hcju/proofwikiqa huggingface dataset:
Let $\mathbb D^3$ be centered at the origin, and $D^3$ be some other unit ball in $\R^3$ such that $\mathbb D^3 \cap D^3 = \O$. Let $\mathbb S^2 = \partial \mathbb D^3$. By the Hausdorff Paradox, there exists a decomposition of $ \mathbb S^2$ into four sets $A, B, C, D$ such that $A, B, C$ and $B \cup C$ are congruent, and $D$ is countable. For $r \in \R_{>0}$, define a function $r^*: \R^3 \to \R^3$ as $\map {r^*} {\mathbf x} = r \mathbf x$, and define the sets: {{begin-eqn}} {{eqn | l = W       | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} A }} {{eqn | l = X       | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} B }} {{eqn | l = Y       | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} C }} {{eqn | l = Z       | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} D }} {{end-eqn}} Let $T = W \cup Z \cup \set {\mathbf 0}$. $W$ and $X \cup Y$ are clearly congruent by the congruency of $A$ with $B \cup C$, hence $W$ and $X \cup Y$ are equidecomposable. Since $X$ and $Y$ are congruent, and $W$ and $X$ are congruent, $X \cup Y$ and $W \cup X$ are equidecomposable.   $W$ and $X \cup Y$ as well as $X$ and $W$ are congruent, so $W \cup X$ and $W \cup X \cup Y$ are equidecomposable.   Hence $W$ and $W \cup X \cup Y$ are equidecomposable, by Equidecomposability is Equivalence Relation. So $T$ and $\mathbb D^3$ are equidecomposable, from Equidecomposability Unaffected by Union. Similarly we find $X$, $Y$, and $W \cup X \cup Y$ are equidecomposable. Since $D$ is only countable, but $\map {\operatorname {SO} } 3$ is not, we have: :$\exists \phi \in \map {\operatorname {SO} } 3: \map \phi D \subset A \cup B \cup C$ so that $I = \map \phi D \subset W \cup X \cup Y$. Since $X$ and $W \cup X \cup Y$ are equidecomposable, by Subsets of Equidecomposable Subsets are Equidecomposable, $\exists H \subseteq X$ such that $H$ and $I$ are equidecomposable. Finally, let $p \in X - H$ be a point and define $S = Y \cup H \cup \set p$. Since: :$Y$ and $W \cup X \cup Y$ :$H$ and $Z$ :$\set 0$ and $\set p$ are all equidecomposable in pairs, $S$ and $\mathbb B^3$ are equidecomposable by Equidecomposability Unaffected by Union. Since $D^3$ and $\mathbb D^3$ are congruent, $D^3$ and $S$ are equidecomposable, from Equidecomposability is Equivalence Relation.   By Equidecomposability Unaffected by Union, $T \cup S$ and $\mathbb D^3 \cup D^3$ are equidecomposable. Hence $T \cup S \subseteq \mathbb D^3 \subset \mathbb D^3 \cup D^3$ are equidecomposable and so, by the chain property of equidecomposability, $\mathbb D^3$ and $\mathbb D^3 \cup D^3$ are equidecomposable. {{qed}} {{AoC|Hausdorff Paradox}}
=== Relexivity === A set is necessarily equidecomposable with itself; the same decomposition and set of isometries suffice for $A$ as for $A$. === Symmetry === There is no order to the relation of being equidecomposable; symmetry follows. === Transitivity === Suppose $A, B, C \subset \R^n$ are sets such that $A, B$ are equidecomposable and $B, C$ are equidecomposable. Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\R^n \to \R^n$ such that: :$\displaystyle A = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)$  and  :$\displaystyle B = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)$ Further let $Y_1, \dots, Y_p$ together with $\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p$ be sets and isometries such that: :$\displaystyle B = \bigcup_{i \mathop = 1}^p \xi_i \left({Y_i}\right)$  and: :$\displaystyle C = \bigcup_{i \mathop = 1}^p \tau_i \left({Y_i}\right)$ Consider the sets :$Z_{i,j} = \nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)$ where $1 \le i \le m$ and $1 \le j \le p$. We have: :$\displaystyle \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \left({\mu_i \circ \nu_i^{-1}}\right) \left({Z_{i,j}}\right) = \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \left({\mu_i \circ \nu_i^{-1}}\right) \left({\nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)}\right) = \bigcup_{i \mathop = 1}^m \left({\mu_i \circ \nu_i^{-1} \circ \nu_i}\right) \left({X_i}\right) = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right) = A$ :$\displaystyle \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \left({\tau_j \circ \xi_j^{-1}}\right) \left({Z_{i,j}}\right) = \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \left({\tau_j \circ \xi_j^{-1}}\right) \left({\nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)}\right) = \bigcup_{j \mathop = 1}^p \left({\tau_j \circ \xi_j^{-1} \circ \xi_j}\right) \left({Y_j}\right) = \bigcup_{j \mathop = 1}^p \tau_j \left({Y_j}\right) = C$ so $Z_{i, j}$ together with the isometries $\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1}$ is a decomposition of $A$ and $C$. Hence these two are equidecomposable. {{qed}} Category:Topology 7ic97xbu5p8wuwze6h1ejtag6kjdbly
We have for each $k \in \left\{{1, \dots, m}\right\}$ a decomposition $\left\{{A_{k, 1}, \cdots, A_{k, l_k}}\right\}$ and set of isometries $\phi_{i, j}: \R^n \to \R^n$ such that: :$\displaystyle S_k = \bigcup_{a \mathop = 1}^{l_k} \phi_{k, a} \left({A_{k, a} }\right)$ and similarly for $T_k$ and some isometries $\theta_{i, j}: \R^n \to \R^n$: :$\displaystyle T_k = \bigcup_{a \mathop = 1}^{l_k} \theta_{k, a} \left({A_{k, a} }\right)$ Thus: :$\displaystyle S = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \phi_{k, i} \left({A_{k, i} }\right)$ and: :$\displaystyle T = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \theta_{k, i} \left({A_{k, i} }\right)$. This yields equivalent decompositions of $S$ and $T$. {{qed}} Category:Topology khsd67uh7ztk2vsb1ngqax6ihrywa2e
Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \ldots, \mu_m, \nu_1, \ldots, \nu_m: \R^n \to \R^n$ such that: :$\displaystyle A = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)$  and  :$\displaystyle B = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)$ Define: :$Y_i = \mu_i^{-1} \left({S \cap \mu_i \left({X_i}\right)}\right)$ Then: {{begin-eqn}} {{eqn | l = \bigcup_{i \mathop = 1}^m \mu_i \left({Y_i}\right)       | r = \bigcup_{i \mathop = 1}^m \left({S \cap \mu_i \left({X_i}\right)}\right)       | c =  }} {{eqn | r = S \cap \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)       | c =  }} {{eqn | r = S \cap A       | c =  }} {{eqn | r = S       | c =  }} {{end-eqn}} and so $\left\{{Y_i}\right\}_{i \mathop = 1}^m$ forms a decomposition of $S$. But for each $i$: :$\left({S \cap \mu_i \left({X_i}\right)}\right) \subseteq \mu_i \left({X_i}\right)$ and so: {{begin-eqn}} {{eqn | l = Y_i       | r = \mu_i^{-1} \left({S \cap \mu_i \left({X_i}\right)}\right)       | c =  }} {{eqn | o = \subseteq       | r = \mu_i^{-1} \left({\mu_i \left({X_i}\right)}\right)       | c =  }} {{eqn | r = X_i       | c =  }} {{end-eqn}} Hence: :$\nu_i \left({Y_i}\right) \subseteq \nu_i \left({X_i}\right)$ and so: {{begin-eqn}} {{eqn | l = \bigcup_{i \mathop = 1}^m \nu_i \left({Y_i}\right)       | o = \subseteq       | r = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)       | c =  }} {{eqn | r = B       | c =  }} {{end-eqn}} Define: : $\displaystyle \bigcup_{i \mathop = 1}^m \nu_i \left({Y_i}\right) = T$ Hence the result. {{qed}} Category:Topology d4nf9og3qpnwksv2n4mhypdck2ylcb4
