hcju/proofwikiqa huggingface dataset:
Let $n$ be triangular. Then: :$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$ So: {{begin-eqn}} {{eqn | l = 25 n + 3       | r = 25 \frac {k \paren {k + 1} } 2 + 3       | c =  }} {{eqn | r = \frac {25 k^2 + 25 k + 6} 2       | c =  }} {{eqn | r = \frac {\paren {5 k + 2} \paren {5 k + 3} } 2       | c =  }} {{end-eqn}} which is triangular. {{qed}}
Let $n$ be triangular. Then: :$\exists k \in \Z: n = \dfrac {k \paren {k + 1} } 2$ So: {{begin-eqn}} {{eqn | l = 49 n + 6       | r = 49 \frac {k \paren {k + 1} } 2 + 6       | c =  }} {{eqn | r = \frac {49 k^2 + 49 k + 12} 2       | c =  }} {{eqn | r = \frac {\paren {7 k + 3} \paren {7 k + 4} } 2       | c =  }} {{end-eqn}} which is triangular. {{qed}}
From Sum of Sequence of n Choose 2 we have: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 2}^n \dbinom j 2       | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2       | c =  }} {{eqn | r = \dbinom {n + 1} 3       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 2}^{n + 1} \dbinom j 2       | r = \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2 + \dbinom {n + 1} 2       | c =  }} {{eqn | r = \dbinom {n + 2} 3       | c =  }} {{end-eqn}} But we have that: :$\dbinom {n + 2} 3 = \dfrac {\paren {n + 2} \paren {n + 1} n} {3 \times 2 \times 1}$ and from Binomial Coefficient with Two: :$T_n = \dfrac {\paren {n + 1} n} 2 = \dbinom {n + 1} 2$ The result follows. {{qed}}
First let $n$ be even. Thus we have: :$n = 2 m$ Then: {{begin-eqn}} {{eqn | l = T_1 + T_2 + T_3 + \dotsb + T_{2 m}       | r = \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} }       | c =  }} {{eqn | r = 2^2 + 4^2 + \dotsb + \paren {2 m}^2       | c = Sum of Consecutive Triangular Numbers is Square }} {{eqn | r = 2^2 \times 1^2 + 2^2 \times 2^2 + \dotsb + 2^2 \times m^2       | c = Sum of Consecutive Triangular Numbers is Square }} {{eqn | r = 2^2 \paren {1^2 + 2^2 + \dotsb + m^2}       | c =  }} {{eqn | r = 4 \frac {m \paren {m + 1} \paren {2 m + 1} } 6       | c =  }} {{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6       | c =  }} {{end-eqn}} Now let $n$ be odd. Thus we have: :$n = 2 m + 1$ Then: {{begin-eqn}} {{eqn | l = T_1 + T_2 + T_3 + \dotsb + T_{2 m}       | r = \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} } + T_{2 m + 1}       | c =  }} {{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + T_{2 m + 1}       | c = from above }} {{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + \frac {\paren {2 m + 1} \paren {2 m + 2} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \frac {2 m \paren {2 m + 1} \paren {2 m + 2} + \paren {6 m + 3} \paren {2 m + 2} } 6       | c =  }} {{eqn | r = \frac {\paren {2 m + 2} \paren {2 m \paren {2 m + 1} + \paren {6 m + 3} } } 6       | c =  }} {{eqn | r = \frac {\paren {2 m + 2} \paren {4 m^2 + 8 m + 3} } 6       | c =  }} {{eqn | r = \frac {\paren {2 m + 2} \paren {2 m + 1} \paren {2 m + 3} } 6       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = T_{9 n + 4} - T_{3 n + 1}       | r = \dfrac {\paren {9 n + 4} \paren {9 n + 5} } 2 - \dfrac {\paren {3 n + 1} \paren {3 n + 2} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \dfrac {\paren {81 n^2 + 81 n + 20} - \paren {9 n^2 + 9 n + 2} } 2       | c =  }} {{eqn | r = \dfrac {72 n^2 + 72 n + 18} 2       | c =  }} {{eqn | r = 36 n^2 + 36 n + 9       | c =  }} {{eqn | r = 9 \paren {4 n^2 + 4 n + 1}       | c =  }} {{eqn | r = 9 \paren {2 n + 1}^2       | c =  }} {{eqn | r = 9 m^2       | c =  }} {{eqn | r = \paren {3 m}^2       | c =  }} {{end-eqn}} {{qed}}
