hcju/proofwikiqa huggingface dataset:
Let $n \in \Z: n \ne 0$. Then: {{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod m       | c = }} {{eqn | ll= \leadstoandfrom       | l = a \bmod m       | r = b \bmod m       | c = {{Defof|Congruence Modulo Integer}} }} {{eqn | ll= \leadstoandfrom       | l = n \left({a \bmod n}\right)       | r = n \left({b \bmod n}\right)       | c = Left hand implication valid only when $n \ne 0$ }} {{eqn | ll= \leadstoandfrom       | l = \left({a n}\right) \bmod \left({m n}\right)       | r = \left({b n}\right) \bmod \left({m n}\right)       | c = Product Distributes over Modulo Operation }} {{eqn | ll= \leadstoandfrom       | l = a n       | o = \equiv       | r = b n       | rr= \pmod {m n}       | c = {{Defof|Congruence Modulo Integer}} }} {{end-eqn}} Hence the result. Note the invalidity of the third step when $n = 0$. {{qed}}
Define $\displaystyle \map G x = \int_a^b \map f {x, y} \rd y$. The continuity of $f$ ensures that $G$ exists. Then by linearity of the integral: :$\dfrac {\Delta G} {\Delta x} = \dfrac {\map G {x + \Delta x} - \map G x} {\Delta x} = \displaystyle \int_a^b \frac {\map f {x + \Delta x, y} - \map f {x, y} } {\Delta x} \rd y$ We want to find the limit of this quantity as $\Delta x$ approaches zero. For each $y \in \closedint a b$, we can consider $\map {f_y} x = \map f {x, y}$ as a separate function of the single variable $x$, with $\dfrac {\d f_y} {\d x} = \dfrac {\partial f} {\partial x}$. Thus by the Mean Value Theorem, there is a number $c_y \in \openint x {x + \Delta x}$ such that: :$\map {f_y} {x + \Delta x} - \map {f_y} x = \map {\dfrac {\d f_y} {\d x} } {c_y} \Delta x$ That is: :$\map f {x + \Delta x, y} - \map f {x, y} = \map {\dfrac {\partial f} {\partial x} } {c_y, y} \Delta x$ Therefore: :$\dfrac {\Delta G} {\Delta x} = \displaystyle \int_a^b \map {\frac {\partial f} {\partial x} } {c_y, y} \rd y$ Now, pick any $\epsilon > 0$ and set $\epsilon_0 = \dfrac {\epsilon} {b - a}$. Since $\dfrac {\partial f} {\partial x}$ is continuous on the compact set $D$, it is uniformly continuous on $D$. Hence for each $x$ and $y$: :$\size {\map {\dfrac {\partial f} {\partial x} } {x + h, y} - \map {\dfrac {\partial f} {\partial x} } {x, y} } < \epsilon_0$ whenever $h$ is sufficiently small. Since $x < c_y < x + \Delta x$, it follows that for sufficiently small $\Delta x$: :$\size {\map {\dfrac {\partial f} {\partial x} } {c_y, y} - \map {\dfrac {\partial f} {\partial x} } {x, y} } < \epsilon_0$ regardless of our choice of $y$. So we can say: {{begin-eqn}} {{eqn | l = \size {\lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} - \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y}       | r = \lim_{\Delta x \mathop \to 0} \size {\int_a^b \map {\frac {\partial f} {\partial x} } {c_y, y} - \frac {\partial f}{\partial x} \rd y} }} {{eqn | o = \le       | r = \lim_{\Delta x \mathop \to 0} \int_a^b \size {\map {\frac {\partial f} {\partial x} } {c_y, y} - \map {\frac {\partial f} {\partial x} } {x, y} } \rd y }} {{eqn | o = \le       | r = \int_a^b \epsilon_0 \rd y }} {{eqn | r = \epsilon }} {{end-eqn}} But since $\epsilon$ was arbitrary, it follows that: :$\displaystyle \lim_{\Delta x \mathop \to 0} \frac {\Delta G} {\Delta x} = \int_a^b \map {\frac {\partial f} {\partial x} } {x, y} \rd y$ and the theorem is proved. {{qed}}
