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#1




Can a prime number be negative?
Well?

#2




Last edited by TriPolar; 06292015 at 11:02 PM. 
#3




#4




If we defined prime numbers to include negative numbers, then we'd lose some important properties of primes, e.g., the unique factorisation of natural numbers into primes. If 3 were a prime, then 9 could be factorised both as 3 x 3 and as 3 x 3.



#5




The answer to this question is yes, if we take it to mean: “Are there negative numbers that are prime elements of the ring of integers (that is, the set of numbers . . ., − 3, − 2, − 1, 0, 1, 2, 3, . . ., combined with the usual operations of addition and multiplication)?”
Why this is the case is linked to the question of why 1 is considered to be neither prime nor composite. And this is because 1 is a unit. It is a number that can divide all the other integers and still have an integer left over every time. In the world of positive integers, 1 is the only unit. But in the world of all the integers, there are two of them: 1 and − 1. Units are different from prime and composite numbers, precisely because they can divide all the other numbers. Because of this, numbers that get divided by one or the other of them are, in some sense, the same number. And we see this with say, 6 and − 6. We can cut them up in a whole bunch of ways: 6 = 2 × 3 6 = − 2 × − 3 − 6 = − 2 × 3 − 6 = 2 × − 3 What all of these have in common is the 2 and the 3. There’s no way to cut up a 6 and end up with a 4 or a 9 or a 5 or anything like that. So, what we’re seeing here is that the negative sign is just kind of there. It’s not really affecting the properties of the numbers in terms of whether they’re prime or not, or what their factors are if they’re composite, like 6 and − 6 are. Now, with the integers, it’s easy to see the pairs of similar numbers that the two units give rise to, and so, it’s easier to just deal with the positive one, and regard the negative one as essentially a subtraction of the positive one. One of them just looks more of a “canonical” prime. And so, you can say that only positive numbers are prime, because it’s easy enough to account for negative primality. Not that this is required, though, but it can make things easier. This is harder to do with the complex numbers, which have four units, 1, − 1, i, and − i. With the complex numbers, it’s harder to pick one of them as the “canonical” prime. For example, one set of 4 equivalent primes is 1 + i, − 1 − i, − 1 + i, and 1 − i. So, these sort of questions don’t really arise in the complex numbers. Last edited by RadicalPi; 06292015 at 11:57 PM. 
#6




Sure
First of all, let's define a unit. A number N is a unit if and only if there exists an m such that Nm = 1. A prime p is thus defined as a nonunit, nonzero number such that for every factorization p = ab that a or b is a unit. In the natural & whole numbers, the only unit is 1 which gives us our typical definition of a prime number. In the rational and real numbers, every nonzero number is a unit therefore there are no primes. In the integers, the only units are 1 and 1 and so the primes are the primes we customarily think of and their additive inverses so if 3 is prime, so is 3. In the complex numbers a+bi  a,b are real, every nonzero number is a unit since (a+bi) x ((abi)/(a^{2}+b^{2})) = 1. Thus like the rational numbers, there are no primes in the complex numbers. Last edited by Saint Cad; 06302015 at 02:58 AM. 
#7




That's the Fundamental Theorem of Arithmetic which by necessity is limited to the factorization of natural numbers and would then exclude negatives through the closure property. Like my above post points out, it is very important that you are specific about which number system you are limiting yourself to.

#8




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Now if you were talking about the extension field Z(i) then it is prime but Z(i) is not the same as C. 
#9




Z(i) is not a field.
__________________
RayMan 


#10




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But, as TriPolar's second link, and the "Generalizations" section of the Wikipedia article, and SaintCad's posts, all note, there are other contexts in which the "prime numbers" of that particular set of numbers are something else. Whether or not there's only one way to factor 5, for example, depends on what set of numbers you're allowed to choose your factors from. 5 = 1*5 = (1)*(5) = 4*(1.25) = 10*(0.5) = ... Last edited by Thudlow Boink; 06302015 at 11:44 AM. 
#11




Or 5 = (2 + i)(2  i) so it is weirdly not prime in the Gaussian integers (a.k.a. Z(i)).

#12




As should be clear by now, it's just a matter of definitional convention, and thus perhaps not a particularly interesting question. But what I would say is this:
When discussing whether a number is prime or not, the sense of "number" which is usually most important is such that two numbers ought be considered equivalent if both divide each other. Accordingly: against the background context of the integers, I would happily say that 2, for example, is prime. But I would also say that 2 is the same prime as 2. And similarly, for most purposes where I would want to talk about "primes" and "composites", I would consider 3 and 3 to be the same prime, 4 and 4 to be the same composite, 5 and 5 to be the same prime, and so on. Last edited by Indistinguishable; 07012015 at 01:51 AM. Reason: Just repeating what RadicalPi already said 
#13




But nevermind my last post. Really, for the most substantive discussion, what we should do is this: rather than us all telling you from on high how we happen to use words, instead, you should tell us what you consider the key properties characterizing being a "prime number", or at least give some indication what it is about the notion of "prime number"s and distinguishing these from other numbers which interests you, and then we may all together consider how the ideas you are interested in interface with consideration of negative numbers.
Last edited by Indistinguishable; 07012015 at 02:03 AM. 
#14




At first I was wondering if there was any conjecture/postulate/theory that states: "any prime can be constructed by the sum of two squares."
0+0=0 0+1=1 1+1=2 3? 1+4=5 7? Not working out too well...unless you use negative squares I then asked myself why a prime itself can't be negative. 


