hcju/proofwikiqa huggingface dataset:
Let $ABCD$ be a rectangle. Seeking a contradiction, assume $ABCD$ is ''not'' a parallelogram. {{image wanted|PW gives an error message when uploading images. An email has been sent to Joe.}} {{WLOG}} let line segments $AD$ and $BC$ ''not'' be parallel. Then extend $AD$ and $BC$ into two infinite straight lines. By the Parallel Postulate, the lines will eventually meet at one side or the other. Let their point of intersection be $E$. There are two possibilities: :$ABE$ is a triangle :$CDE$ is a triangle. {{WLOG}}, let $ABE$ be a triangle. If $ABE$ is a triangle, then by Sum of Angles of Triangle equals Two Right Angles, $\angle ABE + \angle BEA + \angle EAB$ add to two right angles. But by the definition of a rectangle, $\angle ABE + \angle EAB$ add to two right angles. As $\angle BEA$ is not a zero angle (else $ABD$ would not be a triangle), this cannot be so. We are forced to conclude that that $ABCD$ is not a rectangle, a contradiction. {{qed}} Category:Rectangles Category:Parallelograms qiwnxy6fxk3n89o4f39a7c37oltaget
=== Lemma === {{:Condition for Denesting of Square Root/Lemma}}{{qed|lemma}} === Necessary condition === {{begin-eqn}} {{eqn | l = \sqrt {a + \sqrt b}       | r = \sqrt p + \sqrt q }} {{eqn | r = \sqrt {p + q + \sqrt {4 p q} }       | c = Sum of Square Roots as Square Root of Sum }} {{eqn | ll= \leadsto       | l = a       | r = p + q       | c = by Lemma }} {{eqn | l = b       | r = 4 p q       | c = by Lemma }} {{eqn | ll= \leadsto       | l = 0       | r = b - 4 p q }} {{eqn | r = b - 4 p \paren {a - p} }} {{eqn | r = b - 4 a p + 4 p^2 }} {{eqn | ll= \leadsto       | l = p       | r = \frac {4 a \pm \sqrt {16 a^2 - 16 b} } 8       | c = Quadratic Formula }} {{eqn | r = \frac a 2 \pm \frac {\sqrt {a^2 - b} } 2 }} {{end-eqn}} This shows that, in order for $p \in \Q$, we must have $\sqrt {a^2 - b} \in \Q$ as well. The result follows from $\paren {\sqrt {a^2 - b} }^2 = a^2 - b$. {{qed|lemma}} === Sufficient condition === Let $n^2 = a^2 - b$. As $a^2 = b + n^2$ it follows that: :$a \ge \sqrt b$ Then: {{begin-eqn}} {{eqn | l = \sqrt {a + \sqrt b}       | r = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b} } 2}       | c = Square Root of Sum as Sum of Square Roots }} {{eqn | r = \sqrt {\dfrac a 2 + \dfrac {\sqrt {n^2} } 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {n^2} } 2}       | c = substituting $n^2 = a^2 - b$ }} {{eqn | r = \sqrt {\dfrac a 2 + \dfrac n 2} + \sqrt {\dfrac a 2 - \dfrac n 2} }} {{end-eqn}} By Rational Addition is Closed and Rational Subtraction is Closed: :$\dfrac a 2 + \dfrac n 2$ and $\dfrac a 2 - \dfrac n 2$ are rational. {{qed}} Category:Square Roots Category:Condition for Denesting of Square Root 3sarxzk03ddhfc97y129a34h15q2tui
By Characteristic Function Determined by 1-Fiber, it suffices to show: :$\min \set {\map {\chi_A} s + \map {\chi_B} s, 1} = 1 \iff s \in A \cup B$ By the nature of the min operation, this amounts to showing that: :$\map {\chi_A} s + \map {\chi_B} s \ge 1 \iff s \in A \cup B$ As $\chi_A, \chi_B$ are characteristic functions, the left-hand side amounts to: :$s \in A \lor s \in B$ which is precisely the definition of $s \in A \cup B$. {{qed}}
From Subset of Union: :$A, B \subseteq A \cup B$ From Intersection is Subset of Union: :$A \cap B \subseteq A \cup B$ Thus from Characteristic Function of Subset: :$\map {\chi_{A \mathop \cup B} } s = 0 \implies \map {\chi_A} s = \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s = 0$ Now suppose that $\map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$. That is: :$\map {\chi_A} s + \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s$ Suppose the latter equals $1$, that is $s \in A \cap B$. Then both $s \in A$ and $s \in B$, so by definition of characteristic