From Leibniz's Integral Rule: :$\displaystyle \frac \d {\d y} \int_{\map a y}^{\map b y} \map f {x, y} \rd x = \map f {y, \map b y} \frac {\d b} {\d y} - \map f {y, \map a y} \frac {\d a} {\d y} + \int_{\map a y}^{\map b y} \frac {\partial} {\partial y} \map f {x, y} \rd x$ where $\map a y$, $\map b y$ are continuously differentiable. Setting $\map a y$ and $\map b y$ constant, so that $\exists a, b \in \R: \map a y = a, \map b y = b$: :$\dfrac {\d a} {\d y} = \dfrac {\d b} {\d y} = 0$ from which the result follows immediately. {{qed}}
We have that $d = \gcd \set {m, n}$. So: {{begin-eqn}} {{eqn | ll= \exists a, b \in \Z:       | l = a m + b n       | r = d       | c = Bézout's Identity }} {{eqn | lll=\leadsto       | l = a m       | r = d - b n       | c =  }} {{eqn | lll=\leadsto       | l = d - b n       | o = \equiv       | r = 0       | rr= \pmod m       | c = {{Defof|Congruence (Number Theory)|Congruence}} }} {{eqn | lll=\leadsto       | l = b n       | o = \equiv       | r = d       | rr= \pmod m       | c = Modulo Addition: add $b n$ to both sides of congruence }} {{end-eqn}} So $b$ (in the above) fits the requirement for $n'$ in the assertion to be proved. {{qed}} Category:Modulo Arithmetic Category:Named Theorems fs2nvgpt2b1f59a6qjlos2a63rxpz5i
We have that $d = \gcd \set {a, m}$. From Law of Inverses (Modulo Arithmetic), we have: : $\exists a' \in \Z: a a' \equiv d \pmod m$ Hence: {{begin-eqn}} {{eqn | l = a       | o = \equiv       | r = b       | rr= \pmod m }} {{eqn | ll= \leadsto       | l = a a'       | o = \equiv       | r = b a'       | rr= \pmod m       | c = {{Defof|Modulo Multiplication}} }} {{eqn | o = \equiv       | r = d       | rr= \pmod m       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = a x       | o = \equiv       | r = b y       | rr= \pmod m }} {{eqn | ll= \leadsto       | l =  a a' x       | o = \equiv       | r = b a' y       | rr= \pmod m       | c = {{Defof|Modulo Multiplication}} }} {{eqn | ll= \leadsto       | l = d x       | o = \equiv       | r = d y       | rr= \pmod m       | c = from above }} {{eqn | ll= \leadsto       | l = x       | o = \equiv       | r = y       | rr= \paren {\bmod {\dfrac m d} }       | c = Congruence by Product of Moduli }} {{end-eqn}} Hence the result. {{qed}}
=== Necessary Condition === Suppose $a \equiv b \pmod {r s}$. Then from Congruence by Divisor of Modulus it follows directly that: :$a \equiv b \pmod r$ and: :$a \equiv b \pmod s$ Note that for this result it is not required that $r \perp s$. {{qed|lemma}} === Sufficient Condition === Now suppose that $a \equiv b \pmod r$ and $a \equiv b \pmod s$. We have by definition of congruence: :$a \equiv b \pmod r \implies \exists k \in \Z: a - b = k r$ From $a \equiv b \pmod s$ we also have that: :$k r \equiv 0 \pmod s$ As $r \perp s$, we have from Common Factor Cancelling in Congruence: :$k \equiv 0 \pmod s$ So: :$\exists q \in \Z: a - b = q s r$ Hence by definition of congruence: :$a \equiv b \pmod {r s}$ {{qed}}
