hcju/proofwikiqa huggingface dataset:
Let $\circ$ be associative. It will be shown that any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$ is equal to the '''left-associated expression''': :$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_n}$ The proof proceeds by induction on $n$. === Basis for the Induction === The base case, $n = 3$: The product $a_1 \circ a_2 \circ a_3$ has two parenthesizations: :$P_1: a_1 \circ \paren {a_2 \circ a_3}$ :$P_2: \paren {a_1 \circ a_2} \circ a_3$ $P_2$ is the left-associative parenthesization for $n = 3$. From the associativity condition, $P_1$ and $P_2$ are identical. This is the basis for the induction. === Induction Hypothesis === If for all $m < n$, all parenthesizations of the $m$-product are identical to its left-associated parenthesization: :$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_m}$ then all parenthesizations of the $n$-product are identical to its left-associated parenthesization: :$\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_n}$ This is the induction hypothesis. === Induction Step === Let $P_i$ denote any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$. Then $P_i$ is always the product of two smaller products: {{begin-eqn}} {{eqn | l = P_i       | r = A \circ B       | c = outermost multiplication }} {{eqn | r = \paren {a_1 \circ a_2 \circ \cdots \circ a_m} \circ \paren {a_{m + 1} \circ a_{m + 2} \circ \cdots \circ a_n}       | c = }} {{eqn | r = \paren {\paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_m} \circ \paren {\paren {\paren {\cdots \paren {a_{m + 1} \circ a_{m + 2} } \circ a_{m + 3} } \circ \cdots} \circ a_n} }       | c = Induction Hypothesis }} {{eqn | r = \paren {\paren {\paren {\cdots \paren {a_1 \circ a_2} \circ a_3} \circ \cdots} \circ a_{n - 1} } \circ a_n       | c = {{Defof|Associative Operation}} }} {{end-eqn}} By the Principle of Mathematical Induction, the proof is complete. {{qed|lemma}} It follows that all parenthesizations of $a_1 \circ a_2 \circ \dots \circ a_n$ yield identical results. So the theorem holds. {{qed}}
Let $a$ and $b$ be plane numbers. Let $a = c d$ and $b = e f$. From {{EuclidPropLink|book = VIII|prop = 4|title = Construction of Sequence of Numbers with Given Ratios}} we can find $g, h, k$ such that: :$c : e = g : h$ :$d : f = h : k$ Let $d e = l$. We also have that $a = c d$. So from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}}: :$c : e = a : l$ But we also have: :$c : e = g : h$ and so: :$g : h = a : l$ Since: :$e d = l$ :$e f = b$ from {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}}: :$d : f = l : b$ But we also have: :$d : f = h : k$ and so: :$h : k = l : b$ But we have: :$g : h = a : l$ and so from {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}: :$g : k = a : b$ But $g : k$ is the ratio compounded of the ratios of the sides of $a$ and $b$. Therefore $a : b$ is the ratio compounded of the ratios of their sides. {{qed}} {{Euclid Note|5|VIII}}
