hcju/proofwikiqa huggingface dataset:
First we see that $h$ is well-defined, because on $\closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2}$ we have $\map f 1 = b = \map g 0$. Now $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } } \mathop = f \circ k$ where $k: \closedint 0 {\dfrac 1 2} \to \closedint 0 1$ is given by $\map k x = 2 x$. So by Continuity of Composite Mapping, $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } }$ is continuous. Similarly, $\mathbin h {\restriction_{\closedint {\frac 1 2} 1} }$ is continuous. By Continuity from Union of Restrictions, it follows that $h$ is continuous. Finally, $\map h 0 = \map f 0 = a$ and $\map h 1 = \map g 1 = c$. {{qed}}
:250px Let $\triangle ABC$ and $DEF$ be two triangles in which $AB = DE$, $AC = DF$, and $\angle CAB > \angle FDE$. Construct $\angle EDG$ on $DE$ at point $D$. Place $G$ so that $DG = AC$. Join $EG$ and $FG$. Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, by Triangle Side-Angle-Side Equality: :$BC = GE$ By Euclid's first common notion: :$DG = AC = DF$ Thus by Isosceles Triangle has Two Equal Angles: :$\angle DGF = \angle DFG$ So by Euclid's fifth common notion: :$\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$ Since $\angle EFG > \angle EGF$, by Greater Angle of Triangle Subtended by Greater Side: :$EG > EF$ Therefore, because $EG = BC$, $BC > EF$. {{qed}}
Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$ Let $C = \bigcup \CC_x$ === Lemma === {{:Equivalence of Definitions of 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 Connected Sets === It needs to be shown that $C = C'$. {{:Equivalence of Definitions of Component/Equivalence Class equals Union of Connected Sets}}{{qed|lemma}} === Union of Connected Sets is Maximal Connected Set === {{:Equivalence of Definitions of Component/Union of Connected Sets is Maximal Connected Set}}{{qed|lemma}} === Maximal Connected Set is Union of Connected Sets === {{:Equivalence of Definitions of Component/Maximal Connected Set is Union of Connected Sets}}{{qed}}
:300px Let $\triangle ABC$ and $\triangle DEF$ be two triangles in which: :$AB = DF$ :$AC = DE$ :$BC > EF$ {{AimForCont}} that $\angle BAC \not > \angle EDF$. Then either: :$\angle BAC = \angle EDF$ or: :$\angle BAC < \angle EDF$ Let $\angle BAC = \angle EDF$. Then by Triangle Side-Angle-Side Equality: :$BC = EF$ But we know this is not the case, so by Proof by Contradiction: :$\angle BAC \ne \angle EDF$ Suppose $\angle BAC < \angle EDF$. Then by Greater Angle of Triangle Subtended by Greater Side: :$EF > BC$ But we know this is not the case, so by Proof by Contradiction: :$\angle BAC \not < \angle EDF$ Thus: :$\angle BAC > \angle EDF$ {{qed}}
Follows directly from Closure of Subset of Metric Space by Convergent Sequence. {{qed}} Category:Metric Subspaces Category:Sequentially Compact Spaces mqj2iylshn2871w02l98raps0bmr2uw
Let $\rho'$ be a conformal metric on $Y$ in the sense of the definition of extremal length, given in local coordinates as: :$\map {\rho'} z \size {\d z}$ Let $\rho$ be the metric on $X$ obtained as the pull-back of this metric under $\phi$. That is, $\rho$ is given in local coordinates as: :$\map {\rho'} {\map \phi w} \cdot \size {\map {\dfrac {\d \phi} {\d w} } w} \cdot \size {\d w}$ Then the area of $X$ with respect to $\rho$ and the area of $Y$ with respect to $\rho'$ are equal by definition: :$\map A {\rho'} = \map A \rho$ Furthermore, if $\gamma \in \Gamma$ and $\gamma' := \map \phi \gamma$, then also: :$\map L {\gamma, \rho} = \map L {\gamma', \rho'}$ and hence: :$\map L {\Gamma, \rho} = \map L {\Gamma', \rho'}$ In summary, for any metric $\rho'$ on $Y$, there is a metric $\rho$ on $X$ such that: :$\dfrac {\map L {\Gamma, \rho} } {\map A \rho} = \dfrac {\map L {\Gamma', \rho'} } {\map A {\rho'} }$ It thus follows from the definition of extremal length that: :$\map \lambda \Gamma \ge \map \lambda {\Gamma'}$ The opposite inequality follows by exchanging the roles of $X$ and $Y$. {{qed}}