From the discussion of Sine and Cosine are Periodic on Reals, we know that: :$\cos x \ge 0$ on the closed interval $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ and: :$\cos x > 0$ on the open interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ From the same discussion, we have that: :$\map \sin {x + \dfrac \pi 2} = \cos x$ So immediately we have that $\sin x \ge 0$ on the closed interval $\closedint 0 \pi$, $\sin x > 0$ on the open interval $\openint 0 \pi$. But $\map {D_x} {\cos x} = -\sin x$ from Derivative of Cosine Function. Thus from Derivative of Monotone Function, $\cos x$ is strictly decreasing on $\closedint 0 \pi$. From Derivative of Sine Function it follows that: :$\map {D_{xx} } {\cos x} = -\cos x$ On $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ where $\cos x \ge 0$, therefore, $\map {D_{xx} } {\cos x} \le 0$. From Second Derivative of Concave Real Function is Non-Positive it follows that $\cos x$ is concave on $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$. The rest of the result follows similarly. === Graph of Cosine Function === {{:Shape of Cosine Function/Graph}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \left({x + \pi}\right)       | r = \frac {\sin \left({x + \pi}\right)} {\cos \left({x + \pi}\right)}       | c = {{Defof|Real Tangent Function}} }} {{eqn | r = \frac {-\sin x} {-\cos x}       | c = Sine and Cosine are Periodic on Reals }} {{eqn | r = \tan x       | c= }} {{end-eqn}} From Derivative of Tangent Function, we have that: : $D_x \left({\tan x}\right) = \dfrac 1 {\cos^2 x}$ provided $\cos x \ne 0$. From Shape of Cosine Function, we have that $\cos > 0$ on the interval $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$. From Derivative of Monotone Function, $\tan x$ is strictly increasing on that interval, and hence can not have a period of ''less'' than $\pi$. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\tan x}       | r = \lim_{h \mathop \to 0} \frac {\map \tan {x + h} - \tan x} h       | c = {{Defof|Derivative of Real Function at Point}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h} {1 - \tan x \tan h} - \tan x} h       | c = Tangent of Sum }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h - \tan x + \tan^2 x \tan h} {1 - \tan x \tan h} } h }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\tan h + \tan^2 x \tan h} {h \paren {1 - \tan x \tan h} }  }} {{eqn | r = \lim_{h \mathop \to 0} \frac {1 + \tan^2 x} {1 - \tan x \tan h} \cdot \lim_{h \mathop \to 0} \frac {\tan h} h       | c = Product Rule for Limits of Functions }} {{eqn | r = \frac {1 + \tan^2 x} {1 - \tan x \tan 0} \cdot 1       | c = Limit of Tan X over X }} {{eqn | r = 1 + \tan^2 x       | c = Tangent of Zero }} {{eqn | r = \sec^2 x       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = \frac 1 {\cos^2 x}       | c = Secant is Reciprocal of Cosine ($\cos x \ne 0$) }} {{end-eqn}} {{qed}}
From the definition of the tangent function: :$\tan x = \dfrac {\sin x} {\cos x}$ From Derivative of Sine Function: :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ From Derivative of Cosine Function: :$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$ Then: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\tan x}       | r = \frac {\cos x \cos x - \sin x \paren {-\sin x} } {\cos^2 x}       | c = Quotient Rule for Derivatives }} {{eqn | r = \frac {\cos^2 x + \sin^2 x} {\cos^2 x}       | c =  }} {{eqn | r = \frac 1 {\cos^2 x}       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} This is valid only when $\cos x \ne 0$. The result follows from the Secant is Reciprocal of Cosine. {{qed}}
