question about a composite/prime number theorem?

Question

question about a composite/prime number theorem?

Answer 1

I don't see anything new for the theorem about random number. (or I don't get your meaning) Given a situation that, for a total choices of n, each choice has a same probability to be played, ie 1/n. True, because it is uniform distribution. After t times of plays, which t is a very large value, if not infinity, is it true to says that, ALL choices of n will have same frequency of occurance, ie t/n (t * 1/n) after t plays? Since it is uniform distribution, the chance of each choices (1/n) are same. If you put frequency of occurance (t) * chance of each choices (1/n). You will the number that the choices will the occurance in t times of plays. It is a very basic common sense in applied statistics, isn't it ?

Answer 2

You may prove an important lemma on your way to proving

unique factorization.

Lemma: If p is a prime and p divides ab, then p divides a

or p divides b.

Proof. I would like to construct an ancient greek type of proof

for this to demonstrate that the greeks could have accomplished

this result in the time of Euclid or before.

Suppose that p is the smallest prime such that for some a

and b, p divides ab but p does not divide a and p does not

divide b. Since p|ab we write ab=pn. It follows that solving

this problem for a>p or b>p or both, is equivalent to solving

it for a

principle known to archimedes, that is the location of a

remainder upon division by an integer. Here, the

remainders are a-kp

without any loss of generality that ab=pn , a
a>n and b>n for the ordering: n
with positive integers we first must establish that n>1.

If n=1 then p=ab and one of a or b is 1 since a>1,b>1 means

that p is composite a contradiction(we said above that p is prime)

If a=1 then p=b and p|b which we said can't happen by our

initial conditions at the top of the lemma. The same happens

if b=1. Therefore, to resolve this we must have n>1. We are

now set up with ab=pn, p is the smallest prime such that

p divides some product ab but p does not divide a or b,

and the ordering 1
steps in: n>1 implies that n has a prime divisor, say q.

The ordering now contributes n
q

divides ab, and being smaller it is not subject to the same

limitation, which means that q|a or q|b. We divide q into the

appropriate a or b and since q was an arbitrary prime in n,

which can be broken into a finite product of primes

(Your Theorem), we find that all of the primes of n have been

divided into ab leaving us with a result: AB=p,

where A divides a and B divides b. Again, p prime implies

that A=1 or B=1 since AB>1 makes p composite.

If A=1, then p=B but B|b therefore p|b.

If B=1, then p=A but A|a therefore p|a.

This means that p|a or p|b which contradicts the assumption

about p made at the start of the proof. Then, there is no

smallest p such that p|ab and not p|a and not p|b.

Therefore the lemma holds.



If you add your "first part", this lemma will prove uniqueness.

Answer 3

Not that I know of. It's true, so of course it's a valid theorem, but since it's far more useful to add the uniqueness part to the statement, why would we just leave it half finished like that? By the way, the full statement (with uniqueness) is more commonly known as the fundamental theorem of arithmetic.

Answer 4

This is a difficult question to answer. The "unique factorization theorem" (sometimes referred to as the fundamental theorem of arithmetic) is by itself not all that difficult to prove, knowing the basics about prime numbers and divisibility. But placing the question in a broader context, existence and uniqueness of prime factorizations can be an extremely subtle and delicate matter. In number theory it turns out to be useful (even necessary) to work with number systems more general than just the set of all integers, which I shall denote Z.



As an example, consider the set Z[i] where i is the square root of negative one. Elements of this set, called "Gaussian integers", have the form a + bi where a and b are integers. One can without much trouble add, subtract, and multiply such numbers just as one can with ordinary integers; for example, (2 + 3i) * (4 - 2i) = 8 - 4i + 12i - 6i^2 = 8 + 8i + 6 = 14 + 8i. One can then define concepts like divisibility; for example, the computation above shows that that 14 + 8i is divisible by 2 + 3i. From there, one can even define the notion of a prime number, although the problem is a little subtle. Once one has worked hard and made all of the right definitions, it turns out that the unique factorization theorem is still true for Z[i]; every Gaussian integer can be written uniquely as the product of primes (up to a caveat related to the way primes are defined; basically, you can't have i or -i in a prime factorization, just as you can't have 1 or -1).



Now let's vary this example slightly to see how subtle these issues can be. Instead of Z[i], let's look at Z[sqrt(-5)]. This is the set of all numbers of the form a + b sqrt(-5). These numbers can be added, subtracted, and multiplied just as before; also, if one is clever enough to define notions of divisibility and prime for Z[i] then basically the same definitions work here. But there is a problem. One can prove that the numbers 2, 3, 1 + sqrt(-5), and 1 - sqrt(-5) are all prime in Z[sqrt(-5)] (this is NOT trivial, but it's true), and thus unique factorization fails for the number 6: 2*3 = 6, and (1 + sqrt(-5))(1 - sqrt(-5)) = 1 - sqrt(-5) + sqrt(-5) - sqrt(-5)^2 = 1 + 5 = 6. Very unfortunate.



One can define all kinds of different number systems analogous to the integers in this way. And it is not just an academic exercise; for example, Andrew Wiles' proof of Fermat's Last Theorem made heavy use of Z[w] where w is a primitive (complex) nth root of 1. In all cases one can define divisibility and prime for these generalized number systems, but it is an extremely difficult task to determine whether or not the unique factorization theorem is true. In general, the EXISTENCE of prime factorizations is not in question; for the contexts that arise in number theory, this is guaranteed. The uniqueness is really the hard part.



I'm going to wrap up this answer by throwing a bunch of language at you that you might not understand, because I feel a little bad about how imprecise my answer has been up to this point. If nothing else, it might give you something to look up on Wikipedia if you are still curious. By "number system" I really mean the integral closure of Z in a finite algebraic field extension of Q, the set of all rational numbers (more concisely it is the ring of integers in a number field). Z is a Dedekind domain, and one can prove that the ring of integers in any number field is also a Dedekind domain. Every ideal in a Dedekind domain splits as the product of prime ideals, but the decomposition is not guaranteed to be unique. A ring for which the unique factorization theorem holds is called a "unique factorization domain" (UFD), and for a long time people wondered if given any number field K there is a finite algebraic field extension of K whose corresponding ring of integers is a UFD; it turns out that this is not the case, but it wasn't until the 1960's that anybody could construct a counterexample. There is a compromise, though. If you take a number field K whose ring of integers is A and you take any ideal of A, there is a finite algebraic field extension F of K such that the extension of that ideal to the ring of integers of F is a principal ideal. In other words, you can't guarantee unique factorization of ALL integers by passing to an extension, but you can get unique factorization for any given integer. This theorem is one of the crowning achievements of class field theory, a profoundly beautiful and difficult branch of number theory.