nyuuzyou/wikis huggingface dataset:
Chu-Vandermonde Identity for Gaussian Binomial Coefficients

== Theorem ==
Let $q \in \R_{\ne 1}, r, s \in \R, n \in \Z$.
Then:
where $\dbinom r k_q$ is a Gaussian binomial coefficient.
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $58$
Category:Chu-Vandermonde Identity
Category:Gaussian Binomial Coefficients
Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere

== Theorem ==
=== Lemma ===
== Proof ==
:
Let $AB$ be the diameter of the given sphere.
Let a regular dodecahedron and a regular icosahedron be inscribed within.
Let $CDEFG$ be one pentagonal face of the dodecahedron.
Let $KLH$ be one triangular face of the icosahedron.
It is to be demonstrated that the radii of the circles circumscribing $CDEFG$ and $KLH$ are equal.
Let $DG$ be joined.
From :
:$DG$ is the side of a cube which has been inscribed within the given sphere.
Let $MN$ be a straight line such that:
:$AB^2 = 5 \cdot MN^2$
From :
:the square on the diameter of the given sphere is $5$ times the square on the radius of the circle in which is inscribed the pentagonal base of the pyramidal tip of a regular icosahedron which is inscribed within that sphere.
Therefore $MN$ equals the radius of the circle in which that pentagon is inscribed.
Let $MN$ be cut in extreme and mean ratio at $O$ such that $MO$ is the greater segment.
Then from:
:
and the converse of:
:
it follows that:
:$MO$ is the side of a regular decagon which has been inscribed in the circle whose radius is $MN$.
From :
:$5 \cdot MN^2 = AB^2 = 3 \cdot DG^2$
Let $DG$ be cut in extreme and mean ratio such that $CG$ is the greater segment.
From the :
:if two straight lines have been cut in extreme and mean ratio, their segments are in the same ratio.
So:
:$3 \cdot DG^2 : 3 \cdot CG^2 = 5 \cdot MN^2 : 5 \cdot MO^2$
From :
:$MO$ is the side of a regular decagon which has been inscribed
and
:$KL$ is the side of a regular pentagon which has been inscribed
in the circle whose radius is $MN$.
From 
:$MN$ is the side of a regular hexagon which has also been inscribed in that same circle.
Thus from :
:$5 \cdot MO^2 + 5 \cdot MN^2 = 5 \cdot KL^2$
Therefore:
:$5 \cdot KL^2 = 3 \cdot CG^2 + 3 \cdot DG^2$
But from :
:$5 \cdot KL^2$ equals the square on $15$ times the radius of the circle around $KLH$.
From :
:$3 \cdot DG^2 + 3 \cdot CG^2$ equals the square on $15$ times the radius of the circle around $CDEFG$.
Therefore the two radii are equal.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Regular Dodecahedra
Category:Regular Icosahedra
Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere/Lemma

== Lemma to Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere ==
== Proof ==
:
Let $ABC$ be a circle.
Let $AC$ be the side of a regular pentagon which has been inscribed within $ABC$.
Let $D$ be the center of the circle $ABC$.
Let $DF$ be drawn from $D$ perpendicular to $AC$, and produce it both ways to meet $ABC$ in $B$ and $E$.
Let $AB$ and $AE$ be joined.
It is to be demonstrated that:
:$BA^2 + AC^2 = 5 \cdot DE^2$
We have that:
Also:
But from :
:$AC^2 = DE^2 + EA^2$
Therefore:
:$BA^2 + AC^2 = 5 \cdot DE^2$
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Pentagons
Circle Group is Group

== Theorem ==
The circle group $\struct {K, \times}$ is a group.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Also see ==
* Circle Group is Infinite Abelian Group
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.2$
* : $\S 1$: Some examples of groups: Example $1.08$
Category:Circle Group
Category:Complex Numbers
Category:Circle Group is Group
Circle Group is Group/Proof 1

