hcju/proofwikiqa huggingface dataset:
The ovals of Cassini can be defined by a Cartesian equation as follows: {{:Equation of Ovals of Cassini/Cartesian Form}} The lemniscate of Bernoulli can be defined by a Cartesian equation as follows: {{:Definition:Lemniscate of Bernoulli/Cartesian Definition}} Setting $b = a$: {{begin-eqn}} {{eqn | l = \paren {x^2 + y^2 + a^2}^2 - 4 a^2 x^2       | r = a^4       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2 + 2 \paren {x^2 + y^2} a^2 + a^4 - 4 a^2 x^2       | r = a^4       | c =  }} {{eqn | ll= \leadsto       | l = \paren {x^2 + y^2}^2       | r = 4 a^2 x^2 - 2 a^2 x^2 - 2 a^2 y^2       | c = simplifying and rearranging }} {{eqn | r = 2 a^2 \paren {x^2 - y^2}       | c = further simplification }} {{end-eqn}} It follows that the lemniscate of Bernoulli is an oval of Cassini where $b = a$. {{qed}}
{{tidy|}} Let $C$ be the cofactor matrix of $A$.  Then by definition of adjugate matrix, $C^T$ is the adjugate matrix of $A$. Therefore,  {{begin-eqn}} {{eqn | l = A \cdot C^T        | r = \det{A} \cdot I_n       | c = Matrix Product with Adjugate Matrix. }} {{end-eqn}} Because $A$ is invertible, $A^{-1}$ exists. Because $A$ is invertible, $1/\det{A}$ exists by Matrix is Invertible iff Determinant has Multiplicative Inverse. Therefore: :$A^{-1} = \frac{1}{\det{A}}\cdot C^T.$ Since $A \mathbf x = \mathbf b$ by hypothesis,  :$\mathbf x = A^{-1} \mathbf b.$ Therefore, :$\mathbf x = \left(\frac{1}{\det{A}}\cdot C^T\right) \mathbf b.$ For each $r,s \in \{1, \dots, n\}$, let $A_{rs}$ denote the element of $C$ whose index is $(r,s)$.  Then by the definition of transpose, :$C^T = \begin{bmatrix} A_{11} & A_{21} & \dots & A_{n1} \\ A_{12} & A_{22} & \dots & A_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ A_{1n} & A_{2n} & \dots & A_{nn} \end{bmatrix}.$ By the definition of cofactor matrix, $A_{rs}$ is the cofactor of the element of $A$ whose index is $(r,s)$. Let $i \in \{1, \dots, n\}$.  Then by definition of matrix product, :$x_i = \frac{1}{\det{A}}\left(\sum_{j = 1}^n A_{ji} b_j\right).$ Recall that $A_i$ is the matrix obtained from $A$ by replacing the $i$-th column with $\mathbf{b}$.  Then if :$A = \begin{bmatrix} a_{1,1} & a_{1,2} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,n} \end{bmatrix}$ then :$A_i = \begin{bmatrix} a_{1,1} & a_{1,2} & \dots & a_{1, i-1} & b_1 & a_{1, i+1} & \dots & a_{1,n} \\ a_{2,1} & a_{2,2} & \dots & a_{2, i-1} & b_2 & a_{2, i+1} & \dots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n, i-1} & b_n & a_{n, i+1} & \dots & a_{n,n} \end{bmatrix}.$ For all $j \in \{1, \dots, n\}$ the matrix obtained by deleting the $j$-th row and the $i$-th column of $A$ is equal to the matrix obtained by deleting the $j$-th row and $i$-th column of $A_i$.  Therefore, by the definition of a cofactor, the cofactor of $(A_i)_{j,i}$ is equal to $A_{ji}$. By the Expansion Theorem for Determinants, we now have :$\det{A_i} = \sum_{j=1}^n A_{ji} b_j.$ Therefore, :$x_i = \frac{\det{A_i}}{\det{A}}$ as desired. {{Namedfor|Gabriel Cramer|cat = Cramer}}
