hcju/proofwikiqa huggingface dataset:
By definition of a circuit we have: === Lemma 1 === {{:Element of Matroid Base and Circuit has a Substitute/Lemma 1}}{{qed|lemma}} From Independent Subset of Matroid is Augmented by Base: :$\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X$ is a base of $M$ By definition of independent and dependent subsets we have: === Lemma 2 === {{:Element of Matroid Base and Circuit has a Substitute/Lemma 2}} {{qed|lemma}}   From matroid axiom $(\text I 3')$ we have: === Lemma 3 === {{:Element of Matroid Base and Circuit has a Substitute/Lemma 3}} {{qed|lemma}}   From Independent Subset is Base if Cardinality Equals Cardinality of Base: :$\paren{B \setminus \set x} \cup \set y$ is a base of $M$ Because $x \notin \paren{ C \setminus \set x} \cup X$: :$y \ne x$ By definition of set difference: :$y \notin B \setminus \set x$: By definition of set union: :$y \notin \paren{B \setminus \set x } \cup \set x = B$ By the definition of a subset: :$y \notin X$ By definition of set union: :$y \in C \setminus \set x \subseteq C$ By definition of set difference: :$y \in C \setminus B$ {{qed}} Category:Matroid Theory ksu25n4mqulynitqsuv00d79tt9mrx8
From Set Difference is Subset: :$S \setminus T \subseteq S$ From Set Difference with Proper Subset: :$S \setminus T \ne \O$ By definition of proper subset: :$T \ne \O$ From Intersection with Subset is Subset: :$S \cap T = T$ Hence: :$S \cap T \ne \O$ From the contrapositive statement of Set Difference with Disjoint Set: :$S \setminus T \ne S$ It follows that $S \setminus T$ is a proper subset by definition. {{qed}} Category:Set Difference Category:Subsets d3yuafomz3kochodppm5iwa12u8pl6n
By definition of circuit: :$C$ is a minimal dependent subset From Singleton of Element is Subset: :$\set x \subseteq C$ and $\set x \subseteq B$ From matroid axiom $(\text I 2)$: :$\set x \in \mathscr I$ Because $C \notin \mathscr I$: :$\set x \ne C$ Hence: :$\set x \subsetneq C$ From Set Difference with Proper Subset is Proper Subset: :$C \setminus \set x$ is a proper subset of $C$ As $C$ is a minimal dependent subset: :$C \setminus \set x \in \mathscr I$ {{qed}} Category:Matroid Theory a6zgx051bkt13r45ikrt73rhyuvwvce
=== Positive definiteness === By definition of induced norm: :$\forall y \in Y : \norm {y}_Y = \norm {y}_X > 0$ Suppose $y \in Y : \norm {y}_Y = 0$. Since $\norm {\, \cdot \,}_Y$ is an induced norm in $\struct {X, \norm {\, \cdot \,}_X}$: :$\norm {y}_X = 0$. Therefore, $y = \mathbf 0 \in X$. By definition of a vector subspace, $Y$ has the additive identity: :$\mathbf 0 \in Y$. Thus: :$y = \mathbf 0 \in Y \subseteq X$. === Positive homogeneity === Let $y \in Y$. Let $\alpha \in \R$. Then we have that: {{begin-eqn}} {{eqn | l = \norm {\alpha \cdot y}_Y        | r = \norm {\alpha \cdot y}_X        | c = {{defof|Induced Norm}} }} {{eqn | r = \size \alpha \norm {y}_X        | c =  Norm axiom (N2) }} {{eqn | r = \size \alpha \norm {y}_Y       | c = {{defof|Induced Norm}}; $y \in Y$ }} {{end-eqn}} === Triangle inequality === Let $y_1, y_2 \in Y$. By closure axiom of vector space:  :$y_1 + y_2 \in Y$. Furthermore: {{begin-eqn}} {{eqn | l = \norm {y_1 + y_2}_Y       | r = \norm {y_1 + y_2}_X        | c = {{defof|Induced Norm}} }} {{eqn | o = \le       | r = \norm {y_1}_X + \norm {y_2}_X        | c = Norm axiom (N2) }} {{eqn | r = \norm {y_1}_Y + \norm {y_2}_Y       | c = {{defof|Induced Norm}}; $y_1, y_2 \in Y$ }} {{end-eqn}} {{qed}}
