Construction of Geometric Sequence in Lowest Terms

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Theorem

It is possible to find a geometric sequence of integers $G_n$ of length $n + 1$ with a given common ratio such that $G_n$ is in its lowest terms.


In the words of Euclid:

To find numbers in continued proportion, as many as may be prescribed, and the least of those that are in a given ratio.

(The Elements: Book $\text{VIII}$: Proposition $2$)


Porism

In the words of Euclid:

From this it is manifest that, if three numbers in continued proportion be the least of those which have the same ratio with them, the extremes of them are squares, and if four numbers, cubes.

(The Elements: Book $\text{VIII}$: Proposition $2$ : Porism)


Proof 1

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 Proposition $17$ of Book $\text{VII} $: Multiples of Ratios of Numbers that:

$\dfrac a b = \dfrac c d$

As:

$a b = d$
$b^2 = e$

it follows from Proposition $18$ of Book $\text{VII} $: Ratios of Multiples of Numbers that:

$\dfrac a b = \dfrac d e$

As:

$a c = f$
$a d = g$

it follows from Proposition $17$ of Book $\text{VII} $: Multiples of Ratios of Numbers that:

$\dfrac c d = \dfrac f g$

As:

$a d = g$
$a e = h$

it follows from Proposition $17$ of Book $\text{VII} $: Multiples of Ratios of Numbers that:

$\dfrac d e = \dfrac g h$

As:

$a e = h$
$b e = k$

it follows from Proposition $18$ of Book $\text{VII} $: 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 Proposition $22$ of Book $\text{VII} $: 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 Proposition $27$ of Book $\text{VII} $: Powers of Coprime Numbers are Coprime:

$c \perp e$
$f \perp k$

But from Proposition $1$ of Book $\text{VIII} $: Geometric Sequence with Coprime Extremes is in Lowest Terms, these are the least of those with the same common ratio.

Hence the result.

$\blacksquare$


Proof 2

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:

$P = \tuple {q^n, p q^{n - 1}, p^2 q^{n - 2}, \ldots, p^{n - 1} q, p^n}$

$\blacksquare$