Is zero a prime number?

Question

Is zero a prime number?

When talking about prime numbers, it seems like the examples given $(2,3,5,7,11,13,...)$ have the property that they have no factors less than themselves and greater than one. But $0$ also has this property, so is it prime? If not, why not?

Is zero odd or even?

When talking about even numbers, it seems like the examples given $(2,4,6,8,...)$ have the property that, when dividing them by $2,$ have a non-zero quotient and a zero remainder; odd numbers $(1,3,5,7,...)$ have a non-zero quotient and a remainder of $1.$ So, is $0$ odd? even? neither odd nor even?

Is zero a number?

I've heard that every number is either odd or even, but if $0$ is neither odd nor even, does that mean it isn't even a number?

Answer 1

If you are willing to accept the integers as numbers, then you should have no trouble considering $0$ a number. For one willing to define even numbers as "integer multiples of $2$" then it's similarly clear that $0$ should be considered even. I don't want to spend a lot of space here rehashing the evenness of $0$ since there are already questions dedicated to that problem, but fortunately that makes it easy to direct you to the answer: Is zero odd or even?

I've also found some more discussions on the "numberness of zero" that you might find useful: What's the hard part of zero? , Why do some people state that 'Zero is not a number'?

The question as to whether or not it should be considered prime is more interesting.

What should primes be?

After you learn about divisibility and factorization, this idea arises about breaking numbers down into smaller parts (sort of like describing matter with smaller and smaller parts). Divisibility makes a partial order on the nonegative integers. This just means that since $12=3\cdot 4$, the "smaller parts" 3 and 4 dividing 12, we can record this as $3\prec 12$ and $4\prec 12$. Furthermore $2\prec 4$ because $2$ divides $4$, and so on. Since $1$ divides everything, we would say that $1\prec n$ for any nonegative integer $n$.

In physics, we are interested in the smallest things from which everything is built from (the "atoms"!). The idea of atoms has two parts:

• they should all be "small"
• they should build everything else

Well, we can't let $1$ be such a thing, because it would be the only smallest thing, and moreover you can't build anything from $1$ alone. So it is in a sense, too simple.

The next best candidates are those things just above $1$. What just above means becomes clearer if you draw a picture:

This is a sort of Hasse diagram for the nonnegative integers partially ordered by divisibility. Since the diagram is infinite it's not really a Hasse diagram, and the lines to zero don't really come from any numbers, but this is good for our purposes.

From the diagram you can easily see that the primes lie in the first row above $1$, and so they are "as small as possible" without being $1$, and moreover, everything above them (excepting zero) is built out of various combinations of the primes. The gradeschool definition of prime number basically amounts to the fact that nothing lies between $1$ and $p$ for each prime.

Zero, paradoxically, is really aloof and nowhere near the rest of the primes: he doesn't seem very small after all. Moreover he is pretty useless for building numbers since $0n=0$ for any $n$.

So for reasons like these, $0$ is not considered as a prime: he doesn't make a good "atom."

Answer 2

Zero is not a prime number out of almost every definition of prime numbers:

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1. Prime numbers are those natural numbers that are divisible solely by the unity ($1$) and themselves. $0$ is divided by every natural number! $5\cdot0=0,n\cdot0=0,\ldots$ (or not divided at all if you want to exclude zero from the definition of (natural) number).
2. Prime numbers are the cornerstone of arithmetics from its fundamental theorem: every (non-zero) natural number has a unique representation in prime numbers. $0$ only represents itself and that representation is not unique: $0=0^1=0^2=0^{12807}\cdot2^{9987}\cdot97^1\ldots$

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Zero is indeed even as even are defined as those numbers divisible by $2$, and $2\cdot0=0$, therefor $0$ is divisible by $2$. (Unless you want to exclude zero from the definition of (natural) number)

So, the final question: Is zero a number?

In most sets defined as numbers: Integers ($\mathbb Z$), Rationals ($\mathbb Q$), Reals ($\mathbb R$), Complex ($\mathbb C$), then $0$ is an important element. It cannot be excluded from the definition of number in those sets.

The question is about natural numbers ($\mathbb N$), and you can construct a number theory with $0$ as part of the natural numbers and without $0$ as part of the natural numbers. I like to include $0$ as a natural number as it seem to solve some problems more easily than excluding it, yet it adds some complexities.

Note that in the definitions of “prime number” it is better to explicitly exclude $0$, as zero has no purpose in the fundamental theorem of arithmetics, all arithmetics based on that theorem can be done in the set $\mathbb N\setminus\{0\}$ (this is the Natural numbers explicitly excluding $0$).

