# way of converting an approximate rational number of radical combination into original form

Suppose that there is a radical combination $a\sqrt{b}+c\sqrt{d} ...$. where $a,b,c,d\in \mathbb{N}$, for which each term part $\sqrt{b}$ cannot be transformed into the form of $s\sqrt{q}$.

The question is,

1) Suppose that we convert this combination into a rational number approximation. Is there any quick way to know the number of terms that cannot or can be reduced to the form of $x\sqrt{z}$ in the original square root combination using an approximate value? (This would mean that an approximate value would be unique to a particular combination.)

Edit: For example, $12\sqrt{13} + 15\sqrt{17} + \sqrt{19}$. We do addition operation and convert it into a decimal approximation. Using the approximation value how would we be able to know the term that is not of form $x\sqrt{z}$?

2) What restrictions would be needed if there is no way to figure this out in the general case?

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It's quite hard to decipher what you mean. Are you just saying that the sum of the two radicals is not equal to some third number of the same form, i.e. $e\sqrt{f}$ where $e,f$ are positive integers? Without knowing what you mean in the first paragraph there's not much hope of answering the questions you put. – coffeemath Oct 25 '12 at 16:00
I guess the first sentence just wants to say that each number under a radical sign is square-free. – Marc van Leeuwen Aug 18 '13 at 17:39

This may not be what you want, but if there are three (or more) radicals, one can decide to drop one and get another number which is as close as you want to the original. For example take the case of $a=5\sqrt{2}+4\sqrt{3}+6\sqrt{7}.$
There are integers $x,y$ for which $x\sqrt{2}+y\sqrt{3}$ is as close as one wants to any real number. There may be an efficient way to find such $x,y$, but they exist, given any closeness we desire. [Perhaps a computer search here.]
So we may choose some particular $x,y$ for which $x\sqrt{2}+y\sqrt{3}$ is very close to $6\sqrt{7}$. Next we can replace the $6\sqrt{7}$ by the very near value $x\sqrt{2}+y\sqrt{3}$, so that $(5+x)\sqrt{2}+(4+y)\sqrt{3}$ is very close to the original number $a$, but only uses the two radicals $\sqrt{2}$ and $\sqrt{3}$.