Question:
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Let $A\in\mathbb{R}^{n \times n}$ be an orthogonal matrix and let $\varepsilon>0$. Then does there exist a rational orthogonal matrix $B\in\mathbb{R}^{n \times n}$ such that $\|A-B\|<\varepsilon$?


Definitions:
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* A matrix $A\in\mathbb{R}^{n \times n}$ is an *orthogonal matrix* if $A^T=A^{-1}$
* A matrix $A\in\mathbb{R}^{n \times n}$ is a *rational matrix* if every entry of it is rational.



Existence of Rational Orthogonal Matrices
Yes. It is a theorem of Cayley that the mapping $S \rightarrow (S-I)^{-1}(S+1)$ gives a correspondence between the set of $n\times n$ skew-symmetric matrices over $\mathbb{Q}$ and the set of $n\times n$ orthogonal matrices which do not have one as an eigenvalue. Since the mapping is nice, and rational skew-symmetric matrices are dense in the set of all skew-symmetric matrices, you have your result. For more, see [the very nice paper by Liebeck and Osborne](http://www.math.upenn.edu/~pemantle/Summer2007/Library/liebeck-osborne.pdf)


