Basic math: x-y versus sqrt(x)-sqrt(y) First, sorry if this has been asked before. I searched and found lots of info on transformations and back-transformations, but none that offers insight into how you get into the forest and back out
My question: you have 16 - 10 = 6
I want to understand how to get back to that 6 after performing various statistical transforms. For example
$\sqrt{16}-\sqrt{10}\approx 0.837722$ . While each number can be back-transformed (e.g., $\sqrt{16}={4}^2=16$), the same does not hold for the subtracted value. I understand something is going on that changes the subtraction (as in the log transform, which I kinda understand), but I want to extend my understanding to the other transformations... if possible. 
As you can see, while my question is basic math, the application directly relates to statistics and transformations. I also know some folks don't like transformed results to be back-transformed, but that's a philosophical discussion for another day. Even if doing it is "considered wrong," I'd like to understand the "how" and maybe understanding that will let me arrive at the "why" by myself. Examples of transformed values that I'd like to understand include:
$\sqrt{16}- \sqrt{10} = x \overset{trans}{\rightarrow} 6 $
$\frac{1}{16}^2-\frac{1}{10}^2= x \overset{trans}{\rightarrow} 6$
$\frac{1}{16}^3-\frac{1}{10}^3= x \overset{trans}{\rightarrow} 6$
I'd appreciate any resources you can point me to, but please be patient: I'm a bit slow with math and resources full of overly general abstractions might be a bit much. Thank you!
 A: There's no general way to calculate $x-y$ from $\sqrt{x}-\sqrt{y}$ alone.
Two different $(x,y)$ pairs with the same $\sqrt{x}-\sqrt{y}$ can have different $x-y$ values.
For example, consider $x_1=9, y_1=4$, $x_2=64, y_2=49$ and $x_3=169, y_3=144$. Then $\sqrt{x_1}-\sqrt{y_1}=\sqrt{x_2}-\sqrt{y_2}=\sqrt{x_3}-\sqrt{y_3}=1$ but the $x-y$ differences are $5,\,15$ and $25$ respectively.
So for example if you see that the difference in square roots is $1$, you can't tell which of an infinite number of possible $x-y$ values produced it. 
If you knew $x$ or $y$ (or their square roots or something from which they could be computed) or if you knew $\sqrt{x}+\sqrt{y}$ (or any number of other quantities that would allow you to uniquely identify an $(x,y)$ pair), then you could compute $x-y$ readily enough, in essence it's solving a pair of (usually simple) simultaneous equations.
If, however, you know that your original values must be small non-negative integers (as the numbers in your question might suggest), then you may be able to tell what $x-y$ must be (at least some of the time). 
For example, in the case you discuss, where $\sqrt{x}-\sqrt{y}=4-\sqrt{10}\approx 0.83722$ the only pair of small integer solutions (and I expect -- but haven't yet attempted to prove it -- the only non-negative integer solutions at all) is $16,10$. 
However, even if that were the only integer solution there will be an infinity of solutions with difference arbitrarily close to $0.837722...$ (though there may be no nearby solutions that are not much larger than any feasible data values). For example  $\sqrt{3831}-\sqrt{3728}=0.837722...$ comes very close, but these may well be far more (perhaps even a hundred times larger) than a possible value you would expect to observe in some particular problem.
If you only have the difference in square roots stored it will be to some finite precision and then there may be an infinity of integer-pairs that would be indistinguishable (have the same difference of square roots to the available floating point accuracy) -- but you may be able to rule out all but one pair as not consistent with feasible data values.
A: Let $y = f(x)$. To uniquely determine $x$ given $y$, the function $f$ must be injective. A function $f$ is called injective if for any value $y$, there is at most one value $x$ such that $f(x) = y$.
Examples:


*

*On the set of positive real numbers, the function $\sqrt{x}$ is injective and hence you can "undo" it. 

*On the other hand, $f(x_1, x_2) = x_1 - x_2$ is not injective. Trivially, $5-4$ and $3 - 2$ both equal 1. If you're told the difference between two numbers is 1, there's no way to undo the transformation. There's no way to say what those two numbers originally were! The function $f(x_1, x_2) = x_1 - x_2$ is not injective.
To gain intuition, look at the pictures on the Wikipedia article for an injective function.
