hcju/proofwikiqa huggingface dataset:
{{WLOG}}, there are two cases to consider: :$(1): \quad b \ge c$ :$(2): \quad b < c$ First let $b \ge c$. We have: {{begin-eqn}} {{eqn | l = b       | o = \ge       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = a + b       | o = \ge       | r = a + c       | c = Addition of Real Numbers is Compatible with Usual Ordering }} {{eqn | ll= \leadsto       | l = \max \set {a + b, a + c}       | r = a + b       | c = {{Defof|Maximum Element}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a + \max \set {b, c}       | r = a + b       | c = {{Defof|Maximum Element}} }} {{eqn | r = \max \set {a + b, a + c}       | c =  }} {{end-eqn}} {{qed|lemma}} Now let $b < c$. We have: {{begin-eqn}} {{eqn | l = b       | o = <       | r = c       | c =  }} {{eqn | ll= \leadsto       | l = a + b       | o = <       | r = a + c       | c = Addition of Real Numbers is Compatible with Usual Ordering }} {{eqn | ll= \leadsto       | l = \max \set {a + b, a + c}       | r = a + c       | c = {{Defof|Maximum Element}} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a + \max \set {b, c}       | r = a + c       | c = {{Defof|Maximum Element}} }} {{eqn | r = \max \set {a + b, a + c}       | c =  }} {{end-eqn}} {{qed|lemma}} Thus the result holds in both cases. Hence the result, by Proof by Cases. {{qed}}
{{improve|The results upon which this depends need to be enhanced to encompass the full supremum, not just the max}} {{begin-eqn}} {{eqn | l = \left({\sup_{R \left({i}\right)} a_i}\right) + \left({\sup_{S \left({j}\right)} b_j}\right)       | r = \sup_{R \left({i}\right)} \left({a_i + \sup_{S \left({j}\right)} b_j}\right)       | c = Sum with Maximum is Maximum of Sum }} {{eqn | r = \sup_{R \left({i}\right)} \left({\sup_{S \left({j}\right)} \left({a_i + b_j}\right)}\right)       | c = Sum with Maximum is Maximum of Sum }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sup_{R \left({\pi \left({j}\right)}\right)} a_{\pi \left({j}\right)}       | r = \sup_{j \mathop \in I} a_{\pi \left({j}\right)} \left[{R \left({\pi \left({j}\right)}\right)}\right]       | c =  }} {{eqn | r = \sup_{j \mathop \in I} \left({\sup_{i \mathop \in I} a_i \left[{R \left({i}\right)}\right] \left[{i = \pi \left({j}\right)}\right]}\right)       | c =  }} {{eqn | r = \sup_{i \mathop \in I} a_i \left[{R \left({i}\right)}\right] \sup_{j \mathop \in I} \left[{i = \pi \left({j}\right)}\right]       | c =  }} {{eqn | r = \sup_{i \mathop \in I} a_i \left[{R \left({i}\right)}\right]       | c =  }} {{eqn | r = \sup_{R \left({i}\right)} a_i       | c =  }} {{eqn | r = \sup_{R \left({j}\right)} a_j       | c = Change of Index Variable of Supremum }} {{end-eqn}} {{qed}}
{{MissingLinks|relationship between holomorphic on s.c. domain and having a primitive}} Because $V$ is simply connected, $f$ has a primitive. Let one such primitive $F$ be established. By the Fundamental Theorem of Calculus for Contour Integrals: {{begin-eqn}} {{eqn | l = I_1       | r = \int_\gamma \map f z \rd z       | c =  }} {{eqn | r = \map F {z_2} - \map F {z_1}       | c =  }} {{end-eqn}} Next, observe that $\map f {\map \phi u} \map {\phi'} u$ also has a primitive, that is $\map F {\map \phi u}$. Again by the Fundamental Theorem of Calculus for Contour Integrals: {{begin-eqn}} {{eqn | l = I_2       | r = \int_\omega \map f {\map \phi u} \, \map {\phi'} u \rd u       | c =  }} {{eqn | r = \map F {\map \phi {u_2} } - \map F {\map \phi {u_1} }       | c =  }} {{end-eqn}} Because $\map \phi {u_1} = z_1$, we have that: :$\map F {\map \phi {u_1} } = \map F {z_1}$ and likewise for $\map \phi {u_2}$.  