Let $C$ be a circle of diameter $a$ whose circumference passes through the origin $O$. Let the diameter of $C$ which passes through $O$ lie on the polar axis. Let $OQ$ be a chord of $C$. Let $b$ be a real constant. :400px Let $P = \polar {r, \theta}$ denote an arbitrary point on a limaçon of Pascal $L$. We have that: {{begin-eqn}} {{eqn | l = OP       | r = OQ \pm QP       | c = {{Defof|Limaçon of Pascal}} }} {{eqn | ll= \leadsto       | l = r       | r = a \cos \theta \pm b       | c = {{Defof|Cosine of Angle}} }} {{end-eqn}} When $-\dfrac \pi 2 \le \theta \le \dfrac \pi 2$, we have: :$r = b + a \cos theta$ When $\dfrac \pi 2 \le \theta \le \dfrac {3 \pi} 2$, we have: {{begin-eqn}} {{eqn | l = -r       | r = -b + a \, \map \cos {\theta + \pi}       | c =  }} {{eqn | r = -b - a \cos \theta       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = b + a \cos \theta       | c =  }} {{end-eqn}} Hence the result. {{qed}}
:400px By construction: {{begin-eqn}} {{eqn | l = OS       | r = 2 a \sec \theta       | c = {{Defof|Secant Function}} }} {{eqn | l = OR       | r = 2 a \cos \theta       | c = {{Defof|Cosine}} }} {{eqn | l = OP       | r = RS       | c = {{Defof|Cissoid of Diocles}} }} {{eqn | ll= \leadsto       | l = OP       | r = OS - OR       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = 2 a \paren {\sec \theta - \cos \theta}       | c =  }} {{eqn | r = 2 a \sin \theta \tan \theta       | c = Secant Minus Cosine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = r       | r = 2 a \sin \theta \tan \theta       | c = Equation of Cissoid of Diocles: Polar Form }} {{eqn | ll= \leadsto       | l = r^2       | r = 2 a r \sin \theta \tan \theta       | c =  }} {{eqn | ll= \leadsto       | l = x^2 + y^2       | r = 2 a y \dfrac y x       | c = Conversion between Cartesian and Polar Coordinates in Plane }} {{eqn | ll= \leadsto       | l = x \paren {x^2 + y^2}       | r = 2 a y^2       | c = rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = r       | r = 2 a \sin \theta \tan \theta       | c = Equation of Cissoid of Diocles: Polar Form }} {{eqn | ll= \leadsto       | l = r \cos \theta       | r = 2 a \cos \sin \theta \paren {\dfrac {\sin \theta} {\cos \theta} }       | c =  }} {{eqn | l = r \sin \theta       | r = 2 a \sin \theta \sin \theta \paren {\dfrac {\sin \theta} {\cos \theta} }       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = 2 a \sin^2 \theta       | c = Conversion between Cartesian and Polar Coordinates in Plane }} {{eqn | l = y       | r = \dfrac {2 a \sin^3 \theta} {\cos \theta}       | c =  }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn	| l = \int \frac {x^2 + 1} {x^4 + 1} \rd x 	| r = \int \frac {1 + \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x }} {{eqn	| r = \int \frac {1 + \frac 1 {x^2} } {\paren {x - \frac 1 x}^2 + 2} \rd x 	| c = Completing the Square }} {{end-eqn}} Note that, by Derivative of Power:  :$\dfrac \d {\d x} \paren {x - \dfrac 1 x} = 1 + \dfrac 1 {x^2}$ So, we have: {{begin-eqn}} {{eqn	| r = \int \frac 1 {u^2 + 2} \rd u 	| c = substituting $u = x - \dfrac 1 x$ }} {{eqn	| r = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} u} + C 	| c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn	| r = \frac 1 {\sqrt 2} \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + C }} {{end-eqn}} {{qed}} Category:Primitives involving Reciprocals 3qaf3meifuscsjq7r1xd4isj2pp0y51
