# Geometric Progression with Coprime Extremes is in Lowest Terms

## Theorem

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ 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 1

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ 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:

- $G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$

with the same common ratio where:

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

From Proposition $14$ of Book $\text{VII} $: Proportion of Numbers is Transitive:

- $a_0 : a_n = b_0 : b_n$

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.

$\blacksquare$

## Proof 2

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be natural numbers in geometric progression such that $a_0 \perp a_n$.

Suppose $G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$ 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:

- $\dfrac {a_0} {a_n} = \dfrac {b_0} {b_n}$

- $a_0 \perp a_n$

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.

$\blacksquare$