---
book: 9
number: 15
id: "IX.15"
kind: "theorem"
uses: ["[[book-8/proposition-2]]", "[[book-7/proposition-22]]", "[[book-7/proposition-28]]", "[[book-7/proposition-24]]", "[[book-7/proposition-25]]", "[[book-2/proposition-3]]", "[[book-2/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.15"
license: "CC-BY-SA-4.0"
---

# IX.15

*If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number.*

## Proof

Let *A*, *B*, *C*, three numbers in continued proportion, be the least of those which have the same ratio with them; I say that any two of the numbers *A*, *B*, *C* whatever added together are prime to the remaining number, namely *A*, *B* to *C*; *B*, *C* to *A*; and further *A*, *C* to *B*.

For let two numbers *DE*, *EF*, the least of those which have the same ratio with *A*, *B*, *C*, be taken. [[book-8/proposition-2|VIII. 2]]

It is then manifest that *DE* by multiplying itself has made *A*, and by multiplying *EF* has made *B*, and, further, *EF* by multiplying itself has made *C*. [[book-8/proposition-2|VIII. 2]]

Now, since *DE*, *EF* are least, they are prime to one another. [[book-7/proposition-22|VII. 22]]

But, if two numbers be prime to one another, their sum is also prime to each; [[book-7/proposition-28|VII. 28]] therefore *DF* is also prime to each of the numbers *DE*, *EF*.

But further *DE* is also prime to *EF*; therefore *DF*, *DE* are prime to *EF*.

But, if two numbers be prime to any number, their product is also prime to the other; [[book-7/proposition-24|VII. 24]] so that the product of *FD*, *DE* is prime to *EF*; hence the product of *FD*, *DE* is also prime to the square on *EF*. [[book-7/proposition-25|VII. 25]]

But the product of *FD*, *DE* is the square on *DE* together with the product of *DE*, *EF*; [[book-2/proposition-3|II. 3]] therefore the square on *DE* together with the product of *DE*, *EF* is prime to the square on *EF*.

And the square on *DE* is *A*, the product of *DE*, *EF* is *B*, and the square on *EF* is *C*; therefore *A*, *B* added together are prime to *C*.

Similarly we can prove that *B*, *C* added together are prime to *A*.

I say next that *A*, *C* added together are also prime to *B*.

For, since *DF* is prime to each of the numbers *DE*, *EF*, the square on *DF* is also prime to the product of *DE*, *EF*. [[book-7/proposition-24|VII. 24, 25]]

But the squares on *DE*, *EF* together with twice the product of *DE*, *EF* are equal to the square on *DF*; [[book-2/proposition-4|II. 4]] therefore the squares on *DE*, *EF* together with twice the product of *DE*, *EF* are prime to the product of *DE*, *EF*.

Separando, the squares on *DE*, *EF* together with once the product of *DE*, *EF* are prime to the product of *DE*, *EF*.

Therefore, separando again, the squares on *DE*, *EF* are prime to the product of *DE*, *EF*.

And the square on *DE* is *A*, the product of *DE*, *EF* is *B*, and the square on *EF* is *C*.

Therefore *A*, *C* added together are prime to *B*. Q. E. D.