# Do we have negative prime numbers?

Do we have negative prime numbers?

$..., -7, -5, -3, -2, ...$

• looks like there are some directly below your question ;) – jorst Nov 2 '14 at 12:29
• These are just the additive inverses of the (positive) prime numbers. Usually, we only consider the positive ones, most especially when dealing with prime factorizations of (positive) integers. But of course, if you want to factor negative integers as well, then that's a different matter altogether. – Arnie Bebita-Dris Nov 2 '14 at 12:29
• These are, indubitably, prime elements of the ring $\mathbb{Z}$. Some people reserve the term "prime number" for positive prime elements of $\mathbb{Z}$, and then these are not "prime numbers". With the broader terminology of calling all prime elements of $\mathbb{Z}$ prime numbers, they are. – Daniel Fischer Nov 2 '14 at 12:33
• It's not obvious that the person who asked the question has heard of "rings" and will appreciate answers that refer to them. From that perspective, I can imagine all of the existing answers being useless. – Ben Millwood Oct 31 '15 at 8:44
• There is some care to be taken in the definition of a prime. In more advances work irreducibles are distinguished from primes and the core property of a prime is taken to be if $p$ divides $mn$ then it divides one of $m$ and $n$. – Mark Bennet Jan 14 '18 at 17:00

I don't know why this question has a down vote, because it identifies a subtle point about arithmetic which becomes particularly significant when the notion of "prime" is extended to other contexts, and is relevant so far as the integers are concerned when looking at issues like unique factorisation.

When we first encounter prime numbers, we do so in the context of the positive integers. The significant point about this context is that $$+1$$ is the only unit (the only positive integer with a multiplicative inverse). So the question here does not really arise. Often, when the main focus of work is the positive integers, the word prime will be used to imply a positive integer.

As soon as we start to extend this to the integers, and in particular, to consider the integers as having the structure of a ring, we add in a second unit $$-1$$ with $$(-1)^2=1$$. Even in this context it is possible to define the prime numbers as positive integers without too much inconvenience.

But if we extend further and add $$i$$ with $$i^2=-1$$ as another unit - note that $$i\cdot -i=1$$, we are in a different world. For example, $$2=(1+i)(1-i)$$ and $$(1+i)=i(1-i)$$ so that $$2=i(1-i)^2$$. Are these factorisations of $$2$$ to be taken as the same or different?

So very soon, in the context of ring theory and the theory of algebraic integers, we start talking about prime ideals (initially thought of as all the multiples of some prime $$p$$ - but extended beyond that idea too - an ideal which consists of all the multiples of a single element is called principal). And it is somewhat natural, if the ideal is principal, to call the generator a prime element of the ring. However, the primes are then only identified up to multiplication by units - $$1+i$$ generates the same ideal as $$1-i$$. One of the reasons for using ideals is that the uniqueness of factorisation can be maintained in this larger context. In $$\mathbb Z$$ both $$2$$ and $$-2$$ generate the same ideal.

• This question is easy to downvote if you know the answer, but we really should be judging the question based on the perspective of someone who doesn't know. I be that's WHY the question was downvoted, but I disagree with that downvote. – corsiKa Nov 3 '14 at 17:34
• When you say that 1 is "the only positive integer with an multiplicative inverse", do you mean an integer multiplicative inverse? – Marc.2377 Apr 12 '18 at 6:59
• @Marc.2377 Indeed, working within the integers. – Mark Bennet Apr 12 '18 at 7:46

A prime element of a ring is a nonunit $$p$$ with the property that if $$p$$ divides a product $$ab$$ then it divides $$a$$ or $$b$$. In the ring $$\mathbb Z$$ of integers, this property is shared by the (positive) primes $$2,3,5,7,11,\ldots$$ and also the negative primes $$-2,-3,-5,-7,\ldots$$, and even by $$0$$. However, the term prime number is conventionally used only for the positive prime elements of $$\mathbb Z$$, and there are good reasons for this convention, for example $$15=3\cdot 5=(-3)\cdot(-5)$$ shows that the prime factorization would be less unique than we are used to.

