Geometric Sequence with Coprime Extremes is in Lowest Terms/Proof 2

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 $a_0 \perp a_n$.

Suppose $G\,'_n = \left\langle{b_0, b_1, \cdots, b_n }\right\rangle$ be another set of natural numbers in geometric progression with the same common ratio where:


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

By definition of geometric progression:
 * $a_0 = r^n a_n$
 * $b_0 = r^n b_n$

Hence:
 * $\dfrac {a_0} {a_n} = \dfrac {b_0} {b_n}$

By hypothesis:
 * $a_0 \perp a_n$

Thus $\dfrac {a_0} {a_n}$ is in canonical form.

From Canonical Form of Rational Number is Unique it follows that $a_0$ and $a_n$ are the only two integers fulfilling these conditions.

Thus:
 * $\forall p, q: \dfrac p q = \dfrac {a_0} {a_n}: a_0 \le p, a_n \le q$

But it was supposed that:
 * $b_0 < a_0$
 * $b_n < a_n$

From this contradiction it follows that there can be no such $b_0, \ldots, b_n$.

Hence the result.