From the definition of the modulo operation, we have that: :$x = \left \lfloor {x}\right \rfloor + \left({x \bmod 1}\right)$ from which: {{begin-eqn}} {{eqn | l = \left \lfloor {x + y} \right \rfloor       | r = \left \lfloor {\left \lfloor {x}\right \rfloor + \left({x \bmod 1}\right) + \left \lfloor {y}\right \rfloor + \left({y \bmod 1}\right)} \right \rfloor       | c =  }} {{eqn | r = \left \lfloor {x}\right \rfloor + \left \lfloor {y}\right \rfloor + \left \lfloor {\left({x \bmod 1}\right) + \left({y \bmod 1}\right)} \right \rfloor       | c = Floor of Number plus Integer }} {{end-eqn}} Hence the inequality. The equality holds {{iff}}: :$\left \lfloor {\left({x \bmod 1}\right) + \left({y \bmod 1}\right)} \right \rfloor = 0$ that is, {{iff}}: :$x \bmod 1 + y \bmod 1 < 1$ {{qed}}
From the definition of the modulo operation, we have that: :$x = \left \lfloor {x}\right \rfloor + \left({x \bmod 1}\right)$ from which we obtain: :$x = \left \lceil {x}\right \rceil - \left[{x \notin \Z}\right] + \left({x \bmod 1}\right)$ where $\left[{x \notin \Z}\right]$ uses Iverson's convention. {{begin-eqn}} {{eqn | l = \left \lceil {x + y} \right \rceil       | r = \left \lceil {\left \lfloor {x}\right \rfloor + \left({x \bmod 1}\right) + \left \lfloor {y}\right \rfloor + \left({y \bmod 1}\right)} \right \rceil       | c =  }} {{eqn | r = \left \lceil {\left \lceil {x}\right \rceil - \left[{x \notin \Z}\right] + \left({x \bmod 1}\right) + \left \lceil {y}\right \rceil - \left[{y \notin \Z}\right] + \left({y \bmod 1}\right)} \right \rceil       | c =  }} {{eqn | r = \left \lceil {x}\right \rceil + \left \lceil {y}\right \rceil + \left \lceil {\left({x \bmod 1}\right) + \left({y \bmod 1}\right)} \right \rceil - \left[{x \notin \Z}\right] - \left[{y \notin \Z}\right]       | c = Ceiling of Number plus Integer }} {{end-eqn}} We have that: :$x \notin \Z \implies x \bmod 1 > 0$ As $0 \le x \bmod 1 < 1$ it follows that: :$\left[{x \notin \Z}\right] \ge x \bmod 1$ Hence the inequality. The equality holds {{iff}}: :$\left \lceil {\left({x \bmod 1}\right) + \left({y \bmod 1}\right)} \right \rceil = \left[{x \notin \Z}\right] + \left[{y \notin \Z}\right]$ that is, {{iff}} one of the following holds: : $x \in \Z$, in which case $x \bmod 1 = 0$ : $y \in \Z$, in which case $y \bmod 1 = 0$ : both $x, y \in \Z$, in which case $\left({x \bmod 1}\right) + \left({y \bmod 1}\right) = 0$ : both $x, y \notin \Z$ and $\left({x \bmod 1}\right) + \left({y \bmod 1}\right) > 1$. {{qed}}
By definition of congruence modulo $p$: :$\forall x, y \in \R: x \equiv y \pmod p \iff x \bmod p = y \bmod p$  We have that: :$b = a^{\frac{\paren {p - 1} } 2}$ and so: :$b^2 = a^{p - 1}$ Let $a \equiv 0 \pmod p$. Then by definition of congruence modulo $p$: :$p \divides a$ and so: :$p \divides a^{\frac{\paren {p - 1} } 2}$ where $\divides$ denotes divisibility. Thus by definition of congruence modulo $p$: :$b \equiv 0 \pmod p$ and so: :$b \bmod p = 0$ {{qed|lemma}} Otherwise, from Fermat's Little Theorem: :$b^2 \equiv 1 \pmod p$ That is: :$b^2 - 1 \equiv 0 \pmod p$ From Difference of Two Squares: :$b^2 - 1 = \paren {b + 1} \paren {b - 1}$ So either: :$p \divides b + 1$ or: :$p \divides b - 1$ {{AimForCont}} both $p \divides b + 1$ and $p \divides b - 1$. Then by Modulo Subtraction is Well-Defined: :$p \divides \paren {b + 1} - \paren {b - 