Let $P_a = \tuple {a_0, a_1, \ldots, a_n}$ be a geometric sequence of natural numbers such that $a_0 \nmid a_1$. {{AimForCont}} $a_0 \divides a_k$ for some $k: 2 \le k \le n$. Let $b_0, b_1, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_0, a_1, \ldots, a_k$. These can be found by means of {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}. From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}} :$a_0 : a_k = b_0 : b_k$ Also: :$a_0 : a_1 = b_0 : b_1$ and so as $a_0 \nmid a_1$ it follows by {{EuclidDefLink|VII|20|Proportional}}: :$b_0 \nmid b_1$ From One Divides all Integers it follows that: :$b_0 \ne 1$ From {{EuclidPropLink|book = VIII|prop = 3|title = Construction of Sequence of Numbers with Given Ratios}}: :$b_0 \perp b_k$ But as: :$a_0 : a_k = b_0 : b_k$ it follows that: :$a_0 \nmid a_k$ Now suppose $a_j \divides a_k$ such that $0 < j < k$. Let $b_j, \ldots, b_k$ be the least natural numbers which have the same common ratio as $a_j, \ldots, a_k$. These can be found by means of {{EuclidPropLink|book = VII|prop = 33|title = Least Ratio of Numbers}}. From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}: :$a_j : a_k = b_j : b_k$ The other cases follow similarly. {{qed}}
From Form of Geometric Sequence of Integers, the terms of $P$ are in the form: :$\paren 1: \quad: a_j = k q^j p^{n - j}$ where $p \perp q$. Let $a_0 \nmid a_1$. That is: {{begin-eqn}} {{eqn | l = a_0       | o = \nmid       | r = a_1       | c =  }} {{eqn | ll= \leadsto       | l = k q^n       | o = \nmid       | r = k p q^{n - 1}       | c = from $\paren 1$ }} {{eqn | ll= \leadsto       | l = \paren {k q^{n - 1} } q       | o = \nmid       | r = \paren {k q^{n - 1} } p       | c = rearranging }} {{eqn | ll= \leadsto       | l = q       | o = \nmid       | r = p       | c = Contrapositive of Multiple of Divisor Divides Multiple }} {{end-eqn}} Aiming for a contradiction, suppose that: : $\exists i, j \in \set {0, 1, \ldots, n}, i \ne j: a_i \divides a_j$ {{WLOG}} let $i < j$. {{begin-eqn}} {{eqn | l = a_i       | o = \divides       | r = a_j       | c =  }} {{eqn | ll= \leadsto       | l = k q^i p^{n - i}       | o = \divides       | r = k q^j p^{n - j}       | c = from $\paren 1$ }} {{eqn | ll= \leadsto       | l = m \paren {k q^j p^{n - i} } q^{j - i}       | r = \paren {k q^j p^{n - i} } p^{j - i}       | c = for some $m \in \Z$ }} {{eqn | ll= \leadsto       | l = m q^{j - i}       | r = p^{j - i}       | c = Contrapositive of Multiple of Divisor Divides Multiple }} {{eqn | ll= \leadsto       | l = q^{j - i}       | o = \divides       | r = p^{j - i}       | c =  }} {{end-eqn}} But we have that $p \perp q$. From Powers of Coprime Numbers are Coprime: :$q^{j - i} \perp p^{j - i}$ This can happen only when $q = 1$. This is incompatible with $q \nmid p$. From this contradiction, the result follows. {{qed}}
By hypothesis, let $a_0$ be a divisor of $a_n$. {{AimForCont}} $a_0$ is not a divisor of $a_2$. From First Element of Geometric Sequence not dividing Second it would follow that $a_0$ does not divide $a_n$. From this contradiction follows the result. {{qed}} {{Euclid Note|7|VIII}}
Let $S = \sequence {c_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$ such that: :$r$ is the common ratio of $S$ :$S$ is in its lowest terms. From Geometric Sequence in Lowest Terms has Coprime Extremes, $c_0$ is coprime to $c_n$. Then: :$\dfrac {c_0} {c_n} = \dfrac {a_0} {a_n} = \dfrac {b_0} {b_m}$ But: :$c_n = r^n c_0$ and so: :$b_m = r^n b_0$ The result follows. {{qed}} {{Euclid Note|8|VIII}}
