# Defining addition of supernatural numbers?

In the comments on this question Bill Dubuque mentions the supernatural numbers. My curiosity was piqued by the statement on Wikipedia that "there is no natural way to add supernatural numbers" and I soon invented this example:

Let $a$ be the supernatural product of all primes congruent to 1 mod 4, and let $b$ be the supernatural product of all primes congruent to 3 mod 4. Because GCD is defined for supernatural numbers, and the sum of two relatively prime numbers is relatively prime to each of them, we can say that $2a + b = 1$ and also that $a + 2b = 1$; adding these gives $3a + 3b = 2$ or $a + b = \frac{2}{3}$. The value $\frac{2}{3}$ can apparently be interpreted as a "super-rational" number, a supernatural-like number where negative exponents are permitted. So it seems that I can give a consistent definition of addition at least for some supernatural numbers (although the result in this case is "super-rational").

What is the basis of the claim that "there is no natural way to add supernatural numbers"? Do the assumptions underlying my idea lead to any contradiction? If not, to what extent can it be extended to allow the addition of more general forms?

EDIT: I hadn't read the article closely enough to realize that supernatural numbers are allowed to have exponent values of $\infty$, and also it has been pointed out that my idea does not work in any case. What remains of this question I feel is too unfocused. I am accepting Greg Martin's answer.

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I now see there is indeed a problem here. $4a + b$ and $a + 4b$ must both equal $1$ because the addends are relatively prime in each case, and this yields $a + b = \frac{2}{5}$ instead of $\frac{2}{3}$. But my first question still stands unanswered, and I'm also curious if there is any other way to make it work at least for some subset of supernaturals (or super-rationals for that matter). – Dan Brumleve Jan 9 '12 at 5:11
@DanBrumleve: Thanks for the correction (I'll delete my comment). I forgot to include even factors. I agree that it would be interesting to hear of a basis for claiming that there is no natural way to define addition. – Jonas Meyer Jan 9 '12 at 5:16
The assumption here is that since the sum $2a+b$ can't be divided by any natural number (since every prime is divides either $2a$ or $b$, but not both), it must be $1$. But this is true for, say, $2pa+b$ as well, where $p$ is any prime factor of $a$. This seems like it leads to a contradiction. But really, even just multiplying and dividing supernatural numbers is problematic... you already have cases like $x=2^\infty$, where $2x=x$ would seem to imply that either $x=0$ (not great) or $2=1$ (even worse). – mjqxxxx Jan 9 '12 at 5:19
@mjqxxxx, $2^\infty$ is not a supernatural number. The exponents must be non-negative integers (or arbitrary integers in the "super-rational" case). A supernatural can have an infinite number of prime factors, but must have finite exponents. So multiplication and division are well-defined, as are GCD and LCM. – Dan Brumleve Jan 9 '12 at 5:21
@DanBrumleve: That isn't correct according to the usual definition of supernatural number. The multiplicities of the factors are allowed to take the value $\infty$. If you want to restrict to the finite multiplicity case, that should probably be made clear in the question. – Jonas Meyer Jan 9 '12 at 6:09

You say "Because GCD is defined for supernatural numbers, and the sum of two relatively prime numbers is relatively prime to each of them, we can say that $2a+b=1$". This deduction seems hasty to me. I assume you're thinking "two natural numbers that are relatively prime have a sum that has no prime factors in common with either; and the only natural number that has no prime factors is 1". However, you're assuming to start with that $2a+b$ is a natural (or perhaps supernatural) number, but there's no reason that $2a+b$ has to be well-defined.
Addition cannot be defined on the supernatural numbers. Indeed, assume the contrary. Then we get $$2^{\infty}\neq 3\times 2^{\infty}=2^{\infty}+2^{\infty}+2^{\infty}=(2^{\infty}+2^{\infty})+2^{\infty}= 2\times2^{\infty}+2^{\infty}=2^{\infty}+2^{\infty}=2\times2^{\infty}=2^{\infty},$$ which is the contradiction.