---
book: 9
number: 14
id: "IX.14"
kind: "theorem"
uses: ["[[book-7/proposition-30]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.14"
license: "CC-BY-SA-4.0"
---

# IX.14

*If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it.*

## Proof

For let the number *A* be the least that is measured by the prime numbers *B*, *C*, *D*; I say that *A* will not be measured by any other prime number except *B*, *C*, *D*.

For, if possible, let it be measured by the prime number *E*, and let *E* not be the same with any one of the numbers *B*, *C*, *D*.

Now, since *E* measures *A*, let it measure it according to *F*; therefore *E* by multiplying *F* has made *A*.

And *A* is measured by the prime numbers *B*, *C*, *D*.

But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [[book-7/proposition-30|VII. 30]] therefore *B*, *C*, *D* will measure one of the numbers *E*, *F*.

Now they will not measure *E*; for *E* is prime and not the same with any one of the numbers *B*, *C*, *D*.

Therefore they will measure *F*, which is less than *A*: which is impossible, for *A* is by hypothesis the least number measured by *B*, *C*, *D*.

Therefore no prime number will measure *A* except *B*, *C*, *D*. Q. E. D.