Let the common ratio of $P$ be $r$. By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $P$ is given by: :$a_j = q^j p^{n - j}$ such that: :$a_0 = p^n$ :$a_n = q^n$ For $j \in \set {0, 1, \ldots, n}$, let $r_j = q^j$. Let the finite sequence $Q_2 = \sequence {t_j}_{0 \mathop \le j \mathop \le n}$ be defined as: :$\forall j \in \set {0, 1, \ldots, n}: t_j = p^{n - j}$ Then $Q_1$ is a geometric sequence of integers of length $n$ such that: :$t_0 = 1$ :$t_n = p^n$ Let the finite sequence $Q_1 = \sequence {s_j}_{0 \mathop \le j \mathop \le n}$ be defined as: :$\forall j \in \set {0, 1, \ldots, n}: s_j = q^j$ Then $Q_1$ is a geometric sequence of integers of length $n$ such that: :$s_0 = 1$ :$s_n = q^n$ Hence the result. {{qed}} {{Euclid Note|9|VIII|{{EuclidNoteConverse|title = Product of Geometric Sequences from One|book = VIII|prop = 10}}}}
By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $Q_1$ is given by: :$a_j = a^j$ and of $Q_1$ is given by: :$b_j = b^j$ Let the terms of $P$ be defined as: :$\forall j \in \left\{{0, 1, \ldots, n}\right\}: p_j = b^j a^{n - j}$ Then from Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence. {{qed}} {{Euclid Note|10|VIII|{{EuclidNoteConverse|title = Elements of Geometric Sequence between Coprime Numbers|book = VIII|prop = 9}}}}
Let $a^2$ and $b^2$ be square numbers. Consider the number $a b$. We have: :$\dfrac {a^2} {a b} = \dfrac a b = \dfrac {a b} {b^2}$ By definition, it follows that $a b$ is the mean proportional between $a^2$ and $b^2$. Then: :$\paren {\dfrac a b}^2 = \dfrac {a^2} {b^2}$ By definition, it follows that $a^2$ has to $b^2$ the duplicate ratio that $a$ has to $b$. {{qed}} {{Euclid Note|11|VIII}}
Let $a^3$ and $b^3$ be cube numbers. From the corollary to Form of Geometric Sequence of Integers: :$\left({a^3, a^2 b, a b^2, b^3}\right)$ is a geometric sequence. It follows by definition that $a^2 b$ and $a b^2$ are mean proportionals of $a^3$ and $b^3$. Then: :$\left({\dfrac a b}\right)^3 = \dfrac {a^3} {b^3}$ By definition, it follows that $a^3$ has to $b^3$ the triplicate ratio that $a$ has to $b$. {{qed}} {{Euclid Note|12|VIII}}
From Form of Geometric Sequence of Integers, the $j$th term of $P$ is given by: :$a_j = k q^j p^{n - j}$ Thus the $j$th term of $Q$ is given by: :$b_j = k^m \paren {q^m}^j \paren {p^m}^{n - j}$ From Form of Geometric Sequence of Integers, this is a geometric sequence. {{qed}} {{Euclid Note|13|VIII}}
From Between two Squares exists one Mean Proportional: :$\tuple {a^2, ab, b^2}$ is a geometric sequence. Let $a, b \in \Z$ such that $a^2 \divides b^2$. Then from First Element of Geometric Sequence that divides Last also divides Second: :$a^2 \divides a b$ Thus: {{begin-eqn}} {{eqn | l = a^2       | o = \divides       | r = a b       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \Z:       | l = k a^2       | r = a b       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = k a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \divides       | r = b       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{qed|lemma}} Let $a \divides b$. Then: {{begin-eqn}} {{eqn | l = a       | o = \divides       | r = b       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \Z:       | l = k a       | r = b       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = k a^2       | r = a b       | c =  }} {{eqn | ll= \leadsto       | l = k a b       | r = b^2       | c = {{Defof|Geometric Sequence of Integers}} }} {{eqn | ll= \leadsto       | l = k^2 a^2       | r = b^2       | c =  }} {{eqn | ll= \leadsto       | l = a^2       | o = \divides       | r = b^2       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{qed}} {{Euclid Note|14|VIII}}