Thus $I_1 = I_2$. {{qed}} Category:Complex Analysis ixflcuxndq57awug1er3123dc2fhsmp
By Infimum of Set of Integers equals Smallest Element, $S$ has a smallest element $n \in \Z$, that is equals to the infimum of $S$. {{qed}}
By Set of Integers Bounded Below by Real Number has Smallest Element, $S$ has a smallest element, say $n \in S$. By Smallest Element is Infimum, $n$ is the infimum of $S$. {{qed}}
Let $M$ be the greatest element of $T$. Then by definition: :$\forall x \in T: x \preceq M$ By definition of supremum, it is necessary to show that: :$(1): \quad M$ is an upper bound of $T$ in $S$ :$(2): \quad M \preceq U$ for all upper bounds $U$ of $T$ in $S$. By Greatest Element is Upper Bound, $M$ is an upper bound of $T$ in $S$. It remains to be shown that: :$M \preceq U$ for all upper bounds $U$ of $T$ in $S$. Let $U \in S$ be an upper bound of $T$ in $S$. By definition of upper bound: :$\forall t \in T: t \preceq U$ We have that $M \in T$. Therefore: :$M \preceq U$ Hence the result. {{qed}}
Let $M$ be the smallest element of $T$. Then by definition: :$\forall x \in T: m \preceq x$ By definition of infimum, it is necessary to show that: : $(1): \quad m$ is a lower bound of $T$ in $S$ : $(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$. By Smallest Element is Lower Bound, $m$ is a lower bound of $T$ in $S$. It remains to be shown that: :$L \preceq m$ for all lower bounds $L$ of $T$ in $S$. Let $L \in S$ be a lower bound of $T$ in $S$. By definition of lower bound: :$\forall t \in T: L \preceq t$ We have that $m \in T$. Therefore: :$L \preceq m$ Hence the result. {{qed}}
We have that: :$1 \le 1 \cdotp 1 < 2$ Hence $1$ is the floor of $1 \cdotp 1$ by definition. {{qed}}
We have that: :$-2 \le -1 \cdotp 1 < -1$ Hence $-2$ is the floor of $-1 \cdotp 1$ by definition. {{qed}}
We have that: :$-2 < -1 \cdotp 1 \le -1$ Hence $-1$ is the ceiling of $-1 \cdotp 1$ by definition. {{qed}}
We have that: :$0 \le 0 \cdotp 99999 < 1$ Hence $0$ is the floor of $0 \cdotp 99999$ by definition. {{qed}}
{{begin-eqn}} {{eqn | l = \lg 32       | o = \le       | m = \lg 35       | mo= <       | r = \lg 64       | c =  }} {{eqn | ll= \leadsto       | l = 5       | o = \le       | m = \lg 35       | mo= <       | r = 6       | c =  }} {{end-eqn}} Hence $5$ is the floor of $\lg 35$ by definition. {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {100} 3 = 33 + \dfrac 1 3$ and so: :$\left\lfloor{\dfrac {100} 3}\right\rfloor = 33$ Thus: {{begin-eqn}} {{eqn | l = 100 \bmod 3       | r = 100 - 3 \times \left\lfloor{\dfrac {100} 3}\right\rfloor       | c =  }} {{eqn | r = 100 - 3 \times 33       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {100} 7 = 14 + \dfrac 2 7$ and so: :$\left\lfloor{\dfrac {100} 7}\right\rfloor = 14$ Thus: {{begin-eqn}} {{eqn | l = 100 \bmod 7       | r = 100 - 7 \times \left\lfloor{\dfrac {100} 7}\right\rfloor       | c =  }} {{eqn | r = 100 - 7 \times 14       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {-100} 7 = -15 + \dfrac 5 7$ and so: :$\left\lfloor{\dfrac {-100} 7}\right\rfloor = -15$ Thus: {{begin-eqn}} {{eqn | l = -100 \bmod 7       | r = -100 - 7 \times \left\lfloor{\dfrac {-100} 7}\right\rfloor       | c =  }} {{eqn | r = -100 + 7 \times 15       | c =  }} {{eqn | r = 5       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo $0$: :$\forall x \in \R: x \bmod 0 = x$ Hence: :$-100 \bmod 0 = -100$ {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac 5 {-3} = -2 + \dfrac 1 3$ and so: :$\left\lfloor{\dfrac 5 {-3} }\right\rfloor = -2$ Thus: {{begin-eqn}} {{eqn | l = 5 \bmod -3       | r = 5 - \left({-3}\right) \times \left\lfloor{\dfrac 5 {-3} }\right\rfloor       | c =  }} {{eqn | r = 5 - \left({-3}\right) \times \left({-2}\right)       | c =  }} {{eqn | r = 5 - 3 \times 2       | c =  }} {{eqn | r = -1       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {18} {-3} = -6 + \dfrac 0 3$ and so: :$\left\lfloor{\dfrac {18} {-3} }\right\rfloor = -6$ Thus: {{begin-eqn}} {{eqn | l = 18 \bmod -3       | r = 18 - \left({-3}\right) \times \left\lfloor{\dfrac {18} {-3} }\right\rfloor       | c =  }} {{eqn | r = 18 - \left({-3}\right) \times \left({-6}\right)       | c =  }} {{eqn | r = 18 - 3 \times 6       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$ for $y \ne 0$. We have: :$\dfrac {-2} {-3} = 0 + \dfrac 2 3$ and so: :$\left\lfloor{\dfrac {-2} {-3} }\right\rfloor = 0$ Thus: {{begin-eqn}} {{eqn | l = -2 \bmod -3       | r = -2 - \left({-3}\right) \times \left\lfloor{\dfrac {-2} {-3} }\right\rfloor       | c =  }} {{eqn | r = -2 - \left({-3}\right) \times 0       | c =  }} {{eqn | r = -2       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo $1$: :$x \bmod 1 = x - \left \lfloor {x}\right \rfloor$ Thus: {{begin-eqn}} {{eqn | l = 1 \cdotp 1 \bmod 1       | r = 1 \cdotp 1 - \left\lfloor{1 \cdotp 1}\right\rfloor       | c =  }} {{eqn | r = 1 \cdotp 1 - 1       | c =  }} {{eqn | r = 0 \cdotp 1       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \floor {\dfrac x y}$ for $y \ne 0$. We have: {{begin-eqn}} {{eqn | l = \dfrac {0 \cdotp 11} {-0 \cdotp 1}       | r = \dfrac {1 \cdotp 1} {-1}       | c =  }} {{eqn | r = -1 \cdotp 1       | c =  }} {{end-eqn}} and so: :$\floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} } = -2$ Thus: {{begin-eqn}} {{eqn | l = 0 \cdotp 11 \bmod -0 \cdotp 1       | r = 0 \cdotp 11 - \paren {-0 \cdotp 1} \times \floor {\dfrac {0 \cdotp 11} {-0 \cdotp 1} }       | c =  }} {{eqn | r = 0 \cdotp 11 - \paren {-0 \cdotp 1} \times \paren {-2}       | c =  }} {{eqn | r = 0 \cdotp 11 - 0 \cdotp 2       | c =  }} {{eqn | r = -0 \cdotp 09       | c =  }} {{end-eqn}} {{qed}}