function: :$\map {\chi_A} s + \map {\chi_B} s = 1 + 1 = 2$ Since $2 \ne 1$, it follows that $\map {\chi_{A \mathop \cap B} } s \ne 1$, that is it equals $0$. Thence, it follows that: :$\map {\chi_A} s + \map {\chi_B} s = 0$. This only happens when $\map {\chi_A} s = \map {\chi_B} s = 0$. Thus, $s \notin A$ and $s \notin B$, so $s \notin A \cup B$. It finally follows that $\map {\chi_{A \mathop \cup B} } s = 0$. It is now established that: :$\map {\chi_{A \mathop \cup B} } s = 0 \iff \map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$ and from Characteristic Function Determined by 0-Fiber, the result follows. {{qed}}
Suppose $\map {\chi_{A \mathop \cup B} } s = 0$. Then $s \notin A \cup B$, so $s \notin A$ and $s \notin B$. Hence: :$\map {\chi_A} s = \map {\chi_B} s = 0$ and by definition of max operation: :$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$ Conversely, suppose: :$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$ Then it follows that: :$\map {\chi_A} s = \map {\chi_B} s = 0$ because characteristic functions are $0$ or $1$. Hence $s \notin A$ and $s \notin B$, so $s \notin A \cup B$. That is: :$\map {\chi_{A \mathop \cup B} } s = 0$ Above considerations give: :$\map {\chi_{A \mathop \cup B} } s = 0 \iff \max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$ and applying Characteristic Function Determined by 0-Fiber, the result follows. {{qed}}
By Characteristic Function Determined by 1-Fiber, it suffices to show that: :$\map {\chi_A} s \, \map {\chi_B} s = 1 \iff s \in A \cap B$ Now, both $\chi_A$ and $\chi_B$ are characteristic functions. It follows that, for any $s \in S$: :$\map {\chi_A} s \, \map {\chi_B} s = 1 \iff \map {\chi_A} s = \map {\chi_B} s = 1$ By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$. That is, $s \in A \cap B$, by definition of set intersection. {{qed}}
By Characteristic Function Determined by 1-Fiber, it suffices to show that: :$\min \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 1 \iff s \in A \cap B$ By definition of characteristic function, we have: :$\min \left\{{\chi_A \left({s}\right), \chi_B \left({s}\right)}\right\} = 1$ {{iff}} $\chi_A \left({s}\right) = \chi_B \left({s}\right) = 1$ because $\chi_A, \chi_B$ take values in $\left\{{0, 1}\right\}$. By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$. That is, $s \in A \cap B$, by definition of set intersection. {{qed}}
Let two magnitudes $AB, CD$ be equimultiples of two magnitudes $E, F$. Let $AG, CH$ subtracted from them be equimultiples of the same two $E, F$. We need to show that the remainders $GB, HD$ are either equal to $E, F$ or are equimultiples of them. :350px First let $GB = E$. Let $CK$ be made equal to $F$. We have that $AG$ is the same multiple of $E$ that $CH$ is of $F$, while $GB = E$ and $KC = F$. Therefore from Multiplication of Numbers Distributes over Addition, $AB$ is the same multiple of $E$ that $KH$ is of $F$. But by hypothesis $AB$ is the same multiple of $E$ that $CD$ is of $F$. Since then, each of the magnitudes $KH, CD$ is the same multiple of $F$. Therefore $KH = CD$. Let $CH$ be subtracted from each. Therefore the remainder $KC$ equals the remainder $HD$. But $F = KC$, so $HD = F$. Hence, if $GB = E$ then $HD = F$. Similarly we can prove that, even if $GB$ is a multiple of $E$, then $HD$ is also the same multiple of $F$. {{qed}}
{{begin-eqn}} {{eqn | l = \sqrt a + \sqrt b       | r = \sqrt {\paren {\sqrt a + \sqrt b}^2} }} {{eqn | r = \sqrt {\sqrt a^2 + \sqrt b^2 + 2 \sqrt a \sqrt b}       | c = Square of Sum }} {{eqn | r = \sqrt {a + b + \sqrt {4 a b} }       | c = Exponent Combination Laws: Power of Product }} {{end-eqn}} {{qed}} Category:Algebra dqyudpoaz21as5p0ufny7ers4imbydf