== Theorem ==
== Proof ==
First we note that $K \subseteq \C$.
So to show that $K$ is a group it is sufficient to show that $K$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.
From Complex Multiplication Identity is One, the identity element $1 + 0 i$ is in $K$.
Thus $K \ne \O$.
We now show that $z, w \in K \implies z w \in K$:
Next we see that $z \in K \implies z^{-1} \in K$:
:$\cmod z = 1 \implies \cmod {\dfrac 1 z} = 1$
Thus by the Two-Step Subgroup Test:
:$K \le \C_{\ne 0}$
Thus $K$ is a group.
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
Category:Circle Group is Group
Circle Group is Group/Proof 2

== Theorem ==
== Proof ==
We note that $K \ne \O$ as the identity element $1 + 0 i \in K$.
Since all $z \in K$ have modulus $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the polar form:
:$z = \map \exp {i \theta} = \cos \theta + i \sin \theta$
Conversely, if a complex number has such a polar form, it has modulus $1$.
Observe the following property of the complex exponential function:
:$\forall a, b \in \C: \map \exp {a + b} = \map \exp a \map \exp b$
We must show that if $x, y \in K$ then $x \cdot y^{-1} \in K$.
Let $x, y \in K$ be arbitrary.
Choose suitable $s, t \in \hointr 0 {2 \pi}$ such that:
:$x = \map \exp {i s}$
:$y = \map \exp {i t}$
We compute:
So $y^{-1} = \map \exp {-i t}$.
We note that this lies in $K$.
Furthermore, we have:
We conclude that $x y \in K$.
By the Two-Step Subgroup Test, $K$ is a subgroup of $\C$ under complex multiplication.
Category:Circle Group is Group
Circle Group is Group/Proof 3

== Theorem ==
== Proof ==
Taking the group axioms in turn:
=== ===
So $\struct {K, \times}$ is closed.
=== ===
Complex Multiplication is Associative.
=== ===
From Complex Multiplication Identity is One we have that the identity element of $K$ is $1 + 0 i$.
=== ===
We have that:
:$\cmod z = 1 \implies \dfrac 1 {\cmod z} = \cmod {\dfrac 1 z} = 1$
But:
:$z \times \dfrac 1 z = 1 + 0 i$
So the inverse of $z$ is $\dfrac 1 z$.
All the group axioms are satisfied, and the result follows.
== Sources ==
* : Chapter $2$: Examples of Group Structure: $\S 31$
Category:Circle Group is Group
Circle Group is Group/Proof 4

== Theorem ==
== Proof ==
Consider the complex modulus function $\cmod {\, \cdot \,}: \C \to \R, z \mapsto \cmod z$.
By Complex Modulus is Norm, we have that $\cmod z \ge 0$ for all $z \in \C$, and:
:$\cmod z = 0 \iff z = 0$
Let $\C_{\ne 0} := \C \setminus \set 0$ denote the complex numbers without zero.
From Group of Units of Field and Complex Numbers form Field, we have that $\struct {\C_{\ne 0}, \times}$ is a group.
By the above observation, the modulus has a restriction to $\C_{\ne 0}$:
:$\cmod {\, \cdot \,}: \C_{\ne 0} \to \R_{>0}, z \mapsto \cmod z$
From $\cmod 1 = 1$ and Modulus of Product, it follows that $\phi$ is in fact a group homomorphism:
:$\phi: \struct {C_{\ne 0}, \times} \to \struct {\R_{>0}, \times}, z \mapsto \cmod z$
Now $K$ is by definition the kernel of $\phi$.
Hence, by Kernel of Group Homomorphism is Subgroup, $\struct {K, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.
Category:Circle Group is Group
Circle Group is Infinite Abelian Group

