hcju/proofwikiqa huggingface dataset:
:300px Let $ABC$ be a circle. Let $BC$ be the side of a regular decagon which has been inscribed within $ABC$. Let $BC$ be produced to $D$, where $CD$ is equal to the side of a regular hexagon which can be inscribed within $ABC$. It is to be demonstrated that the straight line $BCD$ is cut in extreme and mean ratio at the point $C$, and that $CD$ is the greater segment. Let $E$ be the center of $ABC$. Let $EB, EC, ED$ be joined. Let $BE$ be produced to $A$. We have that $BC$ is the side of a regular decagon. Therefore the arc $ACB$ is five times the arc $CB$. Therefore the arc $AC$ is five times the arc $CB$. But from {{EuclidPropLink|book = VI|prop = 33|title = Angles in Circles have Same Ratio as Arcs}}: :$\angle AEC = 4 \cdot \angle CEB$ From {{EuclidPropLink|book = I|prop = 5|title = Isosceles Triangle has Two Equal Angles}}: :$\angle EBC = \angle ECB$ Therefore by {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle AEC = 2 \cdot \angle ECB$ From {{EuclidPropLink|book = IV|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: :$EC = CD$ So from {{EuclidPropLink|book = I|prop = 5|title = Isosceles Triangle has Two Equal Angles}}: :$\angle CED = \angle CDE$ Therefore by {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle ECB = 2 \cdot \angle EDC$ But it has been proved that: :$\angle AEC = 2 \cdot \angle ECB$ Therefore: :$\angle AEC = 4 \cdot \angle EDC$ But it has been proved that: :$\angle AEC = 4 \cdot \angle BEC$ Therefore: :$\angle EDC = \angle BEC$ But $\angle EBD$ is common to $\triangle BEC$ and $\triangle BED$. Therefore by {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle BED = \angle ECB$ Therefore $\triangle EBD$ is equiangular with $\triangle EBC$. Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$BD : BE = EB : BC$ But $EB = CD$ Therefore: :$BD : DC = DC : CB$ and: :$BD > DC$ Therefore: :$DC > CB$ Therefore $BD$ is cut in extreme and mean ratio at the point $C$, and $CD$ is the greater segment. {{qed}} {{Euclid Note|9|XIII}}
:400px Let $ABC$ be a circle. Let the equilateral triangle $ABC$ be inscribed within the circle $ABC$. It is to be demonstrated that the square on one side of $\triangle ABC$ is $3$ times the square on the radius of the circle $ABC$. Let $D$ be the center of the circle $ABC$. Let $AD$ be joined and produced to $E$. Let $BE$ be joined. We have that $\triangle ABC$ is equilateral. Therefore the arc $BEC$ is a third part of the circumference of the circle $ABC$. Therefore the arc $BE$ is a sixth part of the circumference of the circle $ABC$. Therefore $BE$ is the side of a regular hexagon inscribed within the circle $ABC$. Therefore from {{EuclidPropLink|book = IV|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: :$BE = DE$ We have that: :$AE = 2 \cdot DE$ So: :$AE^2 = 4 \cdot ED^2 = 4 \cdot BE^2$ But from : :{{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}} and: :{{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}} it follows that: :$AE^2 = AB^2 + BE^2 = 4 \cdot BE^2$ Therefore: :$AB^2 = 3 \cdot BE^2$ But: :$BE = DE$ Therefore: :$AB^2 = 3 \cdot DE^2$ Hence the result. {{qed}} {{Euclid Note|12|XIII}}
:600px Let $AB$ be the diameter of the given sphere. Let $AB$ be bisected at $C$. Let $ADB$ be a semicircle on the diameter $AB$. Let $CD$ be drawn from $C$ perpendicular to $AB$. Let $DB$ be joined. Let the square $EFGH$ be set out whose sides are all equal to $DB$. Let $HF$ and $EG$ be joined. Let $K$ be the point where $HF$ and $EG$ meet. From {{EuclidPropLink|book = XI|prop = 12|title = Construction of Straight Line Perpendicular to Plane from point on Plane}}: :let $KL$ be drawn from $K$ perpendicular to the plane of the square $EFGH$. Let $KM$ be produced through to the other side of the plane of the square $EFGH$. Let $KL$ and $KM$ be equal to any one of the straight lines $EK, FK, GK, HK$. Let $LE, LF, LG, LH, ME, MF, MG, MH$ be joined. We have that: :$KE = KH$ and: :$\angle EKH$ is a right angle. Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$HE^2 = 2 \cdot EK^2$ We have that: :$LK = KE$ and: :$\angle LKE$ is a right angle. Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$EL^2 = 2 \cdot EK^2$ But it has already been proved that: :$HE^2 = 2 \cdot EK^2$ Therefore: :$LE^2 = EH^2$ and so: :$LE = EH$ For the same reason: :$LH = HE$ Therefore $\triangle LEH$ is an equilateral triangle. Similarly we can prove that each of the remaining triangles of which the sides of $EFGH$ are the bases are equilateral. Therefore an octahedron has been constructed which is contained by eight equilateral triangles. {{qed|lemma}} Next it is to be demonstrated that the octahedron $EFGHLM$ can be inscribed in the given sphere. We have that: :$LM = KM = KE$ Therefore the semicircle described on $LM$ also passes through $E$. Let $LM$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began. For the same reason as for $E$ it follows that the semicircle passes through the points $F, G, H$. Thus the octahedron $EFGHLM$ has been inscribed in a sphere. We have that: :$LK = KM$ while $KE$ is common. These also contain right angles. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$LE = EM$ From {{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}}: :$\angle LEM$ is a right angle. Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$LM^2 = 2 \cdot LE^2$ We also have that: :$AC = CB$ and so: :$AB = 2 \cdot BC$ But: :$AB : BC = AB^2 : BD^2$ Therefore: :$AB^2 = 2 \cdot BD^2$ But it has already been proved that: :$LM^2 = 2 \cdot LE^2$ and because $EH = DB$: :$DB^2 = LE^2$ Therefore: :$AB^2 = LM^2$ and so: :$AB = LM$ But $AB$ is the diameter of the given sphere. Therefore $LM$ equals the diameter of the given sphere. Therefore the (regular) octahedron $EFGHLM$ has been inscribed in the given sphere. At the same time it has been demonstrated that the square on the diameter of the given sphere is double the square on the edge of the (regular) octahedron $EFGHLM$ inscribed within it. {{qed}} {{Euclid Note|14|XIII}}
:500px Let $AB$ be the diameter of the given sphere. Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$. Let $ADB$ be a semicircle on the diameter $AB$. Let $CD$ be drawn from $C$ perpendicular to $AB$. Let $DB$ be joined. Let the square $EFGH$ be set out whose sides are all equal to $DB$. From {{EuclidPropLink|book = XI|prop = 12|title = Construction of Straight Line Perpendicular to Plane from point on Plane}}: :let $EK, FL, GM, HN$ be drawn from $E, F, G, H$ respectively perpendicular to the plane of the square $EFGH$. Let $EK, FL, GM, HN$ be made equal to one of the straight lines $EF, FG, GH, HE$. Let $KL, LM, MN, NK$ be joined. The cube $FN$ can be seen to have been constructed which is contained by six squares. {{qed|lemma}} Next it is to be demonstrated that the cube $FN$ can be inscribed in the given sphere. Let $KG, EG$ be joined. We have that $KE$ is perpendicular to the plane of the square $EFGH$. So from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$KE$ is perpendicular to the straight line $EG$. Therefore $\angle KEG$ is a right angle. Therefore also if $FK$ is joined, then $GF$ is perpendicular to $FK$. For this reason, the semicircle described on $GK$ will also pass through $F$. Similarly it will also pass through the remaining vertices of the cube. Let $GK$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began. Thus the cube $FN$ has been inscribed in a sphere. We have that: :$GF = FE$ and that $\angle F$ is a right angle. Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$EG^2 = 2 EF^2$ But: :$EF = EK$ Therefore: :$EG^2 = 2 \cdot EK^2$ Therefore from {{EuclidPropLink|book = I|prop = 47|title = Pythagoras's Theorem}}: :$GK^2 = EG^2 + EK^2 = 3 EK^2$ We have that: :$AB = 3 BC$ while: :$AB : BC = AB^2 : BD^2$ Therefore: :$AB^2 = 3 \cdot BD^2$ But it has been proved that: :$GK^2 = 3 EK^2$ while: :$KE = DB$ Therefore: :$KG = AB$ But $AB$ is the diameter of the given sphere. Therefore: :the cube $FN$ can be inscribed in the given sphere and: :the square on the diameter of the given sphere is three times the square on the side of the cube. {{qed}} {{Euclid Note|15|XIII}}
:300px Let $ABC$ be a circle. Let $BC$ be the side of a regular pentagon which has been inscribed within $ABC$. Let $D$ be the center of the circle $ABC$. Let $DE$ be drawn from $D$ perpendicular to $BC$, and produce it both ways to meet $ABC$ in $A$ and $F$. It is to be demonstrated that $DE$ is half the sum of: :the side of a regular hexagon which has been inscribed within $ABC$ and: :the side of a regular decagon which has been inscribed within $ABC$ Let $DC$ and $CF$ be joined. Let $G$ be the point on $AE$ such that $GE = EF$. Let $GC$ be joined. We have that: :the circumference of $ABC$ is $5$ times $\smile BFC$ and: :half the circumference of $ABC$ is $\smile ACF$ where $\smile BFC$ and $\smile ACF$ are used to denote the arc $BFC$ and arc $ACF$. while: :$\smile FC$ is half of $\smile BFC$. Thus: {{begin-eqn}} {{eqn | l = \smile ACF       | r = 5 \cdot \smile FC       | c =  }} {{eqn | ll= \therefore       | l = \smile AC       | r = 4 \cdot \smile CF       | c =  }} {{eqn | ll= \therefore       | l = \angle ADC       | r = 4 \cdot \angle CDF       | c =  }} {{eqn | ll= \therefore       | l = \angle AFC       | r = 2 \cdot \angle CDF       | c =  }} {{eqn | ll= \therefore       | l = \angle GCF       | r = \angle AFC       | c =  }} {{eqn | r = 2 \cdot \angle CDF       | c =  }} {{eqn | ll= \therefore       | l = \angle CDG       | r = \angle DCG       | c = Sum of Angles of Triangle equals Two Right Angles }} {{eqn | ll= \therefore       | l = DG       | r = CG       | c =  }} {{eqn | r = CF       | c =  }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = GE       | r = EF       | c =  }} {{eqn | ll= \therefore       | l = DE       | r = EF + FC       | c =  }} {{eqn | ll= \therefore       | l = 2 \cdot DE       | r = DF + FC       | c = adding $DE$ to each }} {{end-eqn}} But from {{EuclidPropLink|book = VI|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: :$DE$ is the side of a regular hexagon which has been inscribed within $ABC$ and from the construction: :$FC$ is the side of a regular decagon which has been inscribed within $ABC$. Hence the result. {{qed}} {{Euclid Note|1|XIV}}
:300px Let $ABCDE$ be the regular pentagon which is the face of a regular dodecahedron. Let the circle $ABCDE$ be circumscribed around the pentagon $ABCDE$. Let $F$ be the center of the circle $ABCDE$. Let $FG$ be the perpendicular dropped from $F$ to $CD$. Let $CF$ and $FD$ be joined. Then: {{begin-eqn}} {{eqn | l = CD \cdot FG       | r = 2 \cdot \triangle CDF       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | ll= \therefore       | l = 30 \cdot CD \cdot FG       | r = 60 \cdot \triangle CDF       | c =  }} {{eqn | r = 12 \cdot \map {\mathcal A} {ABCDE}       | c =  }} {{end-eqn}} where $\map {\mathcal A} {ABCDE}$ denotes the area of the regular pentagon $ABCDE$. The result follows from the definition of a regular dodecahedron as having $12$ such faces. {{qed}} {{Euclid Note|3|XIV}}
