# “Consistent multiplication” puzzle

Here is the puzzle. A tuple of five numbers $\langle a,b, c, d, Z\rangle$ is called consistent if:

1. The numbers $a, b,c,d, Z$ are all between 0 and 1, inclusive.
2. $a\times b = Z = c \times d$.

We can compare tuples, too: one tuple is smaller than another if each of its entries is less than or equal to the corresponding entry in the other tuple: $a_1 \leq a_2$, $b_1 \leq b_2$, etc. Naturally in that case, the other tuple is called bigger than the first.

The problem I'm trying to solve is this: Given two consistent tuples $T_1$ and $T_2$, find the smallest consistent tuple $T$ that is bigger than both $T_1$ and $T_2$, or prove that there isn't a unique one. I'm looking for a generic rule that I can apply for any pair of consistent tuples.

So far, I've noted that there's always at least one consistent tuple bigger than both $T_1$ and $T_2$, namely $\langle 1,1,1,1,1\rangle$, so the only problem is whether there is a unique smallest consistent tuple.

I also note that an assignment like $a \equiv \max(a_1,a_2)$ and $b \equiv \max(b_1, b_2)$ and $Z = \max(a_1,a_2)\times \max(b_1, b_2)$ works for $a, b, Z$ but might not be extensible to $c$ and $d$, since we have no guarantee that $\max(a_1,a_2)\times \max(b_1,b_2) = \max(c_1,c_2) \times \max(d_1,d_2)$ and in fact that will hardly ever be the case. The question is what to try next.

This is not possible in general. In particular, suppose we fix some $T_1=\langle a_1,b_1,c_1,d_1,Z_1\rangle$ and $T_2=\langle a_2,b_2,c_2,d_2,Z_2\rangle$. We will show that it is usually the case that there are two distinct $T_3,T'_3$ which are both examples of minimal tuples greater than both $T_1$ and $T_2$.
More specifically, define $A=\max(a_1,a_2)$ and $B=\max(b_1,b_2)$ and $C=\max(c_1,c_2)$ and $D=\max(d_1,d_2)$. Unless $AB=CD$, there is not a unique tuple. Let $Z_3=\max(AB,CD)$. Define: $$T_3=\left\langle A,\frac{Z_3}A,C,\frac{Z_3}C,Z_3 \right\rangle$$ $$T'_3=\left\langle\frac{Z_3}B,B,\frac{Z_3}D,D,Z_3 \right\rangle$$ Given that $Z_3$ is as small as it could possibly be, there are no smaller tuples than either of these that are greater than both $T_1$ and $T_2$. Algebraically, it is easy to see that both these tuples are indeed greater than $T_1$ and $T_2$. However, the only way they could be equal would be if $AB=CD$. Thus, to see that such elements do not exist in general, one may choose any example where this condition fails.