#15




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Last edited by Exapno Mapcase; 07012015 at 10:47 AM. 
#16




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Second, what do you mean "negative square"? If you mean something like i^{2}=1 and therefore 1 is a perfect square then you are talking about the complex numbers and there are no primes in the complex. 
#17




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As said upthread, when you ask whether a negative number is a square, you have to ask your context. Certainly when you come to Gaussian integers (of the form a + bi where a and b are odinary integers and i is the square root of 1), then you really have to allow a and b to be positive and negative. For instance 2 = (1+i)(1i) is not a Gaussian prime. Nor is 5 = (2+i)(2i) nor any of the integer primes that are sums of two squares, while all the primes that are not sums of two squares remain prime. And so long as you consider p, p, ip, ip to be different representations of the same prime you have unique factorization in the Gaussian integers. 
#18




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Statement:Any two rational nonzero numbers are equivalent. Let a/b and m/n be any two rational numbers not equal to zero (a/b) / (m/n) = (an)/(bm) e Q (m/n) / (a/b) = (am)/(bn) e Q Quod Erat Demonstratum There is actually a term for what you are discussing (a number times a unit) and for the life of me I cannot remember the term although IIRC it starts out with 'co' 
#19




Two numbers which divide each other are called "associates" or "associated" [though this is just terminology; we might just as well say "similar" or "equivalent" or whatever, as the linguistic context permits]. And, yes, within a field, all nonzero numbers are associated with each other. So, if all you've got to think about is a field, there's nothing interesting to say about primes: everything is either zero or a unit.
Last edited by Indistinguishable; 07012015 at 12:15 PM. Reason: Luckily for mathematics, there are plenty of nonfields to think about. 


#20




Quote:
2k + 1 = k^2 + 2k + 1  k^2 = (k+1)^2  k^2 
#21




Quote:

#22




"Equivalent" is perfectly correct, in that "divides each other" is a valid equivalence relation.
But Indistinguishable's real point there is that vocabulary isn't really relevant to the actual mathematics. The mathematics wouldn't be any different if what we now call "associated" we instead called "similar", and what we now call "similar" we instead called "associated". 
#23




Only for natural numbers unless you are claiming 1=1 or 1/3 = 7/13

#24




"Equivalent" is not the same thing as "equals". An equivalence relation is any relationship with the properties of reflexivity, commutativity, and transitivity. That is to say, a~a, if a~b then b~a, and if a~b and b~c, then a~c, for all a, b, and c. "Equals" is one example of an equivalence relationship, but so is "is similar to", "is proportional to", "has the same magnitude as", or "divides each other".



#25




If you like, I'm saying primality and related concepts are for many purposes cleanest considered not at the level of the integers per se, but rather at the level of the integers (or whatever integral domain) modulo the "divides each other" equivalence relation; that is, passing to the coarsergrained structure where "divides each other" is now taken to be the salient notion of equality because the finer distinctions are not for our purpose of importance. Just as much of rational arithmetic is most naturally considered not at the level of numeratordenominator pairs, but, rather, at the level of numeratordenominator pairs modulo an appropriate notion of equivalence (where we no longer care to distinguish between 3/4 and 6/8). Not that there's never any value in drawing the finer distinctions (if you're interested in mediants, say, you may well need to refrain from identifying 3/4 with 6/8), but there are also contexts for which a coarser notion of identity is most apropos.
Last edited by Indistinguishable; 07032015 at 03:04 PM. 
#26




