hcju/proofwikiqa huggingface dataset:
We have that $\angle BCD + \angle ACB$ make a straight angle. As $CD = AB$ by construction, $CD = BC$ by definition of radius of circle. Thus $\triangle BCD$ is isosceles. By Isosceles Triangle has Two Equal Angles: :$\angle CBD = \angle CDB$ From Sum of Angles of Triangle equals Two Right Angles: :$\angle BCD + 2 \angle CBD$ equals two right angles. Thus: :$2 \angle CBD = \angle ACB$ Similarly, by Isosceles Triangle has Two Equal Angles: :$\angle ACB = \angle CAB$ and again from Sum of Angles of Triangle equals Two Right Angles: :$\angle ABC + 2 \angle ACB$ equals two right angles. and so: :$\angle ABC + 4 \angle CBD$ equals two right angles. But $\alpha + \angle ABC + \angle CBD$ make a straight angle. Thus: :$\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$ and so: :$\alpha = 3 \angle CBD$ {{qed}}
Let $OA$ and $OB$ intersect at $O$. It will be shown that there is no general method using a compass and straightedge construction to construct $OC$ such that $\angle AOB = 3 \times \angle AOC$. It is sufficient to demonstrate that this is impossible for one specific angle. Hence we choose $\angle AOB = 60 \degrees$. Let $A$ and $B$ be points on the unit circle whose center is at $\tuple {0, 0}$. Let $A$ lie on the $x$-axis. Thus: :$O$ is the point $\tuple {0, 0}$ :$A$ is the point $\tuple {1, 0}$ :$B$ is the point $\tuple {\cos 60 \degrees, \sin 60 \degrees}$ These all belong to $\Q \sqbrk {\sqrt 3}$. trisection of $AOB$ is equivalent to constructing the point $\tuple {\cos 20 \degrees, \sin 20 \degrees}$. From Triple Angle Formula for Cosine: :$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$ so: :$8 \cos^3 20 \degrees - 6 \cos 20 \degrees = 2 \cos 60 \degrees = 1$ Thus $\cos 20 \degrees$ is a root of the polynomial: :$8 x^3 = 6 x - 1$ which by Irreducible Polynomial: $8 x^3 - 6 x - 1$ in Rationals is irreducible over $\Q$. Thus $\cos 20 \degrees$ is algebraic over $\Q$ with degree $3$. Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos 20 \degrees$ is not an element of any extension of $\Q$ of degree $2^m$. The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2. {{qed}}
Suppose it is possible. Then from a cube of edge length $L$ we can construct a new cube with edge length $\sqrt [3] 2 L$. $\sqrt [3] 2$ is algebraic of degree $3$. This contradicts Constructible Length with Compass and Straightedge. {{qed}}
Squaring the Circle consists of constructing a line segment of length $\sqrt \pi$ of another. From Constructible Length with Compass and Straightedge, any such line segment has a length which is an algebraic number of degree $2$. But $\pi$ is transcendental. Hence $\pi$ and therefore $\sqrt \pi$ is not such an algebraic number. Therefore any attempt at such a construction will fail. {{Qed}}
From Body under Constant Acceleration: Distance after Time: :$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$ Here the body falls from rest, so: :$\mathbf u = \mathbf 0$ Thus: :$\mathbf s = \dfrac {\mathbf g t^2} 2$ and so taking magnitudes: :$s = \dfrac {g t^2} 2$ It follows by multiplying by $\dfrac 2 g$ that: :$t^2 = \dfrac {2 s} g$ whence: :$t = \sqrt {\dfrac {2 s} g}$ {{qed}}
We have $AO = BO$ since they are radii. Therefore $\triangle AOB$ is isosceles. By Isosceles Triangle has Two Equal Angles: :$\angle OAB = \angle OBA$ By Sum of Angles of Triangle equals Two Right Angles: :$\angle OAB + \angle OBA + \theta = 180 \degrees$ Therefore $\angle OAB = \dfrac {180 \degrees - \theta} 2 = 90 \degrees - \dfrac \theta 2$. Thus: {{begin-eqn}} {{eqn | l = \dfrac {AB} {\sin \theta}       | r = \dfrac {BO} {\sin \angle OAB}       | c = Law of Sines }} {{eqn | l = {AB}       | r = \dfrac {BO \sin \theta} {\sin \angle OAB}       | c =  }} {{eqn | r = \dfrac {2 r \sin \frac \theta 2 \cos \frac \theta 2} {\map \sin {90 \degrees - \frac \theta 2} }        | c = Double Angle Formula for Sine }} {{eqn | r = \dfrac {2 r \sin \frac \theta 2 \cos \frac \theta 2} {\cos \frac \theta 2}       | c = Sine of Supplementary Angle }} {{eqn | r = 2 r \sin \dfrac \theta 2       | c =  }} {{end-eqn}} {{qed}}