By definition of modulo operation: :$x \bmod y := x - y \floor {\dfrac x y}$ for $y \ne 0$. We have: {{begin-eqn}} {{eqn | l = \dfrac {0 \cdotp 11} {0 \cdotp 1}       | r = \dfrac {1 \cdotp 1} 1       | c =  }} {{eqn | r = 1 \cdotp 1       | c =  }} {{end-eqn}} and so: :$\floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} } = 1$ Thus: {{begin-eqn}} {{eqn | l = 0 \cdotp 11 \bmod 0 \cdotp 1       | r = 0 \cdotp 11 - 0 \cdotp 1 \times \floor {\dfrac {0 \cdotp 11} {0 \cdotp 1} }       | c =  }} {{eqn | r = 0 \cdotp 11 - 0 \cdotp 1 \times 1       | c =  }} {{eqn | r = 0 \cdotp 01       | c =  }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \dfrac {x - z} y       | r = k       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = x       | r = z - k y       | c =  }} {{end-eqn}} We also have: :$0 \le z < y$ Hence: :$0 \le \dfrac z y < 1$ and so by definition of floor function: :$(2): \quad \floor {\dfrac z y} = 0$ Thus: {{begin-eqn}} {{eqn | l = x \bmod y       | r = x - y \floor {\dfrac x y}       | c = {{Defof|Modulo Operation}}: note that $y \ne 0$ by hypothesis }} {{eqn | r = \paren {z - k y} - y \floor {\dfrac {z - k y} y}       | c = from $(1)$ }} {{eqn | r = z - k y - y \floor {\dfrac z y - k}       | c =  }} {{eqn | r = z - k y - y \paren {\floor {\dfrac z y} - k}       | c = Floor of Number plus Integer }} {{eqn | r = z - k y + k y - y \floor {\dfrac z y}       | c =  }} {{eqn | r = z - y \times 0       | c = from $(2)$ }} {{end-eqn}} The result follows. {{Qed}}
Proof by Counterexample: Let $a = 6, b = 21, x = 7, y = 12, m = 15$. We note that $\gcd \left({6, 15}\right) = 3$ and so $6$ and $15$ are not coprime. We have that: {{begin-eqn}} {{eqn | l = 6       | o = \equiv       | r = 6       | rr= \pmod {15}       | c =  }} {{eqn | l = 21       | o = \equiv       | r = 6       | rr= \pmod {15}       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \equiv       | r = b       | rr= \pmod {15}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = 6 \times 7       | r = 42       | c =  }} {{eqn | o = \equiv       | r = 12       | rr= \pmod {15}       | c =  }} {{eqn | l = 21 \times 12       | r = 252       | c =  }} {{eqn | o = \equiv       | r = 12       | rr= \pmod {15}       | c =  }} {{eqn | ll= \leadsto       | l = a x       | o = \equiv       | r = b y       | rr= \pmod {15}       | c =  }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = 7       | o = \equiv       | r = 7       | rr= \pmod {15}       | c =  }} {{eqn | l = 12       | o = \equiv       | r = 12       | rr= \pmod {15}       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \not \equiv       | r = y       | rr= \pmod {15}       | c =  }} {{end-eqn}} {{qed}}
Proof by Counterexample: Let $a = 30, b = 40, r = 2, s = 10$. We have that: {{begin-eqn}} {{eqn | l = 30       | o = \equiv       | r = 40       | rr= \pmod 2       | c =  }} {{eqn | l = 30       | o = \equiv       | r = 40       | rr= \pmod {10}       | c =  }} {{eqn-intertext|But note that:}} {{eqn | l = 30       | o = \not \equiv       | r = 40       | rr= \pmod {20}       | c =  }} {{end-eqn}} {{qed}}
From Euler Phi Function of Prime: :$\map \phi p = p - 1$ As $2$ is a prime number it follows that: :$\map \phi 2 = 2 - 1 = 1$ {{qed}}