From Independent Subset is Contained in Maximal Independent Subset : :$\exists B' \subseteq S : X \subseteq B'$ and $B'$ is a maximal independent subset of $S$ By definition of a base: :$B'$ is a base of $M$ From Cardinality of Subset of Finite Set: :$\card X \le \card {B'}$ From All Bases of Matroid have same Cardinality: :$\card{B'} = \card B$ Hence: :$\card X \le \card B$ {{qed}} Category:Matroid Theory tqan50fl5k15k8ca13rj1fevjxdijq7
From Cardinality of Independent Set of Matroid is Smaller or Equal to Base: :$\card X \le \card B$ === Case 1: $\card X < \card B$ === From Independent Set can be Augmented by Larger Independent Set: :$\exists Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$ {{qed|lemma}} === Case 2: $\card X = \card B$ === Let $Z = \O$. Then: :$Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$ {{qed|lemma}} From Independent Subset is Base if Cardinality Equals Cardinality of Base: :$X \cup Z$ is a base of $M$ {{qed}} Category:Matroid Theory tnbbb5wfpvvleyboe6ehakzvdof7wuh
From All Bases of Matroid have same Cardinality: :$\card B = \map \rho S$ where $\rho$ denotes the rank function on $M$. Hence: :$\card X = \map \rho S$ From Independent Subset is Base if Cardinality Equals Rank of Matroid: :$X$ is a base of $M$. {{qed}} Category:Matroid Theory rt8h42qguk3o0v1qbyp50oaql4nlpmz
{{AimForCont}}: :$x \in \paren{ C \setminus \set x} \cup X$ Then: :$\set x, C \setminus \set x \subseteq \paren{ C \setminus \set x} \cup X$ From Union of Subsets is Subset: :$C \subseteq \paren{ C \setminus \set x} \cup X$ From matroid axiom $(\text I 2)$: :$C \in \mathscr I$ This contradicts: :$C \notin \mathscr I$ It follows that: :$x \notin \paren{ C \setminus \set x} \cup X$ {{qed}} Category:Matroid Theory jjgqizigeavgbqieg5kl8mtvsbhg3kh
From Set Difference is Subset: :$B \setminus \set x \subseteq B$ From matroid axiom $(\text I 2)$: :$B \setminus \set x \in \mathscr I$ We have {{begin-eqn}} {{eqn | l = \card {B \setminus \set x} + 1       | r = \paren{\card B - \card{\set x} } + 1       | c = Cardinality of Set Difference with Subset }} {{eqn | r = \paren{\card B - 1 } + 1       | c = Cardinality of Singleton }} {{eqn | r = \card B }} {{eqn | r = \card {\paren{ C \setminus \set x} \cup X}       | c = All Bases of Matroid have same Cardinality }} {{end-eqn}} From matroid axiom $(\text I 3')$: :$\exists y \in \paren{\paren{C \setminus \set x} \cup X} \setminus \paren{B \setminus \set x} : \paren{B \setminus \set x} \cup \set y \in \mathscr I : \card{\paren{B \setminus \set x} \cup \set y} = \card {\paren{ C \setminus \set x} \cup X}$ {{qed}} Category:Matroid Theory shmd8jquqhipde4n19gs7glgxqunhci
From Dependent Subset Contains a Circuit: :there exists a circuit $C$ such that $C \subseteq X \cup \set x$ From Dependent Subset of Independent Set Union Singleton Contains Singleton: :$x \in C$ {{AimForCont}} $C'$ is circuit of $M$ such that: :$C' \ne C$ :$C' \subseteq X \cup \set x$ From Dependent Subset of Independent Set Union Singleton Contains Singleton: :$x \in C'$ By the definition of minimal dependent subset: :$C \not \subseteq C'$ By the definition of a subset: :$\exists y \in C \setminus C'$ By the definition of minimal dependent subset: :$C \setminus \set y \in \mathscr I$ From Independent