Geometric Sequence with Coprime Extremes is in Lowest Terms/Proof 1

Theorem
Let $G_n = \left\langle{a_0, a_1, \ldots, a_n}\right\rangle$ be a geometric progression of integers.

Let:
 * $a_0 \perp a_n$

where $\perp$ denotes coprimality.

Then $G_n$ is in its lowest terms.

Proof
Let $G_n = \left\langle{a_0, a_1, \ldots, a_n}\right\rangle$ be natural numbers in geometric progression such that $a_0 \perp a_n$.

Aiming for a contradiction, suppose there were to exist another set of natural numbers in geometric progression:
 * $G\,'_n = \left\langle{b_0, b_1, \cdots, b_n }\right\rangle$

with the same common ratio where:


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

From :
 * $a_0 : a_n = b_0 : b_n$

But by hypothesis:
 * $a_0 \perp a_n$

and so from:

and:

it follows that:
 * $a_0 \mathop \backslash b_0$

However, this contradicts the assumption that $b_0 < a_0$.

Therefore $a_0, a_1, \cdots, a_n$ are the least of those with the same common ratio.