avewright/proofwiki-math huggingface dataset:
Theorem: Number divides Number iff Cube divides Cube

Let $a, b \in \Z$.
Then:
:$a^3 \divides b^3 \iff a \divides b$
where $\divides$ denotes integer divisibility.
{{:Euclid:Proposition/VIII/15}}

Proof:
Let $a^3$ and $b^3$ be cube numbers.
From {{Corollary|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
      | q = \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
      | q = \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}}
Theorem: Number does not divide Number iff Square does not divide Square

Let $a, b \in \Z$ be integers.
Then:
:$a \nmid b \iff a^2 \nmid b^2$
where $a \nmid b$ denotes that $a$ is not a divisor of $b$.
{{:Euclid:Proposition/VIII/16}}

Proof:
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}}
Theorem: Number does not divide Number iff Cube does not divide Cube

Let $a, b \in \Z$ be integers.
Then:
:$a \nmid b \iff a^3 \nmid b^3$
where $a \nmid b$ denotes that $a$ is not a divisor of $b$.
{{:Euclid:Proposition/VIII/17}}

Proof:
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}}
Theorem: Numbers between which exists one Mean Proportional are Similar Plane

Let $a, b \in \Z$ such that the geometric mean is an integer.
Then $a$ and $b$ are similar plane numbers.
{{:Euclid:Proposition/VIII/20}}

Proof:
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}}
Theorem: Numbers between which exist two Mean Proportionals are Similar Solid

Let $a, b \in \Z$ be the extremes of a geometric sequence of integers whose length is $4$:
:$\tuple {a, m_1, m_2, b}$
That is, such that $a$ and $b$ have $2$ mean proportionals.
Then $a$ and $b$ are similar solid numbers.
{{:Euclid:Proposition/VIII/21}}

Proof:
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}}
Theorem: If First of Three Numbers in Geometric Sequence is Square then Third is Square

Let $P = \tuple {a, b, c}$ be a geometric sequence of integers.
Let $a$ be a square number.
Then $c$ is also a square number.
{{:Euclid:Proposition/VIII/22}}

Proof:
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}}
Theorem: If First of Four Numbers in Geometric Sequence is Cube then Fourth is Cube

Let $P = \tuple {a, b, c, d}$ be a geometric sequence of integers.
Let $a$ be a cube number.
Then $d$ is also a cube number.
{{:Euclid:Proposition/VIII/23}}

Proof:
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}}
Theorem: If Ratio of Square to Number is as between Two Squares then Number is Square

Let $a, b, c, d \in \Z$ be integers such that:
:$\dfrac a b = \dfrac {c^2} {d^2}$
Let $a$ be a square number.
Then $b$ is also a square number.
{{:Euclid:Proposition/VIII/24}}

Proof:
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}}
Theorem: If Ratio of Cube to Number is as between Two Cubes then Number is Cube

Let $a, b, c, d \in \Z$ be integers such that:
:$\dfrac a b = \dfrac {c^3} {d^3}$
Let $a$ be a cube number.
Then $b$ is also a cube number.
{{:Euclid:Proposition/VIII/25}}

Proof:
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}}
Theorem: Square of Cube Number is Cube

Let $a \in \N$ be a natural number.
Let $a$ be a cube number.
Then $a^2$ is also a cube number.
{{:Euclid:Proposition/IX/3}}

Proof:
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}}
Theorem: Square of Cube Number is Cube

Let $a \in \N$ be a natural number.
Let $a$ be a cube number.
Then $a^2$ is also a cube number.
{{:Euclid:Proposition/IX/3}}

Proof:
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}}
Theorem: Cube Number multiplied by Cube Number is Cube

Let $a, b \in \N$ be natural numbers.
Let $a$ and $b$ be cube numbers.
Then $a b$ is also a cube number.
{{:Euclid:Proposition/IX/4}}

Proof:
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 = 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.}}
Theorem: Number Squared making Cube is itself Cube

Let $a \in \Z$ be an integer.
Let $a^2$ be a cube number.
Then $a$ is a cube number.
{{:Euclid:Proposition/IX/6}}

Proof:
$a^2$ 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}}
Theorem: Product of Composite Number with Number is Solid Number

Let $a, b \in \Z$ be positive integers.
Let $a$ be a composite number.
Then $a b$ is a solid number.
{{:Euclid:Proposition/IX/7}}

Proof:
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}}
Theorem: Square Numbers are Similar Plane Numbers

Let $a$ and $b$ be square numbers.
Then $a$ and $b$ are similar plane numbers.