Let's write the fundamental theorem of arithmetics if $0\notin\mathbb N$ (by number then we mean any natural number except 0)

FTA with no zero

Every number $n$ has a unique representation in prime numbers $n=p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}$ (for a finite set of primes $p_j$).

And let's formulate it for $0\in\mathbb N$

FTA with zero

Every number $n$ (with $n\ne0$) has a unique representation in prime numbers $n=p_1^{k_1}\cdot p_2^{k_2}\cdots p_j^{k_j}\cdots$ (where $p_j$ is the $j$^th prime number (and $k_j$ can be $0$)).

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In the first formulation we cannot have $0$ as exponent, so we would strip all primes that don't divide $n$. We just won't include them in the formulation. In the second formulation we must except $0$ from every number, but we can, on the other hand, include all prime numbers and if $p_j$ does not divide $n$ then we set the exponent $k_j=0$.

Answer 3

A: No.

A: Even.

A: Yes.

• Because of the Peano axioms.

Answer 4

By definition, a prime number is a non-$0$, non-unit integer $p$ such that for any integers $m,n$, if $mn$ is a multiple of $p,$ then $m$ or $n$ is a multiple of $p$. Aside from the whole non-$0$ condition, it fits perfectly.

It is even, as it is a multiple of $2$ (as it is a multiple of every number).

It is a (real/complex/integer) number.

Answer 5

Q. Is zero a prime number?

No; but, this is basically just by convention. Here's why.

First, define

$$\mathbb{N} = \{0,1,2,3,\ldots\}$$

so that in particular, $0$ is a natural number.

Preliminary chitchat. A prime number could be defined as a natural number $p$ satisfying the following two conditions:

$$(0) \;\;\mathop{\forall}_{a,b \in \mathbb{N}}\;\;\;p \mid ab \rightarrow (p \mid a) \vee (p \mid b), \qquad (1) \;\;p \mid 1 \rightarrow \mathrm{FALSE}$$

Note that $\mathrm{FALSE}$ is the identity element with respect to Logical OR, which explains to some extent where condition (1) comes from.

If we accept this definition, then $0$ is prime, but $1$ is not.

The usual conventions. Despite the above discussion, we usually declare that $0$ is not prime. There's a couple of reasons for this:

1. We want every non-zero natural number to have a unique factorization into primes.
2. We want the primes to be precisely those elements of $\mathbb{N}$ that cover $1$ with respect to the divisibility order.
3. We want primes to form an antichain with respect to divisibility.

So the definition of "$p$ is prime" becomes:

$$(0) \;\;\mathop{\forall}_{a,b \in \mathbb{N}}\;\;\;p \mid ab \rightarrow (p \mid a) \vee (p \mid b), \qquad (1) \;\;p \mid 1 \rightarrow \mathrm{FALSE}, \qquad (2)\;\; p = 0 \rightarrow \mathrm{FALSE}$$

Under these conventions, $0$ is not a prime.

Why somedays I disagree with the convention that $0$ isn't prime.

By a commutative monoid with $0$, I mean a commutative monoid $M$ together with an element $0 \in M$ satisfying $0a = 0$ and $a0 = 0$.

The way I like to think about the set $P$ of prime numbers is as follows: $P$ is the unique subset of $\mathbb{N}$ such that if $F(P)$ is the commutative monoid with $0$ freely generated by $P$, then the obvious monoid-with-$0$ homomorphism $$F(P) \rightarrow \mathbb{N}$$ is an isomorphism.

Under this definition, $0 \in P$.

Answer 6

Zero is not prime, since it has more than $2$ divisors.

Zero is even, since $0 = 2 \cdot 0$, and $0$ is an integer.

If we use "number" in essentially any of the usual senses (integer, real number, complex number), yes, zero is a number.

Answer 7

• Zero is not a prime number as prime numbers are defined for integers greater than 1.

• Zero is an even number. Definition of an even number with modular arithmetic:

$\forall x\in \mathbb{Z},\, x$ is even if and only if $x\equiv 0 \pmod 2$

As $0$ satisfies the definition, then it is an even number.

• Of course $0$ is a number, because it is a member of some sets who contains only numbers (such as integers, real numbers, complex numbers etc.). If your question is "Is $0$ a natural number?", it's controversial. In some definitions $0$ is a natural number, but in some of them not. Mathematicians do not have an agreement on that, but I'm with the ones who do not accept it as a natural number, because some theorems which are satisfied by all natural numbers are not satisfied by $0$.