---
book: 9
number: 20
id: "IX.20"
kind: "theorem"
uses: ["[[book-7/proposition-31]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.20"
license: "CC-BY-SA-4.0"
---

# IX.20

*Prime numbers are more than any assigned multitude of prime numbers.*

## Proof

Let *A*, *B*, *C* be the assigned prime numbers; I say that there are more prime numbers than *A*, *B*, *C*.

For let the least number measured by *A*, *B*, *C* be taken, and let it be *DE*; let the unit *DF* be added to *DE*.

Then *EF* is either prime or not.

First, let it be prime; then the prime numbers *A*, *B*, *C*, *EF* have been found which are more than *A*, *B*, *C*.

Next, let *EF* not be prime; therefore it is measured by some prime number. [[book-7/proposition-31|VII. 31]]

Let it be measured by the prime number *G*.

I say that *G* is not the same with any of the numbers *A*, *B*, *C*.

For, if possible, let it be so.

Now *A*, *B*, *C* measure *DE*; therefore *G* also will measure *DE*.

But it also measures *EF*.

Therefore *G*, being a number, will measure the remainder, the unit *DF*: which is absurd.

Therefore *G* is not the same with any one of the numbers *A*, *B*, *C*.

And by hypothesis it is prime.

Therefore the prime numbers *A*, *B*, *C*, *G* have been found which are more than the assigned multitude of *A*, *B*, *C*. Q. E. D.