Let $a^3$ and $b^3$ be cube numbers. From the corollary to Form of Geometric Sequence of Integers: :$\tuple {a^3, a^2 b, a b^2, b^3}$ is a geometric sequence. Let $a, b \in \Z$ such that $a^2 \divides b^2$. Then from First Element of Geometric Sequence that divides Last also divides Second: :$a^3 \divides a^2 b$ Thus: {{begin-eqn}} {{eqn | l = a^3       | o = \divides       | r = a^2 b       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \Z:       | l = k a^3       | r = a^2 b       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = k a       | r = b       | c =  }} {{eqn | ll= \leadsto       | l = a       | o = \divides       | r = b       | c = {{Defof|Divisor of Integer}} }} {{end-eqn}} {{qed|lemma}} Let $a \divides b$. Then: {{begin-eqn}} {{eqn | l = a       | o = \divides       | r = b       | c =  }} {{eqn | ll= \leadsto       | lo= \exists k \in \Z:       | l = k a       | r = b       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = k a^3       | r = a^2 b       | c =  }} {{eqn | ll= \leadsto       | l = a^3       | o = \divides       | r = a^2 b       | c = {{Defof|Divisor of Integer}} }} {{eqn | ll= \leadsto       | l = a^3       | o = \divides       | r = b^3       | c = Divisibility of Elements in Geometric Sequence of Integers }} {{end-eqn}} {{qed}} {{Euclid Note|15|VIII}}
Let $a \nmid b$. {{AimForCont}}: :$a^2 \divides b^2$ where $\divides$ denotes divisibility. Then by Number divides Number iff Square divides Square: :$a \divides b$ From Proof by Contradiction it follows that $a^2 \divides b^2$ is false. Thus $a^2 \nmid b^2$. {{qed|lemma}} Let $a^2 \nmid b^2$. {{AimForCont}}: :$a \divides b$ Then by Number divides Number iff Square divides Square: :$a^2 \divides b^2$ From Proof by Contradiction it follows that $a \divides b$ is false. Thus $a \nmid b$. {{qed}} {{Euclid Note|16|VIII}}
Let $a \nmid b$. {{AimForCont}}: :$a^3 \divides b^3$ where $\divides$ denotes divisibility. Then by Number divides Number iff Cube divides Cube: :$a \divides b$ From Proof by Contradiction it follows that $a^2 \divides b^2$ is false. Thus $a^3 \nmid b^3$. {{qed|lemma}} Let $a^3 \nmid b^3$. {{AimForCont}} :$a \divides b$ Then by Number divides Number iff Cube divides Cube: :$a^3 \divides b^3$ From Proof by Contradiction it follows that $a \divides b$ is false. Thus $a \nmid b$. {{qed}} {{Euclid Note|17|VIII}}
Let $m$ and $n$ be similar plane numbers. Then for some $p_1, p_2, q_1, q_2 \in \Z$ such that $p_1 < p_2$ and $q_1 < q_2$: :$m = p_1 p_2$ :$n = q_1 q_2$ such that: :$\dfrac {p_1}{q_1} = \dfrac {p_2}{q_2}$ Thus let: :$r := p_1 q_2 = q_1 p_2$ So: {{begin-eqn}} {{eqn | l = \dfrac m r       | r = \dfrac {p_1 p_2} {p_1 q_2}       | c =  }} {{eqn | r = \dfrac {p_2}{q_2}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dfrac r n       | r = \dfrac {q_1 p_2} {q_1 q_2}       | c =  }} {{eqn | r = \dfrac {p_2}{q_2}       | c =  }} {{end-eqn}} Thus by definition, $\paren {m, r, n}$ is a geometric sequence. By definition, $r$ is a mean proportional between $m$ and $n$. Thus as: :$\dfrac {p_1}{q_1} = \dfrac m r = \dfrac r n = \dfrac {p_2}{q_2}$ it follows that $m$ is in duplicate ratio to $n$ as their sides. {{qed}} {{Euclid Note|18|VIII}}
