I read a little proposition that any quantity obtainable from rational numbers by a finite number of operations $+$, $-$, $\cdot$, $\div$, $\sqrt{{}}$ can be put in a standard form $\frac{p}{q}\cdot A$, where $\frac{p}{q}\in\mathbb{Q}$ and $A$ is an expression of only integers and $+$, $-$, $\cdot$, and $\sqrt{{}}$.
I don't think it's difficult to believe this is the case, but I don't know of a nice, clear way to prove it.
I tried to simplify the argument first by saying that any operation $+$, $-$, $\cdot$, $\div$ can be rewritten entirely in terms of $+$, since $-$ is the addition of a the additive inverse, $\cdot$ is repeated addition, and $\div$ is multiplication of the reciprocal.
I also decided that given a radicand consisting of rationals and operations $+$, $-$, $\cdot$, $\div$, you could then rewrite the radicand as a sum of rationals, and multiply the radical by $\frac{\sqrt{\Delta^2}}{\Delta}$, where $\Delta$ is some multiple of the denominators. This would put the radicand in standard form, with a fraction $\frac{1}{\Delta}$ outside the radical. For example, $ \sqrt{\frac{a}{b}+\frac{c}{d}}=\sqrt{\frac{ad+cb}{bd}}=\frac{\sqrt{(bd)^2}}{bd}\sqrt{\frac{ad+cb}{bd}}=\frac{1}{bd}\sqrt{(bd)(ad+cb)}. $
I suppose for nested radicals, the process could be repeated. Then any sequence of these operations can be put into a sum of terms with rational denominators by rationalizing any denominator, and then one could do the same procedure of pulling a common multiple of all the terms out in front to put the expression in standard form.
However, I wanted to formalize this idea, but it seems like there are so many different things to consider, that I don't know how to go about it. Is there a way to make basic induction work, or perhaps some other method? Thanks.