Let $T_n + T_{n + 1} + T_{n + 2} = m^2$ for some $m \in \Z_{>0}$. We have: {{begin-eqn}} {{eqn | l = T_n + T_{n + 1} + T_{n + 2}       | r = \dfrac {n \paren {n + 1} } 2 + \dfrac {\paren {n + 1} \paren {n + 2} } 2 + \dfrac {\paren {n + 2} \paren {n + 3} } 2       | c =  }} {{eqn | r = \dfrac {n \paren {n + 1} + \paren {n + 1} \paren {n + 2} + \paren {n + 2} \paren {n + 3} } 8       | c =  }} {{eqn | r = \dfrac {n^2 + n + n^2 + 3 n + 2 + n^2 + 5 n + 6} 2       | c = multiplying out }} {{eqn | r = \dfrac {3 n^2 + 9 n + 8} 2       | c = multiplying out }} {{end-eqn}} Thus we need to find $n$ such that $\dfrac {3 n^2 + 9 n + 8} 2$ is a square number. We see that: {{begin-eqn}} {{eqn | l = \dfrac {3 \times 5^2 + 9 \times 5 + 8} 2       | r = \dfrac {75 + 45 + 8} 2       | c =  }} {{eqn | r = \dfrac {128} 2       | c =  }} {{eqn | r = 64       | c =  }} {{end-eqn}} Thus $n$ appears to satisfy the conditions. It remains for us to check: {{begin-eqn}} {{eqn | l = T_5 \times T_6 \times T_7       | r = 15 + 21 + 28       | c =  }} {{eqn | r = 64       | c =  }} {{eqn | r = 8^2       | c =  }} {{end-eqn}} {{qed}}
Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$. We have: {{begin-eqn}} {{eqn | l = T_n \times T_{n + 1} \times T_{n + 2}       | r = \dfrac {n \paren {n + 1} } 2 \dfrac {\paren {n + 1} \paren {n + 2} } 2 \dfrac {\paren {n + 2} \paren {n + 3} } 2       | c =  }} {{eqn | r = \dfrac {n \paren {n + 1}^2 \paren {n + 2}^2 \paren {n + 3} } 8       | c =  }} {{eqn | r = \paren {\dfrac {\paren {n + 1} \paren {n + 2} } 2}^2 \dfrac {n \paren {n + 3} } 2       | c =  }} {{end-eqn}} Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number. We see that: {{begin-eqn}} {{eqn | l = 3 \times \paren {3 + 3}       | r = 3 \times 6       | c =  }} {{eqn | r = 2 \times 3^2       | c =  }} {{end-eqn}} Thus $n = 3$ appears to satisfy the conditions. It remains for us to check: {{begin-eqn}} {{eqn | l = T_3 \times T_4 \times T_5       | r = 6 \times 10 \times 15       | c =  }} {{eqn | r = \paren {2 \times 3} \times {2 \times 5} \times {3 \times 5}       | c =  }} {{eqn | r = \paren {2 \times 3 \times 5}^2       | c =  }} {{eqn | r = 30^2       | c =  }} {{eqn | r = 900       | c =  }} {{end-eqn}} {{qed}}
Let $S$ be a set containing exactly one element. Suppose $S = \set x$ for some object $x$.  Then the power set of $S$ is the set: :$\powerset S = \set {\O, \set x}$ That is: :$\powerset S = \set {\O, S}$ Let $\tau$ be a topology on $S$. $\tau$ is a subset of $\powerset S$, $\tau$ must equal one of the following sets: :$\tau_1 = \O$ :$\tau_2 = \set \O$ :$\tau_3 = \set S$ or :$\tau_4 = \set {\O, S}$. By definition of a topology, $S \in \tau$. Thus $\tau \ne \tau_1$ and $\tau \ne \tau_2$. By Empty Set is Element of Topology, also $\O \in \tau$.  Therefore $\tau \ne \tau_3$. By Indiscrete Topology is Topology, $\tau_4$ is a topology on $S$. So if $S$ is a set containing exactly one element, the only possible topology on $S$ is the indiscrete topology.  {{qed}} Category:Indiscrete Topology Category:Singletons peurip1ml05yqxy25ok7aaanoqbcdpo
