Wed, 01 Feb 2006
The 3n+1 domain
The second proof depends on the (unproved) fact that lowest-term fractions are unique. This is actually a very strong theorem. It is true in the integers, but not in general domains.To fully address this, I need to discuss one of the most fundamental ideas of number theory, the prime factorization. Every positive integer is either a product of two smaller integers, or else is a prime. Primes include 2, 3, 5, and 7. 4 is not prime because it is the product of 2 and 2; 6 is not prime because it is the product of 2 and 3. 221 is the product of 13 and 17; 222 is the product of 6 and 37; 223 is prime. (1 is a special case, and is not considered prime.) This much you probably already know.
The primes were recognized by mathematicians thousands of years ago. Euclid's Elements, written around 2300 years ago, includes a proof that there is no largest prime number. The most important theorem about the positive integers, also known to the Greeks, is that every positive integer has a unique representation as a product of primes. This theorem, fittingly, is known as the "fundamental theorem of arithmetic".
It's quite easy to show that every positive integer can be represented as a product of primes. Suppose that this were not so. Then there would have to be a smallest positive integer n that was not so representable. The number n is either 1, or is prime, or is a product of smaller numbers. If it is 1, it is representable as the "empty product", another special case. If n itself is prime, then it is trivially represented as the product of the single prime n. (Mathematicians customarily allow such things.) Otherwise, n is the product of two smaller numbers, say p and q. But these two numbers are representable as the products of primes, since they are smaller than n, and n is the smallest number not so representable. And then n's representation would be the concatenation of the representations of p and q.
But showing that the representation is unique is trickier. If it seems obvious, you probably haven't thought about it enough. For example, consider the number 24. We can decompose 24 into the product of smaller numbers as 4×6. Then 4 decomposes to 2×2 and 6 to 2×3, so 24 = 2×2×2×3. But what if we had decomposed 24 differently, say as 3×8? Then we decompose 8 as 2×4, and 4 as 2×2, yielding 24 = 3×2×2×2. The two decompositions end up in the same place. But was that guaranteed? What about for a really big number, like 3,628,800? There are a lot of ways to decompose this. Can we be sure that all paths will end at the same place?
For the integers, the answer is yes, they do. But this is a special property of the integers. One can find simple structures, analogous to the integers, where two paths might not end at the same factorization.
The usual example of this is the so-called 3n+1 domain. In this world, we consider only the integers of the form 3n+1. That is, the only numbers of interest are 1, 4, 7, 10, 13, 16, and so on. It's quite easy to show that the product of any collection of these numbers is another one of the form 3n+1. (Briefly, because (3n+1) × (3m+1) = 9mn+3m+3n+1 = 3·(3mn+m+n) + 1. But if you're not convinced by this, just try some examples.)
The 3n+1 domain has its own idea of what is prime and what isn't. 1, as usual, is a special case. 4 is prime, because it is not a product of smaller numbers. (2×2 doesn't work, because the 3n+1 domain has no 2.) 7 is prime, as usual. 10 is prime. (Again, because the 3n+1 domain has neither 2 nor 5.) 13 is prime, as usual. 16 is not prime; it is 4×4, as usual.
The proof that every number has a prime factorization goes through in the 3n+1 domain just as in the regular integers. But the proof that the factorization is unique does not go through, because in fact the 3n+1 domain does not have this property. The smallest example for which it fails is 100, which has not one but two factorizations into primes: 100 = 4 × 25 = 10 × 10.
The failure of the fundamental theorem of arithmetic in the 3n+1 domain triggers the failure of a long series of related results, like the first domino in a series knocking over the others. For example, in the regular integers, if p is prime, and ab is a multiple of p, then either a or b (or both) is a multiple of p. In particular, if the product of two integers is even, then at least one of the two integers must have been even. This does not hold true in the 3n+1 domain. A counterexample is that neither 4 nor 25 is a multiple of the prime 10, but their product is.
Yet another failure is the one with which I opened the article. We say that two integers have a common factor d if they are both multiples of d. (d = 1 is conventionally excluded, since all integers are multiples of 1.) For example, 28 and 18 have a common factor of 2 since they are both multiples of 2; 36 and 63 have a common factor of 9. 14 and 15 have no common factor. When we write fractions, we customarily forbid the numerator and denominator to have a common factor, since the fraction can then be written more simply by canceling the common factor. For example, we never write 36/90; we always cancel the common factor of 18 and write 2/5 instead. When the numerator and denominator have no common factor, we say that the fraction is "in lowest terms".
Each rational number has a unique representation as a lowest-terms fraction, as the quotient of two integers that have no common factor. There is no way to write 2/5 as the quotient of two numbers other than 2 and 5, except by multiplying them by some constant factor, to get something uninteresting like 200/500.
This property also fails in the 3n+1 domain. Some rational numbers have more than one representation. 4/10 = 10/25, but neither 4 and 10 nor 10 and 25 have common factors to cancel, since the 3n+1 domain omits both 2 and 5.
Exercise: Does the analogous 4n+1 domain have unique prime factorizations?
Coming eventually: Gaussian integers and Eisenstein integers.
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