Defining a Binary Operation

Let $a$ and $b$ be arbitrary numbers in $[0,1]$ , we have two cases to consider :

1) $a<b$

From the third property by letting $z=b , y=1$ we have : $(bx)\otimes(b)=b^kx$

Now since $a<b$ we can take $x=a/b \Rightarrow \boxed{a \otimes b=b^{k-1}a} (1)$

2) $a \ge b$

This time we take $z=a , x=1 , y=b/a \Rightarrow \boxed{a \otimes b=a^{k-1}b}(2)$

$\mbox{From } (1) \mbox{ and } (2) \Rightarrow a \otimes b=\begin{cases} a^{k-1}b, & b \le a \\ b^{k-1}a, & b>a \end{cases}$ , for every $a,b\in[0,1](3)$

Choosing $x=1/2 , y=1/4 , z=1/8$ in property (B) together with (3) gives us:

$\frac{1}{2^{3k}}= \frac{1}{2^{k+1}} \otimes \frac{1}{2^3}$ , we'll have 2 cases to check depending on whether $k<2$ or $k \ge 2$ . If $k<2$ then $k=1$ , if $k \ge 2$ then $k=2 \Rightarrow$ If there is $k>0$ for which the operation exists then $k \in \{1,2\}$ .

If $k=1$ , $a \otimes b = min(a,b)$ and all 3 conditions are met .For $k=2$ : $a \otimes b = ab$ and again this meets the required properties .

We will show that there are only two such values of $k$ .

Initially, this problem does not seem to have an obvious solution, so it is best to start with experimentation. The first condition, $(A)$ , suggests that the function will involve some sort of multiplication. This is also corroborated by condition $(C)$ . Clearly, binary operations such as $a\otimes b = a \cdot b$ or $a \otimes b = a^{2} \cdot b^{2}$ satisfy these two. Indeed, if we treat the third condition like a functional equation and set $x = y = 1$ , we find that $z \otimes z = z^{k}$ , which implies that the binary operation must map $a,b$ to $a \otimes b\ = a^{n} \cdot b^{n}$ for some integer $n$ .

However, the second condition says that the operation is associative. Simplifying the second condition using $a \otimes b = a^{n} \cdot b^{n}$ , we obtain $x^{n^{2}} \cdot y^{n} = y^{n^{2}} \cdot x^{n}$ for all real numbers $x, y \in [0,1]$ . The only numbers satisfying $n^{2} = n$ are $0$ and $1$ . Thus, there will be $\boxed{2}$ corresponding unique values of $k$ such that the original conditions are satisfied.

Note that the operations +,- are not closed under [0,1]

And / will be undefined for y=0

From 1 and 2 , we get the two possible operations * and min(x,y)

The operation * gets $z^2xy$ = $z^kxy$ and thus k= 2

The other operation clearly satisfies 1 and 2

Note that since z is in [0,1], if $x \leq y$ then $zx \leq zy$ and thus satisfies 3 when k=1.

I was able to find 2 functions: the minimise function and the multiply function. Those give 1 and 2 respectively as k values, and those are 2 values of k.

Let $\otimes : A \rightarrow A$ where $A = [0,1]$ be a binary operation that satisfies (A) , (B), (C). Now lets try to find the possible values for $k$

From (C) we have that $x \otimes x = x^{k}$

Now lets assume $x < z$ , the case $x > z$ is analogous. Then there exists $y \in [0,1]$ such that $yz = x$

Substituting $x$ for $yz$ in $x\otimes z$ we get $(x\otimes z) = (yz\otimes z )$ (1)

Applying (C) and (A) to the right side of (1) we get

$(x\otimes z) = z^{k}(y\otimes 1 ) = z^{k}y$ , now replacing $y$ for $\frac{x}{z}$ we get

$(x\otimes z) =z^{k-1}x$ in this way $\otimes$ is defined as :

$x\otimes z = \left\{ \begin{array}{lr} x^{k-1}z , x > z\\ x^{k} , x = z\\ z^{k-1}x , z > x \end{array} \right.$

$\otimes$ is well-defined for $k> 0$ .

Lets check property (B) plugging in the values $x = \frac{1}{2}, y = \frac{1}{4}, z = \frac{1}{8}$ ; then we have

$\begin{aligned} \frac{1}{2}\otimes(\frac{1}{4}\otimes\frac{1}{8}) &=& (\frac{1}{2}\otimes\frac{1}{4})\otimes\frac{1}{8} (I)\\ \frac{1}{2}\otimes(\frac{1}{4^{k-1}}\cdot \frac{1}{8}) &=& (\frac{1}{2^{k-1}}\cdot\frac{1}{4})\otimes\frac{1}{8} (II)\\ \frac{1}{2}\otimes\frac{1}{2^{2k+1}} &=& \frac{1}{2^{k+1}}\otimes\frac{1}{2^{3}} (III) \end{aligned}$

Case $k\ge 2$ ,

$\begin{aligned} \frac{1}{2}\otimes\frac{1}{2^{2k+1}} &=& \frac{1}{2^{k+1}}\otimes\frac{1}{2^{3}}\\ \frac{1}{2^{k-1}}\cdot\frac{1}{2^{2k+1}} &=& \frac{1}{2^{k+1}}\cdot\frac{1}{2^{3k-3}}\\ \frac{1}{2^{3k}} &=& \frac{1}{2^{4k-2}} (IV)\\ \end{aligned}$

(IV) holds if and only if $k=2$

Case $k < 2$

$\begin{aligned} \frac{1}{2}\otimes\frac{1}{2^{2k+1}} &=& \frac{1}{2^{k+1}}\otimes\frac{1}{2^{3}}\\ \frac{1}{2^{k-1}}\cdot\frac{1}{2^{2k+1}} &=& \frac{1}{2^{(k+1)(k-1)+3}}\\ \frac{1}{2^{3k}} &=& \frac{1}{2^{k^{2}+2}} (V) \\ \end{aligned}$

(V) holds if and only if $k^{2}-3k+2 = 0$ which is true for $k =1$ and $k =2$ .

Finally there are two values for k such that there exist $\otimes$ that satisfies (A), (B) and (C).

$\otimes$ is as defined above, only replace the respective values of k .