If three numbers in continued proportion are the least of those (numbers) having the same ratio as them then two (of them) added together in any way are prime to the remaining (one). * Let $A$, $B$, $C$ be three numbers in continued proportion (which are) the least of those (numbers) having the same ratio as them. * I say that two of $A$, $B$, $C$ added together in any way are prime to the remaining (one), (that is) $A$ and $B$ (prime) to $C$, $B$ and $C$ to $A$, and, further, $A$ and $C$ to $B$.
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Proofs: 1