Are there any other fraction-to-decimal conversions like $\frac{5}{2} = 2.5$? I noticed that to convert 2.5 to a fraction, it's the number after the decimal divided by the number before the decimal. I wondered if there are any others like these, and I cannot find any nor prove that they don't exist.
So the problem, explicitly, is: are there any pairs $(a,b)$ of positive integers, besides $(2,5)$ such that
$\frac{b}{a} = a + \frac{b}{10^{\lfloor\text{log}_{10}b\rfloor+1}}$
 A: I don't have a complete answer to the problem, but I have shown that $(2, 5)$ is the only coprime pair.

I'll write $d=10$ in the hopes of generalizing it to other $d\in\mathbb{Z}$ with $d\ge 2$.
Multiplying by $d^{\lfloor \log_{d}(b)\rfloor+1}a$ makes it
$$d^{\lfloor \log_{d}(b)\rfloor+1}b=a^2d^{\lfloor \log_{d}(b)\rfloor+1}+ab$$
This then means that $$d^{\lfloor\log_{d}(b)\rfloor+1}(b-a^2)=ab$$
and that we want to find solutions to $d^n(b-a^2)=ab$ where $n\ge 1$ and $d^{n-1}\le b < d^n$.
First, I'll show how to solve the more general equation $k(b-a^2)=ab$ with $k\ge 1$ and $a,b$ positive integers. The following isn't entirely original, as I use this (very useful!) tool a decent amount for this part.  Note that it must be that $a<k$ since otherwise $b=\frac{ka^2}{k-a}<0$. Let $a=k-X$ and $b=Y-2k^2$ with the new restriction that $0<X<k$ and $2k^2<Y$. The equation then simplifies to $kX^2-XY=-k^3\implies X(Y-kX)=k^3$. So $X$ is a small divisor of $k^3$ and $Y=\frac{k^3}{X}+kX$ (as long as $X\not=k$, $Y$ will be more than $2k^2$).
With this question, $k=d^n$. Let $X$ be a divisor of $d^{3n}$ with $0<X<d^n$, and let $Y=\frac{d^{3n}}{X}+d^nX$. Then let $a=d^n-X$ and $b=Y-2\cdot d^{2n}$.
From here, we can show that $(2, 5)$ is the only solution with $\gcd(a, b)=1$ for $d=10$. We want $\gcd\left(10^n-X, \frac{10^{3n}}{X}+10^nX-2\cdot10^{2n}\right)=1$. If $X$ was a multiple of $5$, then $10^n-X$ would also be a multiple of $5$, and the only way $b$ wouldn't be a multiple of $5$ is if $\frac{10^{3n}}{X}$ wasn't, which means that $X=5^{3n}2^m$ with $m\ge 0$, but since $X$ needs to be smaller than $10^n$, this wouldn't work. If $X$ was just $1$, then $b\ge 10^n$, so $X$ must be a power of $2$.  Moreover, it needs to be $2^{3n}$ since otherwise $b$ would be a multiple of $2$. With $X=2^{3n}$, $b=5^{3n}+10^{n}\cdot2^{3n}-2\cdot10^{2n}$ which must be smaller than $10^n$. This only happens when $n$ is less than approximately $1.17$, which in the positive integers, is only $n=1$. Then $n=1$ yields $a=2$ and $b=5$.
If we don't require $\gcd(a, b)=1$, we can continue. We want $d^{n-1}\le b < d^n$, and if we plug in $Y-2\cdot d^{2n}$ for $b$ and $\frac{d^{3n}}{X}+d^nX$ for $Y$ and simplify, we get the following inequality,
$$\left(d^{-1}+2\cdot d^{n}\right)X\le d^{2n}+X^2<\left(1+2\cdot d^{n}\right)X$$
along with the restriction of $0<X<d^n$. Letting $r=d^n$, this means that $X^2-\left(\frac{1}{d}+2r\right)X+r^2\ge0$ and $X^2-(1+2r)X+r^2<0$ and $0<X<r$. Solving these quadratic inequalities yields that $$X+\sqrt{\frac{X}{d}}\le r < X+\sqrt{X}$$ or equivalently $\frac{1+2r-\sqrt{4r+1}}{2}<X\le\frac{d^{-1}+2r-\sqrt{d^{-2}+4r\cdot d^{-1}}}{2}$. This condition and the fact that $X$ divides $d^{3n}=r^3$ determines what $X$ can be. Although $X=r$ would be a divisor of $r^3$, it wouldn't work since $\frac{d^{-1}+2r-\sqrt{d^{-2}+4r\cdot d^{-1}}}{2}<r$. I think (but am not sure!) that there'll be other solutions (and I might write a program to check if there are), but they will likely have very large values of $n$. For $d=10$, it also might be easier to generate $X$ of the form $2^m5^p$ and see if there's a valid $n$.
Edit: From $X+\sqrt{\frac{X}{d}}\le d^n < X+\sqrt{X}$, we can get that $\log_d\left(X+\sqrt{\frac{X}{d}}\right) \le n < \log_d(X+\sqrt{X})$. The difference between the bounds is less than $1$ for all $d, X>0$, so there is at most one integer in that range. Let $n=\lceil \log_d(X+\sqrt{X})\rceil-1$ so that it is the closest integer that satisfies the upper bound. We then want to see whether $\log_d\left(X+\sqrt{\frac{X}{d}}\right) \le n$ or equivalently $$X\le\frac{d^{-1}+2\cdot d^n-\sqrt{d^{-2}+4\cdot d^n\cdot d^{-1}}}{2}$$ under the restriction $X|d^{3n}$. In the case of $d=10$, this means that $X=2^m 5^p$ with $m,p\le 3n$.