Let $m$ and $n$ be similar solid numbers. Then for some $a, b, c, d, e, f \in \Z$ such that $a \le b \le c$ and $d \le e \le f$: :$m = a b c$ :$n = d e f$ such that: :$\dfrac a d = \dfrac b e = \dfrac c f$ Let: :$r := a e = b d = b f = c e$ So: {{begin-eqn}} {{eqn | l = \dfrac {a b} r       | r = \dfrac {a b} {a e}       | c =  }} {{eqn | r = \dfrac b e       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dfrac r {d e}       | r = \dfrac {b d} {d e}       | c =  }} {{eqn | r = \dfrac b e       | c =  }} {{end-eqn}} Thus by definition, $\tuple {a b, r, d e}$ is a geometric sequence. By definition, $r$ is a mean proportional between $a b$ and $d e$. Now consider the products $r c$ and $r f$. We have:  {{begin-eqn}} {{eqn | l = \frac {a b c} {r c}       | r = \frac {a b c} {c b d}       | c =  }} {{eqn | r =  \frac {a b} {b d}       | c =  }} {{eqn | r =  \frac a d       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \frac {r c} {r f}       | r = \frac {c b d} {f b d}       | c =  }} {{eqn | r =  \frac c f       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \frac {f b d} {d e f}       | r = \frac {b d} {d e}       | c =  }} {{eqn | r =  \frac b e       | c =  }} {{end-eqn}} That is: :$\tuple {a b c, b c d, b d f, d e f}$ is a geometric sequence. Also from the above: :$\dfrac a d = \dfrac b e = \dfrac c f$ showing that $m$ is in triplicate ratio to $n$ as their sides. {{qed}} {{Euclid Note|19|VIII}}
Let the geometric mean of $a$ and $b$ be an integer $m$. Then, in the language of {{AuthorRef|Euclid}}, $m$ is a mean proportional of $a$ and $b$. Thus: :$\left({a, m, b}\right)$ is a geometric sequence. From Form of Geometric Sequence of Integers: :$\exists k, p, q \in \Z: a = k p^2, b = k q^2$ So $a$ and $b$ are plane numbers whose sides are: :$k p$ and $p$ and :$k q$ and $q$ respectively. Then: :$\dfrac {k p} {k q} = \dfrac p q$ demonstrating that $a$ and $b$ are similar plane numbers by definition. {{qed}} {{Euclid Note|20|VIII}}
From Form of Geometric Sequence of Integers: :$\exists k, p, q \in \Z: a = k p^3, b = k q^3$ So $a$ and $b$ are solid numbers whose sides are: :$k p$, $p$ and $p$ and :$k q$, $q$ and $q$ respectively. Then: :$\dfrac {k p} {k q} = \dfrac p q$ demonstrating that $a$ and $b$ are similar solid numbers by definition. {{qed}} {{Euclid Note|21|VIII}}
From Form of Geometric Sequence of Integers: :$P = \tuple {k p^2, k p q, k q^2}$ for some $k, p, q \in \Z$. If $a = k p^2$ is a square number it follows that $k$ is a square number: $k = r^2$, say. So: :$P = \tuple {r^2 p^2, r^2 p q, r^2 q^2}$ and so $c = r^2 q^2 = \paren {r q}^2$. {{qed}} {{Euclid Note|22|VIII}}
From Form of Geometric Sequence of Integers: :$P = \tuple {k p^3, k p^2 q, k p q^2, k q^3}$ for some $k, p, q \in \Z$. If $a = k p^3$ is a cube number it follows that $k$ is a cube number: $k = r^3$, say. So: :$P = \tuple {r^3 p^3, r^3 p^2 q, r^3 p q^2, r^3 q^3}$ and so $d = r^3 q^3 = \paren {r q}^3$. {{qed}} {{Euclid Note|23|VIII}}
From {{EuclidPropLink|title = Between two Similar Plane Numbers exists one Mean Proportional|book = VIII|prop = 18}}: :$\tuple {c^2, c d, d^2}$ is a geometric sequence. From {{EuclidPropLink|title = Geometric Sequences in Proportion have Same Number of Elements|book = VIII|prop = 8}}: :$\tuple {a, m, b}$ is a geometric sequence for some $m$. We have that $a$ is a square number. From {{EuclidPropLink|title = If First of Three Numbers in Geometric Sequence is Square then Third is Square|book = VIII|prop = 22}}: :$b$ is a square number. {{qed}} {{Euclid Note|24|VIII}}