Let $A, B$ be (natural) numbers which are the least of those which have the same ratio with them. We need to show that $A$ and $B$ are prime to one another. :250px Aiming for a contradiction, suppose $A$ and $B$ are not coprime. Then by definition there exists some (natural) number $C > 1$ which measures them both. As many times as $C$ measures $A$, let that many units be in $D$. As many times as $C$ measures $B$, let that many units be in $E$. So by {{EuclidDefLink|VII|15|Multiply}}: :$A = C \times D$ :$B = C \times E$ Thus by {{EuclidPropLink|book=VII|prop=17|title=Multiples of Ratios of Numbers}}: :$D : E = A : B$ So $D$ and $E$ are in the same ratio with $A$ and $B$. But by Absolute Value of Integer is not less than Divisors: Corollary: :$D < A$ :$E < B$ This contradicts our hypothesis that $A$ and $B$ are the least of those numbers that are in the same ratio with $A$ and $B$. It follows that $A$ and $B$ must be prime to one another. {{qed}} {{Euclid Note|22|VII|{{EuclidNoteConverse|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}}}
Let $a \perp b$ and $c > 1: c \divides a$. {{AimForCont}} $c \not \perp b$. So by definition of not coprime: :$\exists d > 1: d \divides c, d \divides b$ But from Divisor Relation is Transitive: :$d \divides c, c \divides a \implies d \divides a$ So $d$ is a common divisor of both $a$ and $b$. So, by definition, $a$ and $b$ are not coprime. The result follows by Proof by Contradiction. {{qed}}
Let $A, B$ be two numbers which are prime to one another. Let $C$ be any number greater than $1$ which measures $A$. :250px Suppose $C$ and $B$ are not prime to one another. Then some number $D$ will measure them both. We have that $D$ measures $C$ and $C$ measures $A$. So $D$ measures $A$. But $D$ also measures $B$. So $D$ measures $A$ and $B$ which are prime to one another. By {{EuclidDefLink|VII|12|Relatively Prime}}, this is a contradiction. Therefore there can be no such $D$ that measures both $B$ and $C$. That is, $B$ and $C$ are prime to one another. {{qed}}
Proof by induction: Let  $a \perp b$. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$a^n \perp b^n$ $\map P 1$ is true, as this just says: :$a \perp b$ === Basis for the Induction === By {{EuclidPropLink|book = VII|prop = 25|title = Square of Coprime Number is Coprime}}: :$a^2 \perp b$ Again, by {{EuclidPropLink|book = VII|prop = 25|title = Square of Coprime Number is Coprime}}: :$a^2 \perp b^2$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$a^k \perp b^k$ Then we need to show: :$a^{k + 1} \perp b^{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = a^k        | o = \perp       | r = b^k       | c =  }} {{eqn | ll=\leadsto       | l = a^k \times a       | o = \perp       | r = b^k       | c = {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to all Factors is Coprime to Whole}} }} {{eqn | ll=\leadsto       | l = a^{k + 1}       | o = \perp       | r = b^k       | c =  }} {{eqn | ll=\leadsto       | l = a^{k + 1}       | o = \perp       | r = b^k \times b       | c = {{EuclidPropLink|book = VII|prop = 24|title = Integer Coprime to all Factors is Coprime to Whole}} }} {{eqn | ll=\leadsto       | l = a^{k + 1}       | o = \perp       | r = b^{k + 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N_{>0}: a^n \perp b^n$ {{qed}} {{Euclid Note|27|VII|Euclid's proof does not use the full induction process, which is a later mathematical innovation.}}
=== Necessary Condition === Let $a \perp b$. Suppose $a + b$ is not coprime to $a$. Then: :$\exists d \in \Z_{>1}: d \divides a, d \divides \paren {a + b}$ But then: :$d \divides \paren {\paren {a + b} - a}$ and so: :$d \divides b$ and so $a$ and $b$ are not coprime. From this contradiction it follows that $a + b$ is coprime to $a$. {{qed|lemma}} === Sufficient Condition === Let $a + b$ be coprime to $a$. Suppose $a$ is not coprime to $b$. Then: :$\exists d \in \Z_{>1}: d \divides a, d \divides b$ and so: :$d \divides \paren {a + b}$ and so $a$ and $\paren {a + b}$ are not coprime. From this contradiction it follows that $a$ is coprime to $b$. {{qed}} {{Euclid Note|28|VII}}