== Theorem ==
The circle group $\struct {K, \times}$ is an uncountably infinite abelian group.
== Proof ==
From Circle Group is Group, we have that $\struct {K, \times}$ is a group.
As $K \subseteq \C$, it follows by definition that $\struct {K, \times}$ is a subgroup of $\struct {\C, \times}$.
From Complex Multiplication is Commutative, $\times$ is commutative on $\C$.
From Restriction of Commutative Operation is Commutative, it follows that $\times$ is commutative on $K$.
Hence, by definition, $\struct {K, \times}$ is an abelian group.
Finally we have that the Circle Group is Uncountably Infinite.
Category:Circle Group
Category:Complex Numbers
Category:Circle Group is Infinite Abelian Group
Circle Group is Uncountably Infinite

== Theorem ==
The circle group $\struct {K, \times}$ is an uncountably infinite group.
== Proof ==
From Quotient Group of Reals by Integers is Circle Group, $\struct {K, \times}$ is isomorphic to the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$.
But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the half-open interval $\hointr 0 1$.
A real interval is uncountable by Real Numbers are Uncountable.
Hence the result.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38$
Category:Circle Group
Category:Uncountable Sets
Circle is Bisected by Diameter/Historical Note

== Historical Note on Circle is Bisected by Diameter ==
The result that a Circle is Bisected by Diameter was supposedly attributed to by .
 defines the diameter as the line which passes through the center, but then assumes that it necessarily bisects it:
:
According to in his :
:It seems strange that , a disciple of the Egyptians, should bother to demonstrate something which takes as self-evident. suggests that he proved it by folding the circle over the diameter.
== Sources ==
* : Chapter $\text {A}.1$: Thales (ca. B.C.)
Category:Historical Notes
Circle is Bisected by Diameter/Proof 1

== Theorem ==
== Proof ==
Let $AB$ be a diameter of a circle $ADBE$ whose center is at $C$.
By definition of diameter, $AB$ passes through $C$.
 that $AB$ does not bisect $ADBE$, but that $ADBC$ is larger than $AEBC$.
:
Thus it will be possible to find a diameter $DE$ passing through $C$ such that $DC \ne CE$.
Both $DC$ and $CE$ are radii of $ADBE$.
By Euclid's definition of the circle:
:
:
That is, all radii of $ADBE$ are equal.
But $DC \ne CE$.
From this contradiction it follows that $AB$ bisects the circle.
== Historical Note ==
Category:Circle is Bisected by Diameter
Circle is Bisected by Diameter/Proof 2

== Theorem ==
== Proof ==
:
Let $AB$ be a diameter of a circle whose center is at $O$.
By definition of diameter, $AB$ passes through $O$.
$\angle AOB\cong\angle BOA$ because they are both straight angles.
Thus, the arcs are congruent by Equal Angles in Equal Circles:
:$\stackrel{\frown}{AB}\cong\stackrel{\frown}{BA}$
Hence, a circle is split into two equal arcs by a diameter.
== Historical Note ==
Category:Circle is Bisected by Diameter
Circle is Curve of Second Degree

== Theorem ==
The circle is a curve of degree $2$.
== Proof ==
From Equation of Circle in Cartesian Plane, a circle can be expressed in the form:
:$x^2 + y^2 - 2 a x - 2 b y + a^2 + b^2 - R^2 = 0$
where $\tuple {a, b}$ is the center and $R$ is the radius.
This is a quadratic equation in $2$ variables.
Hence the result by definition of degree of curve.
== Sources ==
* : Chapter $\text {III}$. The Circle: $14$. To find the equation of the circle whose centre is $\tuple {\alpha, \beta}$ and radius $r$
Category:Circles
Circle is Ellipse with Equal Major and Minor Axes

== Theorem ==
Let $E$ be an ellipse whose major axis is equal to its minor axis.
Then $E$ is a circle.
== Proof ==
Let $E$ be embedded in a Cartesian plane in reduced form.
Then from Equation of Ellipse in Reduced Form $E$ can be expressed using the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
where the major axis and minor axis are $a$ and $b$ respectively.
Let $a = b$.
Then:
which by Equation of Circle center Origin is the equation of a circle whose radius is $a$.
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $2$. To find the equation of the ellipse in its simplest form
* 
* 
Category:Circles
Category:Ellipses
Circle of Apollonius in Complex Plane