:300px Let $ABC$ be the equilateral triangle which is the face of a regular icosahedron. Let the circle $ABC$ be circumscribed around the equilateral triangle $ABC$. Let $D$ be the center of the circle $ABC$. Let $DE$ be the perpendicular dropped from $D$ to $BC$. Let $BD$ and $CD$ be joined. Then: {{begin-eqn}} {{eqn | l = DE \cdot BC       | r = 2 \cdot \triangle DBC       | c = Area of Triangle in Terms of Side and Altitude }} {{eqn | ll= \therefore       | l = 3 \cdot DE \cdot BC       | r = 6 \cdot \triangle DBC       | c =  }} {{eqn | r = 2 \cdot \triangle ABC       | c =  }} {{eqn | ll= \therefore       | l = 30 \cdot DE \cdot BC       | r = 20 \cdot \triangle ABC       | c =  }} {{end-eqn}} The result follows from the definition of a regular icosahedron as having $20$ such faces. {{qed}} {{Euclid Note|4|XIV}}
:300px $\qquad$ 300px What is meant by ''its perpendicular'' is a perpendicular from the center of the circle circumscribed around it to the side of it. In the above diagram: From {{EuclidPropLink|book = XIV|prop = 3|title = Size of Surface of Regular Dodecahedron}}: :the surface of the dodecahedron is $30$ times its perpendicular times its side. From {{EuclidPropLink|book = XIV|prop = 4|title = Size of Surface of Regular Icosahedron}}: :the surface of the icosahedron is $30$ times its perpendicular times its side. Hence the result. {{qed}} {{Euclid Note|5|XIV}}
:350px Let a regular dodecahedron, a regular icosahedron and a  cube be inscribed in a given sphere. From {{EuclidPropLink|book = XIV|prop = 3|title = Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere}}: :the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron. Let $ABC$ be the circle which circumscribes both that regular pentagon and that equilateral triangle. Let $CD$ be the side of the equilateral triangle. Let $AC$ be the side of the regular pentagon. Let $E$ be the center of the circumscribing circle. Let $EF$ and $EG$ be perpendiculars dropped from $E$ to $CD$ and $AC$ respectively. Let $EG$ be produced past $G$ to meet the circumference of $ABC$ at $B$. Let $BC$ be joined. Let $H$ be equal to the side of the cube. It is to be demonstrated that the ratio of the surface of the regular dodecahedron to the surface of the regular icosahedron is $H : CD$. From {{EuclidPropLink|book = IV|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: :$EB$ equals the side of the regular hexagon that can be inscribed in $ABC$. From {{EuclidPropLink|book = XIII|prop = 9|title = Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio}}: :$EB + BC$ is divided at $B$ in extreme and mean ratio where $BE$ is the greater segment. From {{EuclidPropLink|book = XIV|prop = 1|title = Perpendicular from Center of Circle to Side of Inscribed Pentagon}}: :$EG = \dfrac {EB + BC} 2$ From {{EuclidPropLink|book = XIII|prop = 12|title = Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle}} (indirectly): :$EF = \dfrac BE 2$ So: :$EG : EF = \left({EB + BC}\right) : BE$ Therefore if $EG$ is divided in extreme and mean ratio, the greater segment is $EF$. But from {{EuclidPropLink|book = XIII|prop = 17|qual=Porism|title = Construction of Regular Dodecahedron within Given Sphere}}: :if $H$ is divided in extreme and mean ratio, the greater segment is equal to $CA$. So: {{begin-eqn}} {{eqn | l = H : CA       | r = EG : EF       | c = as both are in extreme and mean ratio }} {{eqn | n = 1       | ll= \therefore       | l = FE \cdot H       | r = CA \cdot EG       | c = Definition of Ratio }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = H : CD       | r = FE \cdot H : FE \cdot CD       | c =  }} {{eqn | ll= \therefore       | l = H : CD       | r = CA \cdot EG : FE \cdot CD       | c = substituting from $(1)$ }} {{end-eqn}} From {{EuclidPropLink|book = XIV|prop = 5|title = Ratio of Sizes of Surfaces of Regular Dodecahedron and Regular Icosahedron in Same Sphere}}: :(surface of dodecahedron) : (surface of icosahedron) = $CA \cdot EG : FE \cdot CD$ Hence the result. {{qed}} {{Euclid Note|6|XIV}}
