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

## Theorem

Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be a geometric sequence of integers.

Let:

$a_0 \perp a_n$

where $\perp$ denotes coprimality.

Then $G_n$ is in its lowest terms.

## Proof

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

Aiming for a contradiction, suppose there were to exist another set of natural numbers in geometric sequence:

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

with the same common ratio where:

$\forall k \in \N_{\le n}: a_k > b_k$
$a_0 : a_n = b_0 : b_n$

But by hypothesis:

$a_0 \perp a_n$

and so from:

Proposition $21$: Coprime Numbers form Fraction in Lowest Terms

and:

Proposition $20$: Ratios of Fractions in Lowest Terms

it follows that:

$a_0 \divides b_0$

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.

$\blacksquare$