== Theorem ==
Let $\C$ be the complex plane.
Let $\lambda \in \R$ be a real number such that $\lambda \ne 0$ and $\lambda \ne 1$.
Let $a, b \in \C$ such that $a \ne b$.
The equation:
:$\cmod {\dfrac {z - a} {z - b} } = \lambda$
decribes a circle of Apollonius $C$ in $\C$ such that:
:if $\lambda < 0$, then $a$ is inside $C$ and $b$ is outside
:if $\lambda > 0$, then $b$ is inside $C$ and $a$ is outside.
If $\lambda = 1$ then $z$ describes the perpendicular bisector of the line segment joining $a$ to $b$.
== Proof ==
By the geometry, the locus described by this equation is a circle of Apollonius.
== Examples ==
== Sources ==
* : $\S 2$. Geometrical Representations: Exercise $8$
Category:Equation of Circle in Complex Plane
Category:Circle of Apollonius
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius in Complex Plane/Examples

== Examples of Circle of Apollonius in Complex Plane ==
=== Example: $\cmod {\dfrac {z - 3} {z + 3} } = 2$ ===
=== Example: $\cmod {\dfrac {z - 3} {z + 3} } < 2$ ===
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius in Complex Plane/Examples/mod z-3 over z+3 = 2

== Example of Circle of Apollonius in Complex Plane ==
The equation:
:$\cmod {\dfrac {z - 3} {z + 3} } = 2$
describes a circle embedded in the complex plane whose radius is $4$ and whose center is $\paren {-5, 0}$.
== Proof 1 ==
== Proof 2 ==
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius in Complex Plane/Examples/mod z-3 over z+3 = 2/Proof 1

== Example of Circle of Apollonius in Complex Plane ==
== Proof ==
A point $P$ on this circle is $2$ times the distance from $z = 3$ as it is from $z = -3$.
:
Let $z = x + i y$.
Hence the result by Equation of Circle in Complex Plane.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $48 \ \text{(a)}$
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius in Complex Plane/Examples/mod z-3 over z+3 = 2/Proof 2

== Example of Circle of Apollonius in Complex Plane ==
== Proof ==
Hence the result by Equation of Circle in Complex Plane.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $48 \ \text{(a)}$
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius in Complex Plane/Examples/mod z-3 over z+3 less than 2

== Example of Circle of Apollonius in Complex Plane ==
The inequality:
:$\cmod {\dfrac {z - 3} {z + 3} } < 2$
describes the exterior of a circle embedded in the complex plane whose radius is $4$ and whose center is $\paren {-5, 0}$.
== Proof ==
Hence the result by Equation of Circle in Complex Plane.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $48 \ \text{(b)}$
Category:Circle of Apollonius in Complex Plane
Circle of Apollonius is Circle

== Theorem ==
Let $A, B$ be distinct points in the plane.
Let $\lambda \in \R_{>0}$ be a strictly positive real number.
Let $X$ be the locus of points in the plane such that:
:$XA = \lambda \paren {XB}$
Then $X$ is in the form of a circle, known as a circle of Apollonius.
:
If $\lambda < 1$, then $A$ is inside the circle, and $B$ is outside.
If $\lambda > 1$, then $B$ is inside the circle, and $A$ is outside.
== Proof 1 ==
== Proof 2 ==
Category:Circle of Apollonius is Circle
Category:Circle of Apollonius
Circle of Apollonius is Circle/Proof 1