:400px Let a regular dodecahedron, a regular icosahedron and a  cube be inscribed in a given sphere. From {{EuclidPropLink|book = XIV|prop = 3|title = Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere}}: :the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron. Let $AHB$ be the circle which circumscribes both that regular pentagon and that equilateral triangle. Let $C$ be the center of the circumscribing circle. Let $CB$ be a radius of $AHB$. Let $CB$ be divided at $D$ in extreme and mean ratio where $CD$ is the greater segment. Then from: :{{EuclidPropLink|book = XIII|prop = 9|title = Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio}} and the converse of: :{{EuclidPropLink|book = XIII|prop = 5|title = Straight Line cut in Extreme and Mean Ratio plus its Greater Segment}} it follows that: :$CD$ is the side of a regular decagon which has been inscribed in $AHB$. Let $E$ be the side of the equilateral triangle that is the face of the regular icosahedron. Let $F$ be the side of the regular pentagon which is the face of the regular dodecahedron. Let $G$ be the side of the square which is the face of the cube. From {{EuclidPropLink|book = XIII|prop = 17|qual=Porism|title = Construction of Regular Dodecahedron within Given Sphere}}: :if $G$ is divided in extreme and mean ratio, the greater segment is equal to $F$. Thus: {{begin-eqn}} {{eqn | l = E^2       | r = 3 \cdot BC^2       | c = {{EuclidPropLink|book = XIII|prop = 12|title = Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle}} }} {{eqn | l = CB^2 + BD^2       | r = 3 \cdot CD^2       | c = {{EuclidPropLink|book = XIII|prop = 4|title = Area of Squares on Whole and Lesser Segment of Straight Line cut in Extreme and Mean Ratio}} }} {{eqn | ll= \therefore       | l = E^2 : CB^2       | r = \left({CB^2 + BD^2}\right) : CD^2       | c =  }} {{eqn | ll= \therefore       | l = E^2 : \left({CB^2 + BD^2}\right)       | r = CB^2 : CD^2       | c =  }} {{eqn | r = G^2 : F^2       | c =  }} {{eqn | ll= \therefore       | l = G^2 : E^2       | r = F^2 : \left({CB^2 + BD^2}\right)       | c =  }} {{eqn | ll= \therefore       | l = G^2 : E^2       | r = \left({BC^2 + CD^2}\right) : \left({CB^2 + BD^2}\right)       | c = {{EuclidPropLink|book = XIII|prop = 10|title = Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle}} }} {{end-eqn}} Hence the result. {{qed}} {{Euclid Note|7|XIV}}
Let a regular dodecahedron, a regular icosahedron and a  cube be inscribed in a given sphere. From {{EuclidPropLink|book = XIV|prop = 3|title = Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere}}: :the circle which circumscribes the regular pentagon which is the face of the regular dodecahedron is the same size as the circle which circumscribes the equilateral triangle which is the face of the regular icosahedron. In a sphere, equal sections are equally distant from the center. {{explain|The above statement needs a link to a proof.}} Thus the perpendiculars from the center to the faces of the regular icosahedron and regular dodecahedron are equal. So the pyramids whose apices are the center of the sphere and whose bases are the faces of these polyhedra are of the same height. Therefore from {{EuclidPropLink|book = XII|prop = 6|title = Sizes of Pyramids of Same Height with Polygonal Bases are as Bases}}: :the ratios of these pyramids are to one another as their bases. Thus: {{begin-eqn}} {{eqn | l = 12 \text { pentagons} : 20 \text { triangles}       | r = 12 \text { pyramids on pentagons} : 20 \text { pyramids on triangles}       | c =  }} {{eqn | ll= \therefore       | l = \text {surface of dodecahedron} : \text {surface of icosahedron}       | r = \text {volume of dodecahedron} : \text{volume of icosahedron}       | c =  }} {{end-eqn}} Therefore from {{EuclidPropLink|book = XIV|prop = 6|title = Ratio of Sizes of Surfaces of Cube and Regular Icosahedron in Same Sphere}}: {{begin-eqn}} {{eqn | l = \text {volume of dodecahedron} : \text{volume of icosahedron}       | r = \text {side of cube} : \text {side of icosahedron}       | c =  }} {{end-eqn}} {{qed}} {{Euclid Note|8|XIV}}