Subset is Contained in Maximal Independent Subset: :$\exists Y \in \mathscr I : C \setminus \set y \subseteq Y \subseteq X \cup \set x : Y$ is a maximal independent subset of $X \cup \set x$ By assumption: :$X$ is a maximal independent subset of $X \cup \set x$ By matroid axiom $(\text I 3''')$: :$\card Y = \card X$ As $x \in C'$ and $y \notin C'$ then: :$x \ne y$ Hence: :$x \in C \setminus \set y$ By matroid axiom $(\text I 2)$: :$C \setminus \set y \cup  \set y = C \not \subseteq Y$  Hence: :$y \notin Y$ Hence: :$Y \subseteq \paren{X \cup \set x} \setminus \set y$ From Cardinality of Subset of Finite Set: :$\card Y \le \card{\paren{X \cup \set x} \setminus \set y}$ We have: {{begin-eqn}} {{eqn | l = \card{\paren{X \cup \set x} \setminus \set y}       | r = \card{\paren{X \cup \set x} } - \card{\set y}       | c = }} {{eqn | r = \card X + \card{\set x} - \card{\set y}       | c = }} {{eqn | r = \card X +  1 - 1       | c = }} {{eqn | r = \card X       | c = }} {{eqn | r = \card Y       | c = }} {{end-eqn}} From Cardinality of Proper Subset of Finite Set: :$Y = \paren{X \cup \set x} \setminus \set y$ Hence: :$\paren{X \cup \set x} \setminus \set y \in \mathscr I$ As $y \notin C'$ and $C' \subseteq X \cup \set x$ then: :$C' \subseteq \paren{X \cup \set x} \setminus \set y$ From Superset of Dependent Set is Dependent: :$\paren{X \cup \set x} \setminus \set y \notin \mathscr I$ This contradicts the previous statement that  :$\paren{X \cup \set x} \setminus \set y \in \mathscr I$ It follows that $C$ is the unique circuit such that: :$C \subseteq X \cup \set x$ {{qed}} Category:Matroid Theory glbbo8kkqqsz457argdnbvn0w14wu1w
Let $x \ge 0$. By definition of absolute value: :$\size x = x$ We have that: :$-x < 0$ and so by definition of absolute value: :$\size {-x} = -\paren {-x} = x$ {{qed|lemma}} Now let $x < 0$. By definition of absolute value: :$\size x = -x$ We have that: :$-x > 0$ and so: and so by definition of absolute value: :$\size {-x} = -x$ {{qed|lemma}} In both cases it is seen that: :$\size x = \size {-x}$ {{qed}} Category:Absolute Value Function n2lpetttt3eo8db79q0jju2cmvn2wzh
Consider the ordered set $\struct {\powerset S \setminus \mathscr I, \subseteq}$. From Element of Finite Ordered Set is Between Maximal and Minimal Elements: :$\exists C \in \mathscr I : C \subseteq A$ and $A$ is minimal in $\struct {\powerset S \setminus \mathscr I, \subseteq}$. By definition of a circuit: :$C \in \mathscr C$ {{qed}} Category:Matroid Theory mj27wuuxqy537wwzcf4zri7aadjq5z8
In order to be a group, it is necessary that $\struct {\Q, /}$ be an associative structure. But consider the elements $2, 6, 12$ of $\Q$. We have: {{begin-eqn}} {{eqn | l = \paren {12 / 6} / 2       | r = 2 / 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} whereas: {{begin-eqn}} {{eqn | l = 12 / \paren {6 / 2}       | r = 12 / 3       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} That is: :$\paren {12 / 6} / 2 \ne 12 / \paren {6 / 2}$ So it is not generally the case that for $a, b, c \in \Q$: :$\paren {a / b} / c = a / \paren {b / c}$ and so $/$ is not associative on $\Q$. Hence $\struct {\Q, /}$ is not a group. