Geometric Sequence with Coprime Extremes is in Lowest Terms
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 1
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.
$\blacksquare$
Proof 2
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 $G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$ be another set of natural numbers in geometric sequence with the same common ratio where:
- $\forall k \in \N_{\le n}: a_k > b_k$
By definition of geometric sequence:
- $a_0 = r^n a_n$
- $b_0 = r^n b_n$
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.
$\blacksquare$