A special case of the Binet-Cauchy Identity. {{qed}} {{Namedfor|Joseph Louis Lagrange|cat = Lagrange}} Category:Algebra 3n9l8g8wpdzondlfga5uyrxs22net8y
Let $r$ be the common ratio of $\sequence {a_k}$. Let $p, q$ be consecutive terms of $r$. By hypothesis $p, q \in \Q$. Then, by definition of geometric sequence: :$q = r p$ It follows that: :$r = \dfrac q p$ From Rational Numbers form Field, $\Q$ is closed under division. Thus $r \in \Q$ and hence the result. {{qed}} Category:Geometric Sequences qoxjqks99ap78b23fvywcsbpiwnffa0
From Integers form Subdomain of Rationals it follows that $a_k \in \Q$ for all $0 \le k \le n$. The result follows from Common Ratio in Rational Geometric Sequence is Rational. {{qed}} Category:Geometric Sequences of Integers 0qft406itmha5ood9xvu2bzzz0xivkq
Let $r$ be the common ratio of $P$. From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number. Let $r = \dfrac q p$ be in canonical form. Thus, by definition: :$p \perp q$ Let $a$ be the first term of $P$. Then the sequence $P$ is: :$P = \paren {a, a \dfrac q p, a \dfrac {q^2} {p^2}, \ldots, a \dfrac {q^n} {p^n} }$ All the elements of $P$ are integers, so, in particular: :$a \dfrac {q^n} {p^n} \in \Z$ We have that:  :$p \perp q$ From Powers of Coprime Numbers are Coprime: :$q^n \perp p^n$ and so from Euclid's Lemma: :$p^n \divides a$ Thus: :$a = k p^n$ for some $k \in \Z$, and so: :$P = \paren {k p^n, k q p^{n - 1}, k q^2 p^{n - 2}, \ldots, k q^{n - 1} p, k q^n}$ {{qed}} Category:Geometric Sequences of Integers bhfwiu2z97dxzp20i8vusbgjraa7agi
=== Necessary Condition === Let $r$ be an integer. By definition of geometric sequence, the terms of $P$ are in the form: :$a_k = b r^{k - 1}$ where $b$ and $k$ are integers. It follows from Integer Multiplication is Closed that $a_k$ is an integer. {{qed|lemma}} === Sufficient Condition === From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number. Let $r$ be expressed in canonical form as: :$r = \dfrac p q$ where, by definition of canonical form, $p \perp q$, that is, $p$ is coprime to $q$. From Form of Geometric Sequence of Integers, the terms of $P$ are in the form: :$(1): \quad a_j = k q^{j - 1} p^{n - j}$ Let $a_i \divides a_j$. From Powers of Coprime Numbers are Coprime: :$(2): \quad q^{j - i} \perp p^{j - i}$ Then: {{begin-eqn}} {{eqn | l = a_i       | o = \divides       | r = a_j       | c =  }} {{eqn | ll= \leadsto       | l = k q^{i - 1} p^{n - i}       | o = \divides       | r = k q^{j - 1} p^{n - j}       | c = from $(1)$ }} {{eqn | ll= \leadsto       | lo= \exists m \in \Z:       | l = m \paren {k q^{j - 1} p^{n - i} } q^{j - i}       | r = \paren {k q^{j - 1} p^{n - i} } p^{j - i}       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = m q^{j - i}       | r = p^{j - i}       | c = dividing both sides by $k q^{j - 1} p^{n - i}$ }} {{eqn | ll= \leadsto       | l = q^{j - i}       | o = \divides       | r = p^{j - i}       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} But from $(2)$: :$q^{j - i} \perp p^{j - i}$ Thus $q^{j - i} \divides p^{j - i}$ can happen only when $q^{j - 1} = q = 1$. That is, when $r$ is an integer. {{qed}} Category:Geometric Sequences of Integers 3k4xydo9q8ax698lep4fppvf96to2eu
The forward implication is shown in Powers of Coprime Numbers are Coprime. The reverse implication is shown by substituting $n = 1$. {{qed}} Category:Coprime Integers 8k2ky39jzctjiltp3egexjushvinn3s