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\sinh^{-1} } {\frac x a}       | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 + a^2} }       | c = {{Defof|Inverse Hyperbolic Sine|subdef = Real|index = 2}} }} {{eqn | r = \map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 + a^2} }       | c =  }} {{eqn | r = \map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 + a^2} } }       | c =  }} {{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } - \ln a       | c = Difference of Logarithms }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 + a^2} } }       | r = \map {\dfrac \d {\d x} } {\map {\sinh^{-1} } {\frac x a} + \ln a }       | c =  }} {{eqn | r = \dfrac 1 {\sqrt {x^2 + a^2} } + \map {\dfrac \d {\d x} } {\ln a}       | c = Derivative of Inverse Hyperbolic Sine of x over a }} {{eqn | r = \dfrac 1 {\sqrt {x^2 + a^2} } + 0       | c = Derivative of Constant }} {{end-eqn}} We have that $\sqrt {x^2 + a^2} > x$ for all $x$. Thus: {{begin-eqn}} {{eqn | l = x + \sqrt {x^2 + a^2}       | o = >       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \size {x + \sqrt {x^2 + a^2} }       | r = x + \sqrt {x^2 + a^2}       | c = {{Defof|Absolute Value}} }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\size {x + \sqrt {x^2 + a^2} } }       | r = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 + a^2} } }       | c =  }} {{end-eqn}} and the result follows. {{qed}}
We have that $\sqrt {x^2 + a^2} > x$ for all $x$. Thus: :$x + \sqrt {x^2 + a^2} > 0$ and so $\map \ln {x + \sqrt {x^2 + a^2} }$ is defined for all $x$. Then we have: {{begin-eqn}} {{eqn | l = -\map \ln {x + \sqrt {x^2 + a^2} }       | r = \map \ln {\dfrac 1 {x + \sqrt {x^2 + a^2} } }       | c = Logarithm of Reciprocal }} {{eqn | r = \map \ln {\dfrac {x - \sqrt {x^2 + a^2} } {\paren {x + \sqrt {x^2 + a^2} } \paren {x - \sqrt {x^2 + a^2} } } }       | c = multiplying top and bottom by $x - \sqrt {x^2 + a^2}$ }} {{eqn | r = \map \ln {\dfrac {x - \sqrt {x^2 + a^2} } {x^2 - \paren {x^2 + a^2} } }       | c = Difference of Two Squares }} {{eqn | r = \map \ln {\dfrac {-x + \sqrt {x^2 + a^2} } {a^2} }       | c = simplifying }} {{eqn | r = \map \ln {-x + \sqrt {x^2 + a^2} } - \map \ln {a^2}       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = -\map {\sinh^{-1} } {\frac x a}       | r = \map {\sinh^{-1} } {-\frac x a}       | c = Inverse Hyperbolic Sine is Odd Function }} {{eqn | r = \map \ln {-\paren {\frac x a} + \sqrt {\paren {-\frac x a}^2 + a^2} }       | c = {{Defof|Inverse Hyperbolic Sine|subdef = Real|index = 2}} }} {{eqn | r = \map \ln {-\frac x a + \dfrac 1 a \sqrt {x^2 + a^2} }       | c =  }} {{eqn | r = \map \ln {-\dfrac 1 a \paren {x - \sqrt {x^2 + a^2} } }       | c =  }} {{eqn | r = \map \ln {-x + \sqrt {x^2 + a^2} } - \ln a       | c = Difference of Logarithms }} {{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } - \ln a       | c = as $\sqrt {x^2 + a^2} > -x$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\ln \size {x - \sqrt {x^2 + a^2} } }       | r = \map {\dfrac \d {\d x} } {-\map {\sinh^{-1} } {\frac x a} + \ln a}       | c = Sum of Logarithms }} {{eqn | r = -\dfrac 1 {\sqrt {x^2 + a^2} } + \map {\dfrac \d {\d x} } {\ln a}       | c = Derivative of Inverse Hyperbolic Sine of x over a }} {{eqn | r = -\dfrac 1 {\sqrt {x^2 + a^2} } + 0       | c = Derivative of Constant }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }       | r = \map \ln {x + \sqrt {x^2 + a^2} } + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ in Logarithm Form }} {{eqn | l = \int \frac {\d