From the Division Theorem, we have that: :$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$ So, with a view to where this is going, let $q$ and $r$ be renamed such that $a = q' b + r'$ with $0 \le r' < b$. Then let $q' = q + 2$. We have: {{begin-eqn}} {{eqn | l = a       | r = q' b + r'       | c =  }} {{eqn | r = \paren {q + 2} b + r'       | c =  }} {{eqn | r = q b + 2 b + r'       | c =  }} {{end-eqn}} Setting $r = 2 b + r'$, it follows that: :$2 b \le r < 3 b$ and the result follows. {{qed}}
By the Division Theorem, $n$ can be expressed as: :$n = 4 k + r$ where: :$k, r \in \Z$ :$0 \le r < 4$ That is, one of the following holds: {{begin-eqn}} {{eqn | l = n       | r = 4 k }} {{eqn | l = n       | r = 4 k + 1 }} {{eqn | l = n       | r = 4 k + 2 }} {{eqn | l = n       | r = 4 k + 3 }} {{end-eqn}} Of these: {{begin-eqn}} {{eqn | l = n       | r = 4 k }} {{eqn | r = 2 \paren {2 k} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = n       | r = 4 k + 2 }} {{eqn | r = 2 \paren {2 k + 1} }} {{end-eqn}} and so are even by definition. Then: {{begin-eqn}} {{eqn | l = n       | r = 4 k + 1 }} {{eqn | r = 2 \paren {2 k} + 1 }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = n       | r = 4 k + 3 }} {{eqn | r = 2 \paren {2 k + 1} + 1 }} {{end-eqn}} and so are odd by definition. {{qed}}
Let $x$ be an integer. There are three cases to consider: :$(1): \quad x \equiv 0 \pmod 3$: we have $x = 3 k$ :$(2): \quad x \equiv 1 \pmod 3$: we have $x = 3 k + 1$ :$(3): \quad x \equiv 2 \pmod 3$: we have $x = 3 k + 2$ Then: {{begin-eqn}} {{eqn | n = 1       | l = x       | r = 3 k       | c =  }} {{eqn | ll= \leadsto       | l = x^3       | r = \paren {3 k}^3       | c =  }} {{eqn | r = 3^3 k^3       | c =  }} {{eqn | r = 9 \times \paren {3 k^3}       | c =   }} {{eqn | ll= \leadsto       | l = x^3       | o = \equiv       | r = 0 \pmod 9       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 2       | l = x       | r = 3 k + 1       | c =  }} {{eqn | ll= \leadsto       | l = x^3       | r = \paren {3 k + 1}^3       | c =  }} {{eqn | r = \paren {3 k}^3 + 3 \paren {3 k}^2 + 3 \paren {3 k} + 1       | c =  }} {{eqn | r = 9 \times \paren {3 k^3 + 3 k^2 + k} + 1       | c =   }} {{eqn | ll= \leadsto       | l = x^3       | o = \equiv       | r = 1 \pmod 9       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 3       | l = x       | r = 3 k + 2       | c =  }} {{eqn | ll= \leadsto       | l = x^3       | r = \paren {3 k + 2}^3       | c =  }} {{eqn | r = \paren {3 k}^3 + 3 \times 2 \paren {3 k}^2 + 3 \times 2^2 \paren {3 k} + 2^3       | c =  }} {{eqn | r = 9 \times \paren {3 k^3 + 6 k^2 + 4 k} + 8       | c =   }} {{eqn | ll= \leadsto       | l = x^3       | o = \equiv       | r = 8 \pmod 9       | c =  }} {{end-eqn}} {{qed}}
Let $n = r^2 = s^3$ for some $r, s \in \Z$. Then: :$n = \paren {m^2}^3 = \paren {m^3}^2 = m^6$ for some $m \in \Z$ There are $7$ cases to consider: :$(0): \quad m \equiv 0 \pmod 7$: we have $m = 7 k$ :$(1): \quad m \equiv 1 \pmod 7$: we have $m = 7 k + 1$ :$(2): \quad m \equiv 2 \pmod 7$: we have $m = 7 k + 2$ :$(3): \quad m \equiv 3 \pmod 7$: we have $m = 7 k + 3$ :$(4): \quad m \equiv 4 \pmod 7$: we have $m = 7 k + 4$ :$(5): \quad m \equiv 5 \pmod 7$: we have $m = 7 k + 5$ :$(6): \quad m \equiv 6 \pmod 7$: we have $m = 7 k + 6$ Using Congruence of Powers throughout, we make use of: :$x \equiv y \pmod 7 \implies x^j \equiv y^j \pmod 7$ for $j \in \Z_{>0}$. First