From Real Number Line is Metric Space, the distance function defined as $\map d {x, y} = \size {x - y}$ is a metric on $\R$. It remains to be shown that the metric space $\struct {\R, d}$ is complete. By definition, this is done by demonstrating that every Cauchy sequence of real numbers has a limit. This is demonstrated in Cauchy Sequence Converges on Real Number Line. Hence the result. {{qed}}
Let $\rho_1 = \rho_1 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ and $\rho_2 = \rho_2 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ be conformal metrics as in the definition of extremal length. It can be assumed that these are normalized: :$A \left({\rho_j}\right) = L \left({\Gamma_j, \rho_j}\right)$ for $j \in \left\{ {1, 2}\right\}$.  We define another metric $\rho = \rho \left({z}\right) \left\vert{\mathrm d z}\right\vert$ by: :$\rho \left({z}\right) := \max \left({\rho_1 \left({z}\right), \rho_2 \left({z}\right)}\right)$ Note that this is a well-defined metric. By definition, the area form $\rho^2 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ satisfies: :$\rho^2 \left({z}\right) \left\vert{\mathrm d z}\right\vert^2 = \max \left({\rho_1 \left({z}\right)^2, \rho_2 \left({z}\right)^2}\right) \left\vert{\mathrm d z}\right\vert^2 \le \left({\rho_1 \left({z}\right)^2 + \rho_2 \left({z}\right)^2}\right) \left\vert{\mathrm d z}\right\vert^2$ Hence: :$A \left({\rho}\right) \le A \left({\rho_1}\right) + A \left({\rho_2}\right) = L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$ On the other hand, let $\gamma \in \Gamma$. Let $\gamma_1$, $\gamma_2$ be as in the assumption. Then: {{begin-eqn}} {{eqn | l = L \left({\gamma, \rho}\right)       | o = \ge       | r = L \left({\gamma_1, \rho}\right) + L  \left({\gamma_2, \rho}\right)       | c = as $\gamma_1$ and $\gamma_2$ are disjoint }} {{eqn | o = \ge       | r = L \left({\gamma_1, \rho_1}\right) + L \left({\gamma_2, \rho_2}\right)       | c = Definition of $\rho$ }} {{eqn | o = \ge       | r = L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)       | c = Definition of $L \left({\Gamma_j, \rho_j}\right)$ }} {{end-eqn}} Thus  :$ L \left({\Gamma, \rho}\right) \ge L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$ Combining this with the inequality for the area: :$\dfrac {L \left({\Gamma, \rho}\right)^2} {A \left({\rho}\right)} \ge \dfrac {\left({L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)}\right)^2} {L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)} = L  \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$ Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above: :$ L \left({\Gamma}\right) \ge L \left({\Gamma_1}\right) + L \left({\Gamma_2}\right)$ as claimed. {{qed}}
Set $\Gamma := \Gamma_1\cup \Gamma_2$. Let $\rho_1$ and $\rho_2$ be conformal metrics as in the definition of extremal length, normalized such that: :$ L \left({\Gamma_1, \rho_1}\right) = L \left({\Gamma_2, \rho_2}\right) = 1$ We define a new metric by: : $\rho := \max \left({\rho_1, \rho_2}\right)$. {{explain|Prove that $\rho$ is a metric}} Then: : $L \left({\Gamma, \rho}\right) \ge 1$ and: : $A \left({\rho}\right) \le A \left({\rho_1}\right) + A \left({\rho_2}\right)$ {{explain|What is $A$?