Consider the sequence $\sequence {d_n}$ defined as: :$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$ From Lemma 2 it is seen that $\sequence {d_n}$ is a decreasing sequence. From Lemma 3 it is seen that the sequence: :$\sequence {d_n - \dfrac 1 {12 n} }$ is increasing. In particular: :$\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 = \dfrac 1 {12}$ and so $\sequence {d_n}$ is bounded below. From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {d_n}$ is convergent. Let $d_n \to d$ as $n \to \infty$. From Exponential Function is Continuous, we have: :$\exp d_n \to \exp d$ as $n \to \infty$ Let $C = \exp d$. Then: :$\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$ It remains to be shown that $C = \sqrt {2 \pi}$. Let $I_n$ be defined as: :$\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$ Then from Lemma 4: :$\displaystyle \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$ So: {{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \dfrac {n!} {n^n \sqrt n e^{-n} }       | r = \sqrt {2 \pi}       | c = Lemma 5 }} {{eqn | ll= \leadsto       | l = \lim_{n \mathop \to \infty} \dfrac {n!} {\sqrt {2 \pi n} \paren {\frac n e}^n}       | r = 1 }} {{end-eqn}} Hence the result. {{qed}}
From the definition of the cotangent function: :$\cot x = \dfrac {\cos x} {\sin x}$ From Derivative of Sine Function: :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ From Derivative of Cosine Function: :$\map {\dfrac \d {\d x} } {\cos x}= -\sin x$ Then: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\cot x}       | r = \frac {\sin x \paren {-\sin x} - \cos x \cos x} {\sin^2 x}       | c = Quotient Rule for Derivatives }} {{eqn | r = \frac {-\paren {\sin^2 x + \cos^2 x} } {\sin^2 x}       | c =  }} {{eqn | r = \frac {-1} {\sin^2 x}       | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} This is valid only when $\sin x \ne 0$. The result follows from the definition of the real cosecant function. {{qed}}
From the discussion of Sine and Cosine are Periodic on Reals, we have that: : $\sin \paren {x + \dfrac \pi 2} = \cos x$ The result then follows directly from the Shape of Cosine Function. === Graph of Sine Function === {{:Shape of Sine Function/Graph}} {{qed}}
From the definition of the secant function: :$\sec x = \dfrac 1 {\cos x} = \paren {\cos x}^{-1}$ From Derivative of Cosine Function: :$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$ Then: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\sec x}       | r = \map {\dfrac \d {\d x} } {\paren {\cos x}^{-1} }       | c = Exponent Laws }} {{eqn | r = \paren {-\sin x} \paren {-\cos^{-2} x}       | c = Chain Rule for Derivatives, Power Rule }} {{eqn | r = \frac 1 {\cos x} \frac {\sin x} {\cos x}       | c = Exponent Laws }} {{eqn | r = \sec x \tan x       | c = Definitions of secant and tangent }} {{end-eqn}} This is valid only when $\cos x \ne 0$. {{qed}}
From the definition of the cosecant function: :$\csc x = \dfrac 1 {\sin x}$ From Derivative of Sine Function: :$\map {\dfrac \d {\d x} } {\sin x} = \cos x$ Then: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\csc x}       | r = \cos x \frac {-1} {\sin^2 x}       | c = Chain Rule for Derivatives }} {{eqn | r = \frac {-1} {\sin x} \frac {\cos x} {\sin x}       | c =  }} {{eqn | r = -\csc x \cot x       | c = {{Defof|Real Cosecant Function}} and {{Defof|Real Cotangent Function}} }} {{end-eqn}} This is valid only when $\sin x \ne 0$. {{qed}}
* $\tan x$ is continuous and strictly increasing on $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$: Continuity follows from the Quotient Rule for Continuous Functions: :$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$ from Real Sine Function is Continuous and Cosine Function is Continuous :$(2): \quad$ $\cos x > 0$ on this interval. The fact of $\tan x$ being strictly increasing on this interval has been demonstrated in the discussion on Tangent Function is Periodic on Reals. * $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$: From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 \,.\,.