x} {-\sqrt {x^2 + a^2} }       | r = -\map \ln {x + \sqrt {x^2 + a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } + C + \map \ln {a^2}       | c = Negative of Logarithm of x plus Root x squared plus a squared: Corollary }} {{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } + C       | c = subsuming $\map \ln {a^2}$ into the constant }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\cosh^{-1} } {\frac x a}       | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - a^2} }       | c = {{Defof|Inverse Hyperbolic Cosine|subdef = Real|index = 2}} }} {{eqn | r = \map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 - a^2} }       | c =  }} {{eqn | r = \map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 - a^2} } }       | c =  }} {{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a       | c = Difference of Logarithms }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 - a^2} } }       | r = \map {\dfrac \d {\d x} } {\map {\cosh^{-1} } {\frac x a} - \ln a }       | c =  }} {{eqn | r = \dfrac 1 {\sqrt {x^2 - a^2} } + \map {\dfrac \d {\d x} } {\ln a}       | c = Derivative of Inverse Hyperbolic Cosine of x over a }} {{eqn | r = \dfrac 1 {\sqrt {x^2 - a^2} } + 0       | c = Derivative of Constant }} {{end-eqn}} When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined. When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined. When $x < -a$ we have that $x + \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined. Hence the restriction on the domain. {{qed}}
{{begin-eqn}} {{eqn | l = -\map {\cosh^{-1} } {\frac x a}       | r = -\map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - a^2} }       | c = {{Defof|Inverse Hyperbolic Cosine|subdef = Real|index = 2}} }} {{eqn | r = -\map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 - a^2} }       | c =  }} {{eqn | r = -\map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 - a^2} } }       | c =  }} {{eqn | r = -\map \ln {x + \sqrt {x^2 - a^2} } + \ln a       | c = Difference of Logarithms }} {{eqn | r = \map \ln {x - \sqrt {x^2 - a^2} } + \ln a^2 + \ln a       | c = Negative of Logarithm of x plus Root x squared minus a squared }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\map \ln {x - \sqrt {x^2 - a^2} } }       | r = \map {\dfrac \d {\d x} } {-\map {\cosh^{-1} } {\frac x a} - \ln a^3}       | c = Sum of Logarithms }} {{eqn | r = -\dfrac 1 {\sqrt {x^2 - a^2} } - \map {\dfrac \d {\d x} } {\ln a^3}       | c = Derivative of Inverse Hyperbolic Cosine of x over a }} {{eqn | r = -\dfrac 1 {\sqrt {x^2 - a^2} } + 0       | c = Derivative of Constant }} {{end-eqn}} When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined. When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined. When $x < -a$ we have that $x - \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined. Hence the restriction on the domain. {{qed}}