the easy cases: {{begin-eqn}} {{eqn | n = 0       | l = m       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = m^6       | o = \equiv       | r = 0 \pmod 7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 1       | l = m       | o = \equiv       | r = 1 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = m^6       | o = \equiv       | r = 1 \pmod 7       | c =  }} {{end-eqn}} It is sufficient to investigate the congruence modulo $7$ of the integers from $2$ to $6$. Then we have: {{begin-eqn}} {{eqn | n = 2       | l = 2^3       | o = \equiv       | r = 8 \pmod 7       | c =  }} {{eqn | o = \equiv       | r = 1 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = 2^6       | o = \equiv       | r = 1 \pmod 7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 3       | l = 3^2       | o = \equiv       | r = 9 \pmod 7       | c =  }} {{eqn | o = \equiv       | r = 2 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = 3^6 = \paren {3^2}^3       | o = \equiv       | r = 1 \pmod 7       | c = from $(2)$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 4       | l = 4^2       | o = \equiv       | r = 16 \pmod 7       | c =  }} {{eqn | o = \equiv       | r = 2 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = 4^6 = \paren {4^2}^3       | o = \equiv       | r = 1 \pmod 7       | c = from $(2)$ }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 5       | l = 5^2       | o = \equiv       | r = 25 \pmod 7       | c =  }} {{eqn | o = \equiv       | r = 4 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = 5^6 = \paren {4^2}^3       | o = \equiv       | r = 2^3 = 8 \pmod 7       | c = from $(4)$ }} {{eqn | o = \equiv       | r = 1 \pmod 7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | n = 6       | l = 6^2       | o = \equiv       | r = 36 \pmod 7       | c =  }} {{eqn | o = \equiv       | r = 1 \pmod 7       | c =  }} {{eqn | ll= \leadsto       | l = 6^6 = \paren {6^2}^3       | o = \equiv       | r = 1 \pmod 7       | c =  }} {{end-eqn}} Hence the result. {{Qed}}
Let $\sigma$ be the standard deviation of $P$. By the definition of convergence in probability, we aim to show that:  :$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} = 1$ for all real $\varepsilon > 0$. Let $\varepsilon > 0$ be a real number. By Variance of Sample Mean: :$\var {{\overline X}_n} = \dfrac {\sigma^2} n$ By Chebyshev's Inequality, we have for real $k > 0$:  :$\map \Pr {\size {{\overline X}_n - \mu} \ge \dfrac {k \sigma} {\sqrt n}} \le \dfrac 1 {k^2}$ As $\sigma > 0$ and $n > 0$, we can set:  :$k = \dfrac {\sqrt n} {\sigma} \varepsilon$ This gives:  :$\map \Pr {\size {{\overline X}_n - \mu} \ge \varepsilon} \le \dfrac {\sigma^2} {n \varepsilon^2}$ We therefore have:  {{begin-eqn}} {{eqn	| l = \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} 	| r = 1 - \map \Pr {\size { {\overline X}_n - \mu} \ge \varepsilon} }} {{eqn	| o = \ge 	| r = 1 - \frac {\sigma^2} {n \varepsilon^2} }} {{end-eqn}} So:  :$1 - \dfrac {\sigma^2} {n \varepsilon^2} \le \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} \le 1$ We have:  :$\ds \lim_{n \mathop \to \infty} \paren {1 - \dfrac {\sigma^2} {n \varepsilon^2} } = 1$ and:  :$\ds \lim_{n \mathop \to \infty} 1 = 1$ So by the Squeeze Theorem:  :$\ds \lim_{n \mathop \to \infty} \map \Pr {\size { {\overline X}_n - \mu} < \varepsilon} = 1$ for all real $\varepsilon > 0$. {{qed}}