1} = 2$ But $p$ is an odd prime. So it cannot be the case that $p \divides 2$. From this contradiction it follows that  Note that $p$ cannot divide both $b + 1$ and $b - 1$. So either: {{begin-eqn}} {{eqn | l = \paren {b - 1}       | o = \equiv       | r = 0       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \equiv       | r = 1       | rr= \pmod p       | c = Modulo Addition is Well-Defined }} {{end-eqn}} or: {{begin-eqn}} {{eqn | l = \paren {b + 1}       | o = \equiv       | r = 0       | rr= \pmod p       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \equiv       | r = -1       | rr= \pmod p       | c = Modulo Subtraction is Well-Defined }} {{eqn | o = \equiv       | r = p - 1       | rr= \pmod p       | c = Negative Number is Congruent to Modulus minus Number }} {{end-eqn}} Hence the result. {{qed}}
From the definition of a filter we have to prove four things: : $(1): \quad f \left[{\mathcal F}\right] \subset \mathcal P \left({Y}\right)$ : $(2): \quad Y \in f \left[{\mathcal F}\right], \varnothing \notin f \left[{\mathcal F}\right]$ : $(3): \quad U, V \in f \left[{\mathcal F}\right] \implies U \cap V \in f \left[{\mathcal F}\right]$ : $(4): \quad U \in f \left[{\mathcal F}\right], U \subseteq V \subseteq Y \implies V \in f \left[{\mathcal F}\right]$ By construction we have: :$f \left[{\mathcal F}\right] \subseteq \mathcal P \left({Y}\right)$ Since $f^{-1} \left[{\varnothing}\right] = \varnothing \notin \mathcal F$ we know that: :$\varnothing \notin f \left[{\mathcal F}\right]$ Therefore: :$f \left[{\mathcal F}\right] \ne \mathcal P \left({Y}\right)$ which implies $(1)$. Because $f^{-1} \left[{Y}\right] = X \in \mathcal F$, we have: :$Y \in f \left[{\mathcal F}\right]$ Since we've already shown $\varnothing \notin f \left[{\mathcal F}\right]$, this implies $(2)$. Let $U, V \in f \left[{\mathcal F}\right]$. From Preimage of Intersection under Mapping: :$f^{-1} \left[{U \cap V}\right] = f^{-1} \left[{U}\right] \cap f^{-1} \left[{V}\right] \in \mathcal F$ (since $\mathcal F$ is a filter). Thus $U \cap V \in f \left[{\mathcal F}\right]$, and so $(3)$ holds. Finally, let $U \in f \left[{\mathcal F}\right]$ and $V \subseteq Y$ such that $U \subseteq V$. Then from Image of Subset under Relation is Subset of Image/Corollary 3: :$f^{-1} \left[{U}\right] \subseteq f^{-1} \left[{V}\right]$ Since $f^{-1} \left[{U}\right] \in \mathcal F$ and $\mathcal F$ is a filter it follows that: :$f^{-1} \left[{V}\right] \in \mathcal F$ which implies: :$V \in f \left[{\mathcal F}\right]$ and thus $(4)$. {{qed}} Category:Set Theory Category:Filter Theory ixdd9t22ebhxrplrw5qqewfpyfc61t0
=== $(1) \implies (2)$ === Assume $f$ is continuous at $x$ and that $\mathcal F$ is a filter on $X$ that converges to $x$. Suppose $U \subseteq Y$ is a neighborhood of $f \left({x}\right)$. Then, since $f$ is continuous at $x$, $f^{-1} \left({U}\right)$ is a neighborhood of $x$. Since $\mathcal F$ converges to $x$, this implies that $f^{-1} \left({U}\right) \in \mathcal F$. By the definition of $f \left({\mathcal F}\right)$ it follows that $U \in f \left({\mathcal F}\right)$. Thus $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$. === $(2) \implies (1)$ === The set $\mathcal U_x := \left\{{U \subseteq X: U \text{ is a neighborhood of } x}\right\}$ is a filter on $X$. We set $\mathcal F := \mathcal U_x$. By the definition of convergence for filters, $\mathcal F$ converges to $x$. By assumption this implies that $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$. Now let $U \subseteq Y$ a neighborhood of $f \left({x}\right)$. Then $U \in f \left({\mathcal F}\right)$ and by definition this implies $f^{-1} \left({U}\right) \in \mathcal F = \mathcal U_x$. Thus $f^{-1} \left({U}\right)$ is a neighborhood of $x$. Therefore $f$ is continuous at $x$. {{qed}} Category:Continuous Mappings Category:Filter Theory 3muo00vegcsvdg8mu0li73facjxeais
=== Necessary Condition === Let $\FF$ converge to $x$. Let $i \in I$. From Projection from Product Topology is Continuous, $\pr_i : X \to X_i$ is continuous. Thus $\map {\pr_i} \FF$ converges to $\map {\pr_i} x$ as claimed. {{qed|lemma}} === Sufficient Condition === For all $i \in I$, let $\map {\pr_i} \FF$ converge to $\map {\pr_i} x$. Let $V \subseteq X$ a neighborhood of $x$. We have to show that $V \in \mathcal F$. Then there is a set $U$ from the natural basis of $X$ such that $x \in U \subseteq V$. By the definition of the natural basis, there is a finite set $J \subseteq I$ such that: : $\displaystyle U = \bigcap_{j \mathop \in J} \map {\pr_j^{-1}} {U_j}$ where $U_j \subseteq X_j$ is an open set for all $j \in J$. Thus $U_j$ is an open neighborhood of $x_j$ for all $j \in J$. Since $\map {\pr_j} \FF$ converges to $x_j$, this implies $U_j \in \map {\pr_j} \FF$ for all $j \in J$. By the definition of $\map {\pr_j} \FF$, this implies $\map {\pr_j^{-1}} {U_j} \in \mathcal F$ for all $j \in J$. Since $J$ is finite, it follows that: : $\displaystyle U = \bigcap_{j \mathop \in J} \map {\pr_j^{-1}} {U_j} \in \FF$ Recall that $U \subseteq V$. As $\FF$ is a filter, this implies that also $V \in \FF$. Thus $\FF$ converges to $x$ as claimed. {{qed}}
From Image Filter is Filter, we have that $\mathcal F$ is a filter on $Y$. Let $\mathcal G$ be a filter on $Y$ such that $f \left({\mathcal F}\right) \subseteq \mathcal G$. We have to show that $f \left({\mathcal F}\right) = \mathcal G$. Let $U \in \mathcal G$. Assume that $U \notin f \left({\mathcal F}\right)$. By the definition of $f \left({\mathcal F}\right)$ this implies that $f^{-1}(U) \not \in \mathcal F$. By definition of ultrafilter: : $V := X \setminus f^{-1} \left({U}\right) \in \mathcal F$ Because $V \subseteq f^{-1} \left({f \left({V}\right)}\right)$ it follows that $f^{-1}\left({f \left({V}\right)}\right) \in \mathcal F$ and thus also $f \left({V}\right) \in f \left({\mathcal F}\right)$. By assumption we have $f \left({\mathcal F}\right) \subseteq \mathcal G$, thus $f \left({V}\right) \in \mathcal G$. But: : $f \left({V}\right) \cap U = f \left({X \setminus f^{-1} \left({U}\right)}\right) \cap U = \varnothing \notin \mathcal G$ Thus $\mathcal G$ is not a filter, a contradiction to our assumptions. Thus: : $U \in f \left({\mathcal F}\right)$ and therefore $f \left({\mathcal F}\right) = \mathcal G$ Hence $f \left({\mathcal F}\right)$ is an ultrafilter. {{qed}} Category:Set Theory Category:Filter Theory nfsqlqz09qajt8b0sdsksbil8ueflgd