:300px Let $O$ be the center of $C$. Let $AB$ be bisected by $OD$. Consider the pair of triangles $\triangle AOE$ and $\triangle BOE$. We see that: :$AE = ED$ since $AB$ is bisected by $OD$ :$AO = BO$ since they are radii :$OE = OE$ since they are common sides. By Triangle Side-Side-Side Equality, $\triangle AOE = \triangle BOE$. Then we have: :$\angle AOE = \angle BOE = \dfrac \theta 2$ :$\angle OEA = \angle OEB = \dfrac {180 \degrees} 2 = 90 \degrees$ By {{Defof|Sine Function}}: :$\sin \dfrac \theta 2 = \dfrac {AE} {AO} = \dfrac {\frac 1 2 AB} r$ Rearranging, we get: :$AB = 2 r \sin \dfrac \theta 2$ as desired. {{qed}}
:450px Let $ABCD$ be a cyclic quadrilateral. By Angles in Same Segment of Circle are Equal: :$\angle BAC = \angle BDC$ and: :$\angle ADB = \angle ACB$ By Construction of Equal Angle, construct $E$ on $AC$ such that: :$\angle ABE = \angle CBD$ Since: :$\angle ABE + \angle CBE = \angle ABC = \angle CBD + \angle ABD$ it follows that: :$\angle CBE = \angle ABD$ By Equiangular Triangles are Similar: :$\triangle ABE$ is similar to $\triangle DBC$ and: :$\triangle ABD$ is similar to $\triangle EBC$ Thus: :$\dfrac {AE} {AB} = \dfrac {CD} {BD}$ and: :$\dfrac {CE} {BC} = \dfrac {DA} {BD}$ Equivalently: :$AE \times BD = AB \times CD$ and: :$CE \times BD = BC \times DA$ Adding: :$AE \times BD + CE \times BD = AB \times CD + BC \times DA$ Factorizing: :$\paren {AE + CE} \times BD = AB \times CD + BC \times DA$ But: :$AE + CE = AC$ so: :$AC \times BD = AB \times CD + BC \times DA$ {{qed}} {{Namedfor|Claudius Ptolemy|cat = Ptolemy}}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$. Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively. From Spherical Triangle is Polar Triangle of its Polar Triangle we have that: :not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$ :but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. We have: {{begin-eqn}} {{eqn | l = \cos a'       | r = \cos b' \cos c' + \sin b' \sin c' \cos A'       | c = Spherical Law of Cosines }} {{eqn | ll= \leadsto       | l = \map \cos {\pi - A}       | r = \map \cos {\pi - B} \, \map \cos {\pi - C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \map \cos {\pi - a}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = -\cos A       | r = \paren {-\cos B} \paren {-\cos C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \paren {-\cos a}       | c = Cosine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = -\cos A       | r = \paren {-\cos B} \paren {-\cos C} + \sin B \sin C \paren {-\cos a}       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \cos A       | r = -\cos B \cos C + \sin B \sin C \cos a       | c = simplifying and rearranging }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan \dfrac {A + B} 2       | r = \dfrac {\cos \frac {a - b} 2} {\cos \frac {a + b} 2} \cot \dfrac C 2       | c = Napier's Analogies }} {{eqn | n = 1       | ll= \leadsto       | l = \tan \frac {A + B} 2 \cos \frac {a + b} 2       | r = \cos \frac {a - b} 2 \cot \frac C 2       | c = more manageable in this form }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \tan \dfrac {A - B} 2       | r = \dfrac {\sin \frac {a - b} 2} {\sin \frac {a + b} 2} \cot \dfrac C 2       | c = Napier's Analogies }} {{eqn | n = 2       | ll= \leadsto       | l = \tan \frac {A - B} 2 \sin \frac {a + b} 2       | r = \sin \frac {a - b} 2 \cot \frac C 2       | c = more manageable in this form }} {{end-eqn}} Hence we have: {{begin-eqn}} {{eqn | l = \dfrac {\tan \frac {A + B} 2 \cos \frac {a + b} 2} {\tan \frac {A - B} 2 \sin \frac {a + b} 2}       | r = \dfrac {\cos \frac {a - b} 2 \cot \frac C 2} {\sin \frac {a - b} 2 \cot \frac C 2}       | c = dividing $(1)$ by $(2)$ }} {{eqn | ll= \leadsto       | l = \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2} \dfrac 1 {\tan \frac {a + b} 2}       | r = \dfrac 1 {\tan \frac {a - b} 2}       | c = simplifying }} {{eqn | ll= \leadsto       | l = \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2}       | r = \dfrac {\tan \frac {a + b} 2} {\tan \frac {a - b} 2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos a       | r = \cos b \cos c + \sin b \sin c \cos A       | c = Spherical Law of Cosines }} {{eqn | r = \cos b \cos c + \sin b \sin c \paren {2 \cos^2 \dfrac A 2 - 1}       | c = Double Angle Formula for Cosine: Corollary 1 }} {{eqn | r = \map \cos {b + c} + 2 \sin b \sin c \cos^2 \dfrac A 2       | c = Cosine of Sum }} {{eqn | ll= \leadsto       | l = \cos a - \map \cos {b + c}       | r = 2 \sin b \sin c \cos^2 \dfrac A 2       | c = rearranging }} {{eqn | ll= \leadsto       | l = 2 \sin \dfrac {a + \paren {b + c} } 2 \sin \dfrac {\paren {b + c} - a} 2       | r = 2 \sin b \sin c \cos^2 \dfrac A 2       | c = Prosthaphaeresis Formula for Cosine minus Cosine }} {{eqn | ll= \leadsto       | l = \map \sin {\dfrac {a + b + c} 2} \, \map \sin {\dfrac {a + b + c} 2 - a}       | r = \sin b \sin c \cos^2 \dfrac A 2       | c =  }} {{eqn | ll= \leadsto       | l = \sin s \, \map \sin {s - a}       | r = \sin b \sin c \cos^2 \dfrac A 2       | c = setting $s = \dfrac {a + b + c} 2$ and simplifying }} {{end-eqn}} The result follows. {{qed}}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$. Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively. From Spherical Triangle is Polar Triangle of its Polar Triangle we have that: :not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$ :but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$. Let $s' = \dfrac {a' + b' + c'} 2$. We have: {{begin-eqn}} {{eqn | l = \sin \dfrac {A'} 2       | r = \sqrt {\dfrac {\sin \paren {s' - b'} \sin \paren {s' - c'} } {\sin b' \sin c'} }       | c = Sine of Half Angle for Spherical Triangles }} {{eqn | ll= \leadsto       | l = \sin \dfrac {\pi - a} 2       | r = \sqrt {\dfrac {\map \sin {s' - b'}  \, \map \sin {s' - c'} } {\map \sin {\pi - B} \, \map \sin {\pi - C} } }       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle }} {{eqn | ll= \leadsto       | l = \map \sin {\dfrac \pi 2 - \dfrac a 2}       | r = \sqrt {\dfrac {\map \sin {s' - b'}  \, \map \sin {s' - c'} } {\sin B \sin C} }       | c = Sine of Supplementary Angle }} {{eqn | ll= \leadsto       | l = \cos \dfrac a 2       | r = \sqrt {\dfrac {\map \sin {s' - b'}  \, \map \sin {s' - c'} } {\sin B \sin C} }       | c = Sine of Complement equals Cosine }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = s' - b'       | r = \dfrac {\paren {\pi - A} + \paren {\pi - B} + \paren {\pi - C} } 2 - \paren {\pi - B}       | c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle  }} {{eqn | r = \dfrac {\pi - \paren {A + B + C} } 2 + B       | c = simplifying }} {{eqn | r = \dfrac \pi 2 - \paren {S - B}       | c = where $S = \dfrac {A + B + C} 2$ }} {{eqn | ll= \leadsto       | l = \map \sin {s' - b'}       | r = \map \sin {\dfrac \pi 2 - \paren {S - B} }       | c =  }} {{eqn | r = \map \cos {S - B}       | c = Sine of Complement equals Cosine }} {{end-eqn}} and similarly: :$\map \sin {s' - c'} = \map \cos {S - C}$ The result follows. {{qed}}
