dfrac d 2 \times \dfrac {2 n} d       | c =  }} {{eqn | r = n       | c =  }} {{end-eqn}} It may happen that $\dfrac n d - m \le 0$. In that case, notice that: :$\dfrac n d - m \le m - \dfrac n d \le \dfrac n d + m$ and so $m - \dfrac n d$ is also (somewhere) in our sequence of $2 m + 1$ consecutive integers. So all negative integers in our sequence of $2 m + 1$ consecutive integers cancel out with their positive counterparts. Hence we can remove all numbers from $\dfrac n d - m$ to $m - \dfrac n d$. As a result of that process, the sum remains unchanged. We see that the number of integers eliminated is $2 \paren {m - \dfrac n d} + 1$. Since $\dfrac n d$ is a factor of $n$, we have $\dfrac n d \ge 1$. Hence the number of integers remaining is: {{begin-eqn}} {{eqn | l = 2 m + 1 - \paren {2 \paren {m - \dfrac n d} + 1}       | r = \dfrac {2 n} d }} {{eqn | o = \ge       | r = 2 }} {{end-eqn}} This finishes our proof. {{qed}}
By definition of abundance: :$A \left({n}\right) = \sigma \left({n}\right) - 2 n$ By definition of $\sigma$ function: :$\sigma \left({n}\right)$ is the sum of all the divisors of $n$. Thus $\sigma \left({n}\right) - n$ is the aliquot sum of $n$. Hence the result. {{qed}} Category:Deficient Numbers q3ova4ukts2hfa70p7nhtz8jb0ikmaf
By definition of abundance: :$A \left({n}\right) = \sigma \left({n}\right) - 2 n$ By definition of $\sigma$ function: :$\sigma \left({n}\right)$ is the sum of all the divisors of $n$. Thus $\sigma \left({n}\right) - n$ is the aliquot sum of $n$. The result follows. {{qed}} Category:Abundant Numbers no4nk3jiu5f3quesw0kui7z18j5s74x
From Sigma Function of Power of Prime: :$\map \sigma n = \dfrac {p^{k + 1} - 1} {p - 1}$ Thus: {{begin-eqn}} {{eqn | l = \map A n       | r = \map \sigma n - 2 n       | c = {{Defof|Abundance}} }} {{eqn | r = \dfrac {p^{k + 1} - 1} {p - 1} - 2 p^k       | c =  }} {{eqn | r = \dfrac {p^{k + 1} - 1 - \paren {2 p^{k + 1} - 2 p^k} } {p - 1}       | c =  }} {{eqn | r = \dfrac {2 p^k - p^{k + 1} - 1} {p - 1}       | c =  }} {{eqn | r = \dfrac {p^k \paren {2 - p} - 1} {p - 1}       | c =  }} {{eqn | r = \dfrac {p^k - 1} {p - 1} - p^k       | c =  }} {{eqn | r = \dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1}       | c =  }} {{end-eqn}} If $p = 2$ then $\dfrac {p^k} {p - 1} - p^k = 0$ and so: :$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} = - \dfrac 1 {p - 1} < 0$ If $p > 2$ then $\dfrac {p^k} {p - 1} < p^k$ and so: :$\dfrac {p^k} {p - 1} - p^k - \dfrac 1 {p - 1} < 0$ Thus in all cases: :$\map A n < 0$ and so $n = p^k$ is deficient by definition. {{qed}}
{{begin-eqn}} {{eqn | l = \map A n       | r = \dfrac {2^k} {2 - 1} - 2^k - \dfrac 1 {2 - 1}       | c = Power of Prime is Deficient }} {{eqn | r = 2^k - 2^k - 1       | c =  }} {{eqn | r = - 1       | c =  }} {{end-eqn}} and so $n = p^k$ is almost perfect by definition. {{qed}}
Let $x \in X$. By Infimum of Singleton: :$\left\{ {x}\right\}$ admits an infimum and $\inf \left\{ {x}\right\} = x$ By definitions of subset and singleton: :$\left\{ {x}\right\} \subseteq X$ By Singleton is Finite: :$\left\{ {x}\right\}$ is a finite set. Thus by definition of finite infima set: :$x \in \operatorname{fininfs}\left({X}\right)$ {{qed}}
Let $x \in X$. By definition of reflexivity: :$x \preceq x$ Thus by definition of upper closure: :$x \in X^\succeq$ {{qed}}
Since $n^z = n^x + n^y > n^x$ and $n^y$, $z > x,y$. {{WLOG}} assume that $x \le y < z$. {{begin-eqn}} {{eqn | l = n^x + n^y        | r = n^z }} {{eqn | ll = \leadsto       | l = 1 + n^{y - x}       | r = n^{z - x}       | c = Divide both sides by $n^x$ }} {{eqn | ll = \leadsto       | l = 1       | r = n^{y - x} \paren {n^{z - y} - 1} }} {{end-eqn}} Since both $n^{y - x}$ and $n^{z - y} - 1$ are positive integers, both are equal to $1$. This gives: :$y = x$ and $n^{z - y} = 2$ which gives the integer solution: :$n = 2$, $z - y = 1$ Thus the solutions are: :$\tuple {n, x, y, z} = \tuple {2, x, x, x + 1}, x \in \Z$ {{qed}}