Proof:
Let $a = c^2$ and $b = d^2$.
We have (trivially) that:
:$c : d = c : d$
The result follows by definition of similar plane numbers.
{{qed}}
Category:Square Numbers
Category:Euclidean Number Theory
04hgpyj8d33y1qzsa7zr8c6p917xs8x
Theorem: Elements of Geometric Sequence from One which are Powers of Number

Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Then:
:$\forall m, k \in \set {1, \ldots, n}: k \divides m \implies a_m$ is a power of $k$
where $\divides$ denotes divisibility.
{{:Euclid:Proposition/IX/8}}

Proof:
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}}
Theorem: Elements of Geometric Sequence from One where First Element is Power of Number

Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $m \in \Z_{> 0}$.
Let $a_1$ be the $m$th power of an integer.
Then all the terms of $G_n$ are $m$th powers of  integers.
{{:Euclid:Proposition/IX/9}}

Proof:
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}}
Theorem: Elements of Geometric Sequence from One where First Element is not Power of Number

Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $k \in \Z_{> 1}$.
Let $a_1$ not be a power of $k$.
Then $a_m$ is not a power of $k$ except for:
:$\forall m, k \in \set {1, 2, \ldots, n}: k \divides m$
where $\divides$ denotes divisibility.
{{:Euclid:Proposition/IX/10}}

Proof:
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 $k \nmid m$.
Then by the Division Theorem there exists a unique $q \in \Z$ such that:
:$m = k q + b$
for some $b$ such that $0 < b < k$.
Thus:
:$a_m = a^{k q} a^b$
which is not a power of $k$.
{{qed}}
{{Euclid Note|10|IX}}
Theorem: Elements of Geometric Sequence from One which Divide Later Elements

Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $m \in \Z_{> 0}$.
Then:
:$\forall r \in \set {0, 1, \ldots, m}: a_k \divides a_m$
where $\divides$ denotes divisibility.
{{:Euclid:Proposition/IX/11}}

Proof:
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 $q \in \Z$.
Hence the result from Divisors of Power of Prime.
{{qed}}
{{Euclid Note|11|IX}}
Theorem: Elements of Geometric Sequence from One Divisible by Prime

Let $G_n = \sequence {a_n}_{0 \mathop \le i \mathop \le n}$ be a geometric sequence of integers.
Let $a_0 = 1$.
Let $p$ be a prime number such that:
:$p \divides a_n$
where $\divides$ denotes divisibility.
Then $p \divides a_1$.
{{:Euclid:Proposition/IX/12}}

Proof:
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 $q \in \Z$.
Thus by hypothesis:
:$p \divides q^n$
From Euclid's Lemma for Prime Divisors: General Result:
:$p \divides q$
Hence the result.
{{qed}}
{{Euclid Note|12|IX}}
Theorem: Divisibility of Elements of Geometric Sequence from One where First Element is Prime

Let $Q_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of length $n$ consisting of integers only.
Let $a_0 = 1$.
Let $a_1$ be a prime number.
Then the only divisors of $a_n$ are $a_j$ for $j \in \set {1, 2, \ldots, n}$.
{{:Euclid:Proposition/IX/13}}

Proof:
From Form of Geometric Sequence of Integers from One, the elements of $Q_n$ are given by:
:$Q_n = \tuple {1, a, a^2, \ldots, a^n}$
From Divisors of Power of Prime, each of $a_j$ for $j \in \set {1, 2, \ldots, n}$ are the only divisors of $a_n$.
{{qed}}
{{Euclid Note|13|IX}}
Theorem: Sum of Pair of Elements of Geometric Sequence with Three Elements in Lowest Terms is Coprime to other Element

Let $P = \tuple {a, b, c}$ be a geometric sequence of integers in its lowest terms.
Then $\paren {a + b}$, $\paren {b + c}$ and $\paren {a + c}$ are all coprime to each of $a$, $b$ and $c$.
{{:Euclid:Proposition/IX/15}}