• And perhaps it's worth making the point that in a ring like say the Gaussian integers ${\Bbb Z}[i]$, we can't separate out the positive primes because "positive" is meaningless. – David Nov 2 '14 at 12:35
• I think 0 is usually explicitly excluded from being prime, even in general integral domains. (Just like units certainly are, even though they trivially have the property you cite). – hmakholm left over Monica Nov 2 '14 at 12:47
• Units are excluded for the same reason cited in the answer: if they were allowed every element could be factorized in an infinite number of ways. Having a unique factorization is a very strong property, so it makes sense to restrict the notion of "prime number" to allow it in common circumstances. – Bakuriu Nov 2 '14 at 13:43
• This answer is a bit too terse since most students who haven't already taken an advanced undergraduate course in algebra would never think to define a prime number that way - they would take a prime number to be what an algebraist calls an irreducible number. – djechlin Nov 2 '14 at 18:37
• @HenningMakholm Ah, you're right - I viewed this from the perspective of prime ideals of coure ... – Hagen von Eitzen Nov 2 '14 at 18:47

I just wanted to supply a specific quote from a text that defines prime numbers for negatives in the way that several of the other answers already affirm. This is from Hungerford's Abstract Algebra: An Introduction (sec. 1.3):

DEFINITION. An integer $$p$$ is said to be prime if $$p \ne 0, \pm1$$ and the only divisors of $$p$$ are $$\pm1$$ and $$\pm p$$.

EXAMPLE. $$3, -5, 7, -11, 13,$$ and $$-17$$ are prime, but $$15$$ is not (because $$15$$ has divisors other than $$\pm1$$ and $$\pm15$$, such as $$3$$ and $$5$$). The integer $$4567$$ is prime; to prove this from the definition requires a tedious check of all its possible divisors.

It is not difficult to show that there are infinitely many distinct primes (Exercise 25). Because an integer $$p$$ has the same divisors as $$-p$$, we see that

$$p$$ is prime if and only if $$-p$$ is prime.

In this case the Fundamental Theorem of Arithmetic is stated in terms of both positive and negative prime factors (Theorem 1.1), with the standard natural-number statement given thereafter as a corollary (Corollary 1.2).

Note that this is before rings (et al.) appear in this text (although of course they do later), so the definition is not restricted to that domain.

Some more context: In general ring theory one looks at prime elements or irreducible elements as the analogues of prime numbers.

A (nonzero) element $a$ of a ring $R$ is called prime if it generates a prime ideal, or equivalently, if it is not a unit and $a|cb$ implies $a|b$ or $a|c$, i.e. if $a$ divides a product, it divides one of the factors.

An element $a\in R$ is called irreducible if it is not a unit and $a=bc$ implies $b$ or $c$ is a unit, i.e. it cannot be factored into two other nonunits.

In nice rings (such as $\mathbb{Z}$), these coincide and one has a unique factorisation up to units ($\{-1,1\}$ in $\mathbb{Z}$) of every element into prime/irreducible elements.

• technicality; in Z 0 is prime but not irreducible. I might add to your answer some info on why 0 breaks stuff (it's prime but not maximal, is one way). – djechlin Nov 2 '14 at 18:39
• @djechlin See Henning Makholm's comment on Hagen's answer; $0$ is generally explicitly excluded from being prime. But the ideal $\{0\}$ is prime... – 6005 Nov 3 '14 at 3:50
• @Goos not according to the definition here. – djechlin Nov 3 '14 at 11:08
• @djechlin Of course. But one often wants to exclude such cases, just like you exclude units from being irreducible and the entire ring from being a maximal ideal, despite them both satisfying the definition stated naively. – 6005 Nov 3 '14 at 21:13
• @Goos Right, so you shouldn't state definitions naively. If you want to exclude a case you should say "excluding X." Ideally you give a reason for excluding X. But I would highly prioritize not writing something that is plainly incorrect even if we don't have terribly much respect for the exception. This is mathematics, we don't do that. – djechlin Nov 3 '14 at 21:54

Generally, the definition of prime numbers is all those natural numbers greater than 1, having only two divisiors [sic], the number itself and 1. But, can the negative integers also be thought of in the same way?