We note that the inequality, as stated, is symmetrical in $x, y$ and $z$. {{WLOG}}, we can assume that $x \ge y \ge z \ge 0$. Consider the expression: :$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$ We see that every term in the above is non-negative. So, directly: :$(1): \quad \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z} \ge 0$ If $x = y = z$, all of $x - y$, $x - z$ and $y - z$ are $0$. Thus equality holds. Inspection on a case-by-case basis provides evidence for the other conditions for equality. To show these are the only cases, we suppose $x, y, z$ are not equal. Then $x > y > z \ge 0$. We thus have $x - z > y - z$. Hence: {{begin-eqn}} {{eqn | o =        | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z} }} {{eqn | o = \ge       | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} }       | c = second term is non-negative }} {{eqn | o = >       | r = \paren {x - y} \paren {x^t \paren {y - z} - y^t \paren {y - z} }       | c =  }} {{eqn | r = \paren {x - y} \paren {y - z} \paren {x^t  - y^t}       | c =  }} {{eqn | o = >       | r = 0       | c = $x > y > z$ }} {{end-eqn}} Now we suppose two numbers are equal, but the other is neither the same number or $0$. If $x = y > z > 0$: {{begin-eqn}} {{eqn | o =        | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z} }} {{eqn | r = z^t \paren {x - z} \paren {y - z}       | c = $x - y = 0$ }} {{eqn | o = >       | r = 0       | c = $z > 0$, $x - z = y - z > 0$ }} {{end-eqn}} If $x > y = z \ge 0$: {{begin-eqn}} {{eqn | o =        | r = \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z} } + z^t \paren {x - z} \paren {y - z} }} {{eqn | r = \paren {x - y} \paren {x^t \paren {x - z} }       | c = $y - z = 0$ }} {{eqn | o = >       | r = 0       | c = $x > 0$, $x - y = x - z > 0$ }} {{end-eqn}} This shows the equality conditions is {{iff}}. $(1)$ can then be rearranged to '''Schur's inequality'''. {{qed|lemma}} Now, let $t$ be a positive even integer. {{WLOG}}, we can assume that $x \ge y \ge z$. By Pigeonhole Principle, at least $2$ of them have the same sign. Suppose $x, y$ are positive. Once again we consider the expression: :$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$ The first term is still non-negative. The second term is non-negative, since: :$z^t \ge 0 \quad$ Even Power is Non-Negative :$x \ge z$, $y \ge z$ Thus we can still conclude $(1)$, which can then be rearranged to '''Schur's inequality'''. Suppose $y, z$ are negative. Then $-z, -y$ are positive, and $-z \le -y \le -x$. Substituting $x, y, z$ for $-z, -y, -x$ in the above, the result follows. {{qed}} {{namedfor|Issai Schur|cat = Schur}} Category:Algebra ipp6vt84pgfc4whge9w9ntud8exyplv
Into Euler Formula for Sine Function: {{begin-eqn}} {{eqn | l = \dfrac {\sin x} x       | r = \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots       | c =  }} {{eqn | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }       | c =  }} {{end-eqn}} we substitute $x = \dfrac \pi 2$. From Sine of Half-Integer Multiple of Pi: :$\sin \dfrac \pi 2 = 1$ Hence: {{begin-eqn}}      {{eqn | l = \frac 2 \pi       | r = \prod_{n \mathop = 1}^\infty \paren {1 - \frac 1 {4 n^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \frac \pi 2       | r = \prod_{n \mathop = 1}^\infty \paren {\frac {4 n^2} {4 n^2 - 1} }       | c =  }} {{eqn | r = \prod_{n \mathop = 1}^\infty \frac {\paren {2 n} \paren {2 n} } {\paren {2 n - 1} \paren {2 n + 1} }       | c =  }} {{eqn | r = \frac 2 1 \cdot \frac 2 3 \cdot \frac 4 3 \cdot \frac 4 5 \cdot \frac 6 5 \cdot \frac 6 7 \cdot \frac 8 7 \cdot \frac 8 9 \cdots       | c =  }} {{end-eqn}} {{qed}}
