Proving two functions are monotonically related I've been relying on stackexchange a lot lately but this is my first time asking a question. A lot of searching has yielded no answer so hopefully someone can help out.
I'm trying to find (and prove) whether two functions are monotonically related.
By monotonically related I mean that, for two functions, $f(x)$ and $g(x)$, if $f(x_i)<f(x_j)$ then $g(x_i)<g(x_j)$. In other words: $f(x_i)-f(x_j)$ must have the same sign as $g(x_i)-g(x_j)$ for all values of i and j.
To get more specific, my understanding of this is that two functions could be shown to be monotonically related by proving that the function that relates $f(x)$ and $g(x)$ is a monotonic function. I believe what is meant by this is that $f(x)=h(g(x))$ and $h$ has to be a monotonic function.
This is about as far as I can get. I have no idea how to find a function that relates two other functions let alone how to prove said function is a monotonic function. If anyone could help with either of these issues, either showing how to find a function that relates two other functions or more generally how to show that two functions are monotonically related, it would be greatly appreciated.
For reference, the functions I am looking at are the following three:
$$f(x)=\frac{(a-x)^2}{a(2a-x)(b-x)}$$
$$f(x)=\frac{(a-x)}{a(b-x)}$$
$$f(x)=\frac{(a-x)}{a(2a-x)}$$
where x, a and b are all integers greater than 0 and x<=a and x<=b
 A: There are a few questions in this post, so I'll just try to answer one of them. Hopefully I've interpreted it correctly. The quesiton: say $f$ and $g$ are monotonically related. Then there is a monotonic function $h$ such that $f = h∘g$, where $∘$ is function composition.
First, define monotonically related. The function $f$ if monotonically related to $g$ if $f(x) < f(y)$ implies $g(x) < g(y)$. The definition as I've given it may be strengthened to an equivalence instead of an implication but it's all we need for now. I.e. note that the order of $f$ and $g$ matters according to this definition.
Now, it is easy to see that $g(x) = g(y)$ implies $f(x) = f(y)$. Otherwise, $f(x) < f(y)$ or $f(y) < f(x)$, and in either case by our definition of monotonic relatedness then $g(x) \neq g(y)$. Call this result (1).
OK, now for any function we can always define a set valued "inverse". In other words $g^{-1}(y) = \{x | g(x) = y\}$.
Now, we can also define a function $f_s$ as follows. The domain of $f_s$ is the set of all sets whose image is the same under $f$. That is, $f_s$ takes as its argument a set $S_y = \{x | f(x) = y\}$. The value of $f_s(S_y) = y$.
Now, we can define our function $h$. Define $h = f_s∘g^{-1}$. To show that this definition is always possible, consider the result (1). This says that whenever $g(x) = g(y)$ then $f(x) = f(y)$. Now, for all values $u, v$ in the set $g^{-1}(y)$, we know by definition that $g(u) = g(v)$ and so by (1) $f(u) = f(v)$, so the set is a valid input argument for $f_s$ and the definition of $h$ makes sense.
Now, we can easily show that $h∘g = f$. We know by definition of $h$ that $h∘g(x) = f_s∘g^{-1}∘g(x)$. Now, we know that $x$ must be in the set $g^{-1}∘g(x)$, and we know that the value of $f_s$ is the value of $f$ for any value in that set, so after applying $f_s$ we get $f(x)$, as required.

Now that we've defined such a $h$, we need to show that it's monotonic. Well for this we need to strengthen the definition of monotonically related from the earlier one to an equivalence. So now we have also that $g(x) < g(y)$ implies $f(x) < f(y)$. I have a feeling this is the standard definition anyway, because the word "related" has no hint of direction in it.
Now, let's show the monotonicity of $h$. We note that the domain of $h$ is the range of $g$. So to show $h$ is monotonic we just need to show for some $g(x) < g(y)$ that $h(g(x)) < h(g(y))$. But by our definition of $h$, we know that this amounts to $f(x) < f(y)$, which follows directly from monotonic relatedness.

I'm not sure how useful this is. It looks like you're trying to show something specific, involving the particular functions you've posted. Hopefully it helped in some way.

EDIT: We can also prove the converse of the above. That is, given a monotonic $h$ such that $f = h∘g$ then $f$ and $g$ are monotonically related. To prove this, we need to show two things: that $f(x) < f(y)$ implies $g(x) < g(y)$ and vice versa.
So let's prove $f(x) < f(y)$ implies $g(x) < g(y)$. Well from our assumptions $f = h∘g$ and $f(x) < f(y)$, then we get $h(g(x)) < h(g(y))$. Now we know that $g(x) \neq g(y)$, otherwise $h(g(x)) < h(g(x))$, which is clearly false. So, assuming a total ordering, we have either $g(x) < g(y)$ or $g(x) > g(y)$. Well, by monotonicity of $h$, $g(x) > g(y)$ implies $h(g(x)) > h(g(y))$ which contradicts our earlier $h(g(x)) < h(g(y))$. So we're left to conclude $g(x) < g(y)$, finishing this case.
The second case is to show $g(x) < g(y)$ implies $f(x) < f(y)$. This is simpler. We want to show $f(x) < f(y)$, which by our earlier assumption is that same as $h(g(x)) < h(g(y))$. Now this follows directly from $g(x) < g(y)$ and the monotonicity of $h$.
A: From the comments here is another question that I'll address: given any $f$ and $g$, can we always define a $h$ such that $f = h∘g$? Then answer is no. We demonstrate this with a counterexample. Take $f(x) = x$ and $g(x) = x^2$. Notice that $f$ is invertible but strictly speaking $g$ is not. This is the key property underlying the counterexample.
Now, we assume towards a contradiction that there is some $h$ such that $f = h∘g$. Equivalently this is $f(x) = h(g(x))$ for all $x$. Then $f(1) = h(g(1))$ so $1 = h(1)$. But also $f(-1) = h(g(-1))$ so $-1 = h(-1^2) = h(1)$. So $h(1) = -1 = 1$, which is a contradiction of the fact that $-1 \neq 1$.