We have: {{begin-eqn}} {{eqn	| l = \int \frac {x^2 - 1} {x^4 + 1} \rd x 	| r = \int \frac {1 - \frac 1 {x^2} } {x^2 + \frac 1 {x^2} } \rd x }} {{eqn	| r = \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x 	| c = Completing the Square }} {{end-eqn}} Note that, by Derivative of Power:  :$\dfrac \d {\d x} \paren {x + \dfrac 1 x} = 1 - \dfrac 1 {x^2}$ So, we have: {{begin-eqn}} {{eqn	| l = \int \frac {1 - \frac 1 {x^2} } {\paren {x + \frac 1 x}^2 - 2} \rd x 	| r = \int \frac 1 {u^2 - 2} \rd u 	| c = substituting $u = x + \dfrac 1 x$ }} {{eqn	| r = \frac 1 {2 \sqrt 2} \ln \size {\frac {u - \sqrt 2} {u + \sqrt 2} } + C 	| c = Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form }} {{eqn	| r = \frac 1 {2 \sqrt 2} \ln \size {\frac {x + \frac 1 x - \sqrt 2} {x + \frac 1 x + \sqrt 2} } + C }} {{eqn	| r = \frac 1 {2 \sqrt 2} \ln \size {\frac {x^2 - \sqrt 2 x + 1} {x^2 + \sqrt 2 x + 1} } + C }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {x^2 - 1} {x^4 + 1} \rd x       | r = \int \frac {x^2} {x^4 + 1} \rd x - \int \frac 1 {x^4 + 1} \rd x       | c = Linear Combination of Primitives }} {{eqn | r = \frac 1 {4 \sqrt 2} \map \ln {\frac {x^2 - x \sqrt 2 + 1} {x^2 + x \sqrt 2 + 1} } - \frac 1 {2 \sqrt 2} \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} }       | c = Primitive of $\dfrac {x^2} {x^4 + a^4}$, setting $a = 1$ }} {{eqn | o =        | ro= -       | r = \frac 1 {2 \sqrt 2} \paren {\map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } + \frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } }       | c = Primitive of $\dfrac 1 {1 + x^4}$ }} {{end-eqn}} {{finish|Resolve the arctangent equality}}
:400px There exists a unique circle $C$ tangent to $PQ$ passing through $A$ and $B$. From Angles in Same Segment of Circle are Equal, the angle subtended by $AB$ from any point on $C$ is the same as angle $\angle AQB$. All other points on $PQ$ that are not $Q$ itself are outside $C$. Hence the angle subtended by $AB$ to such points is smaller. Hence the result. {{qed}} {{Namedfor|Regiomontanus|cat = Regiomontanus}}
We have by definition of first projection that: :$\map {\pr_1} w = \map {\pr_1} {a, b} = a$ Then: {{begin-eqn}} {{eqn | l = \bigcup \bigcap w       | r = \bigcup \bigcap \tuple {a, b}       | c = Definition of $w$ }} {{eqn | r = \bigcup \bigcap \set {\set a, \set {a, b} }       | c = {{Defof|Kuratowski Formalization of Ordered Pair|Ordered Pair}} }} {{eqn | r = \map \bigcup {\set a \cap \set {a, b} }       | c = Intersection of Doubleton }} {{eqn | r = \bigcup \set a       | c = {{Defof|Set Intersection}} }} {{eqn | r = a       | c = Union of Singleton }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = z \in \bigcap \set {x, y}       | c =  }} {{eqn | ll= \leadstoandfrom       | o =        | r = \forall w \in \set {x, y}: z \in w       | c = {{Defof|Intersection of Class}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \forall w: \paren {\paren {w = x \land w = y} \land z \in w}       | c = {{Defof|Doubleton Class}} }} {{eqn | ll= \leadstoandfrom       | o =        | r = \paren {z \in x \land z \in y}       | c = Equality implies Substitution }} {{eqn | ll= \leadstoandfrom       | o =        | r = z \in \paren {x \cap y}       | c = {{Defof|Class Intersection}} }} {{end-eqn}} {{qed}}
