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Suppose that there is a radical combination $a\sqrt{b}+c\sqrt{d}$ where a,b,c,d are natural. 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 $x\sqrt{z}$ - in this case $\sqrt{19}$?

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

Note: i lost my unregistered account - so unable to edit, so I posted this again as a separate question. Can anyone close the first one?

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1 Answer 1

Given $n$ different squarefree numbers $a_1,\ldots, a_n\in \mathbb N$ and $n$ coefficients $c_1,\ldots , c_n\in\mathbb N$ and a rational approximation $$\sum_{i=1}^n c_i\sqrt{a_i}\approx q\in\mathbb Q$$ determine the number of $i$ such that $c_i=1$. Of course this requires some specification of the meaning of "$\approx$", say we have instead $$p<\sum_{i=1}^n c_i\sqrt{a_i}<q$$ with $p,q\in\mathbb Q$. One could solve this problem by trial and error: There are only $\lfloor q^2\rfloor$ candidates for the number $c_1^2a_1$, which may or may not be $>p$. For each combination of $c_1, \ldots, c_k$, and $a_1,\ldots, a_k$ there are again only finitely many candidates $c_{k+1}^2a_{k+1}^2$ such that the $(k+1)$th partial sum is still $<q$ (and also such that $a_{k+1}\notin\{a_1,\ldots,a_k\}$). Since $n\le q$, this procedure will terminate.


If $n$ is not a square and $1\le k<\sqrt n < k+1$, then $\sqrt n$ cannot be too close to $k$. Indeed, $$\left(k+\frac1{2k+1}\right)^2=k^2+\frac{2k(2k+1)+1}{(2k+1)^2}<k^2+1$$ implies $\sqrt n >k+\frac1{2k+1}$. Thus an irrational square root cannot be too close to an integer. however, with several summands this difference may get smaller.

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