From Variance of Linear Combination of Random Variables, we have:  :$\var {a X + b Y} = a^2 \, \var X + b^2 \, \var Y + 2 a b \, \cov {X, Y}$ where $\cov {X, Y}$ is the covariance of $X$ and $Y$.  From Covariance of Independent Random Variables is Zero:  :$2 a b \, \cov {X, Y} = 0$ The result follows. {{qed}} Category:Variance 8jc0sfnr7tjd01olvha76a5y2a6lz98
Let $A = \set{x}$.   From definition $3$ of a connected set, $A$ is connected in $T$ {{iff}} the subspace $\struct {A, \tau_A}$ is a connected space. From Topology on Singleton is Indiscrete Topology, $\tau_A$ is the indiscrete topology. From Indiscrete Space is Connected, $\struct {A, \tau_A}$ is a connected space. {{qed}} Category:Connected Spaces Category:Singletons 3w8kqpzocosqu4rypn04ewrq9hyrnim
From Singleton is Connected in Topological Space, $\set{x}$ is a connected set of $T$ containing $x$. It follows that $x \in C$. From Union of Connected Sets with Common Point is Connected, $C$ is a  connected set of $T$. Hence $C \in \mathcal C_x$. {{qed}} a6r0x60mfyo94nrjbiounoi7tg99by2
{{begin-eqn}} {{eqn	| l = \var {\overline X} 	| r = \var {\frac 1 n \sum_{i \mathop = 1}^n X_i} }} {{eqn	| r = \frac 1 {n^2} \sum_{i \mathop = 1}^n \var {X_i} 	| c = repeated application of Variance of Linear Combination of Random Variables: Corollary }} {{eqn	| r = \frac 1 {n^2} \sum_{i \mathop = 1}^n \sigma^2 }} {{eqn	| r = \frac {\sigma^2 n} {n^2} 	| c = as $\displaystyle \sum_{i \mathop = 1}^n 1 = n$ }} {{eqn	| r = \frac {\sigma^2} n }} {{end-eqn}} {{qed}} Category:Variance Category:Inductive Statistics 08a68ngd75g37ms3m6i96r0r2s07phh
By definition of divisibility: {{begin-eqn}} {{eqn | lo= \exists k_1 \in \Z:       | l = b       | r = a k_1 }} {{eqn | lo= \exists k_2 \in \Z:       | l = d       | r = c k_2 }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = b d       | r = \paren {a k_1} \paren {c k_2}       | c =  }} {{eqn | r = k_1 k_2 \paren {a c}       | c =  }} {{end-eqn}} and the result follows by definition of divisibility. {{Qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition: :$\forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$ === Basis for the Induction === $\map P 2$ is the case: :$\forall x \in \set {a_1, a_2}: c \divides x \implies \forall x_1, x_2 \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2}$ This is demonstrated to be true in Common Divisor Divides Integer Combination. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$ from which it is to be shown that: :$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$ === Induction Step === This is the induction step: Let : :$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x$ We have that: :$c \divides a_{k + 1} \implies \forall x_{k + 1} \in \Z: c \divides a_{k + 1} x_{k + 1}$ and we have that: :$\forall x \in \set {a_1, a_2, \dotsc, a_k}: c \divides x \implies \forall x_1, x_2, \dotsc, x_k \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k}$ Let $x_1, x_2, \dotsc, x_k \in \Z$ be arbitrary. Let $d = a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k$. Then: {{begin-eqn}} {{eqn | l = c       | o = \divides       | r = cd       | c =  }} {{eqn | lo= \land       | l = c       | o = \divides       | r = a_{k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = c       | o = \divides       | r = \paren {1 \times d + x_{k + 1} a_{k + 1} }       | c =  }} {{eqn | r = \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }       | c =  }} {{end-eqn}} But $x_1, x_2, \dotsc, x_k \in \Z$ are arbitrary, and so: :$\forall x \in \set {a_1, a_2, \dotsc, a_k, a_{k + 1} }: c \divides x \implies \forall x_1, x_2, \dotsc, x_k, x_{k + 1} \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_k x_k + a_{k + 1} x_{k + 1} }$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 2}: \forall x \in \set {a_1, a_2, \dotsc, a_n}: c \divides x \implies \forall x_1, x_2, \dotsc, x_n \in \Z: c \divides \paren {a_1 x_2 + a_2 x_2 + \dotsb + a_n x_n}$ {{qed}}
From One Divides all Integers: :$1 \divides a$ and: :$1 \divides b$ where $\divides$ denotes divisibility. The result follows by definition of common divisor. {{Qed}}
From One is Common Divisor of Integers: :$1$ is a common divisor of $a$ and $b$. Thus, whatever $a$ and $b$ are: :$1 \in S$ The result follows by definition of empty set. {{Qed}}
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text { is path-connected in } T}$ Let $C = \bigcup \CC_x$. === Lemma === {{:Equivalence of Definitions of Path Component/Lemma 1}}{{qed|lemma}} Let $C'$ be the equivalence class containing $x$ of the equivalence relation $\sim$ defined by: :$y \sim z$ {{iff}} $y$ and $z$ are connected in $T$. === Equivalence Class equals Union of Path-Connected Sets === It needs to be shown that $C = C'$. {{:Equivalence of Definitions of Path Component/Equivalence Class equals Union of Path-Connected Sets}}{{qed|lemma}} === Union of Path-Connected Sets is Maximal Path-Connected Set === {{:Equivalence of Definitions of Path Component/Union of Path-Connected Sets is Maximal Path-Connected Set}}{{qed|lemma}} === Maximal Path-Connected Set is Union of Path-Connected Sets === {{:Equivalence of Definitions of Path Component/Maximal Path-Connected Set is Union of Path-Connected Sets}}{{qed}}
From Integer Divisor Results, the divisors of $a$ include $a$ itself. From Integer Divides its Negative, $a \divides \paren {-a}$. Thus we have: :$a \divides a$ and: :$a \divides -a$ and so: :$\gcd \set {a, -a} \ge \size a$ From Absolute Value of Integer is not less than Divisors, there is no divisor of $a$ which is greater than $a$. That is: :$\gcd \set {a, -a} \le \size a$ Hence the result. {{qed}} Category:Greatest Common Divisor 68esbzc2bdbir4akn0p2a4dpg5258u2
From Point is Path-Connected to Itself, $\set x$ is a path-connected of $T$ containing $x$. It follows that $x \in C$. From Union of Path-Connected Sets with Common Point is Path-Connected, $C$ is a  path-connected of $T$. Hence $C \in \CC_x$. {{qed}} Category:Equivalence of Definitions of Path Component l7pjkqyt9ps2ou2m3yshgvg09jqxose
{{begin-eqn}} {{eqn | l = y \in C'       | o = \leadstoandfrom       | r = x \text{ is path-connected to } y \text{ in } T       | c = Definition of $\sim$ }} {{eqn | o = \leadstoandfrom       | r = \exists B \text{ a connected set of } T, x \in B, y \in B       | c = Points are Path-Connected iff Contained in Path-Connected Set }} {{eqn | o = \leadstoandfrom       | r = \exists B \in \CC_x : y \in B       | c = Equivalent definition }} {{eqn | o = \leadstoandfrom       | r = y \in \bigcup \CC_x       | c = {{Defof|Set Union|subdef = General Definition}} }} {{eqn | o = \leadstoandfrom       | r = y \in C       | c = Definition of $C$ }} {{end-eqn}} The result follows.
