Is there a function $f(A,B,C,D)$, using only $+$, $-$, $\times$, $\div$, that can indicate when $ABasically, there's this website where I can create some custom calculation by describing a function based on certain variables. However, its computation is very limited and I can only use basic operators ($+, -, \times, \div$) and numerical constants ($0,1,2$, etc.). It's a restricted company website, so I can't share it.
I have four input variables $A,B,C,D$ within the range $(-\infty, \infty)$. I need to create a function $f$ (using only the operators above) that outputs $E=f(A,B,C,D)$ such that I will know what values of $E$ will make the condition,
$$
A<B \text{ and } C<D
$$
true. $E$ should be a single decimal number that I can compare (e.g. if $E>=0$, then the condition is True, else False).
Is it possible to create such a function $f$?
As noted above, the function $f$ can only be defined using the operators $+,-,\times,\div$, and numerical constants. You can also use parentheses. For example, $(A+B)/(C-8.99-D)$, $A/C+9$, $A/D/C$, and $A-(B-4-(D/C))+C$ are valid functions. You're not allowed to use any other operators: absolute value $|A|$, exponents, logs, mod (%), logic comparators (AND, OR, XOR, etc.), normal comparators (<, >, <=, >=), IF statements, WHILE loops, creating other functions, MAX/MIN functions, SQRT(), basically any other stuff.
 A: The functions that you're able to create with these limited operations are precisely the rational functions over $\mathbb{Q}$ in variables $A,B,C,D$. Indeed, any rational function can be created with the operations $+,-,\times,\div$ and numerical constants, and you can prove that any function created this way is rational by strong induction on the number of operations used.
So, you want to know whether there exists a rational function $f : \mathbb{R}^4 \rightarrow \mathbb{R}$ such that $f(a,b,c,d)>0$ implies $a<b$ and $c<d$. (I have assumed for simplicity that we want a condition of the form $f>0$. A more general condition is described below.) We can simplify this question further by reducing it to two variables.
Indeed, this question is equivalent to asking if there exists a rational function $g : \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $g(x,y)>0$ implies $x,y>0$. Indeed, if such a $g$ exists, then $f(a,b,c,d) = g(b-a,d-c)$ satisfies the condition: if $g(b-a,d-c)>0$, then $b-a>0$ and $d-c>0$, so $a<b$ and $c<d$. Conversely, if $f$ exists, then we can take $g(x,y) = f(0,x,0,y)$, and $f(0,x,0,y)>0$ implies $x>0$ and $y>0$.
So: does such a rational function $g$ exist? I'm not sure, but I suspect the answer is no; I've tried and failed to come up with an example. Proving this false will probably be tricky, though, so I've asked a separate question on the site and will update this answer when I find out more.

What if we want to consider more general conditions than just $f>0$? Well, you want your function $f$ to be such that the value $f(a,b,c,d)$ can be used to infer whether the condition "$a<b$ and $c<d$" is true. That is, if the condition is true for $(a,b,c,d)$ but false for $(a',b',c',d')$, then we should never have $f(a,b,c,d)=f(a',b',c',d')$. If we let
$$
U = \{(a,b,c,d) \in \mathbb{R}^4 : a<b \text{ and } c<d\},
$$
then this is equivalent to saying that $f(U)$ and $f(\mathbb{R}^4 \backslash U)$ are disjoint. Then, we can use our same trick to reduce this to a function $g(x,y)$, and we're now asking whether $g(V)$ and $g(\mathbb{R}^2 \backslash V)$ are disjoint, where $V = \{(x,y) \in \mathbb{R}^2 : x,y>0\}$.