Let the point $\tuple {x, y}$ satisfy the equation: :$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 = R^2$ By the Distance Formula, the distance between this $\tuple {x, y}$ and $\tuple {a, b}$ is: :$\sqrt {\paren {x - a}^2 + \paren {y - b}^2}$ But from equation $(1)$, this quantity equals $R$. Therefore the distance between points satisfying the equation and the center is constant and equal to the radius. Thus $\tuple {x, y}$ lies on the circumference of a circle with radius $R$ and center $\tuple {a, b}$. Now suppose that $\tuple {x, y}$ does not satisfy the equation: :$\paren {x - a}^2 + \paren {y - b}^2 = R^2$ Then by the same reasoning as above, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ does not equal $R$. Therefore $\tuple {x, y}$ does not lie on the circumference of a circle with radius $R$ and center $\tuple {a, b}$. Hence it follows that the points satisfying $(1)$ are exactly those points which are the circle in question. {{qed}}
Let the point $\tuple {x, y}$ satisfy the equations: :$x = a + R \cos t$ :$y = b + R \sin t$ By the Distance Formula, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ is: :$\sqrt {\paren {\paren {a + R \cos t} - a}^2 + \paren {\paren {b + R \sin t} - b}^2}$ This simplifies to: :$\sqrt {R^2 \cos^2 t + R^2 \sin^2 t} = R \sqrt {\cos^2 t + \sin^2 t}$ Then by Sum of Squares of Sine and Cosine, this distance equals $R$. Therefore the distance between points satisfying the equation and the center is constant and equal to the radius. {{qed}}
Let the point $\polar {r, \theta}_\text {Polar}$ satisfy the equation: :$r^2 - 2 r r_0 \map \cos {\theta - \varphi} + \paren {r_0}^2 = R^2$ Let the points $\polar {r, \theta}$ and $\polar {r_0, \varphi}$ be rewritten in Cartesian coordinates: :$\polar {r, \theta}_\text {Polar} = \tuple {r \cos \theta, r \sin \theta}_\text{Cartesian}$ :$\polar {r_0, \varphi}_\text{Polar} = \tuple {r_0 \cos \varphi, r_0 \sin \varphi}_\text{Cartesian}$ Thus the distance between $\polar {r, \theta}_\text {Polar}$ and $\polar {r_0, \varphi}_\text{Polar}$ is: {{begin-eqn}} {{eqn | o =        | r = \sqrt {\paren {r \cos \theta - r_0 \cos \varphi}^2 + \paren {r \sin \theta - r_0 \sin \varphi}^2} }} {{eqn | r = \sqrt {r^2 \cos^2 \theta + \paren {r_0}^2 \cos^2 \varphi - 2 r r_0 \cos \theta \cos \varphi + r^2 \sin^2 \theta + \paren {r_0}^2 \sin^2 \varphi - 2 r r_0 \sin \theta \sin \varphi} }} {{eqn | r = \sqrt {r^2 \paren {\cos^2 \theta + \sin^2 \theta} + \paren {r_0}^2 \paren {\cos^2 \varphi + \sin^2 \varphi} - 2 r r_0 \paren {\cos \theta \cos \varphi + \sin \theta \sin \varphi} } }} {{end-eqn}} {{begin-eqn}} {{eqn | r = \sqrt {r^2 + \paren {r_0}^2 - 2 r r_0 \map \cos {\theta - \varphi} }       | c = Cosine of Difference and Sum of Squares of Sine and Cosine }} {{end-eqn}} But from the equation, this quantity equals $R$. Therefore the distance between points satisfying the equation and the center is constant and equal to the radius $R$. {{qed}} Category:Equation of Circle fl8juwo98qgj9lsefc4c4of07uedhad
{{begin-eqn}} {{eqn | l = A \paren {x^2 + y^2} + B x + C y + D       | r = 0       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x^2 + y^2 + \frac B A x + \frac C A y        | r = - \frac D A       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x^2 + 2 \frac B {2 A} x + \frac {B^2} {4 A^2} + y^2 + 2 \frac C {2 A} y + \frac {C^2} {4 A^2}       | r = \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac D A       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x + \frac B {2 A} }^2 + \paren {y + \frac C {2 A} }^2       | r = \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac D A       | c =  }} {{eqn | r = \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac {4 A D} {4 A^2}       | c =  }} {{eqn | r = \frac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}       | c =  }} {{end-eqn}} This last expression is non-negative {{iff}} $B^2 + C^2 \ge 4 A D$. In such a case $\dfrac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}$ is in the form $R^2$ and so: :$\paren {x + \dfrac B {2 A} }^2 + \paren {y + \dfrac C {2 A} }^2 = \dfrac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}$ is in the form: :$\paren {x - a}^2 + \paren {y - b}^2 = R^2$ Hence the result from Equation of Circle: Cartesian. {{qed}}