\, \dfrac \pi 2}\right)$. We have that: : $(1): \quad \cos x \to 0$ as $x \to \dfrac \pi 2 ^-$ : $(2): \quad \sin x \to 1$ as $x \to \dfrac \pi 2 ^-$ Thus it follows that: * $\tan x = \dfrac {\sin x} {\cos x} \to + \infty$ as $x \to \dfrac \pi 2 ^-$. * $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$: From Sine and Cosine are Periodic on Reals, we have that $\sin x < 0$ and $\cos x > 0$ on $\left({-\dfrac \pi 2 \,.\,.\, 0}\right)$. We have that: : $(1): \quad \cos x \to 0$ as $x \to -\dfrac \pi 2 ^+$ : $(2): \quad \sin x \to -1$ as $x \to -\dfrac \pi 2 ^+$ Thus it follows that $\tan x = \dfrac {\sin x} {\cos x} \to - \infty$ as $x \to -\dfrac \pi 2 ^+$. * $\tan x$ is not defined and discontinuous at $x = \left({n + \dfrac 1 2}\right) \pi$: From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi \implies \cos x = 0$. As division by zero is not defined, it follows that at these points $\tan x$ is not defined either. Now, from the above, we have: : $(1): \quad \tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$ : $(2): \quad \tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$ As $\tan \left({x + \pi}\right) = \tan x$ from Tangent Function is Periodic on Reals, it follows that $\tan x \to - \infty$ as $x \to \dfrac \pi 2 ^+$. Hence the left hand limit and right hand limit at $x = \dfrac \pi 2$ are not the same. From the periodic nature of $\tan x$, it follows that the same applies $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi$. The fact of its discontinuity at these points follows from the definition of discontinuity. * $\tan \left({n \pi}\right) = 0$: Follows directly from Sine and Cosine are Periodic on Reals: $\forall n \in \Z: \sin \left({n \pi}\right) = 0$. === Graph of Tangent Function === {{:Shape of Tangent Function/Graph}} {{qed}}
Let $y = \arcsin x$ where $-1 < x < 1$. Then: :$x = \sin y$ Then from Derivative of Sine Function: :$\dfrac {\d x} {\d y} = \cos y$ Hence from Derivative of Inverse Function: :$\dfrac {\d y} {\d x} = \dfrac 1 {\cos y}$ From Sum of Squares of Sine and Cosine, we have: :$\cos^2 y + \sin^2 y = 1 \implies \cos y = \pm \sqrt {1 - \sin^2 y}$ Now $\cos y \ge 0$ on the image of $\arcsin x$, that is: :$y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ Thus it follows that we need to take the positive root of $\sqrt {1 - \sin^2 y}$. So: {{begin-eqn}} {{eqn | l = \frac {\d y} {\d x}       | r = \frac 1 {\sqrt {1 - \sin^2 y} }       | c =  }} {{eqn | r = \frac 1 {\sqrt {1 - x^2} }       | c =  }} {{end-eqn}} {{qed}}
Let $y = \arccos x$ where $-1 < x < 1$. Then: :$x = \cos y$ Then from Derivative of Cosine Function: :$\dfrac {\d x} {\d y} = -\sin y$ Hence from Derivative of Inverse Function: :$\dfrac {\d y} {\d x} = \dfrac {-1} {\sin y}$ From Sum of Squares of Sine and Cosine, we have: :$\cos^2 y + \sin^2 y = 1 \implies \sin y = \pm \sqrt {1 - \cos^2 y}$ Now $\sin y \ge 0$ on the range of $\arccos x$, that is, for $y \in \closedint 0 \pi$. Thus it follows that we need to take the positive root of $\sqrt {1 - \cos^2 y}$. So: :$\dfrac {\d y} {\d x} = \dfrac {-1} {\sqrt {1 - \cos^2 y} }$ and hence the result. {{qed}}
Let $y \in \R$ such that: : $\exists x \in \left[{-1 \,.\,.\, 1}\right]: x = \cos \left({y + \dfrac \pi 2}\right)$ Then: {{begin-eqn}} {{eqn | l = x       | r = \cos \left({y + \frac \pi 2}\right)       | c =  }} {{eqn | r = -\sin y       | c = Cosine of Angle plus Right Angle }} {{eqn | r = \sin \left({-y}\right)       | c = Sine Function is Odd }} {{end-eqn}} Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$. Then we can write $-y = \arcsin x$. But then $\cos \left({y + \dfrac \pi 2}\right) = x$. Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$. Hence $y + \dfrac \pi 2 = \arccos x$. That is, $\dfrac \pi 2 = \arccos x + \arcsin x$. {{qed}}
