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.


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