How to define the '*' operation in the ring

How to define the '*'(multiplication) operation in the ring, where the set is real positive numbers and the '+' operation is multiplication, so that the ring will be without zero divisors? The same question for the ring, where the set is rational positive numbers.

I understand that the zero element in this ring is '1', so we need the equation a*b=1 to be unsolvable if a and b are not 1. But I've tried all the typical operations such as addition or exponentiation and haven't succeeded.

Let's look at the abelian group structure of your set $\mathbb R _+$ - it is isomorphic to $\mathbb R$ with usual addition, via

$$f : \mathbb R \rightarrow \mathbb R _+$$ $$f(x) = e^x$$

since

$$e^{x+y} = e^x \cdot e^y$$

Now, just transfer the usual multiplication of $\mathbb R$ using this isomorphism, that is, define multiplication $*$ on $\mathbb R_+$ as

$$e^x * e^y = e^{xy}$$

which is

$$x * y = e^{\ln x \ln y}$$

The case of positive rationals is a bit subtler. Again, look at the additive structure of our to-be ring. Every positive rational can be uniquely represented as $p_1^{n_1} p_2^{n_2} \dots p_k ^{n_k}$, where $p_i$ are primes, and $n_i \in \mathbb Z$. Multiplication of rationals corresponds to adding the $n_i$'s, which provides an isomorphism of abelian groups

$$\mathbb Q_+ \cong \bigoplus\limits_{i=0}^{\infty}\mathbb Z$$

There are many possible ring structures on this abelian group. As an example, notice that $\mathbb Z[X]$, the ring of polynomials with coefficients in $\mathbb Z$, has the same additive structure, and contains no zero divizors. Thus, we can transfer the multiplication from $\mathbb Z[X]$ to $\mathbb Q _+$.

• Thanks! Everything is clear. And what about the case with rational positive numbers set? As I understand, now we can't use 'e' or 'ln'. Or the operation still could contain irrational parts? – SilverLight Apr 17 '18 at 15:47
• @SilverLight My apologies - I missed the rationals part of your question. I've updated the answer. – lisyarus Apr 17 '18 at 15:59
• Wow! That's amazing! Thank you very much! – SilverLight Apr 17 '18 at 16:03
• @SilverLight Always glad to be of help! You can accept the answer if it suits you. – lisyarus Apr 17 '18 at 16:05