Geometric Sequence with Coprime Extremes is in Lowest Terms

Theorem
Let $G_n = \left\langle{a_1, a_2, \ldots, a_n}\right\rangle$ be a geometric progression where $a_1, \ldots, a_n$ are all natural numbers.

Let:
 * $a_1 \perp a_n$

where $\perp$ denotes coprimality.

Then the terms of $G_n$ are the lowest possible natural numbers with the same common ratio.


 * If there be as many numbers whatsoever of continuously proportional numbers, and the extremes of them be prime to one another, then the numbers are the least of those which have the same ratio as them.

Proof
Let $G_n = \left\langle{a_1, a_2, \ldots, a_n}\right\rangle$ be natural numbers in geometric progression such that $a_1 \perp a_n$.

Let $G\,'_n = \left\langle{b_1, b_2, \cdots, b_n }\right\rangle$ be another set of natural numbers in geometric progression with the same common ratio where:


 * $\forall k \in \N_{\le n}: a_k > b_k$
 * $\forall k \in \N_{\le n}: a_k : b_k = a_1 : b_1$

From :
 * $a_1 : a_n = b_1 : b_n$

But by hypothesis:
 * $a_1 \perp a_n$

and so from:

and:

it follows that:
 * $a_1 \mathop \backslash b_1$

However, this contradicts the assumption that $b_1 < a_1$.

Therefore $a_1, a_2, \cdots, a_n$ are the least of those with the same common ratio.