Is there a simple closed form of $|\alpha(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor) + \beta(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor)|$? Let $d_n(x)$ denote the $n$'th digit after the decimal point in $x$. 
Examples:
$d_8(e) = 2,\;d_5(\pi) = 9$
$\alpha(x)$ and $\beta(x)$ are defined this way:
$$d_n(\alpha(x))  = \left\{
\begin{array}{c l}     
    0 & \;d_n(x)\leq5\\
    d_n(x)-5 & \text{otherwise}
\end{array}
\right.$$
$$d_n(\beta(x))  = \left\{
\begin{array}{c l}     
    0 & \;d_n(x)\geq5\\
    5-d_n(x) & \text{otherwise}
\end{array}
\right.$$
Now consider$|\alpha(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor) - \beta(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor)|$, which yield numbers in the form of
$a\sqrt{b} - \frac{c}{9}$ or
$\frac{d}{9} - e \sqrt{f}$, where $a,b,c,d,e,f,n$ all are integers greater than $0$.
Or , in more detail: $$0<\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor < \frac{1}{2} \implies \sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor = \frac{d}{9} - e \sqrt{f}$$
$$\frac{1}{2}<\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor \implies \sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor = a\sqrt{b} - \frac{c}{9}$$
$$\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor = 0 \implies \sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor = \frac{d}{9} - e \sqrt{n}$$
The question:
Does $|\alpha(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor) + \beta(\sqrt{n}-\left\lfloor \sqrt{n} \right \rfloor)|$ have similary simple forms?

Please excuse any misuse of special characters, as my knowledge of what special characters usually denote is lacking. I simply used $\alpha$ and $\beta$ as I started out using $a$ and $b$ in their place.
 A: Just to make sure everything's formal and nice, we should probably  state somewhere that $\alpha$ and $\beta$ have no digits to the left of the decimal point - we could extend them somehow, but for simplicity, let's just consider them maps $(0,1)\rightarrow(0,1)$ - so we're staying in the unit interval. Also, since some rationals do not have a unique decimal representation, we'll only work with irrational numbers.
The first thing to note is that, as I think you are trying to get at in your post, for any $x\in(0,1)$ the difference $\alpha(x)-\beta(x)=x-\frac{5}9$. This is fairly easy to prove - in particular, note that, if $S$ is the set of positions $n$ such that $d_n(x)<5$ and $S^{C}$ is the complement thereof, we get
$$\alpha(x)=\sum_{n\in S}10^{-n}(d_n(x)-5)$$
$$\beta(x)=\sum_{n\in S^C}10^{-n}(5-d_n(x))$$
which is just summing up the decimal expansion of each. This means that
$$\alpha(x)-\beta(x)=\left(\sum_{n\in S}10^{-n}(d_n(x)-5)\right)+\left(\sum_{n\in S^C}10^{-n}(d_n(x)-5)\right)$$
where we have negated the expression for $\beta$ by switching $(5-d_n(x))$ to $(d_n(x)-5)$. However, the same term is being summed in both sides, so really, we can take the same sum over $S\cup S^{C}=\mathbb{N}$ - this gives
$$\alpha(x)-\beta(x)=\sum_{n=1}^{\infty}10^{-n}(d_n(x)-5)$$
$$\alpha(x)-\beta(x)=\left(\sum_{n=1}^{\infty}10^{-n}d_n(x)\right)-\left(\sum_{n=1}^{\infty}5\cdot 10^{-n}\right)$$
where the sum on the left is clearly just $x$, being the decimal expansion therefore, and the sum on the right is a geometric series and converges to $\frac{5}9$, leaving
$$\alpha(x)-\beta(x)=x-\frac{5}9.$$
Notice that this applies to every $x$ - there's nothing special about the numbers $\sqrt{n}-\lfloor\sqrt{n}\rfloor$ that makes it easier to write your difference. If you can write $x$ easily, you can write $\alpha(x)-\beta(x)$ equally easily. (And obviously taking the absolute value hardly changes anything).
The expression $\alpha(x)+\beta(x)$, however, we would expect to be much harder to write out explicitly, except by computing it. In particular, notice that $\alpha(x)-\beta(x)$ is easy to write - meaning that, if $\alpha(x)+\beta(x)$ were easy to write, so must their difference, which is $2\beta(x)$. But it is somewhat absurd to think that $\beta(x)$ alone would have a nice closed form! It would require deep results about the digits of $x$ to create a closed form, and mathematics simply doesn't have the tools to say much about the digits of numbers even as simple as $\sqrt{n}$. We don't even know if its digits behave somewhat randomly or not. Moreover, if every digit of $\sqrt{n}$ occurs infinitely often for any non-square $n$ - which is generally believed to be true - the value $\beta(\sqrt{n}-\lfloor\sqrt{n}\rfloor)$, which has only digits $0$ through $5$ could not possibly be expressed as a square root of any other number plus a rational, implying that, since $\alpha+\beta$ would have no nice form.
Perhaps the only thing we can easily say is that, if we wrote $\alpha(x)-\beta(x)$ in base $10$, but using digits $-5$ through $4$ instead of the usual digits (meaning that the number $15$ would be represented, for instance, as the two digits $2,-5$ in this base system, since $2\cdot 10^1 + (-5)\cdot 10^0=15$. Check out this page on Wikipedia for a similar base system), then $\alpha(x)+\beta(x)$, in typical decimal, is just $\alpha(x)-\beta(x)$ in the modified decimal, where we ignore the negative signs in the digits. This is because, the digits from $-\beta(x)$ turn out to be $d_n(x)-5$, which are the digits $-5$ through $0$ and the digits from $\alpha(x)$ are those from $0$ to $4$ - so we're basically ensuring that we never have to carry. For instance, if
$$\alpha(x)-\beta(x)=.3572\ldots$$
then, in our modified decimal system, we would write it out as
$$\alpha(x)-\beta(x)=.4,-4,-3,2,\ldots$$
which would mean that $x$ is just those digits, plus $5$, so $x=.9127\ldots$.
(check it for yourself that $.3572 = 4\cdot 10^{-1}+(-4)\cdot 10^{-2}+(-3)\cdot 10^{-3}+2\cdot 10^{-4}$, if you doubt my weird-base-conversion-skills). Then, $\alpha(x)+\beta(x)$ is just the freaky decimal conversion without its signs, and is hence
$$\alpha(x)+\beta(x)=.4432\ldots$$
This gives, at the least, a procedure for obtaining $\alpha(x)+\beta(x)$ (and also for obtaining $x$) given knowledge of $\alpha(x)-\beta(x)$, but I doubt any better relations can be found.
If you have questions, I'd be happy to answer them; I can understand that my techniques may be confusing, since they are very much non-standard.