}} Hence: {{begin-eqn}} {{eqn | l=\frac 1 {\lambda \left({\Gamma}\right)}       | o=\le       | r=\frac {A \left({\rho}\right)} {L \left({\Gamma, \rho}\right)}       | c= }} {{eqn | o=\le       | r=A \left({\rho}\right)       | c= }} {{eqn | o=\le       | r=A \left({\rho_1}\right) + A \left({\rho_2}\right)       | c= }} {{eqn | r=\frac 1 {L \left({\Gamma_1, \rho_1}\right)} + \frac 1 {L \left({\Gamma_2, \rho_2}\right)}       | c= }} {{end-eqn}} Taking the infimum over all metrics $\rho_1$ and $\rho_2$, the claim follows. Now suppose that the disjointness assumption holds, and let $\rho$ again be a Borel-measurable conformal metric, normalized such that $L \left({\Gamma, \rho}\right)= 1$.  We can define $\rho_1$ to be the restriction of $\rho$ to $A_1$, and likewise $\rho_2$ to be the restriction of $\rho$ to $A_2$. By this we mean that, in local coordinates, $\rho_j$ is given by :$ \rho_j \left({z}\right) \ \left|{\mathrm d z}\right| = \begin{cases} \rho \left({z}\right) \ \left|{\mathrm d z}\right| & : z \in A_j \\ 0 \ \left|{\mathrm d z}\right| & : \text{otherwise} \end{cases}$ {{explain|The above section from "By this we mean" needs considerably more explanation, as none of the concepts introduced here can be understood without reference to links from elsewhere.}} Then: : $A \left({\rho}\right) = A \left({\rho_1}\right) + A \left({\rho_2}\right)$ and: : $L \left({\Gamma_1, \rho_1}\right), L \left({\Gamma_2, \rho_2}\right) \ge 1$ {{explain|How do these two statements follow from what went before?}} Hence: {{begin-eqn}} {{eqn | l=A \left({\rho}\right)       | r=A \left({\rho_1}\right) + A \left({\rho_2}\right)       | c= }} {{eqn | o=\ge       | r=\frac {A \left({\rho_1}\right)} {L \left({\Gamma_1, \rho}\right)} + \frac {A \left({\rho_2}\right)} {L \left({\Gamma_2, \rho}\right)}       | c= }} {{eqn | o=\ge       | r=\frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}       | c= }} {{end-eqn}} Taking the infimum over all metrics $\rho$, we see that: :$\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$ Together with the first part of the Proposition, this proves the claim. {{qed}} Category:Geometric Function Theory 4c236pxipdbjkcuxt6h6zrtidkaqptr
The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$. Hence: {{begin-eqn }} {{eqn | l =\frac 1 {\lambda \left({\Gamma}\right)}       | o = \ge       | r = \frac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)}       | c = Comparison Principle for Extremal Length }} {{eqn | r = \frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}       | c = Extremal Length of Union }} {{end-eqn}} {{qed}} {{MissingLinks|Some results and definitions need to be linked.}}
Let $M = \struct {X, d}$ be a metric space. By the triangle inequality, we have: :$\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$ By subtracting $\map d {y, z}$ from both sides: :$\map d {x, y} \ge \map d {x, z} - \map d {y, z}$ Now we consider 2 cases.  ;Case $1$: Suppose $\map d {x, z} - \map d {y, z} \ge 0$. Then: :$\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$ and so: :$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$ ;Case 2: Suppose $\map d {x, z} - \map d {y, z} < 0$. Applying the triangle inequality again, we have: :$\forall x, y, z \in X: \map d {y, x} + \map d {x, z} \ge \map d {y, z}$ Hence: :$\map d {x, y} \ge \map d {y, z} - \map d {x, z}$ Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that: :$\map d {y, z} - \map d {x, z} > 0$ and so: :$\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$ Thus we obtain: :$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$ Since these cases are exhaustive, we have shown that: :$\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$ {{qed}} Category:Triangle Inequality Category:Named Theorems Category:Inequalities qymhy24nk2xkx1f0gpt8zzppyct76rb
