If we have two rectangles and their dimensions, is there a mathematical test to simultaneously compare the two numbers. For examples, if we wanted to compare the area, we would compare the products, but that doesn't tell us if both dimensions are larger in one rectangle or just their product.
Another way to phrase this problem would be, given two pairs of numbers $(a,b)$ and $(p,q)$, is there a function $f(x,y)$ such that if $f(a,b)<f(p,q)$ then $a<p$ and $b<q$? That is, can we use a single comparison to determine whether the rectangle with dimensions $(a,b)$ can be contained in one with dimensions $(p,q)$?
It is impossible to determine whether $(a,b)$ is a smaller rectangle than $(p,q)$ using just a single comparison. Let us write $(a,b)<(p,q)$ whenever $a<p$ and $b<q$ and never any other time. Essentially, the problem we have is that, in the real numbers, the following is true:
However, we find that we have pairs like $(2,2)$ and $(1,3)$ where neither can be said to be greater. We will call such pairs incomparable. In a technical sense, we are saying that the real numbers are totally ordered and rectangles are only partially ordered. This can quickly be exploited to show the non-existence of such an $f$.
Let us consider the four rectangles $(4,1),\,(1,4),\,(2,2),\,(3,3)$. We can see that we need $f(1,4)=f(4,1)$ since, if they differed, our lemma above would say one was greater and either $(1,4)<(4,1)$ or $(4,1)<(1,4)$, neither of which is true. In fact, the only comparison that exists in our set of rectangles is $(2,2)=(3,3)$, but to get the rest to be incomparable we need $f(4,1)=f(2,2)$ and $f(4,1)=f(3,3)$ implying that $f$ is constant on this set - meaning it cannot be used to compare anything.
In fact, in more generality, we can prove that to satisfy the condition $f(a,b)<f(p,q)$ implies $(a,b)<(p,q)$, we need to set $f$ to a constant, in which case it's not very useful for comparing anything.