Let $S$ be the set of all complex numbers $s$ such that $\map f s$ converges. By hypothesis, there is some $s_0 = \sigma_0 + it_0 \in \C$ such that $\map f {s_0}$ converges, so $S$ is not empty. {{questionable|Which hypothesis?}} Moreover, $S$ is bounded below, for otherwise it follows from Dirichlet Series Convergence Lemma that $\map f s$ converges for all $s \in \C$, a contradiction of our assumptions. Therefore the infimum: :$\sigma_c = \inf \set {\sigma: s = \sigma + i t \in S} \in \R$ is well defined. Now if $s = \sigma + it$ with $\sigma > \sigma_c$, then there is $s' = \sigma' + i t' \in S$ with $\sigma' < \sigma$, and $\map f {s'}$ is convergent. Then it follows from Dirichlet Series Convergence Lemma that $\map f s$ is convergent. If $s = \sigma + it$ with $\sigma < \sigma_c$, and $\map f s$ is convergent then $s$ contradicts the definition of $\sigma_c$. Therefore, $\sigma_c$ has the claimed properties. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\begin{cases} 1 & : n = 0 \\ n \left({n - 1}\right)! & : n > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^n k$ === Basis for the Induction === $P \left({0}\right)$ is the case: :$\displaystyle \prod_{k \mathop = 1}^0 k = 1$ which holds by definition of vacuous product. Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$\begin{cases} 1 & : r = 0 \\ r \left({r - 1}\right)! & : r > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^r k$ from which it is to be shown that: :$\begin{cases} 1 & : r = 0 \\ \left({r + 1}\right) r! & : r > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^{r + 1} k$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \prod_{k \mathop = 1}^{r + 1} k       | r = \left({r + 1}\right) \prod_{k \mathop = 1}^r k       | c =  }} {{eqn | r = \left({r + 1}\right) r!       | c = Induction Hypothesis }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\begin{cases} 1 & : n = 0 \\ n \left({n - 1}\right)! & : n > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^n k$ {{qed}} Category:Factorials srborupz7w0nhjdkg5c49ygatcsxskk
{{begin-eqn}} {{eqn | l = \paren {\dfrac 1 2}!       | r = \map \Gamma {1 + \dfrac 1 2}       | c = Gamma Function Extends Factorial }} {{eqn | r = \dfrac 1 2 \map \Gamma {\dfrac 1 2}       | c = Gamma Difference Equation }} {{eqn | r = \dfrac 1 2 \sqrt {\pi}       | c = Gamma Function of One Half }} {{end-eqn}} {{qed}}
From Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta: {{:Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta}} Setting: : $k = 2$ : $\alpha_1 = \alpha_2 = 0$ : $\beta_1 = -\dfrac 1 2, \beta_2 = \dfrac 1 2$ we see that: :$\alpha_1 + \alpha_2 = \beta_1 + \beta_2$ So this reduces to: {{begin-eqn}} {{eqn | l = \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2}       | r = \dfrac {\Gamma \left({1 - \frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)} {\Gamma \left({1}\right) \Gamma \left({1}\right)}       | c =  }} {{eqn | r = \Gamma \left({\frac 1 2}\right) \Gamma \left({1 + \frac 1 2}\right)       | c = as $\Gamma \left({1}\right) = 0! = 1$ }} {{eqn | r = 2 \Gamma \left({1 + \frac 1 2}\right)^2       | c = Gamma Difference Equation }} {{end-eqn}} We then note that by Wallis's Product: :$\displaystyle \prod_{n \mathop \ge 1} \dfrac n {n - \frac 1 2} \dfrac n {n + \frac 1 2} = \frac \pi 2$ Thus: {{begin-eqn}} {{eqn | l = \frac \pi 2       | r = 2 \Gamma \left({1 + \frac 1 2}\right)^2       | c =  }} {{eqn | ll= \leadsto       | l = \frac \pi 4       | r = \left({\left({\frac 1 2}\right)!}\right)^2       | c = Gamma Function Extends Factorial }} {{eqn | ll= \leadsto       | l = \left({\frac 1 2}\right)!       | r = \frac {\sqrt \pi} 2       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = n^{\underline {n - 1} }       | r = \dfrac {n!} {\left({n - \left({n - 1}\right)}\right)!}       | c = Falling Factorial as Quotient of Factorials }} {{eqn | r = \dfrac {n!} {1!}       | c =  }} {{eqn | r = n!       | c = Factorial of $1$ }} {{end-eqn}} {{qed}} Category:Falling Factorials bh4xpr99b4kgrrvf6xuijumw5kx1z4v
