# Is zero a prime number?

Q: Zero is it a prime number?

Q: Zero is odd or even?

Q: Zero is a number?

• If yes or no, then why?
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I don't think this question should be closed. The OP has asked quite a few questions on prime numbers today, and is asking them out of interest. Closing it because it lacks motivation is petty! –  user1729 Oct 25 '13 at 11:16

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

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 and that representation is not unique: $0=0^1=0^2=0^{12807}\cdot2^{9987}\cdot97^1\ldots$

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 $0$)).

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 $k_j=0$.

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aded some ....... to validate post: tx u, Carlos. –  Vitalie Ghelbert Oct 25 '13 at 20:52
You can well upvote the answer for validation, @VitalieGhelbert. Or accept it. –  Carlos Eugenio Thompson Pinzón Oct 25 '13 at 20:56
At the moment could only accept, not sufficient days on the Earth to UpVote :p, first day on Math :) –  Vitalie Ghelbert Oct 25 '13 at 20:58
mmm, sorry, I'm laying! neet 15 points, to UpVote, lost some points on some recent posts. If the remuneration was in negatives, i'll be first one :p –  Vitalie Ghelbert Oct 25 '13 at 20:59
An integer number is prime when it has ${\bf\mbox{only two }}$ divisors and they are not equal. It excludes ONE. –  Felix Marin Oct 31 '13 at 8:18

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. 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."

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+1 Very nice! The partial ordering argument is one of the best I've seen. –  Cameron Buie Oct 29 '13 at 15:02
Nice answer +1. However when you wrote (about $1$) that "All of its multiples are just $1$", you must have been thinking of something different. Maybe "powers" instead of "multiples"? –  Marc van Leeuwen Feb 7 at 8:05
@MarcvanLeeuwen your guess as to my state of mind at the time is as good as mine at this point :) I decided on just removing the line. Thanks for the careful reading! –  rschwieb Feb 7 at 10:59

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.

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If 0 were a prime we would not have uniqueness in the fundamental theorem of arithmetic. Pretty good reason to leave 0 out. –  M.B. Oct 25 '13 at 10:54
@M.B. That, plus a whole load of very similar cases, is one reason $1$ isn't considered a prime either. –  Arthur Oct 25 '13 at 11:52

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.

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Dear @Bongers : I don't disagree with the observation in the first line, but in almost all sources, prime numbers are defined to be nonzero, and I think that kind of trumps anything you might say about divisors. I'm not suggesting you need to change anything: I just wanted to bring this to future readers' attention. Regards! –  rschwieb Oct 25 '13 at 12:46
@rschwieb I find the definition of prime numbers to be non-zero an exceptionally boring reason as to why zero is not a prime number. It implies that "zero is not a prime number because someone defined it that way", which isn't true. Zero is not a prime number because it has more than two divisors, and this allows people define it out. –  user1729 Oct 28 '13 at 20:19
@user1729 Aesthetically I totally agree with you :) I was just merely stating the reality of the printed situation. Since the "real" definition of prime requires you to exclude zero, I imagine that the "nonzero" condition appears in the "gradeschool" definition as an artifact, and they mostly just don't notice this convenience possible with their definition. What annoys me more is the fact that so many geometry texts insist a trapezoid has exactly two parallel sides... –  rschwieb Oct 29 '13 at 13:27
• 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 $0$.
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No. Its factors are not just 1 and itself, but any number and itself.

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Zero is not a prime number. By definition a prime number has to be greater than $1$.

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