Quote:
First, among the complex numbers, let's say those of the form a + b * i for ordinary integers a and b are "Gaussian integers". Also, as usual, we'll say the size a + b * i of a complex number a + b * i is sqrt(a^2 + b^2) [equivalently, the squared size a + b * i^2 of a complex number a + b * i is obtained by multiplying it by its conjugate a  b * i]. So the question is, which primes are squared sizes of Gaussian integers?; i.e., which primes are products of two conjugate Gaussian integers? The benefit of looking at things this way is that the Gaussian integers have a structure very similar to the ordinary integers. In particular, they come with a notion of prime factorization very similar to that for the ordinary integers. To see this, let's review how prime factorization works: Let's call two integers "similar" if each is a multiple of the other (so, e.g., x and x are similar; note that similar values have the same size). From now on, except where otherwise specified, I will not distinguish between similar values; they're as good as equal to me. We say a nonzero integer p is "prime" if any way of expressing it as a product of integers involves a factor of p. (Note that 1 (or any value similar to 1; aka, any "unit") is not considered prime because of the empty product.) By repeatedly breaking any particular nonzero integer into a product of smaller integers, if possible, and stopping when reaching primes, we find that every nonzero integer has some representation as a product of primes. (Note that this factorization process cannot go on forever, by consideration of how sizes get smaller at each stage) Not only that, but this representation is unique. (This follows from the fact that if a prime divides a product, it divides one of the product's factors. [This in turn can be seen as so: if prime p divides ab but does not divide a, we have that p is a common divisor of ab and pb, so p must divide gcd(ab, pb) = gcd(a, p) * b = 1 * b = b. (This reasoning in turn uses basic facts about gcd [specifically, that any common divisor divides the gcd, and that gcd scales up as its inputs scale up] which can be seen from, for example, the Euclidean algorithm to compute gcds, which makes use of a "division and remainder (smaller than the divisor)" operator, whose existence just depends on the fact that any ratio of integers has distance less than 1 from some integer. Again, this algorithm cannot go on forever by consideration of how sizes get smaller at each stage.)]) All of the above was phrased in terms of integers, but, in fact, it works just as well replacing "integer" by "Gaussian integer" throughout! So we have that each nonzero Gaussian integer uniquely factorizes into a product of Gaussian primes. In particular, each ordinary integer also has a factorization as a product of Gaussian primes, which we can obtain by first factoring it into ordinary primes, and then further factoring those ordinary primes into Gaussian primes. So let's try to understand how ordinary primes factor into Gaussian primes. Let p be an ordinary prime and let q be a Gaussian prime factor of p. By symmetry, q's conjugate must also be a prime factor of p. It may be that q is similar to its conjugate or not. If q and its conjugate aren't similar, then their product is a nonunit integer factor of p, and thus must be p itself. In this case, let's say p is a "bifurcating" prime; it's prime within the integers, but splits into two dissimilar conjugate primes within the Gaussian integers. On the other hand, if q and its conjugate are similar, then q is either similar to an integer or similar to an integer multiple of 1 + i. In the former case, as p is prime within the integers, we must have that q is similar to p itself, and thus p is already a Gaussian prime (let's call such p "atomic"). In the latter case (with q similar to a multiple of 1 + i), we have that p is divisible by 1 + i, and thus p^2 = p^2 is divisible by 1 + i^2 = 2; this can only occur when p itself is even, which is to say, when p is 2. And, indeed, 2 has the Gaussian prime factorization (1 + i) * (1  i), a product of two similar conjugates. In summary, every odd ordinary prime is either itself a Gaussian prime (in which case we'll call it "atomic"), or the product of two dissimilar conjugate Gaussian primes (in which case we'll call it "bifurcating"). The one other ordinary prime is 2, which has the Gaussian prime factorization (1 + i) * (1  i), a product of two similar conjugates. Phew! That's quite a bit already, but we're almost done. Now we just need to figure out which odd primes are atomic/bifurcating. The bifurcating ones will be products of conjugates (i.e., sums of two squares) and the atomic ones will not be. If ordinary prime p equals (A + iB) * (A  iB) = A^2 + B^2, then B won't be divisible by p, and so A/B will be a welldefined square root of 1 in the field of integers mod p. Conversely, if there is a square root r of 1 in the field of integers mod p, then we have that 1 + r^2 = (1 + ir) * (1  ir) is a multiple of p, even though p divides neither factor, and thus, p cannot be a Gaussian prime itself. Thus, p is atomic just in case there is no square root of 1 modulo p. How do we tell if 1 has a square root modulo p? More generally, how do we tell if a value has a square root modulo p? Well, note that two values have the same square just in case they are each other's negations; restricting attention to the case where p is odd, no nonzero value is its own negation, so the p  1 nonzero values square to precisely (p  1)/2 distinct nonzero values. Furthermore, Fermat's Little Theorem tells us r^(p  1) = (r^2)^((p  1)/2) is 1 modulo p for any nonzero r; accordingly, the nonzero values with square roots modulo p are precisely the (p  1)/2 many solutions to x^((p  1)/2) = 1. Does (1)^((p  1)/2) = 1? Well, by the nature of 1, this is true just in case (p  1)/2 is even, which is to say, just in case p is 1 mod 4. Thus, we see that the bifurcating odd primes are precisely those which are 1 mod 4 (with the primes which are 3 mod 4 being atomic, and the even prime 2 being its own odd man out), completing our investigation. Last edited by Indistinguishable; 07032015 at 03:18 PM. Reason: "similar", "associate", "equivalent", "equal"... Who cares? 
#27




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(As for why this fact about fractions is true, you may enjoy puzzling it over yourself. Note that, for our present purposes, we need reasoning that establishes this in such a general way as to show that it holds not just for fractions over the integers, but just as well for fractions over the Gaussian integers. [Spoiler available here.]) Last edited by Indistinguishable; 07032015 at 03:34 PM. 
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