Proof:
Let the common ratio of $P$ in canonical form be $\dfrac q p$.
By Form of Geometric Sequence of Integers in Lowest Terms:
:$P = \tuple {p^2, p q, q^2}$
Then:
{{begin-eqn}}
{{eqn | l = p
      | o = \perp
      | r = q
      | c = {{Defof|Canonical Form of Rational Number}}
}}
{{eqn | ll= \leadsto
      | l = q
      | o = \perp
      | r = \paren {p + q}
      | c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | ll= \leadsto
      | l = q
      | o = \perp
      | r = p \paren {p + q}
      | c = Integer Coprime to Factors is Coprime to Whole
}}
{{eqn | ll= \leadsto
      | l = q^2
      | o = \perp
      | r = p \paren {p + q}
      | c = Square of Coprime Number is Coprime
}}
{{eqn | ll= \leadsto
      | l = q^2
      | o = \perp
      | r = p^2 + p q
      | c = factorising
}}
{{eqn | ll= \leadsto
      | l = c
      | o = \perp
      | r = a + b
      | c = 
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = p^2
      | o = \perp
      | r = p q + q^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a
      | o = \perp
      | r = b + c
      | c = factorising
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = p + q
      | o = \perp
      | r = p
      | c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | lo= \land
      | l = p + q
      | o = \perp
      | r = q
      | c = Numbers are Coprime iff Sum is Coprime to Both
}}
{{eqn | ll= \leadsto
      | l = \paren {p + q}^2
      | o = \perp
      | r = p q
      | c = as above
}}
{{eqn | ll= \leadsto
      | l = p^2 + q^2 + 2 p q
      | o = \perp
      | r = p q
      | c = factorising
}}
{{eqn | ll= \leadsto
      | l = p^2 + q^2
      | o = \perp
      | r = p q
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b
      | o = \perp
      | r = a + c
      | c = 
}}
{{end-eqn}}
{{qed}}
{{Euclid Note|15|IX}}
Theorem: Two Coprime Integers have no Third Integer Proportional

Let $a, b \in \Z_{>0}$ be integers such that $a$ and $b$ are coprime.
Then there is no integer $c \in \Z$ such that:
:$\dfrac a b = \dfrac b c$
{{:Euclid:Proposition/IX/16}}

Proof:
Suppose such a $c$ exists.
From Coprime Numbers form Fraction in Lowest Terms, $\dfrac a b$ is in canonical form.
From Ratios of Fractions in Lowest Terms:
:$a \divides b$
where $\divides$ denotes divisibility.
This contradicts the fact that $a$ and $b$ are coprime.
Hence such a $c$ cannot exist.
{{qed}}
{{Euclid Note|16|IX}}
Theorem: Last Element of Geometric Sequence with Coprime Extremes has no Integer Proportional as First to Second

Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers such that $a_0 \ne 1$.
Let $a_0 \perp a_n$, where $\perp$ denotes coprimality.
Then there does not exist an integer $b$ such that:
:$\dfrac {a_0} {a_1} = \dfrac {a_n} b$
{{:Euclid:Proposition/IX/17}}

Proof:
{{AimForCont}} there exists $b$ such that $\dfrac {a_0} {a_1} = \dfrac {a_n} b$.
Then:
:$\dfrac {a_0} {a_n} = \dfrac {a_1} b$
By Ratios of Fractions in Lowest Terms:
:$a_0 \divides a_1$
where $\divides$ denotes divisibility.
From Divisibility of Elements in Geometric Sequence of Integers:
:$a_0 \divides a_n$
But $a_0 \perp a_n$.
From this contradiction it follows that there can be no such $b$.
{{qed}}
{{Euclid Note|17|IX}}
Theorem: Condition for Existence of Third Number Proportional to Two Numbers

Let $a, b, c \in \Z$ be integers.
Let $\tuple {a, b, c}$ be a geometric sequence.
In order for this to be possible, both of these conditions must be true:
:$(1): \quad a$ and $b$ cannot be coprime
:$(2): \quad a \divides b^2$
where $\divides$ denotes divisibility.
{{:Euclid:Proposition/IX/18}}

Proof:
Let $P = \tuple {a, b, c}$ be a geometric sequence.
Then by definition their common ratio is:
:$\dfrac b a = \dfrac c b$
From Two Coprime Integers have no Third Integer Proportional it cannot be the case that $a$ and $b$ are coprime.
Thus condition $(1)$ is satisfied.
From Form of Geometric Sequence of Integers, $P$ is in the form:
:$\tuple {k p^2, k p q, k q^2}$
from which it can be seen that:
:$k p^2 \divides k^2 p^2 q^2$
demonstrating that condition $(2)$ is satisfied.
{{qed}}
{{Euclid Note|18|IX}}
Theorem: Odd Divisor of Even Number also divides its Half

Let $a, b \in \Z$ be integers.
Let $a$ be odd and $b$ be even.
Let:
:$a \divides b$
where $\divides$ denotes divisibility.
Then:
:$a \divides \dfrac b 2$ 
{{:Euclid:Proposition/IX/30}}

Proof:
By definition of an even number:
:$\exists r \in \Z: b = 2 r$
By definition of an odd number:
:$\exists s \in \Z: a = 2 s + 1$
Thus:
:$a \divides 2 \iff a = \pm 1$
in which case from One Divides all Integers:
:$a \divides \dfrac b 2$
We have that:
:$a \divides 2 r$
and so by Euclid's Lemma:
:$a \divides r = \dfrac b 2$
{{qed}}
{{Euclid Note|30|IX}}
Theorem: Odd Number Coprime to Number is also Coprime to its Double

