Dear Mr. Bogomonly

Somewhere on Chris Caldwell's site, http://www.utm.edu/research/primes/ he discussed whether or not the number one was prime. For some reason, which escapes me, he considered it not to be prime. My problem is if this is true, the all prime would fail the Fundamental Theorem of Arithmetic test.

i.e. 5 is only representable as 1*5 if every N is representable as the product of 2 primes then one must be prime.

Any comments?

Frank V Anzalone

P.S. Love your site

Dear Frank:

> Somewhere on Chris Caldwell's site, > http://www.utm.edu/research/primes/
> he discussed whether or not the number one was prime.

I doubt there's a discussion. It's a matter of definition and its
motivation.

It's a common convention to not consider 1 a prime. Otherwise,
many definitions and theorem would have to deal with a special case.

By definition, a number is prime if, besides 1, it's only divisible by itself.

In this form, 1 is automatically excluded as it can't be "besides 1."

But regardless, a better definition may be the more explicit:

n > 1 is a prime if it's only divisible by 1 and itself.

The Fundamental Theorem of Arithmetic states that any number can be uniquely represent as a product of primes. If 1 is a prime the representation can't be unique, for 1*2 = 1*1*2 = ...

You have to add that there are products that consist of a single term - primes' decomposition into factors. You may like this or not. For whatever reasons, mathematicians like this situation more than accepting 1 as a prime, but than modifying the FTA.

> if every N is representable as the
> product of 2 primes then
> one must be prime.

This is not a prerequisite that there should be at least 2 factors, and in fact contradicts the definitions.

> P.S. Love your site

Thank you

All the best,
Alexander Bogomolny