# 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:

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

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:

- $a_0 \divides b_0$

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.

$\blacksquare$

## Historical Note

This proof is Proposition $1$ of Book $\text{VIII}$ of Euclid's *The Elements*.

It is the converse of Proposition $3$: Geometric Sequence in Lowest Terms has Coprime Extremes.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{VIII}$. Propositions