Let $a, b \in \Z$ be integers.
Let $a$ be odd.
Let:
:$a \perp b$
where $\perp$ denotes coprimality.
Then:
:$a \perp 2 b$
{{:Euclid:Proposition/IX/31}}

Proof:
By definition of odd number:
:$a \perp 2$
The result follows from Integer Coprime to Factors is Coprime to Whole.
{{qed}}
{{Euclid Note|31|IX}}
Theorem: Power of Two is Even-Times Even Only

Let $a > 2$ be a power of $2$.
Then $a$ is even-times even only.
{{:Euclid:Proposition/IX/32}}

Proof:
As $a$ is a power of $2$ greater than $2$:
:$\exists k \in \Z_{>1}: a = 2^k$
Thus:
:$a = 2^2 2^{k - 2}$
and so has $2^2 = 4$ as a divisor.
Let $b$ be an odd number.
By definition:
:$b \nmid 2$
The result follows by Integer Coprime to Factors is Coprime to Whole.
{{qed}}
{{Euclid Note|32|IX}}
Theorem: Number whose Half is Odd is Even-Times Odd

Let $a \in \Z$ be an integer such that $\dfrac a 2$ is an odd integer.
Then $a$ is even-times odd.
{{:Euclid:Proposition/IX/33}}

Proof:
By definition:
:$a = 2 r$
where $r$ is an odd integer.
Thus:
:$a$ has an even divisor
and:
:$a$ has an odd divisor.
Hence the result by definition of even-times odd integer.
As $r$ is an odd integer it follows that $2 \nmid r$.
Thus $a$ is not divisible by $4$.
Hence $a$ is not even-times even.
{{qed}}
{{Euclid Note|33|IX}}
Theorem: Number neither whose Half is Odd nor Power of Two is both Even-Times Even and Even-Times Odd

Let $a \in \Z$ be an integer such that:
:$(1): \quad a$ is not a power of $2$
:$(2): \quad \dfrac a 2$ is an even integer.
Then $a$ is both even-times even and even-times odd.
{{:Euclid:Proposition/IX/34}}

Proof:
As $\dfrac a 2$ is an even integer it follows that:
:$\dfrac a 2 = 2 r$
for some $r \in \Z$.
That is:
:$a = 2^2 r$
and so $a$ is even-times even by definition.
{{qed|lemma}}
{{AimForCont}} $a$ is not even-times odd.
Then $a$ does not have an odd divisor.
Thus in its prime decomposition there are no odd primes.
Thus $a$ is in the form:
:$a = 2^k$
for some $k \in \Z_{>0}$.
That is, $a$ is a power of $2$.
This contradicts condition $(1)$.
From this contradiction it is deduced that $a$ is even-times odd.
{{qed}}
{{Euclid Note|34|IX}}
Theorem: Scaled Real Function that Increases Without Bound

Let $f: \R \to \R$ be a real function.
Let $\lambda \in \R_{\ne 0}$ be a nonzero constant.
Then:
For $\lambda > 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = + \infty$
For $\lambda < 0$:
:$\ds \lim_{x \mathop \to +\infty} \map f x = +\infty \implies \lim_{x \mathop \to +\infty} \lambda \map f x = -\infty$
:$\ds \lim_{x \mathop \to -\infty} \map f x = +\infty \implies \lim_{x \mathop \to -\infty} \lambda \map f x = -\infty$

Proof:
Let $\ds \lim_{x \mathop \to +\infty} \map f x = +\infty$.
From the definition of infinite limit at infinity, this means that:
:$\forall M > 0: \exists N > 0: x > N \implies \map f x > M$.
Suppose $\lambda > 0$.
Then $M > 0 \iff \lambda^{-1} M > 0$.
Also, $\map f x > \lambda^{-1} M \iff \lambda f\map f x > M$.
So:
:$\forall M > 0: \exists N > 0: x > N \implies \lambda \map f x > M$
From the definition of infinite limit at infinity:
:$\ds \lim_{x \mathop \to +\infty} \lambda \map f x = +\infty$.
The proof for $\ds \lim_{x \mathop \to -\infty} \map f x = +\infty$ is analogous.
Now, suppose $\lambda < 0$.
Then $-\lambda f > 0$, and so $-\lambda f \to + \infty$, from above.
Write $\lambda f = -\paren {-\lambda f}$ and the result follows from Negative of Real Function that Increases Without Bound.
{{qed}}