I'll consider the problem of computing $x^\frac1{q}, \; q > 0$; as I've already mentioned in the comments, one can decompose any positive rational number as $m+\dfrac{p}{q}$, where $m,p$ are nonnegative integers, $q$ is a positive integer, and $p < q$. Thus for computing $x^{m+\frac{p}{q}}$, one could use binary exponentiation on $x^m$ and $\left(x^\frac1{q}\right)^p$ and multiply the results accordingly.
A.N. Khovanskiĭ, in his book on continued fractions, displays a continued fraction representation for the binomial function:
$$(1+z)^\alpha=1+\cfrac{2\alpha z}{2+(1-\alpha)z-\cfrac{(1-\alpha^2)z^2}{3(z+2)-\cfrac{(4-\alpha^2)z^2}{5(z+2)-\cfrac{(9-\alpha^2)z^2}{7(z+2)-\cdots}}}}$$
which converges for $|\arg(z+1)| < \pi$.
Letting $z=x-1$ and $\alpha=\dfrac1{q}$, one can then evaluate this continued fraction (with, say, Lentz-Thompson-Barnett) to generate a "seed" that can be subsequently polished with Newton-Raphson, Halley, or any of a number of iterations with high-order convergence. You'll have to experiment with how accurate a seed you need to start up the iteration, by picking a not-too-small tolerance when evaluating the continued fraction.
Here's some Mathematica code demonstrating what I've been saying earlier, for computing $\sqrt[3]{55}$:
With[{q = 3, t = 55, prec = 30},
y = N[2 + (1 - 1/q) (t - 1), prec];
c = y; d = 0; k = 1;
While[True,
u = (k^2 - q^-2) (t - 1)^2; v = (2 k + 1) (t + 1);
c = v - u/c; d = 1/(v - u d);
h = c*d; y *= h;
If[Abs[h - 1] <= 10^-4, Break[]];
k++];
FixedPoint[
Function[x, x ((1 + q) t - x^q (1 - q))/(x^q (1 + q) - (1 - q) t)],
1 + 2 (t - 1)/q/y]]
Here, I've arbitrarily chosen to stop when the continued fraction has already converged to $\approx 4$ digits, and then polished the result with Halley's method. The result here is good to $\approx 28$ digits. Again, you'll have to experiment on the accuracy versus expense of evaluating the "seed", as well as picking the appropriate iteration method for polishing the seed.
When I learned division in elementary school, I learned "remainder" at the same time. I think it is mostly terminological that this is not called an "operation", because the division algorithm produces both the whole number quotient and the remainder of division of two natural numbers, at the same time.
On the other hand, when we move to the rationals, "remainder of division" is no longer a very interesting operation, because the rationals are a field. Students are taught to stop using the division algorithm and start using a different algorithm to divide fractions. This is perhaps a reason that the remainder operation is de-emphasized. But students are certainly still able to compute remainders if they are asked to; they just don't describe it as an "arithmetical operation".
Best Answer
Let $$f_1(x)=x+c_1$$ $$f_2(x)=x\cdot c_2$$ where $c_1,c_2$ are some constants, and $f_1$ represents addition and subtraction, and $f_2$ represents multiplication and division. Functions $f_1$ and $f_2$ have inverse function, but $\lfloor{x}\rfloor$ does not have an inverse function. So, if we assume it is possible to find $\lfloor{x}\rfloor$ from $x$ using $f_1$ and $f_2$, then it is possible to find $x$ from $\lfloor{x}\rfloor$. Contradiction.