Is -1 a Prime Number? Exploring the Smallest Prime Number Debate

[Sven Tingblad]

I'm sorry if this is a stupid question, it certainly seems like someone must have have thought of it before. Anyway, here it is: is -1 a prime number? After all, primes are divisible only by the number 1 and by themselves. -1 is divisible only by 1 and itself. Is it therefore the smallest prime number?

 

[Muzza]

No. See the definition of a prime number. A prime must be strictly greater than 1. There are reasons for this, for example, it's nice to be able to say that every natural number can be written uniquely (up to order of factors) as the product of primes. If you allow 1 to be a prime, this no longer holds (since, say, 6 = 2 * 3 * 1 = 2 * 3 * 1 * 1). If you allow negative primes, you get a similar situation (since 6 = 2 * 3 = (-1)(-1)(2)(3), etc).

 

[Sven Tingblad]

Thanks for clearing that up. I knew something had to be wrong with it, or it would have been discovered centuries ago. I guess nothing is as simple as it seems.

 

[mee]

No. See the definition of a prime number. A prime must be strictly greater than 1. There are reasons for this, for example, it's nice to be able to say that every natural number can be written uniquely (up to order of factors) as the product of primes. If you allow 1 to be a prime, this no longer holds (since, say, 6 = 2 * 3 * 1 = 2 * 3 * 1 * 1). If you allow negative primes, you get a similar situation (since 6 = 2 * 3 = (-1)(-1)(2)(3), etc).

Is it that the extra digits of 1 or -1 are not as efficient or what?

 

[HallsofIvy]

The main reason why neither 1 nor -1 is a prime number is the "prime factorization" theorem: every positive integer can be factore into a product of primes in a unique way.

 

[Janitor]

... the "prime factorization" theorem- HallsofIvy

Also known as the "fundamental theorem of arithmetic," if I am not mistaken.

 

[shmoe]

Is it that the extra digits of 1 or -1 are not as efficient or what?

It's not a matter of efficiency but uniqueness. As it is now, prime factorization says "every natural number can be written uniquely (up to order of factors) as the product of primes.".

If you allowed 1 or -1 to be primes you'd have to change this to say "every natural number can be written uniquely (up to order of factors) as the product of primes (up to possibly some added 1's and -1's thrown into the mix for fun)".

It wouldn't fundamentally affect things, but it would make life much messier.

 

[HallsofIvy]

Also known as the "fundamental theorem of arithmetic," if I am not mistaken.

Yep! Completely correct (last added to go over 10 characters).

 

[matt grime]

-1 of course is a unit (it is invertible), and units are by definition excluded from being considered primes.

however, there is another point: if we allow -1 to be considered we must consider all negatives, surely, and there is no smallest prime, but of course there is a smallest positive prime. And by prime we use the definition of prime relevant to rings such as Z, since the notion of having exactly two factors is not the one we want to consider (-1 divides 2 for instance, so 2 might appear to have three factors: -1,1,2).

 

[CRGreathouse]

-1 of course is a unit (it is invertible), and units are by definition excluded from being considered primes.

however, there is another point: if we allow -1 to be considered we must consider all negatives, surely, and there is no smallest prime, but of course there is a smallest positive prime. And by prime we use the definition of prime relevant to rings such as Z, since the notion of having exactly two factors is not the one we want to consider (-1 divides 2 for instance, so 2 might appear to have three factors: -1,1,2).

Obviously, as you mentioned, -1 is a unit and as such can't be called prime. I have seen negatives considered primes under some definitions, though; uniqueness was by associate classes ([tex]x[/tex] is associate to [tex]y[/tex] if [tex]x=yu[/tex] for a unit [tex]u[/tex]). Thus [tex]12=2^2\cdot3=(-2)^2(-3)[/tex] is a unique factorization, since -2 is associate to 2 (and so forth).

 

[humanino]

Check this line of reasoning.
Let [tex]n[/tex] denote the largest of all natural integers. We have [tex]n^2\leq n[/tex]. Therefore [tex]n=1[/tex] or [tex]n=0[/tex]. Thus [tex]n=1[/tex].

Whence the importance of not assuming existence without care.
Why did I came to think about that ?

 

[Gokul43201]

Check this line of reasoning.
Let [tex]n[/tex] denote the largest of all natural integers. We have [tex]n^2\leq n[/tex]. Therefore [tex]n=1[/tex] or [tex]n=0[/tex]. Thus [tex]n=1[/tex].

Whence the importance of not assuming existence without care.
Why did I came to think about that ?

This baffles me. :

 

[humanino]

Sorry Gokul. I'm pointless as usual

 

[mathwonk]

factorization into primes is in fact not usually said to be absolkutely unique, as one does generally consider as primes all negatives of primes. in all rings factorization into primes is unique only up to (order) and associates. hence in rings where there are a lot of units, hence a lot of associates, factorization is more complicated.

of course one could restrict attention in afctoring only to positive integers, but this is not wise. for example the usual theorem on rational roots reminds us that the possible rational roots of a monic polynomial with integer coefficients, has as possible rational roots only those positive or negative factors of its constant term.

as son as we consider the simplest generalization of integers, say gaussian integers {of form a+bi where a,b are integers}, one has not the luxury of distinguishing positive from negative factors.

so the first thing to determine in questions of uniqueness of factorization, is what are the units in a ring. e.g. in the gaussian integers they are 1,-1,i,-i. then units are never allowed to be primes, for the rerasons given above. still factorization is unique only up to associates.

nonetheless, this at least makes the number of prime factors, and the number of non associate prime factors, and the multiplicity of each prime factor (along with occurrences of its associates) uniquely determined.

moreover if one considers all multiples of the prime as the "ideal" it generates, then the prime ideals are indeed unique (at least whenever factorization is unique up to associates).