Let $X \subseteq Y$. Let $x \in X^\succeq$. By definition of upper closure of subset: :$\exists y \in X: y \preceq x$ By definition of subset: :$y \in Y$ Thus by definition of upper closure of subset: :$x \in Y^\succeq$ {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} {n!} {n^n \sqrt n e^{-n} }       | r = \sqrt {2 \pi}       | c = Lemma for Stirling's Formula }} {{eqn | ll= \leadsto       | l = e       | r = \lim_{n \mathop \to \infty} \dfrac {n \paren {2 \pi n}^{1 / 2 n} } {\sqrt [n] {n!} }       | c =  }} {{eqn | ll= \leadsto       | l = e       | r = \lim_{n \mathop \to \infty} \dfrac {n \sqrt {\paren {2 \pi n}^{1 / n} } } {\sqrt [n] {n!} }       | c =  }} {{end-eqn}} From Limit of Root of Positive Real Number: :$\displaystyle \lim_{n \mathop \to \infty} \paren {2 \pi}^{1 / 2 n} = 1$ and from Limit of Integer to Reciprocal Power: :$n^{1 / n} = 1$ Hence the result. {{qed}}
Firstly, we will prove that $\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$, where $n \in \N$. Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: : $\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$ === Basis for the Induction === $\map P 1$ is true, as this says $\displaystyle \frac {\sin z} z = \frac {\sin z} z$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \frac {\sin z} z = \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}$ Then we need to show: :$\displaystyle \frac {\sin z} z = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \frac {\sin z} z       | r = \paren {\frac {2^k} z} \sin \frac z {2^k} \prod_{i \mathop = 1}^k \cos \frac z {2^i}       | c = Induction Hypothesis }} {{eqn | r = \paren {\frac {2^k} z} \paren {2 \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } } \prod_{i \mathop = 1}^k \cos \frac z {2^i}       | c = Double Angle Formula for Sine }} {{eqn | r = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \cos \frac z {2^{k + 1} } \prod_{i \mathop = 1}^k \cos \frac z {2^i}       | c =  }} {{eqn | r = \paren {\frac {2^{k + 1} } z} \sin \frac z {2^{k + 1} } \prod_{i \mathop = 1}^{k + 1} \cos \frac z {2^i}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \frac {\sin z} z = \paren {\frac {2^n} z} \sin \frac z {2^n} \prod_{i \mathop = 1}^n \cos \frac z {2^i}$ And then: {{begin-eqn}} {{eqn | l = \frac {\sin z} z       | r = \lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } \prod_{i \mathop = 1}^n \cos \frac z {2^i}       | c =  }} {{eqn | r = \paren {\lim_{n \mathop \to \infty} \paren {\frac {2^n} z} \paren {\sin \frac z {2^n} } } \prod_{i \mathop = 1}^{\infty} \cos \frac z {2^i}       | c =  }} {{eqn | r = \paren 1 \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}       | c = Limit of Sine of X over X }} {{eqn | r = \prod_{i \mathop = 1}^\infty \cos \frac z {2^i}       | c =  }} {{end-eqn}} {{qed}}
By assumption and popular belief, ordinary space (on the measurable local level) is an instance of a $3$-dimensional Euclidean space. Hence the results of {{ElementsLink}} can be applied. Let there be $4$ straight lines $a$, $b$, $c$ and $d$ in space. Let $a$ be perpendicular to each of $b$, $c$ and $d$. From Three Intersecting Lines Perpendicular to Another Line are in One Plane, $b$, $c$ and $d$ are all in the same plane. Let $b$ be perpendicular to $c$ and $d$. Either $c$ and $d$ are the same line or not. If $c = d$ then $c$ is not perpendicular to $d$. Otherwise, from Equal Corresponding Angles implies Parallel Lines, $c$ is parallel to $d$. In either case $c$ is not perpendicular to $d$. Hence the result. {{qed}}
