Definition of a prime number

Cecil's Columns/Staff Reports
21

I looked over this thread and I noticed that the term unit is used, but never defined. Therefore I submit that a unit is a number which has an inverse.

I found another definition of prime. (I can’t provide a link 'cuz Netscape crashed.) A number p is prime if whenever p = a*b, then a or b is a unit. The page also gave Zorblak’s definition and said that when the two sets are different, the numbers I just described are irreducible.

Virtually yours,

DrMatrix
“Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?” - WallyM7

22

So is 1 prime under this definition, or did you leave out the “p must not itself be a unit” part?

It is too clear, and so it is hard to see.

23

Yeah, I think he left out that p must not be a unit. Also, as DrMatrix mentioned in passing, it’s really not the definition of a prime, but the definition of an irreducible. It turns out that, in a principal ideal domain, primes and irreducibles are the same things.

24

Oh, and not only in a principal ideal domain, but in a unique factorization domain, they’re the same things, as well.

25

Yes, I forgot to say p must not be a unit. (This after going to the trouble of defining a unit.)

Does anyone have an example where irreducibles and primes are not the same sets? I seem to recal something about sqrt(5) or sqrt(-5), but I might be thinking about something else.

Virtually yours,

DrMatrix
“Feynman was wrong.
I understand Quantum Physics completely.
Anybody seen my drugs?” - WallyM7

26

Yeah, one example is in Z[sqrt(-5)] (Z for integers, of course). In this ring, for example, 2 + sqrt(-5) is irreducible but not prime.

Actually, if you look at the algebraic integers contained in Q[sqrt(d)], they form a unique factorization domain when d = -1, -2, -3, -7, -11, -19, -43, -67, and -163, but not for any other negative integer values for d. I’m not sure about the positive integers.

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(sig line courtesy of WallyM7)

27

OK, I don’t understand this notation. Could someone give a brief description of what it means?

It is too clear, and so it is hard to see.

28

In general, Z is the set of polynomials (over x) with integer coefficients. So Z[sqrt(-5)] would be the polynomials over sqrt(-5) with integer coefficients. Since [sqrt(-5)]^2 is an integer, this really just amounts to numbers of the form a + b[sqrt(-5)], with a and b integers.

As for the other part I mentioned, Q would be the polynomials with rational coefficients. So Q[sqrt(d)] would be numbers of the form a + b[sqrt(d)], with a and b rational. Then I wanted to take the algebraic integers that are in Q[sqrt(d)]. The algebraic integers are all the roots of monic polynomials with integer coefficients (“monic” meaning the coefficient of the highest power of x is just 1).

(The set of algebraic integers in Q[sqrt(d)] will not necessarily be Z[sqrt(d)]).

29

Some mathematical definitions are arbitrary, “Counting Numbers” for instance…

Zero used to be excluded, but now many prominent mathmeticians are point out than if you have five apples and a bully takes five of them away, you have ‘none’ or zero. The concept of apples doesn’t magically disappear when you don’t have any, thus you must have way to express how many you (don’t) have. I personally start counting at zero. If I have no money and you give me a dollar bill then ask me how much I have, I take my old total and add one, $0 + $1, getting one. This is the same operation I’d do if I was starting at $10,276.

Anyways, some definitions are arbitrary, some highly respected mathmeticians think one thing, some think the other, and there isn’t really a right answer, because if nobody counted 13, for superstitious reasons, we could exclude it from the set of “counting numbers”.

Other definitions appear arbitrary, yet aren’t if you examine them in depth.

10^2 which is 10*10 is 100, 10^1 which is 10 is … well … 10. 10^0 which is … is 1… How is that again?

It makes absolutely no sense if N^x just means writing ‘x’ Ns with *s between them. Yet if you understand the operation, especially the inverse operations, it makes sense.

So, is the definition of ‘prime numbers’ arbitrary in its exclusion of 1? 1 does fit the standard definition, having only one and itself as factors, if you don’t assume that the two factors must be different.

Are there any functions which prime numbers can be tested with which we could show that 1 is not a prime number, though it appears to be?

Just saying that “one is the unit, and thus different” isn’t good enough.

There is one piece of evidence against 1 being prime. That is, take the simple method of finding primes, start at a number, cross out all of its multiples, go to the next non-crossed out number, cross out all of its multiples. When you decide to stop, look at the list of numbers you’ve passed. Any that aren’t crossed out are prime.

If you start with 1, it’s trivial to see that this doesn’t work, where if you start at two, it does.

But, does this mean the prime-generation method is flawed, or that one isn’t prime?

What tests for primeness, asside from a definition which arbitrarily excludes it, does 1 fail?

30

The definition of prime integers (only divisors are 1 and itself, excluding one) is not arbitrary because of the idea behind the definition.

Look at the Fundamental Theorem of Arithmetic, for example:

Every nonzero integer can be uniquely written as the product of primes.

The idea behind the definition of primes is that we want them to be thought of as the “multiplicative building blocks” of the nonzero integers. Using multiplication and the primes, there is a unique way to construct any nonzero integer.

The number 1 doesn’t really fit into that idea. 1 is merely the multiplicative identity, it doesn’t actually “build” numbers through multiplication as the primes do.

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(sig line courtesy of WallyM7)

31

For a semi-rigorous algebraic definition of “prime” see my post on the thread about “is zero even”. Perhaps that’s what you’re looking for. In short, 1 is not prime because 1Z = Z is not a proper ideal of Z, the integers. See any general abstract algebra or ring theory text.

Joe.

32

I seem to recall seeing an alternate definition of a prime (which I think only held on the positive real integers) which ivolved a bunch of factorials, but unfortunately, I can’t seem to dig up the source where I saw it… Does anyone happen to remember how that definition went, and what it would say about the primality of 1?

“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein

33

You’re probably thinking of Wilson’s Theorem, which says:

If p is an integer, p > 1, then p is a prime if and only if (p-1)! = -1 (mod p), i.e. (p-1)! + 1 is divisible by p.

If I omitted the part about p > 1, then technically 1 would also satisfy this theorem. (In general, though, congruence mod 1 isn’t really ever used though, since then there’s only one number, zero).

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34

Thanks, Cabbage, I think that was what I was remembering. One wonders if one could use that definition with the gamma function representation of a non-integral factorial, and a generalization of the concept of modularity to the reals, to define a set of “primes” over the reals… I suspect, tho, that it would still only work for the conventional integer primes…

“There are only two things that are infinite: The Universe, and human stupidity-- and I’m not sure about the Universe”
–A. Einstein