# Fundamental Theorem of Arithmetic: why greater than 1?

The theorem, as wikipedia states it, is

Every integer greater than 1[note 1] either is prime itself or is the product of prime numbers, and that this product is unique.

It does have a note there that maybe 1 can be included, but it's still "careful" about 1, and often I see it without a note.

Why exclude 1? 1 is of course the product formed from the empty set, which is the only possible prime factorization of 1 and it doesn't contain anything that isn't a prime, so this seems fine to me.

• You are quite right: $1$ has a unique prime factorisation, just like every other positive integer. The reason the theorem is expressed this way is that your version might confuse people who are not mathematically sophisticated ("Why is the product of all elements in the empty set equal to $1$?"). Personally I think this is a very good reason! Commented Nov 28, 2015 at 15:45
• In short, it is simply easier to way. Admitting an 'empty set' factorization adds a complication with no clear benefit. Commented Nov 28, 2015 at 15:47
• @SimonS not to be contrary, but it seems nice to me to get rid of a special case Commented Nov 28, 2015 at 15:50
• "not to be contrary": surely you mean "to be contrary"? The Wikipedia article covers your objection explicitly. My comment justifies it pedagogically. Everything looks just fine to me. Commented Nov 30, 2015 at 23:22

I'm glad you understand the concept of the empty product. Not everyone does. Wikipedia, despite its many, many flaws (peruse Wikipediocracy sometime), at least recognizes that this has to be dumbed down for the general public, it has to be made less sophisticated. Some people have trouble with the idea of a single integer being a product of one integer, how can they understand an integer being the product of no integers at all?

This is why the article says "either is prime itself or is the product of prime numbers" rather than "every positive integer is a product of primes." 2 is the product of a single prime, itself, while 1 is the product of no primes at all. You understand that and I understand that. But if you're writing for someone who might not necessarily understand these "subtleties," you have to dumb it down.

You also have to consider the history of the subject. 1 has never been a prime number, but it took mathematicians a long time to recognize this. In explaining a unique factorization domain, you have to keep in mind that some people may not know that 1 is really a unit, not a prime.

Also, and I am neither the first nor the last to say this, but too many people have this idea that unique factorization is something that needs protection. It is true that in one sense, $1^3 \times 2 \times 5$ and $(-1)^4 \times 5 \times 2$ are different things. But both expressions evaluate to the same number and involve the same prime numbers. The units and the ordering changed, not the prime numbers.

Study some basic algebraic number theory and the answer is simple: $1$ is neither a prime nor a composite number, it is a unit.

To use the Euclidean algorithm (in those rings where it can be used), you need some kind of norm function that maps the algebraic integers of that ring to real, rational, positive integers. To use the Euclidean algorithm in $\mathbb{Z}$ (which consists of the positive and negative integers, and $0$), the obvious choice of function is the absolute value function.

If $m$ is a nonzero integer and $p$ is a prime, then $|p| < |mp|$ and $|m| < |mp|$, and you're of course rolling your eyes at me and saying "duh!" But $|m| = |1m|$. This also holds true for $-1$.

This does nothing to change whether or not the ring has unique factorization or not. How do you factorize $-49$? Easy: $(-1) \times 7^2$. Or you could say $(-7) \times 7$, same thing.

It's better then to say

In a unique factorization domain, each nonzero non-unit number is the unique product of primes without regard to order or multiplication by units.

$1$ is neither a prime nor a composite number and it cannot be expressed as the product of any prime or composite number. It can only be expressed as the product of infinite $1$'s which is not unique.

All $1$ is composed of is (product of) multiple $1$s which would destroy the product representation (factorization) of all integers if we included $1$ in their factorization. For example, if $1 = 1 \cdot 1 \cdots 1$, then $10 = 2 \cdot 5 \cdot 1 \cdot \cdots 1$ which wouldn't be unique.

• Yes but that's not the proposed factorization Commented Nov 28, 2015 at 20:42