Product of Geometric Sequences from One

Theorem

Let $Q_1 = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ and $Q_2 = \sequence {b_j}_{0 \mathop \le j \mathop \le n}$ be geometric sequences of integers of length $n + 1$.

Let $a_0 = b_0 = 1$.

Then the sequence $P = \sequence {p_j}_{0 \mathop \le j \mathop \le n}$ defined as:

$\forall j \in \set {0, \ldots, n}: p_j = b^j a^{n - j}$

is a geometric sequence.

In the words of Euclid:

If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion.

(The Elements: Book $\text{VIII}$: Proposition $10$)

Proof

By Form of Geometric Sequence of Integers with Coprime Extremes, the $j$th term of $Q_1$ is given by:

$a_j = a^j$

and of $Q_1$ is given by:

$b_j = b^j$

Let the terms of $P$ be defined as:

$\forall j \in \left\{{0, 1, \ldots, n}\right\}: p_j = b^j a^{n - j}$

Then from Form of Geometric Sequence of Integers it follows that $P$ is a geometric sequence.

$\blacksquare$

Historical Note

This proof is Proposition $10$ of Book $\text{VIII}$ of Euclid's The Elements.
It is the converse of Proposition $9$ of Book $\text{VIII} $: Elements of Geometric Sequence between Coprime Numbers.

Sources

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