If there are any multitude whatsoever of numbers in continued proportion, and the first does not measure the second, then no other (number) will measure any other (number) either. * Let $A$, $B$, $C$, $D$, $E$ be any multitude whatsoever of numbers in continued proportion, and let $A$ not measure $B$. * I say that no other (number) will measure any other (number) either.
If $A,B,C,D,E$ are given with $B=An,$ $C=An^2,$ $D=An^3,$ $E=An^4,$ and $n$ is not a positive integer, then $A$ is not a divisor of $B.$ In this geometric progression, none of the numbers will be a multiple of any of the proceeding numbers.
Proofs: 1
Proofs: 1