== Theorem ==
== Proof ==
:
Let $P$ be an arbitrary point such that $\dfrac {AP} {PB} = \lambda$.
Let $\angle APB$ be bisected internally and externally to intersect $AB$ at $X$ and $Y$ respectively.
Then:
:$\dfrac {AX} {XB} = \dfrac {AP} {PB} = \lambda$
and:
:$\dfrac {AY} {YB} = \dfrac {AP} {PB} = \lambda$
Thus $X$ and $Y$ are the points which divide $AB$ internally and externally in the given ratio.
Therefore, for a given $A$, $B$ and $\lambda$, $X$ and $Y$ are fixed.
From Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular, $\angle XPY$ is a right angle.
Hence by Thales' Theorem, $P$ lies on the circumference of a circle of which $XY$ is a diameter.
== Sources ==
* : $\text {IV}$. Pure Geometry: Plane Geometry: Apollonius' Circle
Category:Circle of Apollonius is Circle
Circle of Apollonius is Circle/Proof 2

== Theorem ==
== Proof ==
Let $A = \tuple {x_a, y_a}, B = \tuple {x_b, y_b}$.
Let $X = \tuple {x, y}$.
We have:
From Equation of Circle in Cartesian Plane, this is an equation in the form:
:$A \paren {x^2 + y^2} + B x + C y + D = 0$
with radius $R$ and center $\tuple {a, b}$, where:
:$R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
:$\tuple {a, b} = \tuple {\dfrac B {2 A}, \dfrac C {2 A} }$
Here, we have:
Hence the center $\tuple {a, b}$ can be evaluated:
Then the radius can be evaluated:
Category:Circle of Apollonius is Circle
Circles Touch at One Point at Most

== Theorem ==
== Proof ==
 it is possible for two circles to touch at more points than one.
:
First, let the circle $ABDC$ touch the circle $EBFD$ internally at more than one point, that is, at $B$ and $D$.
Let $G$ be the center of the circle $ABDC$, and $H$ be the center of the circle $EBFD$.
(It is clear that in the diagram these centers are not actually at $G$ and $H$, and in fact $EBFD$ is obviously not a circle - it is the point of this proof to demonstrate that this would not be possible.)
From Line Joining Centers of Two Circles Touching Internally the straight line $GH$ will pass through both $B$ and $D$.
Since $G$ is the center of the circle $ABDC$, we have:
But since $H$ is the center of the circle $EBFD$, we have that $BH = HD$.
But we have just shown that $BH \gg HD$, which is impossible.
Therefore a circle does not touch another circle internally at more than one point.
Next suppose the circle $ACK$ touches the circle $ABDC$ at more points than one, that is, at $A$ and $C$, and join $AC$.
The two points $A$ and $C$ fall on the circumference of both circles $ABDC$ and $ACK$.
So it follows from Chord Lies Inside its Circle that $AC$ lies with both circles.
But from this line would fall inside $ABDC$ and outside $ACK$.
This reveals a contradiction.
Therefore a circle does not touch another circle externally at more than one point.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Circles with Same Poles are Parallel

== Theorem ==
Let $S$ be a sphere.
Let $C$ and $D$ be circles on $S$ (either great circles or small circles).
Let $C$ and $D$ both have the same pair of poles.
Then $C$ and $D$ are parallel.
== Proof ==
Let the poles of $C$ and $D$ be $A$ and $B$.
$C$ and $D$ each lie embedded in a plane.
Both of these planes by definition are perpendicular to $AB$.
The result follows from Planes Perpendicular to same Straight Line are Parallel.
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $2$. The spherical triangle.
Category:Spherical Geometry
Circuit of Simple Graph has Three Edges or More