We have by definition of second projection that: :$\map {\pr_1} w = \map {\pr_1} {a, b} = b$ We consider: {{begin-eqn}} {{eqn | l = \bigcup w       | r = \bigcup \tuple {a, b}       | c = Definition of $w$ }} {{eqn | r = \bigcup \set {\set a, \set {a, b} }       | c = {{Defof|Kuratowski Formalization of Ordered Pair|Ordered Pair}} }} {{eqn | r = \set a \cup \set {a, b}       | c = Union of Doubleton }} {{eqn | n = 1       | r = \set {a, b}       | c = {{Defof|Union of Set of Sets}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \bigcap w       | r = \bigcap \tuple {a, b}       | c = Definition of $w$ }} {{eqn | r = \bigcap \set {\set a, \set {a, b} }       | c = {{Defof|Kuratowski Formalization of Ordered Pair|Ordered Pair}} }} {{eqn | r = \set a \cap \set {a, b}       | c = Intersection of Doubleton }} {{eqn | n = 2       | r = \set a       | c = {{Defof|Intersection of Set of Sets}} }} {{end-eqn}} Suppose $\displaystyle \bigcup w \ne \bigcap w$. Then: {{begin-eqn}} {{eqn | l = \map \bigcup {\bigcup w \setminus \bigcap w}       | r = \map \bigcup {\set {a, b} \setminus \set a}       | c = from $(1)$ and $(2)$ above }} {{eqn | r = \bigcup \set b       | c = {{Defof|Set Difference}}, which holds because $a \ne b$ }} {{eqn | r = b       | c = Union of Singleton }} {{end-eqn}} demonstrating that the first case holds. Now suppose that $\bigcup w = \bigcap w$. Thus: {{begin-eqn}} {{eqn | l = \bigcup w       | r = \bigcap w       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = \set {a, b}       | r = \set a       | c =  }} {{eqn | n = 4       | ll= \leadsto       | l = a       | r = b       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \bigcup \bigcup w       | r = \bigcup \bigcup \tuple {a, b}       | c = Definition of $w$ }} {{eqn | r = \bigcup \bigcup \set {\set a, \set {a, b} }       | c = {{Defof|Kuratowski Formalization of Ordered Pair|Ordered Pair}} }} {{eqn | r = \bigcup \bigcup \set {\set a, \set a}       | c = from $(3)$ above }} {{eqn | r = \map \bigcup {\set a \cup \set a}       | c = Union of Doubleton }} {{eqn | r = \bigcup \set a       | c = Union is Idempotent }} {{eqn | r = a       | c = Union of Singleton }} {{eqn | r = b       | c = from $(4)$ above }} {{end-eqn}} The result follows; {{qed}}
Proof by Counterexample: Let $\Z$ denote the set of integers. We have that Integers under Usual Ordering form Totally Ordered Set. Let $m \in \Z$ be an arbitrary integer. Let $f_m: \Z \to \Z$ be the order isomorphism defined as: :$\forall a \in \Z: \map {f_m} a = a + m$ It follows that there are at least as many order isomorphisms of from $\Z$ to $\Z$ as there are integers. Hence the result {{qed}}
First primitive: {{begin-eqn}} {{eqn | l = \int \dfrac x {x^2 + 1} \rd x       | r = \dfrac 1 2 \int \dfrac {2 x} {x^2 + 1} \rd x       | c =  }} {{eqn | r = \dfrac {\map \ln {x^2 + 1} } 2       | c = Primitive of Function under its Derivative }} {{end-eqn}} The integration constant is not added due to the series never having a constant during its integration. {{qed|lemma}} Second primitive: {{begin-eqn}} {{eqn | l = \int \dfrac {\map \ln {x^2 + 1} } 2 \rd x       | r = \dfrac 1 2 \paren {x \map \ln {x^2 + 1^2} - 2 x + 2 \times 1 \times \arctan \frac x 1}       | c = Primitive of $\ln {x^2 + a^2}$, putting $a = 1$ }} {{eqn | r = \dfrac {x \map \ln {x^2 + 1} } 2 - x + \map \arctan x       | c = }} {{end-eqn}} {{qed|lemma}} Third primitive: {{begin-eqn}} {{eqn | l = \int \paren {\dfrac {x \map \ln {x^2 + 1} } 2 + \map \arctan x - x} \rd x       | r = \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + \int \map \arctan x \rd x - \int x \rd x       | c = }} {{eqn | r = \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + \int \map \arctan x \rd x - \dfrac {x^2} 2       | c = Integral of Power }} {{eqn | r = \int \dfrac {x \map \ln {x^2 + 1} } 2 \rd x + x \map \arctan x - \dfrac {\map \ln {x^2 + 1} } 2 - \dfrac {x^2} 2       | c = Primitive of $\arctan \dfrac x a$ }} {{eqn | r = \frac 1 2 \paren {\frac {\paren {x^2 + 1} \map \ln {x^2 + 1} - x^2} 2} + x \map \arctan x - \dfrac {\map \ln {x^2 + 1} } 2 - \dfrac {x^2} 2       | c = Primitive of $x \map \ln {x^2 + a^2}$ with $a = 1$ }} {{eqn | r = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4       | c = simplifying }} {{end-eqn}} {{qed}} Category:Pi as Sum of Sequence of Reciprocal of Product of Three Consecutive Integers oegt7rnht3rfy34wiqrvyl9x2723lf1
=== Necessary Condition === Let: :$S$ have a supremum. We need to show that: :$(1): \quad S_i$ has a supremum for every $i$ in $\set {1, 2, \ldots, n}$ :$(2): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ By Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S_i$ exists for every $i$ in $\set {1, 2, \ldots, n}$. This proves $(1)$. Also by Supremum of Set of Real Numbers is at least Supremum of Subset, $\sup S \ge \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$. Therefore: :$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ In addition, by Supremum of Subset of Union Equals Supremum of Union: :$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$ Therefore: :$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ Accordingly: :$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ This proves $(2)$. {{qed|lemma}} === Sufficient Condition === Let: :$S_i$ have a supremum for every $i$ in $\set {1, 2, \ldots, n}$. We need to show that: :$(3): \quad S$ has a supremum :$(4): \quad \sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ Every $S_i$ has a supremum. Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ exists as a real number. The number $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ is an upper bound for every $S_i$. Therefore, $\max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ is an upper bound for $S$ as $S = \bigcup S_i$. By the Continuum Property $S$ has a supremum as $S$ is a non-empty real set with an upper bound. This proves $(3)$. We have proved that $S$ has a supremum. By Supremum of Set of Real Numbers is at least Supremum of Subset: :$\sup S \ge \sup S_i$ for every $i$ in $\set {1, 2, \ldots, n}$ Therefore: :$\sup S \ge \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ In addition, by Supremum of Subset of Union Equals Supremum of Union: :$\sup S = \sup S_j$ for a $j$ in $\set {1, 2, \ldots, n}$ Therefore: :$\sup S \le \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ Accordingly: :$\sup S = \max \set {\sup S_1, \sup S_2, \ldots, \sup S_n}$ This proves $(4)$. {{qed}}
