Associative binary operation on the positive reals

Question

Let $*$ be an associative binary law on the interval $(0,\infty)$ of the positive reals. Suppose that for every $a,b,c \in (0, \infty)$ we know $$ a * b * c = \frac{abc}{ab+bc+ca}.$$ Can we show that for all $a,b \in (0, \infty)$ we have $a * b = \frac{ab}{a+b}$?

I suspect that the answer to this question is yes, but was not able to make any meaningful progress towards a full proof in the last few days. Any ideas would be appreciated.

Answer 1

The following is an idea that won't fit into a comment. I'm not sure whether it is helpful, or whether it is merely the kind of thing that people are encouraged to put into the question body to indicate "what you tried."

What if we use the binary operation $*$ to define a new binary operation $\oplus$, and this time our set has an identity element (say "e")?

If for $a > 0$ and $b > 0$ we have $a \oplus b = a * b$ and if $$(a \oplus b \oplus e) = (a \oplus b)$$ then we could use a formula for $$(a \oplus b \oplus c)$$ (that is similar to our formula for $(a * b * c)$ ) and we would have some hope of obtaining a formula for $a \oplus b$, which should itself simply be equal to $(a*b)$, and then we will be done.

Now, one difficulty is that the original expression provided for $(a * b * c)$ has the product $a b$ in the denominator, and we need a generalization of that expression so that, when c is our identity element, the term $a b$ will disappear.

Consider the following:

$$(a \oplus b \oplus c) = \frac{abc}{((a b)(c - 1)) + b c + c a}$$

If the above equation could be relied upon, then we would be able to obtain in particular $$(a \oplus b \oplus 1) = \frac{ab}{b + a}$$ and in that scenario it would seem that 1 serves as an identity element.

Now, one difficulty with the above proposal is that we need to continue to have our original formula:

$$(a * b * c) = \frac{abc}{(a b + b c + c a)}$$ Furthermore, we could run into trouble if $$((a b)(c - 1)) + b c + c a) < 0$$ So it seems that we need to try a somewhat different approach.

Observe that $$\frac{abc}{(a b + b c + c a)} = \frac{1}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

So when $c$ is equal to the identity element, we could evaluate our expression by considering the limit as $c$ approaches infinity. In other words, we could define:

$$(a \oplus b \oplus e) = \lim\limits_{c \to \infty} (a * b * c)$$

Technically, that's not going to be a definition, but a consequence of some other statement that will be a definition, because $\oplus$ is simply going to be a binary operation, and thus it needs to be defined as such.

Now, the above isn't an answer, but perhaps it is the beginning of an attempt to fulfill the request for ideas.

Answer 2

This answer is merely a reduction to a simpler problem and some partial results. The proof is completed in the answer by krm2233.

The assumption is $$a * b * c = \frac{1}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}},$$ that is, $f(a * b * c) = f(a) + f(b) + f(c)$ for $f(x) := 1/x$. The map $f : (0,\infty) \to (0,\infty)$ is bijective (it is inverse to itself) and hence allows us to define an associative binary operation $\oplus$ on $(0,\infty)$ such that $f(a * b) = f(a) \oplus f(b)$. We have therefore reduced the problem to the following one, which looks much easier:

Let $\oplus$ be an associative binary operation on $(0,\infty)$ such that $$x \oplus y \oplus z = x + y + z.$$ The question is if $\oplus$ is necessarily the addition.

The answer is yes if $\oplus$ had an extension to $[0,\infty)$, satisfying the same equation, and has $0$ as a neutral element: Then we simply deduce $x \oplus y = x \oplus y \oplus 0 = x + y$.

As in the answer by Franklin, we can prove that every element of $(0,\infty)$ has the form $x \oplus y$, and that each $x \mapsto x \oplus y$ (for fixed $y$) is injective; likewise $x \mapsto y \oplus x$. It then follows that $\oplus$ is commutative, since for all $x,y,z$ we have $(x \oplus y) \oplus z = x + y + z = y + x + z = (y \oplus x) \oplus z$, hence $x \oplus y = y \oplus x$.

Here is an observation which almost looks like the start of an inductive proof: If $a \oplus b =a+b =: y$, then $x \oplus y = x+y$. (Proof: $x \oplus y = x \oplus a \oplus b = x + a + b = x + y$.)

In particular, it suffices to prove $a \oplus a = a + a$ for all $a$ (since every $y$ has the form $a+a$).

Answer 3

Here’s the beginning of an answer, establishing some useful properties of this operation.

First note all positive real numbers are in the image of this binary operation, since $a \ast a \ast a = a/3$ for all positive real $a$.

Next notice that $\ast$ is an injective function in each argument, for fixed values of the other argument. To see why, we use your identity to notice that $(a\ast b)\ast c$ is an injective function of $c$, and by our above observation, $a\ast b$ can be any number in the domain. By similar reasoning, $a\ast (b\ast c)$ is an injective function of $a$ for fixed positive reals $b,c$, and $b\ast c$ can assume any value in the domain.

From your identity, we can see that the value of $a\ast b\ast c$ is determined by the value of $a$ and the value of $1/b+1/c$. Therefore by injectiveness when the first argument is fixed, we have that $b\ast c$ is a function of $1/b+1/c$. Let this function be $f$.

Now we need to argue $f(x)=1/x$. Maybe someone else can finish?

Answer 4

Thanks to Martin's answer, it suffices to consider the simpler situation of $\oplus \colon (0,\infty)\times(0,\infty)\to (0,\infty)$ an associative binary operation which satisfies $a\oplus b \oplus c = a+b+c$, and determine whether or not it must be the case that $a\oplus b = a+b$.

Theorem: $a\oplus b = a+b$.

Proof:

Step 1: For all $a,b \in (0,\infty)$ we have $a\oplus b \leq a+b$.

Indeed if $a,b \in (0,\infty)$ have $a\oplus b>a+b$ then let $c = a\oplus b-(a+b)>0$.

For any $x \in (0,\infty)$ let $r_x \colon (0,\infty)\to (0,\infty)$ be the map given by $y \mapsto y\oplus x$. Then $r_c(a\oplus b) = a\oplus b\oplus c = a+b+c=a\oplus b$, and hence $r_c^2(a\oplus b) = a\oplus b$. But for any $x$ we have $r_c^2(x) = (x\oplus c)\oplus c = x\oplus c \oplus c = x+2c$, so that $r_c^2(a\oplus b) = a\oplus b+2c$, hence as $c>0$ we have a contradiction.

Step 2 We have $a\oplus b = a+b$

Suppose that $a\oplus b<a+b$. Then if $a_1 = a\oplus b$, using Step 1 we have
$$ 2a+b = a\oplus a\oplus b = a\oplus a_1 \leq a+a_1 <2a+b $$ which is a contradiction.