From {{EuclidPropLink|title = Between two Similar Solid Numbers exist two Mean Proportionals|book = VIII|prop = 19}}: :$\left({c^3, c^2 d, c d^2, d^3}\right)$ is a geometric sequence. From {{EuclidPropLink|title = Geometric Sequences in Proportion have Same Number of Elements|book = VIII|prop = 8}}: :$\left({a, m_1, m_2, b}\right)$ is a geometric sequence for some $m$. We have that $a$ is a cube number. From {{EuclidPropLink|title = If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube|book = VIII|prop = 23}}: :$b$ is a cube number. {{qed}} {{Euclid Note|25|VIII}}
Let $a$ and $b$ be similar plane numbers. From Between two Similar Plane Numbers exists one Mean Proportional, there exists a mean proportional $m$ between them. By definition of mean proportional: :$\left({a, m, b}\right)$ is a geometric sequence. From Form of Geometric Sequence of Integers: :$\exists k, p, q \in Z: a = k p^2, b = k q^2, m = k p q$ Thus: :$\dfrac a b = \dfrac {k p^2} {k q^2} = \dfrac {p^2} {q^2}$ Hence the result. {{qed}} {{Euclid Note|26|VIII}}
Let $a$ and $b$ be similar solid numbers. From Between two Similar Solid Numbers exist two Mean Proportionals, there exists two mean proportionals $m_1$ and $m_2$ between them. By definition of mean proportional: :$\left({a, m_1, m_2, b}\right)$ is a geometric sequence. From Form of Geometric Sequence of Integers: :$\exists k, p, q \in Z: a = k p^3, b = k q^3, m_1 = k p^2 q, m_2 = k p q^2$ Thus: :$\dfrac a b = \dfrac {k p^3} {k q^3} = \dfrac {p^3} {q^3}$ Hence the result. {{qed}} {{Euclid Note|27|VIII}}
Let $a$ and $b$ be similar plane numbers. From {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}}: :$a : b = a^2 : a b$ From {{EuclidPropLink|book = VIII|prop = 18|title = Between two Similar Plane Numbers exists one Mean Proportional}}: :there exists one mean proportional between $a$ and $b$. From {{EuclidPropLink|book = VIII|prop = 8|title = Geometric Sequences in Proportion have Same Number of Elements}}: :there exists one mean proportional between $a^2$ and $a b$. From {{EuclidPropLink|book = VIII|prop = 22|title = If First of Three Numbers in Geometric Sequence is Square then Third is Square}} it follows that: :$a b$ is square. {{qed}} {{Euclid Note|1|IX|{{EuclidNoteConverse|prop = 2|title = Numbers whose Product is Square are Similar Plane Numbers}}}}
Let $a$ and $b$ be natural numbers such that $a b$ is square. From {{EuclidPropLink|book = VII|prop = 17|title = Multiples of Ratios of Numbers}}: :$a : b = a^2 : a b$ We have that $a b$ and $a^2$ are both square. By Square Numbers are Similar Plane Numbers they are similar plane numbers. From {{EuclidPropLink|book = VIII|prop = 18|title = Between two Similar Plane Numbers exists one Mean Proportional}}: :there exists one mean proportional between $a^2$ and $a b$. From {{EuclidPropLink|book = VIII|prop = 8|title = Geometric Sequences in Proportion have Same Number of Elements}}: :there exists one mean proportional between $a$ and $b$. The result follows from {{EuclidPropLink|book = VIII|prop = 20|title = Numbers between which exists one Mean Proportional are Similar Plane}}. {{Qed}} {{Euclid Note|2|IX|{{EuclidNoteConverse|prop = 1|title = Product of Similar Plane Numbers is Square}}}}