From the definition of an integral domain: :All elements of $D^* = D \setminus \set {0_D}$ are cancellable :$\struct {D^*, \circ}$ is a commutative semigroup. So by the Inverse Completion Theorem, there exists an inverse completion of $\struct {D, \circ}$. From Construction of Inverse Completion, this is done as follows: Let $\ominus$ be the congruence relation defined on $D \times D^*$ by: :$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$ The fact that this is a congruence relation is proved in Equivalence Relation on Semigroup Product with Cancellable Elements. Let $\struct {D \times D^*, \otimes}$ be the external direct product of $\struct {D, \circ}$ with $\struct {D^*, \circ}$, where $\otimes$ is the operation on $D \times D^*$ induced by $\circ$. Let the quotient structure defined by $\ominus$ be $\struct {\dfrac {D \times D^*} \ominus, \otimes_\ominus}$. where $\otimes_\ominus$ is the operation induced on $\dfrac {D \times D^*} \ominus$ by $\otimes$. Let us use $D'$ to denote the quotient set $\dfrac {D \times D^*} {\ominus}$. Let us use $\circ'$ to denote the operation $\otimes_\ominus$. Thus $\struct {D', \circ'}$ is the inverse completion of $\struct {D, \circ}$. An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$. As the Inverse Completion is Unique up to isomorphism, it follows that we can ''define'' the structure $\struct {K, \circ}$ which is isomorphic to $\struct {D', \circ'}$. An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$. So an element of $K$ is the isomorphic image of an element $\eqclass {\paren {x, y} } \ominus$ of $\dfrac {D \times D^*} \ominus$. Hence every element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$. Alternatively, from the definition of division product, of the form $\dfrac x y$. Hence we can therefore interpret any element of $\struct {K, \circ}$ as equivalence classes of elements of the form $\dfrac x y$. {{qed}} Category:Integral Domains Category:Inverse Completions l7kiwb0myfc07a0q01ixcexn3b2jvxf
Let $\left({K, \circ}\right)$ be the inverse completion of $\left({D, \circ}\right)$. We define $\left({K, \circ}\right)$ of $\left({D, \circ}\right)$ by Inverse Completion of Integral Domain Exists. The structure of $\left({K, \circ}\right)$ is such that element of $\left({K, \circ}\right)$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$. From Zero of Inverse Completion of Integral Domain, $0_K$ is all elements of $D \times D^*$ of the form $\dfrac {0_D} x$. Therefore the elements of $K^*$ are those of the form $\dfrac x \circ y^{-1}$ where $x, y \in D^*$. By Product of Division Products, $\displaystyle \frac a b \circ \frac c d = \frac {a \circ c} {b \circ d}$. As $\left({D, +, \circ}\right)$ is an integral domain, none of its non-zero elements are zero divisors. Therefore $\forall x, y \in D^*: x \circ y \ne 0_D$. So: : $\displaystyle \forall \frac a b, \frac c d \in K^*: \frac {a \circ c} {b \circ d} \in K^*$ Hence the result. {{qed}} Category:Integral Domains Category:Inverse Completions 6ivccqlw1ptq3qv4ftk49lrdkgm451r
{{expand|A commutative diagram would be nice. Anyone out there good at xyplot?}} From the definition, we know that a mapping is a relation. First, note that from the definition of composition of relations, the following must be the case before the above expression is even to be defined: :$(1): \quad \Dom {f_2} = \Cdm {f_1}$ :$(2): \quad \Dom {f_3} = \Cdm {f_2}$ where $\Cdm f$ denotes the codomain of the mapping $f$. The two '''composite relations''' can be seen to have the same domain, as follows: {{begin-eqn}} {{eqn | l = \Dom {\paren {f_3 \circ f_2} \circ f_1}       | r = \Dom {f_1}       | c = Domain of Composite Relation }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \Dom {f_3 \circ \paren {f_2 \circ f_1} }       | r = \Dom {f_2 \circ f_1}       | c = Domain of Composite Relation }} {{eqn | r = \Dom {f_1}       | c = Domain of Composite Relation }} {{end-eqn}} Also they have the same codomain, as is seen by: {{begin-eqn}} {{eqn | l = \Cdm {\paren {f_3 \circ f_2} \circ f_1}       | r = \Cdm {f_3 \circ f_2}       | c = Codomain of Composite Relation }} {{eqn | r = \Cdm {f_3}       | c = Codomain of Composite Relation }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \Cdm {f_3 \circ \paren {f_2 \circ f_1} }       | r = \Cdm {f_3}       | c = Codomain of Composite Relation }} {{end-eqn}} As a mapping is a relation, we can use that the Composition of Relations is Associative: :$\forall x \in \Dom {f_1}: \map {\paren {f_3 \circ f_2} \circ f_1} x = \map {f_3 \circ \paren {f_2 \circ f_1} } x$ Hence the result. {{qed}}