Denote as $\ln_\R$ the real natural logarithm. {{begin-eqn}} {{eqn | l = \map \ln {r e^{i \theta} }       | r = \set {\map {\ln_\R} r + i \theta + 2 k \pi i: k \in \Z}       | c = {{Defof|Natural Logarithm/Complex|index = 1|Complex Natural Logarithm}} }} {{eqn | ll= \leadsto       | l = \map \ln {1 e^{i \times 0} }       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = substituting $r = 1$ and $\theta = 0$ }} {{eqn | ll= \leadsto       | l = \map \ln {1 e^{i \times \paren {0 + 0 i} } }       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = Real Number as Complex Number }} {{eqn | ll= \leadsto       | l = \map \ln {1 e^0}       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = {{Defof|Complex Multiplication}} }} {{eqn | ll= \leadsto       | l = \map \ln {1 \times \paren {1 + 0 i} }       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = {{Defof|Exponential Function/Complex|subdef = Differential Equation|Complex Exponential Function}} }} {{eqn | ll= \leadsto       | l = \map \ln {1 \times 1}       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = Real Number as Complex Number }} {{eqn | ll= \leadsto       | l = \map \ln 1       | r = \set {\map {\ln_\R} 1 + i \times 0 + 2 k \pi i: k \in \Z}       | c = Real Multiplication Identity is One }} {{eqn | ll= \leadsto       | l = \map \ln 1       | r = \set {0 + i \times 0 + 2 k \pi i: k \in \Z}       | c = Real Logarithm of 1 is 0 }} {{eqn | ll= \leadsto       | l = \map \ln 1       | r = \set {2 k \pi i: k \in \Z}       | c = Complex Addition Identity is Zero }} {{end-eqn}} {{qed}} Category:Logarithms dq6xxfp2b3e0z07i4h1ynn87w1a4ft5
Let $z_0 \in D$ be a point in $D$. {{begin-eqn}} {{eqn | l = \map {f'} {z_0}       | r = \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h       | c = {{Defof|Derivative of Complex Function}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {z_0 + h} + \map k {z_0 + h} } - \paren {\map j {z_0} +\map k {z_0} } } h       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {z_0 + h} + \map k {z_0 + h} - \map j {z_0} - \map k {z_0} } h       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\map j {z_0 + h} - \map j {z_0} } + \paren {\map k {z_0 + h} - \map k {z_0} } } h       | c =  }} {{eqn | r = \lim_{h \mathop \to 0} \paren {\frac {\map j {z_0 + h} - \map j {z_0} } h + \frac {\map k {z_0 + h} - \map k {z_0} } h}       | c = Complex Multiplication Distributes over Addition }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {z_0 + h} - \map j {z_0} } h + \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h       | c = Sum Rule for Limits of Functions }} {{eqn | r = \map {j'} {z_0} + \map {k'} {z_0}       | c = {{Defof|Derivative of Complex Function}} }} {{eqn | ll=\leadsto       | lo= \forall z \in D:       | l = \map {f'} z       | r = \map {j'} z + \map {k'} z       | c = {{Defof|Derivative of Complex Function}} }} {{end-eqn}} {{qed}} Category:Complex Differential Calculus q1pk390l1uyh5gi0a5194llmaqratl5
The function $f_c: D \to \C$ is defined as: :$\forall z \in D: f_c \left({z}\right) = c$ Thus: {{begin-eqn}} {{eqn | l = f_c' \left({z}\right)       | r = \lim_{h \mathop \to 0} \frac {f_c \left({z + h}\right) - f_c \left({z}\right)} h       | c = Definition of Complex Differentiation }} {{eqn | r = \lim_{h \mathop \to 0} \frac {c - c} h       | c = }} {{eqn | r = \lim_{h \mathop \to 0} \frac 0 h       | c = }} {{eqn | r = 0       | c = }} {{end-eqn}} {{qed}} Category:Complex Differential Calculus Category:Constant Mappings mxqh76yge3gg3g7uru7gkaor12euiib
The identity mapping is defined as: :$\forall x \in \R: \map {I_\R} x = x$ Thus: {{begin-eqn}} {{eqn | l = \map {I_\R'} x       | r = \lim_{\delta x \mathop \to 0} \frac {\map {I_\R} {x + \delta x} - \map {I_\R} x} {\delta x}       | c = {{Defof|Differentiation}} }} {{eqn | r = \lim_{\delta x \mathop \to 0} \frac {\paren {x + \delta x} - x} {\delta x}       | c = {{Defof|Identity Mapping}} }} {{eqn | r = \lim_{\delta x \mathop \to 0} \frac {\delta x} {\delta x}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Derivatives Category:Identity Mappings gl1j1aj81eg3xmxml3sygtd5c9v40wq