Let $j \in N$ Suppose $\map N {j + 1} \ge \map N j$ By definition: :$\norm{\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$ Now suppose $\map N j \ge \map N {j + 1}$ Then: {{begin-eqn}} {{eqn | l = \norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p       | o = \le        | r = p^{-\paren {j + 2} } }} {{eqn | o = <        | r = p^{-\paren {j + 1} }       | c = Power Function on Integer between Zero and One is Strictly Decreasing }} {{end-eqn}} In either case: :$\norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren{j + 1} }$ Since $j$ was arbitrary, then: :$\forall j \in \N: \norm {\beta_{\map N {j + 1} } - \beta_{\map N j} }_p \le p^{-\paren {j + 1} }$ {{qed}} Category:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence lijtj2pogl5m9jroyyrmxyiechknxpz
For all $j \in \N$: {{begin-eqn}} {{eqn | l = \norm {\alpha_{j + 1} - \alpha_j}_p       | r = \norm {\alpha_{j + 1} - \gamma_{j + 1} + \gamma_{j + 1} - \gamma_j + \gamma_j - \alpha_j}_p }} {{eqn | o = \le       | r = \max \set {\norm {\alpha_{j + 1} - \gamma_{j + 1} }_p \mathop , \norm {\gamma_{j + 1} - \gamma_j }_p \mathop , \norm{\gamma_j - \alpha_j}_p} }} {{eqn | o = \le       | r = \max \set {p^{-\paren{j + 2} } , \norm{\gamma_{j + 1} - \gamma_j }_p \mathop , p^{-\paren {j + 1} } }       | c = Assumption on $\sequence{\alpha_n}$ }} {{eqn | o = \le       | r = \max \set {p^{-\paren {j + 2} } , p^{-\paren {j + 1} } , p^{-\paren {j + 1} } }       | c = Assumption on $\sequence{\gamma_n}$ }} {{eqn | r = p^{-\paren {j + 1} }       | c =  Power Function on Integer between Zero and One is Strictly Decreasing }} {{end-eqn}} {{qed}} Category:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence qxhjdz1mi46bo8j2ae9dfvjkt5km0li
{{begin-eqn}} {{eqn | l = \dfrac {x^2} {a^2 + \lambda}       | r = 1 - \dfrac {y^2} {b^2 + \lambda}       | c = from $(1)$ }} {{eqn | r = \dfrac {b^2 + \lambda - y^2} {b^2 + \lambda}       | c =  }} {{eqn | n = 2       | ll= \leadsto       | l = a^2 + \lambda       | r = \dfrac {x^2 \paren {b^2 + \lambda} } {b^2 + \lambda - y^2}       | c =  }} {{eqn | l = \dfrac {y^2} {b^2 + \lambda}       | r = 1 - \dfrac {x^2} {a^2 + \lambda}       | c = from $(1)$ }} {{eqn | r = \dfrac {a^2 + \lambda - x^2} {a^2 + \lambda}       | c =  }} {{eqn | n = 3       | ll= \leadsto       | l = b^2 + \lambda       | r = \dfrac {y^2 \paren {a^2 + \lambda} } {a^2 + \lambda - x^2}       | c =  }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \map {\dfrac \d {\d x} } {\dfrac {x^2} {a^2 + \lambda} + \dfrac {y^2} {b^2 + \lambda} }       | r = \map {\dfrac \d {\d x} } 1       | c = Differentiating $(1)$ {{WRT|Differentiation}} $x$ }} {{eqn | n = 4       | ll= \leadsto       | l = \dfrac {2 x} {a^2 + \lambda} + \dfrac {2 y} {b^2 + \lambda} \dfrac {\d y} {\d x}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a^2 + \lambda} y'       | r = -\dfrac {2 x} {2 y} \paren {b^2 + \lambda}       | c = putting $y' = \dfrac {\d y} {\d x}$ }} {{eqn | r = -\dfrac x y \dfrac {y^2 \paren {a^2 + \lambda} } {a^2 + \lambda - x^2}       | c = from $(3)$ }} {{eqn | ll= \leadsto       | l = y'       | r = -\dfrac {x y} {a^2 + \lambda - x^2}       | c =  }} {{eqn | ll= \leadsto       | l = y' \paren {a^2 + \lambda}       | r = x^2 y' - x y       | c =  }} {{eqn | n = 5       | ll= \leadsto       | l = a^2 + \lambda       | r = \dfrac {x^2 y' - x y} {y'}       | c =  }} {{end-eqn}} and similarly: {{begin-eqn}} {{eqn | l = \paren {b^2 + \lambda}       | r = -\dfrac {2 y} {2 x} \paren {a^2 + \lambda} y'       | c = from $(4)$, putting $y' = \dfrac {\d y} {\d x}$ }} {{eqn | r = -\dfrac {y y'} x \dfrac {x^2 \paren {b^2 + \lambda} } {b^2 + \lambda - y^2}       | c = from $(2)$ }} {{eqn | ll= \leadsto       | l = 1       | r = -\dfrac {x y y'} {b^2 + \lambda - y^2}       | c =  }} {{eqn | n = 6       | ll= \leadsto       | l = b^2 + \lambda       | r = y^2 - x y y'       | c =  }} {{end-eqn}} Eliminating $\lambda$: {{begin-eqn}} {{eqn | l = b^2 + \paren {\dfrac {x^2 y' - x y} {y'} - a^2}       | r = y^2 - x y y'       | c = from $(5)$ and $(6)$ }} {{eqn | ll= \leadsto       | l = y' \paren {b^2 - a^2} + x^2 y' - x y - y^2 y + x y \paren {y'}^2       | r = 0       | c = multiplying through by $y'$ and rearranging }} {{eqn | ll= \leadsto       | l = x y \paren {\paren {y'}^2 - 1} + \paren {x^2 - y^2 - a^2 + b^2} y'       | r = 0       | c = multiplying through by $y'$ and rearranging }} {{end-eqn}} {{qed}}
From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \beta_j}$ is a null sequence. Let $\epsilon \in \R_{> 0}$. From Sequence of Powers of Number less than One, the sequence $\sequence {p^n}$ is a null sequence. Then there exists $j \in \N$ such that $p^{-j} < \epsilon$. Then for all $i \ge \map N j$: {{begin-eqn}} {{eqn | l = \norm {\alpha_i - \beta_i}_p        | r = \norm {\alpha_i - \beta_{\map N i} + \beta_{\map N i} - \beta_i}_p }} {{eqn | o = \le        | r = \max \set {\norm {\alpha_i - \beta_{\map N i} }_p, \norm {\beta_{\map N i} - \beta_i}_p} }} {{eqn | o = \le        | r = \max \set {\norm {p^{-\paren{i + 1} } }_p, \norm {\beta_{\map N i} - \beta_i}_p}       | c = }} {{eqn | o = \le        | r = \max \set {\norm {p^{-\paren{i + 1} } }_p, \norm {p^{-\paren {i + 1} } }_p}       | c = }} {{eqn | r = \norm {p^{-\paren {i + 1} } }_p       | c = }} {{eqn | o = \le        | r = \norm {p^{-j } }_p       | c = }} {{eqn | o = <        | r = \epsilon       | c = }} {{end-eqn}} It follows that $\sequence {\alpha_n - \beta_n}$ is a null sequence by definition. {{qed}} Category:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence ce462vsjcz4mtk0kqamekvumjtpkqp0
Let $\sequence {\alpha'_j}$ be a coherent sequence not equal to $\sequence {\alpha_j}$. From Representatives of same P-adic Number iff Difference is Null Sequence, it needs only to be shown that $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence. Since $\sequence {\alpha'_j} \ne \sequence{\alpha_j}$ then: :$\exists i_0 \in \N : \alpha'_{i_0} \ne \alpha_{i_0}$ By definition of coherent sequences: :$0 \le \alpha_{i_0}, \alpha'_{i_0} < p^{i_0 + 1}$ From Reduced Residue System Modulo Prime: :$\alpha_{i_0} \not \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$ By definition of a coherent sequence, for all $i > i_0$: :$\alpha_i \equiv \alpha_{i_0} \pmod {p^{i_0 + 1} }$ :$\alpha'_i \equiv \alpha'_{i_0} \pmod {p^{i_0 + 1} }$ Then: :$\forall i > i_0: \alpha_i \equiv \alpha_{i_0} \not \equiv \alpha'_{i_0}  \equiv \alpha'_i \pmod {p^{i_0 + 1} }$ That is: :$\forall i > i_0: p^{i_0 + 1} \nmid \alpha_i - \alpha'_i$ By definition of the $p$-adic norm on integers: :$\forall i > i_0: \norm {\alpha_i - \alpha'_i} > \dfrac 1 {p^{i_0 + 1} }$ By definition of convergence, $\sequence {\alpha_j - \alpha'_j}$ is not a null sequence. The result follows. {{qed}} Category:Equivalence Class in P-adic Integers Contains Unique Coherent Sequence kesyxphk5qvqissz4ablkm7be3mms6o