From Equation of Circle: Cartesian, the equation of a circle with radius $R$ and center $\tuple {a, b}$ expressed in Cartesian coordinates is: :$\paren {x - a}^2 + \paren {y - b}^2 = R^2$ Setting $a = b = 0$ yields the result. {{qed}}
:600px Let $d$ be the distance to be found between $A = \tuple {x_1, y_1, z_1}$ and $B = \tuple {x_2, y_2, z_2}$. Let the points $C$ and $D$ be defined as: :$C = \tuple {x_2, y_1, z_1}$ :$D = \tuple {x_2, y_2, z_1}$ Let $d'$ be the distance between $A$ and $D$. From Distance Formula, it can be seen that: :$d' = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$ We note that $\triangle ADB$ is a right triangle. Thus by Pythagoras's Theorem: :$AB^2 = AD^2 + DB^2$ Thus: {{begin-eqn}} {{eqn | l = d^2       | r = d'^2 + DB^2       | c =  }} {{eqn | r = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2 + \paren {z_1 - z_2}^2       | c =  }} {{end-eqn}} and so: :$d = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2 + \paren {z_1 - z_2}^2}$ as it was to be proved. {{qed}}
Let the point $\tuple {x, y, z}$ satisfy the equation: :$(1): \quad \paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$ By the Distance Formula in 3 Dimensions, the distance between this $\tuple {x, y, z}$ and $\tuple {a, b, c}$ is: :$\sqrt {\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2}$ But from equation $(1)$, this quantity equals $R$. Therefore the distance between points satisfying the equation and the center is constant and equal to the radius. Thus $\tuple {x, y, z}$ lies on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$. Now suppose that $\tuple {x, y, z}$ does not satisfy the equation: :$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$ Then by the same reasoning as above, the distance between $\tuple {x, y, z}$ and $\tuple {a, b, c}$ does not equal $R$. Therefore $\tuple {x, y, z}$ does not lie on the surface of a sphere with radius $R$ and center $\tuple {a, b, c}$. Hence it follows that the points satisfying $(1)$ are exactly those points which are the sphere in question. {{qed}}
From Equation of Sphere in Rectangular Coordinates, the equation of a sphere with radius $R$ and center $\tuple {a, b, c}$ expressed in Cartesian coordinates is: :$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$ Setting $a = b = c = 0$ yields the result. {{qed}}
By definition, a conic section is the set of points of intersection between a cone and a plane. Let $P = \tuple {\alpha, \beta, \gamma}$ be the apex of the cone. Let $Q = \tuple {x, y, z}$ be a point of intersection between the plane and the cone. From Equation of Right Circular Cone, we have: :$(1): \quad \paren {x - \alpha}^2 + \paren {y - \beta}^2 = \eta \paren {z - \gamma}^2$ for some (strictly) positive real number $\eta$. From Equation of Plane, we have: $A x + B y + C z + D = 0$ for some real numbers $A, B, C, D$. Let $C \ne 0$. Then: :$z = - D - \dfrac {A x + B y} C$ Hence, plugging in $z$ into $(1)$: :$\paren {x - \alpha}^2 + \paren {y - \beta}^2 = \eta \paren {\dfrac {A x + B y} C + D + \gamma}^2$ Opening up the brackets and making the corresponding notation of constants, we get the desired result. {{qed|lemma}} Let $C = 0$. Then: :$A x + B y + D = 0$ In order for the plane to be well-defined, $A$ or $B$ must be non-zero. {{WLOG}}, since the cone equation $(1)$ is symmetric in $x$ and $y$, we can consider $B \ne 0$ (otherwise interchange coordinates). Then: :$y = - D - \dfrac {A x} B$ Hence, plugging  $y$ into $(1)$: :$\paren {x - \alpha}^2 + \paren {-D - \dfrac {A x} B - \beta}^2 = \eta \paren {z - \gamma}^2$ Opening up the brackets and making the corresponding notation of constants and coordinates ($z \leftrightarrow y$), we get the desired result. {{qed}}
