Why is Zero not composite?
Andrew Mclaughlin 
I saw this question:
Vidya wonders why numbers one and zero are neither composite numbers nor prime numbers. Could you please explain to Vidya what can happen if they are either composite numbers or prime numbers? Use an example to punch your points.
Apart from the fact that I cannot comprehend how to answer, I am also unable to understand how zero is not composite.
A number $n$ is composite if $k\mid n$, where $1<k<n$. $k\mid n$ if $kx=n$. Since, $0k=0$ for all $k$, shouldn't $0$ be some sort of "hyper-composite"?
Is it? If not, why not?
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$\begingroup$The answer they're looking for, I think, is that if $1$ is considered a prime or $0$ considered a composite, then we no longer have unique factorization. Or at least the statement of the unique factorization theorem becomes uglier.
For the first point, $6 = 2\cdot 3 = 1^5\cdot 2 \cdot 3$ gives two different prime factorizations of $6$. Ick.
For the second, $0 = 0*3 = 0 *2.$ So $3$ is a factor of the first product, but not the second. Again, ick. And this breaks even more things. What's gcd$(0,0)$, for instance?
$\endgroup$ 12 $\begingroup$While $0$ is indeed divisible by any prime, no product of primes will make $0$. Therefore, $0$ is not composite.
$\endgroup$ 2 $\begingroup$We can of course assume that $0$ is a composite number. It would not break mathematics or anything. However it would make a lot of theorems and statements more tedious. For instance, we know that any composite number can be written as a product of a finite amount of primes. However If we let $0$ be composite then we have to either always say "Every composite except $0$ can be ..." or ignore such theorems.
To put it simply we let $0$ be non-composite because it is convenient.
One could also argue that "composite" numbers multiplied with each other should create a new composite number, and this is not true for $0$. But then this is more of a philosophical stance than a mathematical one I would claim.
$\endgroup$ $\begingroup$A number $n$ is composite if [there is an integer $k$ such that] $k|n$, where $1<k<n$.
I found your statement of the definition of a composite number a little ambiguous, so I have inserted a few extra words emphasizing that there must actually exist one number that can be $k$ in the two conditions given.
As you observed, if $n=0$ then the first condition, $k\mid n,$ is satisfied by many integers--every one of them, in fact.
But according to the definition that you yourself gave, there is also a second condition that $k$ must satisfy: $1 < k < n.$ If $n = 0,$ is there any integer $k$ that satisfies this condition, that is, $1 < k < 0$? No, there is no integer that is both greater than $1$ and less than $0.$
Therefore, according to this definition, $0$ is not composite.
Other answers have indicated why it is desirable that $0$ not be composite, that is, why we would want to include the second condition in the given definition.
$\endgroup$ $\begingroup$For the Fundamental Theorem of Arithmetic we usually assume that $n$ is a positive integer, and consider $n=1$ separate. Indeed, $1$ is neither a prime nor composite, according to your definition. Although zero satisfies $0k=0$ for $1<k<n$, it is not considered to be a positive integer. Hence it is not composite, i.e., not considered for having a prime decomposition. The usual prime numbers then start with $2,3,\ldots $, and the positive integers to be composed into these primes then start with $n=2,3,4,5,\ldots$
Wikipedia says for $n=1$: Using the empty product rule one need not exclude the number $1$, and the theorem can be stated as: every positive integer has unique prime factorization.
However, we exclude $0$.
$\endgroup$ 2 $\begingroup$In more general stuctures, rings, we define units, irreducible elements and prime elements. (What follows is slightly modified quotes from Wolfram MathWorld.)
Unit
A unit in a ring is an element $u$ such that there exists $u^{-1}$ where $u \cdot u^{-1}=1$.
Irreducible Element
An element $a$ of a ring which is nonzero, not a unit, and whose only divisors are the trivial ones (i.e., the units and the products ua, where u is a unit). Equivalently, an element a is irreducible if the only possible decompositions of a into the product of two factors are of the form $a=u^{-1}ua$, where $u^{-1}$ is the multiplicative inverse of $u$.
Prime Element
A nonzero and noninvertible element a of a ring $\mathbf R$ ... characterized by the condition that whenever $a$ divides a product in $\mathbf R$, $a$ divides one of the factors. The prime elements of $\mathbb Z$ are the prime numbers, $\mathbf P$.
Comments
In an integral domain, every prime element is irreducible, but the converse holds only in unique factorization domains. The ring $\mathbb Z[i \sqrt 5]$, where $i$ is the imaginary unit, is not a unique factorization domain, and there the element $2$ is irreducible, but not prime, since $2$ divides the product $(1-i \sqrt 5)(1+i \sqrt 5)=6$, but it does not divide any of the factors.
In the ring of integers, all irreducible elements are also prime elements.
Note that prime and irreducible elements cannot be zero or units.
So why omit zero?
Note that $0$ meets the definition of a prime number since, if $0$ divides $ab$, then zero divides $a$ or zero divides $b$.
But zero is not irreducible since every number divides $0$.
So if we want our prime numbers to also be irreducible, we must require that they not be equal to $0$.
$\endgroup$ 2 $\begingroup$You cannot answer this question without a formal defintion of a composite number. Usual definitions exclude $0$ explicitly and there is nothing to prove.
$1$ is obviously not composite.
The case of primality is easier: you need to have exactly two distinct divisors, which is not achieved by $0$ nor $1$.
Had another convention been used, this would have broken the Fundamental Theorem of Arithmetic, because $0^m=0$ and $1^m=1$, causing indeterminacies.
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