Last Element of Geometric Sequence with Coprime Extremes has no Integer Proportional as First to Second

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $G_n = \sequence {a_j}_{0 \mathop \le j \mathop \le n}$ be a geometric sequence of integers such that $a_0 \ne 1$.

Let $a_0 \perp a_n$, where $\perp$ denotes coprimality.

Then there does not exist an integer $b$ such that:

$\dfrac {a_0} {a_1} = \dfrac {a_n} b$


In the words of Euclid:

If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first is to the second.

(The Elements: Book $\text{IX}$: Proposition $17$)


Proof

Aiming for a contradiction, suppose there exists $b$ such that $\dfrac {a_0} {a_1} = \dfrac {a_n} b$.

Then:

$\dfrac {a_0} {a_n} = \dfrac {a_1} b$

By Ratios of Fractions in Lowest Terms:

$a_0 \divides a_1$

where $\divides$ denotes divisibility.

From Divisibility of Elements in Geometric Sequence of Integers:

$a_0 \divides a_n$

But $a_0 \perp a_n$.

From this contradiction it follows that there can be no such $b$.

$\blacksquare$


Historical Note

This proof is Proposition $17$ of Book $\text{IX}$ of Euclid's The Elements.


Sources