Let us suppress the operation of $G$ for brevity. === $(2) \implies (1)$ === Suppose that $G$ has the cross cancellation property. Then, for all $x, y \in G$: {{begin-eqn}} {{eqn | l=y \left({x y}\right)       | r=\left({y x}\right) y       | c=Associativity }} {{eqn | ll=\implies       | l=x y       | r=y x       | c=Cross Cancellation Property }} {{end-eqn}} Thus, $G$ is abelian. {{qed|lemma}} === $(1) \implies (2)$ === Conversely, suppose $G$ is abelian. Let $a, b, c \in G$ be such that $a b = c a$. Since $G$ is abelian, $c a = a c$. We conclude that: :$a b = c a = a c$ Thus, by left cancellation, $b = c$. {{qed}} Category:Abelian Groups taj0pft4of6zsk03msxwfbb1l3egaff
Let us suppress the operation of $G$ for brevity. === $(2) \implies (1)$ === Suppose that $G$ satisfies the middle cancellation property. Then, for all $g, h \in G$: {{begin-eqn}} {{eqn | l=e h       | r=h e       | c=Definition of identity }} {{eqn | ll=\implies       | l=g g^{-1} h       | r=hg^{-1}g       | c=Definition of inverse }} {{eqn | ll=\implies       | l=g h       | r=h g       | c=Definition of middle cancellation property }} {{end-eqn}} Thus $G$ is abelian. {{qed|lemma}} === $(1) \implies (2)$ === Conversely, suppose $G$ is abelian. Then, for all $a, b, c, d, x \in G$: {{begin-eqn}} {{eqn | l=a x b       | r=c x d }} {{eqn | ll=\implies       | l=a b x       | r=c d x       | c=Definition of Abelian Group }} {{eqn | ll=\implies       | l=a b       | r=c d       | c=Right Cancellation }} {{end-eqn}} Thus the middle cancellation property holds in $G$. {{qed}} Category:Abelian Groups tj0sq9li4dt91qjulmy5oyraictdbln
Recall that by Integers form Ordered Integral Domain, $\struct {\Z, +, \times, \le}$ is an ordered integral domain By Rational Numbers form Field, $\struct {\Q, +, \times}$ is a field. In the formal definition of rational numbers, $\struct {\Q, +, \times}$ is the field of quotients of $\struct {\Z, +, \times, \le}$ By Total Ordering on Field of Quotients is Unique, it follows that $\struct {\Q, +, \times}$ has a unique total ordering on it that is compatible with its ring structure. {{explain|Review the ordering / total ordering question}} Thus $\struct {\Q, +, \times, \le}$ is an ordered field. {{Qed}}
From the definition of rational numbers, we can express $a$ and $b$ as $a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$. Thus from Mediant is Between: :$\dfrac {p_1} {q_1} < \dfrac {p_1 + p_2} {q_1 + q_2} < \dfrac {p_2} {q_2}$ From Rational Numbers form Field: :$\dfrac {p_1 + p_2} {q_1 + q_2} \in \Q$ Hence $c = \dfrac {p_1 + p_2} {q_1 + q_2}$ is an element of $\Q$ between $a$ and $b$. {{qed}} Category:Analysis cvubwxgv3tvpgfvlnsrbxlucyikct61
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real addition, $x + y$ is defined as: :$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$ We have that $\forall i \in \N: x_i \in \Q, y_i \in \Q$, therefore $x_i + y_i \in \Q$. So it follows that $\eqclass {\sequence {x_n + y_n} } {} \in \R$. {{qed}}
