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

Let the number *A* be the least that is measured by the prime numbers *B, C,* and *D.*

I say that *A* is not measured by any other prime number except *B, C,* or *D.*

If possible, let it be measured by the prime number *E,* and let *E* not be the same as any one of the numbers *B, C,* or *D.*

Now, since *E* measures *A,* let it measure it according to *F,* therefore *E* multiplied by *F* makes *A.* And *A* is measured by the prime numbers *B, C,* and *D.*

But, if two numbers multiplied by one another make some number, and any prime number measures the product, then it also measures one of the original numbers, therefore each of *B, C,* and *D* measures one of the numbers *E* or *F.*

Now they do not measure *E,* for *E* is prime and not the same with any one of the numbers *B, C,* or *D.* Therefore they measure *F,* which is less than *A,* which is impossible, for *A* is by hypothesis the least number measured by *B, C,* and *D.*

Therefore no prime number measures *A* except *B, C,* and *D.*

Therefore, *if a number is the least that is measured by prime numbers, then it is not measured by any other prime number except those originally measuring it.*

Q.E.D.

The proof is clear, and it depends on VII.30, that if a prime divides a product, then it divides one of the factors.