Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the (cumbersome) statement: "Every integer greater than 1 can be written as a unique product of integers belonging to a certain subset, $S$ of integers.

When $S$ is the set of primes, this is the Fundamental Theorem of Arithmetic. My question is this: Are there any other types of numbers, for which this is true.

EDIT: As the answers show, this obviously cannot be done. What if we relax the integer condition, i.e. can there be any other canonical representation of positive integers using complex numbers?

share|improve this question
1  
No. Indeed, take any prime, it must be a product of something in $S$, so $S$ has to contain all the primes. Since you want unicity too, it can only contain the primes. –  Najib Idrissi Feb 2 '12 at 19:17

3 Answers 3

up vote 5 down vote accepted

If you mean that every positive integer gets a unique multiplicative factorization, then no, there is no other canonical representation. Why? Because then every prime number $p$ can be factorized, but the only way that's possible is if the components of the factorizations include the primes them-selves. Furthermore, you can't add any other number to the list because then the factorization of this number would be non-unique.

Alternatively, there are non-multiplicative representations of integers. The $p$-adic representation is just writing $n$ in "base $p$": $n=a_0+a_1p+a_2p^2+\cdots+a_rp^r$. Even though the golden ratio is not a rational number, we can write integers in base golden ratio.


Algebraic number theory studies number fields and rings of integers beyond just $\mathbb{Q}$ and $\mathbb{Z}$. Of note, there is not necessarily unique factorization of the elements. For example, in $\mathbb{Z}[\sqrt{-5}]$, we have

$$6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}).$$

This lead to some headaches (I assume anyway), until mathematicians figured out that even though the numbers don't factor uniquely, the ideals of the integers factor uniquely into products of prime ideals, which has lead to other algebraic constructions based off of them designed ultimately to study the structure of numbers. (If you don't understand this section of my answer, don't worry about it. It's for a later time then.)

share|improve this answer
    
Very informative answer, thanks a lot. From my cursory reading of the Wikipedia entry, it seems that that concept I was referring to crudely as $S$ above is called an ideal. The example you cite is precisely what prompted my question: Can it be shown that no such unique factorization exists? –  recipriversexclusion Feb 2 '12 at 19:48
    
Yes. So it sounds like the $S$'s you're looking for might be precisely "sets of prime elements of number fields with class number 1." –  Cam McLeman Feb 2 '12 at 19:51
    
@reci: One can think crudely of ideals as "sets of multiples." (More abstractly: an additive subgroup absorbing ambient multiplication. It takes a good while to develop the machinery to understand these sorts of phrases.) Here we speak of factoring sets of numbers into other sets instead of factoring the numbers themselves. Other constructions, like the class number, provide information on factoring actual numbers in the integers. Also, try finding a nonunit divisor of 2,3,1+√-5,1-√-5 in Z[√-5], maybe find out why it's not possible. –  anon Feb 2 '12 at 20:03
    
@reci: The answer to the edit in your question depends on what ring of integers we're working in. In quadratic extensions Z[√d], we use things like quadratic reciprocity to check when the usual primes are still primes, how they ramify etc. Whether or not it is a UFD (unique factorization is present) is much harder to answer. –  anon Feb 2 '12 at 20:19

No. Such a set $S$ must include the primes (because they have no other factors). If $s \in S$ is not prime, then it can be written as a product of primes, i.e., as a product of other elements of $S$. This contradicts uniqueness. Therefore the only such set $S$ is the set of the primes.

share|improve this answer

When you relax the integrality condition you start to pick up other solutions - for instance, the set of square roots of primes, $S=\{\sqrt{p}:p\in P\}$, is such that every number $n$ has a unique factorization $n=s_1^{e_1}s_2^{e_2}\ldots$ in terms of the elements of $S$; just consider the breakdown $n=p_1^{d_1}p_2^{d_2}\ldots$ and then set $e_1=2d_1$, etc. On the other hand, this gives up existence : it's no longer the case that every sequence $e_1, e_2, \ldots$ of whole numbers corresponds to a unique integer; instead we have the additional requirement that all the $e_i$ be even.

More generally, any breakdown of primes into products of (non-integral) real factors will give you a solution: if $p_1 = t_{11}\cdot t_{12} \cdot \ldots\cdot t_{1m}$, $p_2 = t_{21}\cdot t_{22}\cdot\ldots\cdot t_{2n}$, and so on, then obviously any factorization of a number $n$ into products of $p_i$ extends to a (unique) factorization of $n$ into products of the $t_{ij}$. What's more, any unique factorization of all numbers into real 'factors' $x_i$ has to come from this kind of breakdown, because obviously the prime numbers all need unique factorizations, and the factorization of a number $n$ into primes then 'extends through' the factorization of those primes into a factorization of $n$ in terms of the $x_i$, which by definition must be the factorization of $n$. So you don't get anything particularly interesting by removing the non-integrality condition, just 'refinements' of the factorizations you already had.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.