From Path-Connected iff Path-Connected to Point, $\map \gamma I$ is a path-connected set {{iff}} every point of $\map \gamma I$ is path-connected to a common point. It is shown that every point of $\map \gamma I$ is path-connected to $\map \gamma a$. Let $H = \map \gamma I$. Let $x = \map \gamma a$. From Point is Path-Connected to Itself, $x$ is path-connected to itself. Let $y$ be any element of $H \setminus \set x$. By assumption: :$\exists t \in \hointl a b : \map \gamma t = y$ From Continuity of Composite with Inclusion, $\mathbin \gamma {\restriction_{\closedint a t} } : \closedint a t \to H$ is a continuous mapping. It follows that $\mathbin \gamma {\restriction_{\closedint a t} }$ is a path joining $x$ to $y$. The result follows. {{qed}} Category:Path-Connected Spaces bbtlwcjrgjq5oioz83uirrn3xkzwzec
{{Proofread}} We use Lagrangian mechanics to derive the result. First we remember that $\mathbf r = \map {\mathbf r} t$ is a function of time $t$. Let $R = \size r$ be the constant radius of the circle $C$ for clarity. The constraint can then be written as: :$\forall t: \map f {\mathbf r} := \size {\mathbf r}^2 - R^2 = 0$ The Principle of Stationary Action then states that the first variation of the action $S$ must vanish. :$\ds \delta S = \delta \int L \rd t = 0$ The Lagrangian $L$ is given by the kinetic energy alone because the constraint force is handled implicitly by the constraint and there are no other forces acting on $P$. :$L = \dfrac 1 2 m \size {\mathbf v}^2$ To incorporate the constraint we use a scalar Lagrange multiplier $\lambda$. Because we are dealing with functionals and not functions and the constraint is local and not global $\lambda$ is also a function of time $t$, as is $\mathbf r$: :$\lambda = \map \lambda t$  The augmented action $S^+$ is given by: :$\ds S^+ = S + \int \lambda f \d t$ We have that: :$\dfrac {\d f} {\d \mathbf r} = 2 \mathbf r$ Then: :$\ds \delta S^+ = \int \paren {-\map {\dfrac \d {\d t} } {m \mathbf v} + \map \lambda {2 \mathbf r} } \cdot \delta \mathbf r \rd t + \int f \delta \lambda \rd t = 0$ This means that both the term in the brackets as well as $f$ must vanish. We eliminate $\lambda$ from the term in the brackets by taking its scalar product with $\mathbf r$ and simplifying $\map {\dfrac \d {\d t} } {m \mathbf v}$ to $m \mathbf a$: :$-\map {\dfrac \d {\d t} } {m \mathbf v} \cdot \mathbf r + \map \lambda {2 \mathbf r} \cdot \mathbf r = 0$ This gives: :$2 \lambda = \dfrac {m \mathbf a \cdot \mathbf r} {R^2}$ We have to replace $\mathbf a \cdot \mathbf r$. To do this, we take the second derivative {{WRT|Differentiation}} $t$ of the constraint, which must vanish too: :$\dfrac {\d^2 f} {\d t^2} = 2 \map {\dfrac \d {\d t} } {\mathbf r \cdot \mathbf v} = 2 \paren {\mathbf a \cdot \mathbf r + \size {\mathbf v}^2} = 0$ So :$2 \lambda = \dfrac {-m \size {\mathbf v}^2} {R^2}$ Plugging in $2 \lambda$ in the term with the big brackets yields: :$-m \mathbf a - \dfrac {m \size {\mathbf v}^2} {R^2} \mathbf r = 0$ Division by $m$ completes the proof. {{qed}}