By definition of composite number: :$\exists b \in \Z: b \divides a$ If $b$ is a prime number then the proof is complete. Otherwise $b$ is composite. By definition of composite number: :$\exists c \in \Z: c \divides b$ and so: :$c \divides a$ Again, if $c$ is a prime number then the proof is complete. Continuing in this manner, some prime number will be found which will divide the number before it. This will be a divisor of $a$. Because if not, then an infinite series of numbers will be divisors of $a$, each of which is less than the other. This is impossible in numbers. Hence the result. {{qed}} {{Euclid Note|31|VII|}}
By definition of composite number $a$ is either prime or composite. Let $a$ be prime. Then the statement of the result is fulfilled. Let $a$ be composite. Then by {{EuclidPropLink|book = VII|prop = 31|title = Composite Number has Prime Factor}}: :$\exists p: p \divides a$ where $p$ is a prime number. The result follows by Proof by Cases. {{qed}} {{Euclid Note|32|VII|}}
Let $A, B, C$ be the given numbers, as many as we please. $A, B, C$ are either coprime or not. Let $A, B, C$ be coprime. Then by {{EuclidPropLink|book = VII|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}} they are the least of those which have the same ratio with them. Let $A, B, C$ not be coprime. Then by {{EuclidPropLink|book = VII|prop = 3|title = Greatest Common Divisor of Three Numbers}} their GCD $D$ can be found. As many times as $D$ measures each of $A, B, C$, let those be $E, F, G$ respectively. That is: :$D E = A, D F = B, D G = C$ By {{EuclidPropLink|book = VII|prop = 20|title = Ratios of Fractions in Lowest Terms}}, $E, F, G$ are in the same ratio with $A, B, C$. Suppose $E, F, G$ are not the least of those which have the same ratio with $A, B, C$. Let those numbers be $H, K, L$. Then $H$ measures $A$ the same number of times that $K, L$ measure $B, C$. Let this number of times be $M$. We have that $H$ measures $A$ according to the units of $M$. It follows from {{EuclidPropLink|book=VII|prop=16|title=Natural Number Multiplication is Commutative}} that $M$ also measures $A$ according to the units of $H$. For the same reason, $M$ also measures $B$ and $C$ according to the units of $K$ and $L$. Therefore $M$ measures $A$. That is: :$A = H M$ For the same reason: :$A = E D$ That is: :$E D = H M$ Therefore by {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}: :$E : H = M : D$ But $E > H$ and so $M > D$. Also, $M$ measures $A$, $B$ and $C$. But by hypothesis $D$ is the greatest common measure of $A, B, C$. Therefore there cannot be any numbers less than $E, F, G$ which are in the same ratio with $A, B, C$. Therefore $E, F, G$ are the least of those which are in the same ratio with $A, B, C$. {{qed}} {{Euclid Note|33|VII|}}
Let $d = \lcm \set {a, b}$. This exists from {{EuclidPropLink|book = VII|prop = 34|title = Existence of Lowest Common Multiple}}. Either $c \divides d$ or not, where $\divides$ denotes divisibility. Suppose $c \divides d$. But by definition of lowest common multiple, $a \divides d$ and $b \divides d$ also. Suppose $a, b, c$ are divisors of $e$ where $e < d$. Then $a, b$ are divisors of $e$. That is, $e$ is a common divisor of $a$ and $b$ which is lower than $d$. But by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}: :$d \divides e$ which is impossible. It follows that there can be no such $e$. Therefore $d = \lcm \set {a, b, c}$. Now suppose $c \nmid d$. Let $e = \lcm \set {c, d}$. This exists from {{EuclidPropLink|book = VII|prop = 34|title = Existence of Lowest Common Multiple}}. Since $a$ and $b$ are both divisors of $d$, it follows that: :$a \divides e$ :$b \divides e$ But we have that $c \divides e$ as well. Suppose $a, b, c$ are divisors of $f$ where $f < e$. Then $a, b$ are divisors of $f$. So by {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}: :$d = \lcm \set {a, b} \divides f$ But also $c \divides f$. Therefore $c$ and $d$ are divisors of $f$. By {{EuclidPropLink|book = VII|prop = 35|title = LCM Divides Common Multiple}}: :$e = \lcm \set {c, d} \divides f$ But this is impossible as by hypothesis $f < e$. Therefore $a, b, c$ are divisors of no number smaller than $e$. Therefore $e = \lcm \set {a, b, c}$. {{qed}}
Suppose $\displaystyle  \lim_{x \mathop \to +\infty} \map f x = +\infty$. Then by the definition of infinite limits at infinity: :$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$ But $M > 0 \iff -M < 0$. Likewise $\map f x > M \iff -\map f x < -M$. Putting $M' = -M$: :$\forall M' < 0: \exists N > 0: x > N \implies -\map f x < M'$ The result then follows from the definition of negative infinite limit at infinity. The proof for $\displaystyle \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analagous. {{qed}}
Let $b \divides a$. By definition of divisibilty: :$\exists c \in \N: c \times b = a$ Then also: :$c \times 1 = c$ So by {{EuclidPropLink|book = VII|prop = 15|title = Alternate Ratios of Multiples}}: :$1 : b = c : a$ Hence the result. {{qed}} {{Euclid Note|37|VII|}}
Let $a$ have an aliquot part $b$. Let $c$ be an integer called by the same name as the aliquot part $b$. Then: :$1 = \dfrac 1 c \times c$ and so by {{EuclidPropLink|book = VII|prop = 15|title = Alternate Ratios of Multiples}}: :$ 1 : c = b : a$ Hence the result. {{qed}} {{Euclid Note|38|VII|}}
Let $a, b, c$ be the given aliquot parts. Let $d, e, f$ be the numbers called by the same name as the aliquot parts $a, b, c$. From {{EuclidPropLink|book = VII|prop = 36|title = LCM of Three Numbers}}, let: :$g = \lcm \set {d, e, f}$ So $g$ has aliquot parts called by the same name as $d, e, f$. Therefore $g$ has the aliquot parts $a, b, c$. Suppose there exists $h \in \N: h < g$ which has the aliquot parts $a, b, c$. By {{EuclidPropLink|book = VII|prop = 38|title = Divisor is Reciprocal of Divisor of Integer}}, $h$ will be measured by numbers called by the same name as the aliquot parts $a, b, c$. Therefore $h$ is measured by $d, e, f$. But $h < g$ which is impossible. Therefore there is no number less than $g$ which has the aliquot parts $a, b, c$. {{qed}} {{Euclid Note|39|VII|}}
Suppose $\displaystyle \lim_{x \mathop \to +\infty} \map f x = -\infty$. Then by the definition of negative infinite limit at infinity: :$\forall M < 0: \exists N > 0: x > N \implies \map f x < M$ But $M < 0 \iff -M > 0$. Likewise $\map f x < M \iff -\map f x > -M$. Putting $M' = -M$: :$\forall M' > 0: \exists N > 0: x > N \implies -\map f x > M'$ The result then follows from the definition of infinite limit at infinity. The proof for $\displaystyle \lim_{x \mathop \to -\infty} \map f x = -\infty$ is analogous. {{qed}}
Let $r = \dfrac a b$ be the given common ratio. Let the required geometric sequence have a length of $4$. Let $a^2 = c$. Let $a b = d$. Let $b^2 = e$. Let: :$a c = f$ :$a d = g$ :$a e = h$ and let: :$b e = k$ As: :$a^2 = c$ :$a b = d$ it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that: :$\dfrac a b = \dfrac c d$ As: :$a b = d$ :$b^2 = e$ it follows from {{EuclidPropLink|book = VII|prop = 18|title = Ratios of Multiples of Numbers}} that: :$\dfrac a b = \dfrac d e$ As: :$a c = f$ :$a d = g$ it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that: :$\dfrac c d = \dfrac f g$ As: :$a d = g$ :$a e = h$ it follows from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}} that: :$\dfrac d e = \dfrac g h$ As: :$a e = h$ :$b e = k$ it follows from {{EuclidPropLink|book = VII|prop = 18|title = Ratios of Multiples of Numbers}} that: :$\dfrac a b = \dfrac h k$ Putting the above together: :$c, d, e$ are in geometric sequence with common ratio $\dfrac a b$ :$f, g, h, k$ are in geometric sequence with common ratio $\dfrac a b$ We have that $a$ and $b$ are the smallest numbers with the same ratio. So by {{EuclidPropLink|book = VII|prop = 22|title = Numbers forming Fraction in Lowest Terms are Coprime}}: :$a \perp b$ where $\perp$ denotes coprimality. We also have that: :$a^2 = c, b^2 = e$ and: :$a c = e, b e = k$ so by {{EuclidPropLink|book = VII|prop = 27|title = Powers of Coprime Numbers are Coprime}}: :$c \perp e$ :$f \perp k$ But from {{EuclidPropLink|book = VIII|prop = 1|title = Geometric Sequence with Coprime Extremes is in Lowest Terms}}, these are the least of those with the same common ratio. Hence the result. {{qed}}
Let the required length of the geometric sequence $P$ be $n$. Let $r$ be the given common ratio. From Common Ratio in Integer Geometric Sequence is Rational, $r$ is a rational number. Let $r = \dfrac p q$ be in canonical form. Thus, by definition: :$p \perp q$ Let $a$ be the first term of $P$. Then the sequence $P$ is: :$P = \tuple {a, a \dfrac p q, a \dfrac {p^2} {q^2}, \ldots, a \dfrac {p^n} {q^n} }$ All the elements of $P$ are natural numbers, so, in particular: :$a \dfrac {p^n} {q^n} \in \N$ From Powers of Coprime Numbers are Coprime: :$p^n \perp q^n$ and so from Euclid's Lemma: :$q^n \divides a$ Thus: :$a = k q^n$ for some $k \in \N$, and so: :$P = \tuple {k q^n, k p q^{n - 1}, k p^2 q^{n - 2}, \ldots, k p^{n - 1} q, k p^n}$ From Geometric Sequence with Coprime Extremes is in Lowest Terms: :$k q^n \perp k p^n$ from which it follows that $k = 1$. It follows that the required geometric sequence is: :$P = \tuple {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$ {{qed}}
Let $a_0, a_1, a_2, \ldots, a_n$ be natural numbers. Let $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ be a geometric sequence with common ratio $r$. Let $a_0, a_1, \ldots, a_n$ be the smallest such natural numbers. From {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}, let $d_0, d_1$ be the smallest natural numbers such that $d_1 = r d_0$. From {{EuclidPropLink|book = VIII|prop = 2|title = Construction of Geometric Sequence in Lowest Terms}} one can build a sequence of $3, 4, \ldots, n$ terms with the same property. Let the geometric sequence so constructed with $n$ terms be $\sequence {b_k}_{0 \mathop \le k \mathop \le n}$. From {{EuclidPropLink|book = VII|prop = 22|title = Numbers forming Fraction in Lowest Terms are Coprime}}, $d_0$ and $d_1$ are coprime. From {{EuclidPropLink|book = VIII|prop = 2|qual = Porism|title = Construction of Geometric Sequence in Lowest Terms}}, each of the extreme terms of the intermediate geometric sequences is a power of a natural number. From {{EuclidPropLink|book = VII|prop = 27|title = Powers of Coprime Numbers are Coprime}}, each of those extreme terms is coprime. We have that $a_0, \ldots, a_n$ are the smallest such natural numbers such that $\sequence {a_k}_{0 \mathop \le k \mathop \le n}$ is a geometric sequence with common ratio $r$. We have also constructed $b_0, \ldots, b_n$ to have the same property. Thus: : $a_0 = b_0, a_1 = b_1, \ldots, a_n = b_n$ and the result follows. {{qed}}
Let $P$ be a geometric sequence of natural numbers of length $n$. Let the common ratio of $P$ be expressed in canonical form as $\dfrac p q$. From Construction of Geometric Sequence in Lowest Terms: :$P = \paren {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$ By definition of canonical form: :$p \perp q$ It follows from Powers of Coprime Numbers are Coprime that: :$p^n \perp q^n$ Hence the result. {{qed}}