{{begin-eqn}} {{eqn | l = \frac {\d \Psi} {\d x}       | r = \frac {\partial \Psi} {\partial x} \frac {\d x} {\d x} + \frac {\partial \Psi} {\partial y} \frac {\d y} {\d x}       | c = Chain Rule for Real-Valued Functions }} {{eqn | r = \frac {\partial \Psi} {\partial x} + \frac {\partial \Psi} {\partial y} \frac {\d y} {\d x}       | c = Derivative of Identity Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map f t}       | r = \int_0^{\infty} e^{-s t} \map f t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} \int_{k T}^{\paren {k + 1} T} e^{-s t} \map f t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions: Corollary }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s \paren {t + k T} } \map f {t + k T} \rd t       | c = Change of Limits of Integration }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s \paren {t + k T} } \map f t \rd t       | c = General Periodicity Property }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s t - s k T} \map f t \rd t       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} \int_0^T e^{-s t} e^{- s k T} \map f t \rd t       | c = Product of Powers }} {{eqn | r = \sum_{k \mathop = 0}^{\infty} e^{- s k T} \int_0^T e^{-s t} \map f t \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \paren {\int_0^T e^{-s t} \map f t \rd t} \sum_{k \mathop = 0}^{\infty} e^{- s k T}       | c = Scaling of Summations }} {{eqn | r = \paren {\int_0^T e^{-s t} \map f t \rd t} \sum_{k \mathop = 0}^{\infty} \paren {e^{- s T} }^k       | c = Power of Power }} {{eqn | r = \paren {\int_0^T e^{-s t} \map f t \rd t} \paren {\frac 1 {1 - e^{-s T} } }       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t       | c = Real Multiplication is Commutative }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map f t}       | r = \int_0^{\infty} e^{-s t} \map f t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_T^{\infty} e^{-s t} \map f t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f {t + T} \rd t       | c = Change of Limits of Integration }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s \paren {t + T} } \map f t \rd t       | c = {{Defof|Real Periodic Function}} }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t - s T} \map f t \rd t       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_0^{\infty} e^{-s t} e^{-s T} \map f t \rd t       | c = Exponent Combination Laws: Product of Powers }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \int_0^{\infty} e^{-s t} \map f t \rd t       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + e^{-s T} \laptrans {\map f t}       | c = {{Defof|Laplace Transform}} }} {{eqn | ll= \leadsto       | l = \paren {1 - e^{-s T} } \laptrans {\map f t}       | r = \int_0^T e^{-s t} \map f t \rd t       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map f t}       | r = \int_0^{\infty} e^{-s t} \map f t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^T e^{-s t} \map f t \rd t + \int_T^{2 T} e^{-s t} \map f t \rd t + \int_{2 T}^{3 T} e^{-s t} \map f t \rd t + \dotsb       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^T e^{-s u} \map f u \rd t + \int_T^{2 T} e^{-s \paren {u + T} } \map f {u + T} \rd u + \int_{2 T}^{3 T} e^{-s \paren {u + 2 T} } \map f {u + 2 T} \rd u + \dotsb       | c = Integration by Substitution: $t = u$, $t = u + T$, $t = u + 2 T$, $\ldots$ }} {{eqn | r = \int_0^T e^{-s u} \map f u \rd t + e^{-s T} \int_0^T e^{-s u} \map f u \rd u + e^{-2 s T} \int_0^T e^{-s u} \map f u \rd u + \dotsb       | c = Laplace Transform of Function of t minus a, and adjusting limits of integration }} {{eqn | r = \paren {1 + e^{-s T} + e^{-2 s T} + \dotsb} \int_0^T e^{-s u} \map f u \rd u       | c = simplifying }} {{eqn | r = \frac 1 {1 - e^{-s T} } \int_0^T e^{-s t} \map f t \rd t       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} {{qed}}
By {{EuclidPostulateLink|Fifth}}, it is possible to construct exactly one straight line parallel to each of $AC$, $BC$ and $AC$. :500px From Parallelism implies Equal Corresponding Angles: :$\angle ABC = \angle ECB = \angle DAB$ :$\angle ACB = \angle CBE = \angle CAF$ :$\angle CAB = \angle ABD = \angle ACF$ By Triangle Angle-Side-Angle Equality: :$\triangle ABC = \triangle ABD$ :$\triangle ABC = \triangle CFA$ :$\triangle ABC = \triangle ECB$ Thus: :$FC = CE$ :$DB = BE$ :$FA = AD$ The result follows by definition of medial triangle. {{qed}}