== Theorem ==
Let $G$ be a simple graph.
Let $C$ be a circuit in $G$.
Then $C$ has at least $3$ edges.
== Proof ==
By definition, a circuit contains at least one edge.
That is, a single vertex does not comprise a trivial degenerate circuit with no edges.
Suppose $C$ contains only one edge $e$.
As $G$ is a simple graph it contains no loops.
Therefore $e$ starts and ends at two distinct vertices.
Therefore $C$, consisting entirely of $e$ can not be a circuit.
Suppose $C$ contains exactly $2$ edges $e$ and $f$.
Let $e = u v$ where $u$ and $v$ are vertices of $G$.
Then for $C$ to be a circuit it must follow that $f = v u$.
Thus there are two edges between the same vertices $v$ and $u$.
As $G$ is an undirected graph, it must be that $e$ and $f$ are part of a multiple edge between $u$ and $v$.
But $G$ is not a multigraph, and so such a circuit can not exist.
A circuit containing $3$ edges definitely exists, though.
Let $G = \struct {V, E}$ where $V = \set {u, v, w}$ and $E = \set {\set {u, v}, \set {u, w}, \set {v, w} }$.
Then $\tuple {u, w, v, u}$ is a $3$-edge circuit.
== Sources ==
* : $\S 2.3$: Connected Graphs
Category:Circuits (Graph Theory)
Category:Simple Graphs
Circuits of Matroid iff Matroid Circuit Axioms

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Then:
:$\mathscr C$ is the set of circuits of a matroid on $S$
:$\mathscr C$ satisfies the circuit axioms
== Proof ==
From Equivalence of Definitions of Matroid Circuit Axioms it is sufficient to show:
:$(\text a)\quad$if $\mathscr C$ is the set of circuits of a matroid then $\mathscr C$ satisfies circuit axioms (formulation 1)
and
:$(\text b)\quad$if $\mathscr C$ satisfies circuit axioms (formulation 2) then $\mathscr C$ is the set of circuits of a matroid
=== Circuits of Matroids implies Formulation 1 ===
Let $\mathscr C$ be the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$
It is shown that $\mathscr C$ satisfies the circuit axioms (formulation 1):
=== Formulation 2 implies Circuits of Matroid ===
Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms (formulation 2):
== Also see ==
* Equivalence of Definitions of Matroid Circuit Axioms
Category:Matroid Circuits
Category:Circuits of Matroid iff Matroid Circuit Axioms
Circuits of Matroid iff Matroid Circuit Axioms/Circuits of Matroid implies Formulation 1

== Theorem ==
Let $S$ be a finite set.
Let $\mathscr C$ be a non-empty set of subsets of $S$.
Let $\mathscr C$ be the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$
Then:
:$\mathscr C$ satisfies the circuit axioms:
== Proof ==
==== $\mathscr C$ satisfies $(\text C 1)$ ====
By definition of circuit of a matroid:
:for all $C \in \mathscr C: C$ is a dependent subset
By definition of a dependent subset:
:for all $C \in \mathscr C$, $C \notin \mathscr I$
By definition of a matroid:
:$\O \in \mathscr I$
Hence:
:$\O \notin \mathscr C$
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 1)$.
==== $\mathscr C$ satisfies $(\text C 2)$ ====
Let $C_1, C_2 \in \mathscr C : C_1 \neq C_2$.
By definition of circuit of a matroid:
:$C_2$ is a dependent subset of $S$ which is a minimal dependent subset
and
:$C_1$ is a dependent subset
Hence:
:$C_1 \nsubseteq C_2$
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 2)$.
==== $\mathscr C$ satisfies $(\text C 3)$ ====
Let $\rho$ denote the rank function of the matroid $M$.
 $C_1, C_2 \in \mathscr C:$
:$C_1 \neq C_2$
and
:$\exists z \in C_1 \cap C_2 : \nexists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$
From Proper Subset of Matroid Circuit is Independent:
:$\paren{C_1 \cup C_2} \setminus \set z$ is an independent subset.
Similarly:
:$\paren{C_1 \cap C_2}$ is an independent subset.
We have:
This is a contradiction.
It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.
== Sources ==
* Chapter $1.$ $\S 9.$ Circuits
Category:Circuits of Matroid iff Matroid Circuit Axioms