{{begin-eqn}} {{eqn	| l = \frac {\map \d {\map \arctan x} } {\d x}  	| r = \lim_{h \mathop \to 0} \frac {\map \arctan {x + h} - \map \arctan x} h  	| c = {{Defof|Derivative of Real Function at Point}} }} {{eqn	| r = \lim_{h \mathop \to 0} \frac {\arctan {x + h} + \map \arctan {-x} } h	 	| c = Arctangent Function is Odd }} {{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac {x + h - x} {1 + x \paren {x + h} } } 	| c = Sum of Arctangents }} {{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac h {1 + x^2 + h x} } }} {{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \paren {\frac h {1 + x^2 + h x} - \frac 1 3 \paren {\frac h {1 + x^2 + h x} }^3 + \frac 1 5 \paren {\frac h {1 + x^2 + h x} }^5 + \map \OO {h^7} } 	| c = {{Defof|Real Arctangent}} }} {{eqn	| r = \lim_{h \mathop \to 0} \paren {\frac 1 {1 + x^2 + h x} - \frac {h^2} {3 \paren {1 + x^2 + h x}^3} + \frac {h^4} {5 \paren {1 + x^2 + h x}^5} + \map \OO {h^6} } }} {{eqn	| r = \frac 1 {1 + x^2 + 0 x} - \frac {0^2} {3 \paren {1 + x^2 + 0 x}^3} + \frac {0^4} {5 \paren {1 + x^2 + 0 x}^5} }} {{eqn	| r = \frac 1 {1 + x^2}  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \arctan x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \tan y       | c = {{Defof|Real Arctangent}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = \sec^2 y       | c = Derivative of Tangent Function }} {{eqn | r = 1 + \tan^2 y       | c = Difference of Squares of Secant and Tangent }} {{eqn | r = 1 + x^2       | c = Definition of $x$ }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac 1 {1 + x^2}       | c = Derivative of Inverse Function }} {{end-eqn}} {{qed}}
Since: :$\tuple {x_1, 0} + \tuple {x_2, 0} = \tuple {x_1 + x_2, 0}$ :$\tuple {x_1, 0} \tuple {x_2, 0} = \tuple {x_1 x_2, 0}$ we can identify a complex number (definition 2) $\tuple {x_1, 0}$ with the real number $x_1$. Specifically, we can define an isomorphism between the set of complex numbers (definition 2) of the form $\tuple {x, 0}$ and the field of real numbers. Now, we define $i = \tuple {0, 1}$. Then: {{begin-eqn}} {{eqn | l = x + i y       | r = \tuple {x, 0} + \tuple {0, 1} \tuple {y, 0}       | c =  }} {{eqn | r = \tuple {x, y}       | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}} and {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{end-eqn}} Finally, we see that: {{begin-eqn}} {{eqn | l = i^2       | r = \tuple {0, 1} \tuple {0, 1}       | c =  }} {{eqn | r = \tuple {0 \cdot 0 - 1 \cdot 1, 0 \cdot 1 + 1 \cdot 0}       | c =  }} {{eqn | r = \tuple {-1, 0}       | c =  }} {{eqn | r = -1       | c =  }} {{end-eqn}} Thus we can say that $i = \sqrt {-1}$. {{qed}}
From Complex Numbers under Addition form Abelian Group, we have that $\struct {\C, +}$ forms an abelian group. From Non-Zero Complex Numbers under Multiplication form Abelian Group, we have that $\struct {\C_{\ne 0}, \times}$ forms an abelian group. Finally, we have that Complex Multiplication Distributes over Addition. Thus all the criteria are fulfilled, and $\struct {\C, +, \times}$ is a field. {{qed}}
From the definition of complex numbers, we define the following: {{begin-eqn}} {{eqn | l = z       | o = :=       | r = \tuple {x_1, y_1} }} {{eqn | l = w       | o = :=       | r = \tuple {x_2, y_2} }} {{end-eqn}} where $x_1, x_2, y_1, y_2 \in \R$. Then: {{begin-eqn}} {{eqn | l = z_1 z_2       | r = \tuple {x_1, y_1} \tuple {x_2, y_2}       | c = {{Defof|Complex Number|index = 2}} }} {{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = \tuple {x_2 x_1 - y_2 y_1, x_1 y_2 + x_2 y_1}       | c = Real Multiplication is Commutative }} {{eqn | r = \tuple {x_2 x_1 - y_2 y_1, x_2 y_1 + x_1 y_2}       | c = Real Addition is Commutative }} {{eqn | r = \tuple {x_2, y_2} \tuple {x_1, y_1}       | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}} }} {{eqn | r = z_2 z_1       | c = {{Defof|Complex Number|index = 2}} }} {{end-eqn}} {{qed}}