Product of Geometric Progressions from One

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Theorem

Let $Q_1 = \left\langle{a_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ and $Q_2 = \left\langle{b_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ be geometric progressions of integers of length $n + 1$.

Let $a_0 = b_0 = 1$.

Then the sequence $P = \left\langle{p_j}\right\rangle_{0 \mathop \le j \mathop \le n}$ defined as:

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

is a geometric progression.


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 Progression 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 Progression of Integers it follows that $P$ is a geometric progression.

$\blacksquare$


Historical Note

This theorem 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 Progression between Coprime Numbers.


Sources