Construction of Geometric Sequence in Lowest Terms/Proof 2

Theorem
It is possible to find a geometric progression of a given length of the smallest possible numbers with a given common ratio.

Proof
Let the required length of the geometric progression $P$ be $n$.

Let $r$ be the given common ratio.

From Common Ratio in Integer Geometric Progression 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:
 * $P = \left({a, a \dfrac p q, a \dfrac {p^2} {q^2}, \ldots, a \dfrac {p^{n - 1} } {q^{n - 1} } }\right)$

All the elements of $P$ are natural numbers, so, in particular:
 * $a \dfrac {p^{n - 1} } {q^{n - 1} } \in \N$

From Powers of Coprime Numbers are Coprime:
 * $p^{n - 1} \perp q^{n - 1}$

and so from Euclid's Lemma:
 * $q^{n - 1} \mathop \backslash a$

Thus:
 * $a = k q^{n - 1}$

for some $k \in \N$, and so:
 * $P = \left({k q^{n - 1}, k p q^{n - 2}, k p^2 q^{n - 3}, \ldots, k p^{n - 2} q, k p^{n - 1} }\right)$

From Geometric Progression with Coprime Extremes is in Lowest Terms:
 * $k q^{n - 1} \perp k p^{n - 1}$

from which it follows that $k = 1$.

It follows that the required geometric progression is:
 * $P = \left({q^{n-1}, p q^{n-2}, p^2 q^{n-3}, \ldots, p^{n - 2} q, p^{n - 1} }\right)$