{{begin-eqn}} {{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }       | r = \ln \size {x + \sqrt {x^2 - a^2} } + C       | c = Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ in Logarithm Form }} {{eqn | l = \int \frac {\d x} {-\sqrt {x^2 + a^2} }       | r = -\ln \size {x + \sqrt {x^2 - a^2} } + C       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \ln \size {x - \sqrt {x^2 - a^2} } + C + \map \ln {a^2}       | c = Negative of Logarithm of x plus Root x squared minus a squared }} {{eqn | r = \ln \size {x - \sqrt {x^2 - a^2} } + C       | c = subsuming $\map \ln {a^2}$ into the constant }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 a \map {\tanh^{-1} } {\frac x a}       | r = \dfrac 1 a \cdot \dfrac 1 2 \map \ln {\dfrac {1 + \frac x a} {1 - \frac x a} }       | c = {{Defof|Inverse Hyperbolic Tangent|subdef = Real|index = 2}} }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} }       | c = multiplying top and bottom of argument by $a$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } }       | r = \map {\dfrac \d {\d x} } {\dfrac 1 a \map {\tanh^{-1} } {\frac x a} }       | c =  }} {{eqn | r = \dfrac 1 a \cdot \dfrac a {a^2 - x^2}       | c = Derivative of Inverse Hyperbolic Tangent of x over a }} {{eqn | r = \dfrac 1 {a^2 - x^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac 1 a \map {\coth^{-1} } {\frac x a}       | r = \dfrac 1 a \cdot \dfrac 1 2 \map \ln {\dfrac {\frac x a + 1} {\frac x a - 1} }       | c = {{Defof|Inverse Hyperbolic Cotangent|subdef = Real|index = 2}} }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} }       | c = multiplying top and bottom of argument by $a$ }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } }       | r = \map {\dfrac \d {\d x} } {\dfrac 1 a \map {\coth^{-1} } {\frac x a} }       | c =  }} {{eqn | r = \dfrac 1 a \cdot \dfrac {-a} {x^2 - a^2}       | c = Derivative of Inverse Hyperbolic Cotangent of x over a }} {{eqn | r = \dfrac 1 {a^2 - x^2}       | c = simplifying }} {{end-eqn}} {{qed}}
From Set Difference is Subset: :$C \setminus \set x \subseteq C$ Because $x \in C$ and $x \notin C \setminus \set x$: :$C \setminus \set x \ne C$ Hence: :$C \setminus \set x \subsetneq C$ By definition of circuit: :$C$ is a minimal dependent subset Hence: :$C \setminus \set x \in \mathscr I$ {{qed}} Category:Matroid Theory l0nw0nrmr3yf8agel9n9z7hwfdruvn2
From Set is Subset of Union: :$B \subseteq B \cup \set x$ Because $x \in B \cup \set x$ and $x \notin B$: :$B \ne B \cup \set x$ Hence: :$B \subsetneq B \cup \set x$ By definition of base: :$B$ is a maximal independent subset Hence: :$B \cup \set x \notin \mathscr I$ {{qed}} Category:Matroid Theory 05pwsmu9mo9f7zw2mm4sdq0xb4x0i3q
Let $\size x < a$. Let: {{begin-eqn}} {{eqn | l = u       | r = \tanh^{-1} {\frac x a}       | c = {{Defof|Real Inverse Hyperbolic Tangent}}, which is defined where $\size {\dfrac x a} < 1$ }} {{eqn | ll=\leadsto       | l = x       | r = a \tanh u       | c =  }} {{eqn | ll=\leadsto       | l = \frac {\d x} {\d u}       | r = a \sech^2 u       | c = Derivative of Hyperbolic Tangent }} {{eqn | ll=\leadsto       | l = \int \frac {\rd x} {a^2 - x^2}       | r = \int \frac {a \sech^2 u} {a^2 - a^2 \tanh^2 u} \rd u       | c = Integration by Substitution }} {{eqn | r = \frac a {a^2} \int \frac {\sech^2 u} {1 - \tanh^2 u} \rd u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \int \frac {\sech^2 u} {\sech^2 u} \rd u       | c = Sum of Squares of Hyperbolic Secant and Tangent }} {{eqn | r = \frac 1 a \int \rd u }} {{eqn | r = \frac 1 a u + C       | c = Integral of Constant }} {{eqn | r = \frac 1 a \tanh^{-1} \frac x a + C       | c = Definition of $u$ }} {{end-eqn}}