Consider the focus-directrix property of a parabola $P$. Let the focus of $P$ be the point $\tuple {0, f}$ on a Cartesian plane. Let the directrix of $P$ be the straight line $y = -d$. Let $\tuple {x, y}$ be an arbitrary point on $P$. Then by the focus-directrix property: :$y + d = \sqrt {\paren {x - k}^2 + \tuple {y - f}^2}$ where: :$y + d$ is the distance from $\tuple {x, y}$ to the straight line $y = -d$ :$\sqrt {\paren {x - k}^2 + \paren {y - f}^2}$ is the distance from $\tuple {x, y}$ to the point $\tuple {k, f}$ by the Distance Formula. Hence: {{begin-eqn}} {{eqn | l = \paren {y + d}^2       | r = \paren {x - k}^2 + \paren {y - f}^2       | c =  }} {{eqn | ll= \leadsto       | l = y^2 + 2 y d + d^2       | r = x^2 - 2 k x + k^2 + y^2 - 2 f y + f^2       | c =  }} {{eqn | ll= \leadsto       | l = 2 y \paren {f + d}       | r = x^2 - 2 k x + f^2 + k^2 - d^2       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \frac 1 {2 \paren {f + d} } x^2 - \frac k {\paren {f + d} } x + \frac {f - d} 2       | c =  }} {{end-eqn}} This is in the form $y = a x^2 + b^2 + c$. By setting $k$, $f$ and $d$ appropriately in terms of $a$, $b$ and $c$, the specific focus and directrix can be appropriately positioned. {{qed}} Category:Parabolas Category:Quadratic Equations Category:Graph of Quadratic describes Parabola 0clbumwrj6btk7hwlqwr202ws8raqz7
This is a particular instance of Graph of Quadratic describes Parabola, where: :$y = a x^2 + b x + c$ is the equation of a parabola. The result follows by setting $a = 1, b = 0, c = 0$. {{qed}}
From Graph of Quadratic describes Parabola: Corollary 1, where: :$y = x^2$ is the equation of a parabola. Let $f: \R \to \R$ be the real function defined as: :$\map f x = x^2$ From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$. Also we have from Positive Real Number has Two Square Roots: :$\forall x \in \R: \paren {-x}^2 = x^2$ Thus it is necessary to apply a bijective restriction upon $f$. Let $g: \R_{\ge 0} \to \R_{\ge 0}$ be the bijective restriction of $f$ to $\R_{\ge 0} \times \R_{\ge 0}$: :$\forall x \in \R_{\ge 0}: \map g x = x^2$ From Inverse of Bijection is Bijection, $g^{-1}: \R_{\ge 0} \to \R_{\ge 0}$ is also a bijection. By definition: :$\forall x \in \R_{\ge 0}: \map {g^{-1} } x = +\sqrt x$ Then from Graph of Inverse Mapping, the graph of $g^{-1}$ is the same as the graph of $g$, reflected in the line $x = y$. As the graph of $f$ is a parabola, the graph of $g$ is also a parabola, but because of the restriction to $\R_{\ge 0}$, just half of it. Thus the graph of $g^{-1}$ is also half a parabola. {{qed}}
Consider the natural numbers $\N$ defined as the naturally ordered semigroup. From the definition of the naturally ordered semigroup, it follows that $\left ({\N, +}\right)$ is a commutative semigroup. {{qed}}
This is simply the contrapositive of Gödel's First Incompleteness Theorem. {{qed}} {{Namedfor|Kurt Friedrich Gödel}}
Let $g: D \to C$ be the right inverse of $f$: :$f \circ g = \operatorname{id}_D$ which is guaranteed to exist by definition of split epimorphism. Therefore: :$f \circ g \circ f = \operatorname{id}_D \circ f = f \circ \operatorname{id}_C$ by the property of the identity morphism. Since $f$ is left cancellable, by the definition of monomorphism, we have: :$g \circ f = \operatorname{id}_C$ Hence $f$ is a split monomorphism with left inverse $g$. {{qed}} Category:Category Theory etbcg72blm50bar942619desslok0uv
{{explain|What is a Dual proof}} Dual proof of Monomorphism that is Split Epimorphism is Split Monomorphism. {{qed}} Category:Category Theory peimi58rmzw3im8toqfkj41g1bmf24k