Geometric Sequence with Coprime Extremes is in Lowest Terms

Theorem

 * If there are any multitude whatsoever of continuously proportional numbers, and the outermost of them are prime to one another, then the numbers are the least of those having the same ratio as them.

Proof
Let $a_1, a_2, \cdots, a_n$ be natural numbers, where $a_1 \perp a_n$, where $\perp$ denotes the coprime relation.

Let $b_1, b_2, \cdots, b_n$ be another set of natural numbers, where:


 * $\forall k \in \N_{\le n}: a_k < b_k$
 * $\exists r: \forall k \in \N_{\le n}: \frac{a_k}{b_k} = r$

that is, $b_1, b_2, \cdots, b_n$ are smaller than $a_1, a_2, \cdots, a_n$ respectively while having the same ratio as them.

Since $a_1 \perp a_n$, by Ratios of Fractions in Lowest Terms, $a_1 \backslash b_1$.

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

Therefore $a_1, a_2, \cdots, a_n$ are the least of those having the same ratio as them.