First note that if $x = a$ then $a^2 - x^2 = 0$ and so $\dfrac 1 {a^2 - x^2}$ is undefined. === Proof for $\size x < a$ === {{:Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form/Proof}}{{qed|lemma}} === Proof for $\size x > a$ === {{:Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form/Proof}}{{qed}} Category:Primitive of Reciprocal of a squared minus x squared psko6uv1r93j0d6ykdjnb58lpixw3u0
Let $\size x > a$. Let: {{begin-eqn}} {{eqn | l = u       | r = \coth^{-1} {\frac x a}       | c = {{Defof|Real Inverse Hyperbolic Cotangent}}, which is defined where $\dfrac x a > 1$ }} {{eqn | ll=\leadsto       | l = x       | r = a \coth u       | c =  }} {{eqn | ll=\leadsto       | l = \frac {\d x} {\d u}       | r = -a \csch^2 u       | c = Derivative of Hyperbolic Cotangent }} {{eqn | ll=\leadsto       | l = \int \frac {\rd x} {a^2 - x^2}       | r = \int \frac {-a \csch^2 u} {a^2 - a^2 \coth^2 u} \rd u       | c = Integration by Substitution }} {{eqn | r = -\frac a {a^2} \int \frac {\csch^2 u} {-\paren {\coth^2 u - 1} } \rd u       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 a \int \frac {\csch^2 u} {\csch^2 u} \rd u       | c = Difference of Squares of Hyperbolic Cotangent and Cosecant }} {{eqn | r = \frac 1 a \int \rd u }} {{eqn | r = \frac 1 a u + C       | c = Integral of Constant }} {{eqn | r = \frac 1 a \coth^{-1} \frac x a + C       | c = Definition of $u$ }} {{end-eqn}}
Let $\size x < a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a^2 - x^2}       | r = \frac 1 a \tanh^{-1} {\frac x a} + C       | c = Primitive of $\dfrac 1 {a^2 - x^2}$: $\tanh^{-1}$ form }} {{eqn | r = \frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} } } + C       | c = $\tanh^{-1} \dfrac x a$ in Logarithm Form }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C       | c = simplifying }} {{end-eqn}}
Let $\size x < a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a^2 - x^2}       | r = \int \frac {\d x} {\paren {a + x} \paren {a - x} }       | c = Difference of Two Squares }} {{eqn | r = \int \frac {\d x} {2 a \paren {a + x} } + \int \frac {\d x} {2 a \paren {a - x} }       | c = Partial Fraction Expansion }} {{eqn | r = \frac 1 {2 a} \int \frac {\d x} {a + x} + \frac 1 {2 a} \int \frac {\d x} {a - x}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {2 a} \ln \size {a + x} - \frac 1 {2 a} \ln \size {a - x} + C       | c = Primitive of Reciprocal }} {{eqn | r = \frac 1 {2 a} \map \ln {a + x} - \frac 1 {2 a} \map \ln {a - x} + C       | c = as both $a + x < 0$ and $a - x < 0$ }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C       | c = Difference of Logarithms }} {{end-eqn}} {{qed}}
Let $\size x < a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a^2 - x^2}       | r = -\int \frac {\d x} {x^2 - a^2}       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a - x} {a + x} } + C       | c = Primitive of $\dfrac 1 {x^2 - a^2}$ for $\size x < a$ }} {{eqn | r = \frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C       | c = Logarithm of Reciprocal }} {{end-eqn}} {{qed}}
Let $\size x > a$. Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {a^2 - x^2}       | r = \frac 1 a \coth^{-1} {\frac x a} + C       | c = Primitive of $\dfrac 1 {a^2 - x^2}$: $\coth^{-1}$ form }} {{eqn | r = \frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C       | c = $\coth^{-1} \dfrac x a$  in Logarithm Form }} {{eqn | r = \dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C       | c = simplifying }} {{end-eqn}} {{qed}}