From Limit of Error in Stirling's Formula: :$e^{1 / \paren {12 n + 1} } \le \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } \le e^{1 / 12 n}$ {{Finish}}
By definition of the termial on the real numbers: :$x? = \dfrac {x \left({x + 1}\right)} 2$ Thus: {{begin-eqn}} {{eqn | l = x? - x       | r = \dfrac {x \left({x + 1}\right)} 2 - x       | c = {{Defof|Termial/Real Numbers|Termial}} }} {{eqn | r = \dfrac {x \left({x + 1}\right) - 2 x} 2       | c =  }} {{eqn | r = \dfrac {x^2 + x - 2 x} 2       | c =  }} {{eqn | r = \dfrac {x^2 - x} 2       | c =  }} {{eqn | r = \dfrac {\left({x - 1}\right) x} 2       | c =  }} {{eqn | r = \left({x - 1}\right)?       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{m \mathop \to \infty} \frac {m^n m!} {\left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + m}\right)}       | r = \lim_{m \mathop \to \infty} \frac {m^n m!} {\frac {\left({n + m}\right)!} {n!} }       | c =  }} {{eqn | r = n! \lim_{m \mathop \to \infty} \frac {m^n} {\left({m + 1}\right) \left({m + 2}\right) \cdots \left({m + n}\right)}       | c =  }} {{eqn | r = n!       | c =  }} {{end-eqn}} Now we have from Factorial of Integer plus Reciprocal of Integer that: :$\displaystyle \lim_{m \mathop \to \infty} \dfrac {\left({m + n}\right)!} {m! m^n} = 1$ Now: {{begin-eqn}} {{eqn | l = \dfrac {n! \left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + m}\right)} {m! m^n}       | r = \dfrac {\left({m + n}\right)!} {m! m^n}       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{m \mathop \to \infty} \dfrac {n! \left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + m}\right)} {m! m^n}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{m \mathop \to \infty} \dfrac {m! m^n} {\left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + m}\right)}       | r = n!       | c =  }} {{end-eqn}}
{{begin-eqn}} {{eqn | l = \map \Gamma {-\dfrac 1 2}       | r = \frac {\map \Gamma {\frac 1 2} } {-1/2}       | c = Gamma Difference Equation }} {{eqn | r = -2 \, \map \Gamma {\frac 1 2}       | c =  }} {{eqn | r = -2 \sqrt \pi       | c = Gamma Function of One Half }} {{end-eqn}} {{qed}}
$n$ can be represented as: {{begin-eqn}} {{eqn | l = n       | r = \sum_{j \mathop = 1}^r 2^{e_j}       | c = where $e_1 > e_2 > \cdots > e_r > 0$ }} {{eqn | r = 2^{e_1} + 2^{e_2} + \cdots + 2^{e_r}       | c =  }} {{end-eqn}} where all of $e_1, e_2, \ldots, e_r$ are integers. Then $r$ is the sum of the digits in $n$ as a binary number. From Legendre's Theorem, we have: :$\mu = \dfrac {n - r} {2 - 1}$ Hence the result. {{qed}} {{Namedfor|Adrien-Marie Legendre}}
