I recently came across a binary operation (in a very non-algebraic context - it's a way to organize a certain updating of log-likelihood-ratios) and was idly wondering whether it is any kind of reasonable algebraic object. The answer may well be no but it does satisfy some properties that look like those of a ring.

Let $\boxplus \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ be given by $a \boxplus b = \log\left(\frac{1+e^{a+b}}{e^a+e^b}\right)$.

It's a fun exercise to check the following properties:

  1. $a\boxplus b=b \boxplus a$
  2. $a\boxplus b=0$ iff $a=0$ or $b=0$
  3. $a \boxplus (-b) = (-a) \boxplus b= -(a \boxplus b)$
  4. $a \boxplus \infty = a$, in the sense that $\displaystyle \lim_{x \to \infty} a\boxplus x=a$
  5. $(a\boxplus b) \boxplus c=a \boxplus (b \boxplus c)$

So just looking at these properties I though maybe $(\mathbb{R},+,\boxplus)$ is a commutative ring without identity. But it doesn't satisfy the distributive law. Is there anything that can be said about such a structure?

Edit: After a helpful comment from Bill I would like to point out which of these properties I find 'ringlike'. Property 3 is a statement that is true in a ring and doesn't make sense in just the semigroup $(\mathbb{R},\boxplus)$. Property 2 is the definition of an integral domain (if it was a ring). So it seems to me that having $-$ and $0$ puts me in the mind of connecting $\boxplus$ with $+$. But I don't know if this is necessary: is there some theory of semigroups with an extra unary negation operator? Maybe that's all I need.

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    $\begingroup$ What about other possible pairings with your exotic operation other than addition? Wouldn't multiplication, min, max, or even GCD, LCM distribute? $\endgroup$ Mar 15, 2012 at 19:41
  • $\begingroup$ @TegiriNenashi I don't think you're commenting in the right place. $\endgroup$
    – user23211
    Mar 15, 2012 at 20:01
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    $\begingroup$ A ring without the distributive law consists of a set with a monoid structure and group structure, without the two structures having any relationship. So it has nothing to do with a ring. $\endgroup$ Mar 15, 2012 at 20:05
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    $\begingroup$ Bill - I agree that in general a ring without distributive law has no relationship between the ring and the monoid but I think my example does. Property 3 connects inverses in the additive group with the monoid structure. Or should I not be thinking about minuses that way? Maybe it's just a group with an extra unary operation of negation? $\endgroup$ Mar 15, 2012 at 20:24
  • $\begingroup$ @ymar I have meant that OP bundled his operation with plus while there are other choices. Yes, 0 constant hints to addition, but so does infinity for min (max). I just suggested if other coupling would give rise to genuine (semi)ring. $\endgroup$ Mar 15, 2012 at 21:56

1 Answer 1


Thanks to Tegiri Nenashi for his comment suggesting using operations other than plus. This answer will use max. If we wish to have our $\boxplus$ be the multiplicative operation on a semiring then the additive identity needs to annihilate the semiring multiplicatively, thus the additive identity must be $0$. So the additive operation, $\max$, must be applied on nonnegative reals only. If we want the multiplicative identity, $\infty$, to be in the semiring also then we must take the underlying set to be $[0,\infty]$.

Inspired by reasonable intuitions and limiting arguments define $\max(a,\infty)=\infty$ and $a\boxplus \infty =a$ for all $a \in [0,\infty]$.

Then I claim that $([0,\infty],\max,0,\boxplus,\infty)$ is a commutative semiring. It is easy to check that $([0,\infty],\max,0)$ is a commutative monoid with identity $0$. Properties 1, 4 and 5 show that $([0,\infty],\boxplus,\infty)$ is a commutative monoid with identity $\infty$. Property 2 shows that $0$ annihilates $[0,\infty]$.

Finally we need the distributive law $$a \boxplus \max(b,c)=\max(a\boxplus b,a\boxplus c).$$

Without loss of generality $b \leq c$ so we need to show that $a \boxplus c=\max(a\boxplus b,a\boxplus c)$ i.e. $a\boxplus b \leq a\boxplus c$. If $a=0$ or $a=\infty$ or $b=c$ this is true, otherwise we need $$\frac{1+e^{a+b}}{e^a+e^b}\leq \frac{1+e^{a+c}}{e^a+e^c}$$ i.e. $$e^a+e^c+e^{2a+b}+e^{a+b+c} \leq e^a+e^b+e^{2a+c}+e^{a+b+c}$$ i.e. $$e^{2a}(e^b-e^c) \leq e^b-e^c.$$ But $e^b-e^c < 0$ so our inequality is equivalent to $e^{2a} \geq 1$ which is true.


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