# Zeta function justifying 1 not being prime?

1. Nov 26, 2005

### Jameson

I was trying to explain to my family last night why 1 is not generally defined as a prime number and I thought of the Zeta Function. There is the standard way to write it,

(1)$$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$$

but then there is also the Euler product formula:

(2)$$\prod_{p}\frac{1}{1-p^{-s}}$$

Obtaining the product formula through the sieving method requires one to factor out (1-1/p) from the (1) equation, p being prime numbers starting at 2. If we include 1 as a prime number, this entire method would fail.

Do you think this is a good way of showing why analytically 1 should not be considered prime?

Jameson

2. Nov 26, 2005

### shmoe

The Euler product would not fail in any way, it would just be expressed differently and be the product over primes greater than 1. It's just a definition to exclude 1, changing this would have no effect on the underlying mathematics just how you express it.

It turns out to be more convenient to exclude 1 as it doesn't behave like the usual primes since it divides every other number (it's a unit). Any doubters should pick up a number theory text, change the definition of prime to include 1, then go through the book and modify all the statements to make sense and be valid. Compare which version is more compact.

3. Nov 26, 2005

### matt grime

Don't bother to explain why 1 is excluded. Make them make the case for it to be included. If it helps, remind them the definition of prime in the positive integers is (equivalent to ) the number has exactly two *distinct* positive prime factors. It is just a definition and this is the one we NOW prefer (old fashioned mathematics, say a couple of centuries ago would have had a different definition and allowed 1 as a prime).

One of the key things about primes in the integers is that if p divides ab and p is a prime then it divides one of a and b, this is vacuous if p is allowed to be 1. It doesn't say anything about a or b at all.

There are always degenerate cases, in algebra we tend to include the degenerate cases: as an example subset is allowed to include the empty set and the whole set, but have the label *proper* subset to exclude the degenerate case. As Shmoe said, a result itself does not become false because you redefine the labels, but the statement of the result is incorrect.

I think I will attempt to never again justify why 1 is excluded, but make people justify why they think it should be included. It is after all 'my' definition, and the definition of all those who matter (ie mathematicians), and we don't include 1, we have our reasons, if someone else doesn't like it, tough. If you want a set of elements that includes all the primes and 1 make a name for it; it won't be any use I imagine, since, should I ever need it I can simply state 'take the primes, union 1' and there we are.

Last edited: Nov 26, 2005
4. Nov 26, 2005

### Jameson

Right. And by saying that it's the product over all primes greater than one suggests something about one being prime. I don't see a way to express the product that includes one. Do you know of way?

My main point was that by this sieving method all of the numbers factored out happen to be primes, and one isn't included by this method.

5. Nov 26, 2005

### matt grime

It wouldn't include one... why would it...? it doesn't... if you were to suddenly make one a prime then the statement that the zeta function is a product over all primes is false, because you've changed the meaning of one of the terms! That the zeta function is still a product over 2,3,5,7,.. is still true. it's just that you're no longer calling this the complete set of primes. 1 isn't a prime, shmoe only said what he did to show you how the result is still true with the corrected statement IF YOU (yes, YOU, no one else) alters the definition of prime to include 1. 1 is not considered a prime, end of story.

6. Nov 26, 2005

### matt grime

Oh, and the dirichlet series holds because each integer n in the sum has a unique decomposition into primes (1 is not a prime) that is why the numbers 2,3,5,7,11,.. appear in the product, it really is nothing inexplicable.

7. Nov 26, 2005

### Jameson

I'm on your side. :tongue2: I'm just trying to show that perhaps analytic functions demonstrate that 1 doesn't have a place in the set of primes.

8. Nov 26, 2005

### matt grime

right, cos that's the only statement in mathematics that's made easier because we exclude 1 from being a prime (and is a direct consequence of the fundamental theorem of arithmetic anyway)

9. Nov 26, 2005

### shmoe

The "greater than one" bit is only needed if you allowed 1 to be a prime. If you really wanted, you could include 4 in your definition of prime, and pi (and....) The Euler product would then awkwardly be "over primes not equal to 1, 4 or pi".

As matt says, 'take the primes, union 1'. That's it.

What makes a definition "correct"? Nothing really. Even 100 years or so ago some took "prime" to include 1, most notably the tables of Lehmer. A big consideration is how 'natural' the statement of important results becomes under the definition. You can take this 'naturalness' as a measure of which definition better captures a like group of objects. If you include 1 as a prime, many statements about 'primes' will be excluding 1 (do my suggested text book exercise), but I can't think of an important case of wanting "primes unions 1".

You should compare this with the convention that's sometimes adopted that x^0=1 irregardless of what the symbol x represents. You can then write power series as $\sum_{n=0}^{\infty}a_n x^n$ instead of the more cumbersome $a_0+\sum_{n=1}^{\infty}a_n x^n$. There's no mathematical signifigance here when you define the symbol x^0 to mean "1", but it makes notation simpler.

Last edited: Nov 26, 2005