Write:  :$Q_n = \displaystyle \prod_{i \mathop = 1}^n \closedint {c - R} {c + R}_i$ Let $x, y \in Q_n$ By the definition of the usual metric, the distance between any two points $x$ and $y$ is given by: :$\displaystyle \map d {y - x} = \paren {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2}^{1 / 2}$ By Positive Power Function on Non-negative Reals is Strictly Increasing, this quantity is maximal when each summand is maximal. Consider $x_i, y_i$ in the $i$th interval: :$\closedint {c - R} {c + R}_i$ To maximize $\size {y_i - x_i}$, take $x_i = \min \closedint {c - R} {c + R}_i$ and $y_i = \max \closedint {c - R} {c + R}_i$. Then: :$\size {y_i - x_i} = \size {c + R - \paren {c - R} } = 2$ By the definition of an $n$-cube, each interval is of the same length. Thus: {{begin-eqn}} {{eqn | l = \paren {\map \sup {\map d {y - x} } }^2       | r = \map \sup {\sum_{i \mathop = 1}^n \paren {y_i - x_i}^2 } }} {{eqn | r = \paren {\sum_{i \mathop = 1}^n \paren {2 R}^2} }} {{eqn | r = \paren {2 R}^2 n       | c = Sum of Identical Terms }} {{eqn | r = \paren {2 R \sqrt n}^2 }} {{end-eqn}} {{qed}} Category:Real Analysis 8djg9lndytb6ghltw3d54vcs5hztvzv
$n$ can be represented as: {{begin-eqn}} {{eqn | l = n       | r = \sum_{j \mathop = 0}^m a_j p^j       | c = where $0 \le a_j < p$ }} {{eqn | r = a_0 + a_1 p + a_2 p^2 + \cdots + a_m p^m       | c = for some $m > 0$ }} {{end-eqn}} Using De Polignac's Formula, we may extract all the powers of $p$ from $n!$. :$\mu = \displaystyle \sum_{k \mathop > 0} \floor {\dfrac n {p^k} }$ where $\mu$ is the multiplicity of $p$ in $n!$: :$p^\mu \divides n!$ :$p^{\mu + 1} \nmid n!$ We have that: {{begin-eqn}} {{eqn | l = \floor {\dfrac {n!} p}       | r = \floor {\dfrac {a_0 + a_1 p + a_2 p^2 + a_3 p^3 + \cdots + a_m p^m} p}       | c =  }} {{eqn | r = a_1 + a_2 p + a_3 p^2 + \cdots + a_m p^{m - 1}       | c =  }} {{eqn | l = \floor {\dfrac {n!} {p^2} }       | r = \floor {\dfrac {a_0 + a_1 p + a_2 p^2 +  + a_2 p^2 + \cdots + a_m p^m} {p^2} }       | c =  }} {{eqn | r = a_2 + a_3 p + \cdots + a_m p^{m - 2}       | c =  }} {{eqn | o = \vdots }} {{eqn | l = \floor {\dfrac {n!} {p^m} }       | r = \floor {\dfrac {a_0 + a_1 p + a_2 p^2 +  + a_2 p^2 + \cdots + a_m p^m} {p^m} }       | c =  }} {{eqn | r = a_m       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \mu       | r = a_m \paren {p^{m - 1} + p^{m - 2} + \cdots + p + 1}       | c =  }} {{eqn | o =        | ro= +       | r = a_{m - 1} \paren {p^{m - 2} + p^{m - 3} + \cdots + p + 1}       | c =  }} {{eqn | o = \vdots }} {{eqn | o =        | ro= +       | r = a_2 \paren {p + 1}       | c =  }} {{eqn | o =        | ro= +       | r = a_1       | c =  }} {{eqn | r = a_m \paren {\dfrac {p^m - 1} {p - 1} } + a_{m - 1} \paren {\dfrac {p^{m - 1} - 1} {p - 1} }       | c = Sum of Geometric Sequence }} {{eqn | o = \vdots }} {{eqn | o =        | ro= +       | r = a_2 \paren {\dfrac {p^2 - 1} {p - 1} } + a_1 \paren {\dfrac {p^1 - 1} {p - 1} } + a_0 \paren {\dfrac {p^0 - 1} {p - 1} }       | c = where the last term evaluates to $0$ }} {{eqn | r = \dfrac {\paren {a_m p^m + a_{m - 1} p^{m - 1} + \cdots + a_2 p^2 + a_1 p + a_0} - \paren {a_m + a_{m - 1} + \cdots + a_2 + a_1 + a_0} } {p - 1}       | c =  }} {{eqn | r = \dfrac {n - r} {p - 1}       | c =  }} {{end-eqn}} Hence the result. {{qed}} {{Namedfor|Adrien-Marie Legendre|cat = Legendre}}