First Element of Geometric Sequence that divides Last also divides Second
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Theorem
Let $P = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers of length $n$.
Let $a_0$ be a divisor of $a_n$.
Then $a_0$ is a divisor of $a_2$.
In the words of Euclid:
- If there be as many numbers as we please in continued proportion, and the first measures the last, it will measure the second also.
(The Elements: Book $\text{VIII}$: Proposition $7$)
Proof
By hypothesis, let $a_0$ be a divisor of $a_n$.
Aiming for a contradiction, suppose $a_0$ is not a divisor of $a_2$.
From First Element of Geometric Sequence not dividing Second it would follow that $a_0$ does not divide $a_n$.
From this contradiction follows the result.
$\blacksquare$
Historical Note
This proof is Proposition $7$ of Book $\text{VIII}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VIII}$. Propositions