From the Reduction Formula for Integral of Power of Sine, we have: :$\displaystyle (1): \quad \int \sin^n x \rd x = - \frac 1 n \sin^{n - 1} \cos x + \frac {n - 1} n \int \sin^{n - 2} x \rd x$ Let $I_n$ be defined as: :$\displaystyle I_n = \int_0^{\pi / 2} \sin^n x \rd x$ As $\cos \dfrac \pi 2 = 0$ from Shape of Cosine Function, we have from $(1)$ that: :$(2): \quad I_n = \dfrac {n-1} n I_{n-2}$ To start the ball rolling, we note that: :$\displaystyle I_0 = \int_0^{\pi / 2} \rd x = \frac \pi 2 \qquad \qquad I_1 = \int_0^{\pi / 2} \sin x \rd x = \left[{- \cos x}\right]_0^{\pi / 2} = 1$ We need to separate the cases where the subscripts are even and odd: {{begin-eqn}}      {{eqn | l = I_{2 n}       | r = \frac {2 n - 1} {2 n} I_{2 n - 2}       | c =  }} {{eqn | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} I_{2 n - 4}       | c =  }} {{eqn | r = \cdots       | c =  }} {{eqn | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 I_0       | c =  }} {{eqn | n = A       | r = \frac {2 n - 1} {2 n} \cdot \frac {2 n - 3} {2 n - 2} \cdot \frac {2 n - 5} {2 n - 4} \cdots \frac 3 4 \cdot \frac 1 2 \cdot \frac \pi 2 }} {{end-eqn}} {{begin-eqn}}      {{eqn | l = I_{2 n+1}       | r = \frac {2 n} {2 n + 1} I_{2 n - 1}       | c =  }} {{eqn | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} I_{2 n - 3}       | c =  }} {{eqn | r = \cdots       | c =  }} {{eqn | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 I_1       | c =  }} {{eqn | n = B       | r = \frac {2 n} {2 n + 1} \cdot \frac {2 n - 2} {2 n - 1} \cdot \frac {2 n - 4} {2 n - 3} \cdots \frac 4 5 \cdot \frac 2 3 }} {{end-eqn}} By Shape of Sine Function, we have that on $0 \le x \le \dfrac \pi 2$: :$0 \le \sin x \le 1$ Therefore: :$0 \le \sin^{2 n + 2} x \le \sin^{2 n +1} x \le \sin^{2 n} x$ It follows from Relative Sizes of Definite Integrals that: :$\displaystyle 0 < \int_0^{\pi / 2} \sin^{2 n + 2} x \rd x \le \int_0^{\pi / 2} \sin^{2 n + 1} x \rd x \le \int_0^{\pi / 2} \sin^{2 n} x \rd x$ That is: :$(3): \quad 0 < I_{2 n + 2} \le I_{2 n + 1} \le I_{2 n}$ By $(2)$ we have: :$\dfrac {I_{2 n + 2} } {I_{2 n} } = \dfrac {2 n + 1} {2 n + 2}$ Dividing $(3)$ through by $I_{2n}$ then, we have: :$\dfrac {2 n + 1} {2 n + 2} \le \dfrac {I_{2 n + 1}} {I_{2 n}} \le 1$ By Squeeze Theorem, it follows that: :$\dfrac {I_{2 n + 1} } {I_{2 n} } \to 1$ as $n \to \infty$ which is equivalent to: :$\dfrac {I_{2 n} } {I_{2 n + 1} } \to 1$ as $n \to \infty$ Now we take $(B)$ and divide it by $(A)$ to get: :$\dfrac {I_{2 n + 1} } {I_{2 n} } = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \dfrac 2 \pi$ So: :$\dfrac \pi 2 = \dfrac 2 1 \cdot \dfrac 2 3 \cdot \dfrac 4 3 \cdot \dfrac 4 5 \cdots \dfrac {2 n} {2 n - 1} \cdot \dfrac {2 n} {2 n + 1} \cdot \left({\dfrac {I_{2 n} } {I_{2 n + 1} } }\right)$ Taking the limit as $n \to \infty$ gives the result. {{qed}}