Almost, but not quite. We could say that a negative integer is prime if it has only two divisors ($-1$ and itself) among the negative integers.

This is very similar to the positive integers, for which we say that a positive integer is prime if it has only two divisors among the positive integers.

For example, 163 is divisible by 1 and itself. It's also divisible by $-163$ and $-1$, but we usually don't bother with that when we are trying to see if 163 is prime.

Similarly to check whether $-163$ is prime, we could try dividing it by $-2, -3, -5, -7, -11$. We don't need to try dividing by $-13$ because $$-13 < (-1) \frac{\sqrt{-163}}{i}.$$

More commonly though, what people usually do is multiply the number by $-1$ and check whether that is prime, and that way they sidestep the issue of imaginary numbers. For example, try Divisors[-163] in Wolfram Alpha.

This is a question that is going to get lots of duplicates. That's how I came across it.

Is $-14$ prime?

Of course it isn't! Since $$\frac{-14}{2} = -7,$$ for starters. It is also divisible by $-7, -2, 7$, as well as by its obvious divisors, $-1, 1, 14$ and itself. Clearly $-14$ is a composite number.

But is $-7$ prime? That's the tougher question. It is trivially divisible by itself as well as by $-1, 1, 7$, but it's not divisible by any other numbers (we're operating in the domain of the integers, $\textbf{Z}$).

My terminology "obvious divisor" is of course not standard. What I mean by it is that if $n$ is any integer, we automatically know that it is divisible by itself, by $-n$ and by $1$ and $-1$.

Let's agree that if $n$ is divisible only by these numbers that we automatically know to be divisors but not by any others, then it is a prime number, but if it's divisible by these "automatic divisors" and by other integers (with all the divisors forming a finite set), then it is a composite number.

Invariably this causes people to whine about breaking the fundamental theorem of arithmetic. But if we recognize that $1$ and $-1$ are special numbers in that they are units, then the big and melodramatic problem of the fundamental theorem becomes irrelevant nonsense. In $\textbf{Z}$, each nonunit nonzero number has unique factorization regardless of units.

Don't worry about the prime counting function either. If $x$ is a positive real number, then $\pi(x)$ counts how many numbers are between $0$ and $x$. But if $x$ is negative, then $\pi(x)$ counts how many numbers are between $0$ and $|x|$ or between $0$ and $x$; either way the answer will be the same.

Picture the prime numbers as dots on the real number line. They are symmetrical, with the positive primes reflected in the negative primes on the other side of $0$.

Of course there are negative prime numbers. For a number $p$ in a given domain to be prime, it needs to satisfy $p|ab$ always means $p|a$ or $p|b$ (maybe both). For example, if $p = -3$, $a = -2$, $b = 30$, we see the condition is satisfied with $p|b$. But if $p = -4$, neither condition is satisfied. This means that $-3$ is a negative prime number and $-4$ is a negative composite number.

It also needs to be said that there is a negative unit, $-1$. A unit divides every number in the domain. If $p$ is prime and $p = ab$, either $a$ or $b$ (but not both) must be a unit. For example, $-3 = -1 \times 3 = 1 \times -3$.

This is no problem whatsoever for unique factorization (if it exists in the domain at hand). We simply say that factorization is unique regardless of ordering (e.g., $5 \times 3$ is not a distinct factorization of $15$) and regardless of multiplication by units.

However, for the prime counting function $\pi(x)$, we're generally only concerned with positive primes. After all, $\pi(7) = \infty$ would be kind of useless. Or we can just say that $\pi(x)$ counts how many primes there are between $0$ and $x$. Then it turns out that $\pi(x) = \pi(-x)$, e.g., $\pi(-10) = 4$ just as $\pi(10) = 4$.