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers: :$\eqclass {\sequence {x_n} } {} \equiv \eqclass {\sequence {y_n} } {} \iff \forall \epsilon > 0: \exists n \in \N: \forall i, j > n: \size {x_i - y_j} < \epsilon$ Let $x = \eqclass {\sequence {x_n} } {}$ and $y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real addition, $x + y$ is defined as: :$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$. We need to show that: :$\sequence {x_n}, \sequence {x'_n} \in \eqclass {\sequence {x_n} } {}, \sequence {y_n}, \sequence {y'_n} \in \eqclass {\sequence {y_n} } {} \implies \sequence {x_n + y_n} = \sequence {x'_n + y'_n}$ That is: :$\forall \epsilon > 0: \exists N: \forall i, j > N: \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} } < \epsilon$ Let $\epsilon > 0$, such that: :$\exists N_1: \forall i, j \ge N_1: \size {x_i - x'_j} < \epsilon / 2$ :$\exists N_2: \forall i, j \ge N_2: \size {y_i - y'_j} < \epsilon / 2$ Now let $N = \max \set {N_1, N_2}$. Then we have: :$\forall i, j \ge N: \size {x_i - x'_j} + \size {y_i - y'_j} < \epsilon$ But: {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \size {x_i - x'_j} + \size {y_i - y'_j}       | c =  }} {{eqn | o = >       | r = \size {\paren {x_i - x'_j} + \paren {y_i - y'_j} }       | c = Triangle Inequality }} {{eqn | r = \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} }       | c =  }} {{end-eqn}} So: :$\forall i, j \ge N: \size {\paren {x_i + y_i} - \paren {x'_j + y'_j} } < \epsilon$ Hence the result. {{qed}} Category:Real Addition 8qgf4h4p5nwy6mzoo9i5n9beoijf0e7
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as $\eqclass {\sequence {x_n} } {} \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$. We need to show that: :$\sequence {x_n}, \sequence {x'_n} \in \eqclass {\sequence {x_n} } {}, \sequence {y_n}, \sequence {y'_n} \in \eqclass {\sequence {y_n} } {} \implies \sequence {x_n \times y_n} = \sequence {x'_n \times y'_n}$ That is: :$\forall \epsilon > 0: \exists N: \forall i, j > N: \size {\paren {x_i \times y_i} - \paren {x'_j \times y'_j} } < \epsilon$ As $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are Cauchy sequences, from Cauchy Sequence in Metric Space is Bounded they are bounded. Let $B_x = 2 \map \sup {\sequence {x_n} }$ and $B_y = 2 \map \sup {\sequence {y_n} }$. Let $B = \max \set {B_x, B_y}$. Now let $\epsilon > 0$. Then: :$\exists N_1: \forall i, j > N_1: \size B \size {x_i - x'_j} < \epsilon / 2$ :$\exists N_2: \forall i, j > N_2: \size B \size {y_i - y'_j} < \epsilon / 2$ Now let $N = \max \set {N_1, N_2}$. Then we have: :$\forall i, j \ge N: \size B \size {x_i - x'_j} + \size B \size {y_i - y'_j} < \epsilon$ So: {{begin-eqn}} {{eqn | l = \epsilon       | o = >       | r = \size B \size {x_i - x'_j} + \size B \size {y_i - y'_j}       | c =  }} {{eqn | o = \ge       | r = \size {B \paren {x_i - x'_j} + B \paren {y_i - y'_j} }       | c = Triangle Inequality }} {{eqn | o = \ge       | r = \size {\paren {x_i - x'_j} \paren {y_i + y'_j} + \paren {y_i - y'_j} \paren {x_i + x'_j} }       | c =  }} {{eqn | r = \size {x_i y_i - x'_j y_i + x_i y'_j - x'_j y'_j + x_i y_i + x'_j y_i - x_i y'_j - x'_j y'_j}       | c =  }} {{eqn | r = \size {2 x_i y_i - 2 x'_j y'_j}       | c =  }} {{eqn | r = \size {x_i y_i - x'_j y'_j}       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Real Multiplication dcv2j6vq1ced8nmyqcmdsk1od2poq9d
From the definition, the real numbers are the set of all equivalence classes $\eqclass {\sequence {x_n} } {}$ of Cauchy sequences of rational numbers. Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such equivalence classes. From the definition of real multiplication, $x \times y$ is defined as: :$\eqclass {\sequence {x_n} } {} \times \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n \times y_n} } {}$ We have that: :$\forall i \in \N: x_i \in \Q, y_i \in \Q$ therefore $x_i \times y_i \in \Q$. So it follows that: :$\eqclass {\sequence {x_n \times y_n} } {} \in \R$ {{qed}}