By the definition of cube number: :$\exists k \in \N: k^3 = a$ Thus: {{begin-eqn}} {{eqn | l = a^2       | r = \paren {k^3}^2       | c =  }} {{eqn | r = k^6       | c =  }} {{eqn | r = \paren {k^2}^3       | c =  }} {{end-eqn}} Thus: :$\exists r = k^2 \in \N: a = r^3$ Hence the result by definition of cube number. {{qed}}
From Cube Number multiplied by Cube Number is Cube, if $a$ and $b$ are cube numbers then $a b$ is a cube number. The result follows by setting $b = a$. {{qed}}
By the definition of cube number: :$\exists r \in \N: r^3 = a$ :$\exists s \in \N: s^3 = b$ Thus: {{begin-eqn}} {{eqn | l = a b       | r = \paren {r^3} \paren {s^3}       | c = Exponent Combination Laws: Power of Product }} {{eqn | r = \paren {r s}^3       | c =  }} {{end-eqn}} Thus: :$\exists k = r s \in \N: a = k^3$ Hence the result by definition of cube number. {{qed}} {{Euclid Note|4|IX|The proof as given here is not that given by {{AuthorRef|Euclid}}, as the latter is unwieldy and of limited use.}}
Let $a, b \in \Z$ be integers such that $a^3 b$ is a cube number. By Square of Cube Number is Cube, $a^3 \cdot a^3$ is a cube number. By Multiples of Ratios of Numbers: :$a^3 \cdot a^3 : a^3 b = a^3 : b$ From Between two Cubes exist two Mean Proportionals, there exist two mean proportionals between $a^3 \cdot a^3$ and $a^3 b$. By Geometric Sequences in Proportion have Same Number of Elements, there exist two mean proportionals between $a^3$ and $b$. The result follows from If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube. {{qed}} {{Euclid Note|5|IX}}
$a^2$ and $a^3$ are both cube numbers. From Between two Cubes exist two Mean Proportionals the sequence: :$a^3, m_1, m_2, a^2$ is a geometric sequence of integers for some $m_1, m_2 \in \Z$. By Geometric Sequences in Proportion have Same Number of Elements: :$a^2, m_3, m_4, a$ is a geometric sequence of integers for some $m_3, m_4 \in \Z$. From If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube it follows that $a$ is a cube number. {{qed}} {{Euclid Note|6|IX}}
By definition of composite number: :$\exists p, q \in \Z_{>1}: a = p q$ Then: :$a b = p q b$ Hence the result by definition of solid number. {{qed}} {{Euclid Note|7|IX}}
Let $a = c^2$ and $b = d^2$. We have that: : $c : d = c : d$ The result follows by definition of similar plane numbers {{qed}} Category:Euclidean Number Theory r9mhpqv1z04e43xu2c1t037yxl8odoh
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as: :$a_j = q^j$ for some $q \in \Z$. Let $k, m \in \set {1, 2, \ldots, n}$ such that $k \divides m$. By definition of divisibility: :$\exists r \in \Z: m = r k$ Then: {{begin-eqn}} {{eqn | l = a_m       | r = q^m       | c =  }} {{eqn | r = q^{r k}       | c =  }} {{eqn | r = \paren {q^r}^k       | c =  }} {{end-eqn}} That is, $a_m$ is a power of $k$. {{qed}} {{Euclid Note|8|IX}}
By Form of Geometric Sequence of Integers from One, the general term of $G_n$ can be expressed as: :$a_j = q^j$ for some $q \in \Z$. Let $a_2 = k^m$. By definition of geometric sequence: :$\forall j \in \set {1, 2, \ldots, n}: a_j = r a_{j - 1}$ where $r$ is the common ratio. This holds specifically for $j = 1$: :$k^m = r \cdot 1$ Thus: {{begin-eqn}} {{eqn | l = a_j       | r = \paren {k^m}^j       | c =  }} {{eqn | r = k^{m j}       | c =  }} {{eqn | r = \paren {k^j}^m       | c =  }} {{end-eqn}} Hence the result. {{qed}} {{Euclid Note|9|IX}}