If there are any multitude whatsoever of numbers in continued proportion, and the outermost of them are prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them. * Let $A$, $B$, $C$, $D$ be any multitude whatsoever of numbers in continued proportion. * And let the outermost of them, $A$ and $D$, be prime to one another. * I say that $A$, $B$, $C$, $D$ are the least of those (numbers) having the same ratio as them.
This proposition states that if \(a,p,q\) are numbers (i.e. a positive integers) and \(p/q\) is a common ratio of the geometric progression. \[a,a\frac pq,a \left(\frac pq\right)^2,\ldots, a \left(\frac pq\right)^n\quad\quad( * )\]
for some \(n\) such that \(a \left(\frac pq\right)^n\) is also number (i.e. a positive integer), which is co-prime to \(a\), i.e. \[\gcd\left(a,a \left(\frac pq\right)^n\right)=1,\] then the numbers in the geometric progression \( ( * ) \) are the lowest possible with these properties (i.e. having the same common ratio \(p/q\) and having \(a \left(\frac pq\right)^n\) and \(a\) co-prime to each other).
Proofs: 1
Proofs: 1