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