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Clearly the sum of two irrational numbers is not necessarily irrational. But is it true that it is 'almost always' irrational, in the sense that $$\displaystyle\lim_{x\to\infty}\dfrac{\lambda(P\cap B(x))}{\lambda(R\cap B(x))}=1$$ where $\lambda$ is Lebesgue measure, $R\subset \mathbb{R}^2$ is the set of points with irrational coordinates and $P\subset R$ is the set of those points the sum of whose coordinates is irrational (and $B(x)$ the disc of radius $x$ centred at the origin)? And I guess the same question applies to transcendental numbers. Intuitively it seems true, but I don't know how one would prove this. If not, then there is the question of whether the limit exists and if it does what is it.

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Let $NP$ be the set of pairs whose sum is rational. I think its easier to prove $\lambda(NP\cap B(x))=0$. In fact since $NP\cap B(x)\subset NP$, we just prove $\lambda(NP)=0$ and we are done. Let $$NP_x=\{(x,y)|x+y\in \mathbb{Q}\}$$ Notice that the restriction addition to this set is translation by $x$ which is measure preserving, hence the inverse image of the rational numbers has measure zero. However, there is a weak form of Fubini's theorem that says that if a subset of a product measure space ( which $\mathbb{R}^2$ is) has the property that its intersection with each slice has measure zero then the set has measure zero. Hence $\lambda(NP)=0$.

To bring this back to the specific question you are asking, $NP\cup P=\mathbb{R}^2$, so for any open ball $B(x)$ $\lambda(P\cap B(x))=1$. On the other hand the set of points whose coordinates are irrational is the complement of a set of measure zero, so $\lambda(R\cap B(x))=1$. Hence you are taking the limit of $1/1$.

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  • $\begingroup$ Nice answer (I suppose you mean area $\pi x^2$ of $B(x$) as opposed to 1?). And I suppose the same argument would go through for transcendentals because $\overline{\mathbb{Q}}$ is countable too. $\endgroup$ – AlephNull Jan 6 '19 at 15:03
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    $\begingroup$ Fubini's theorem is a bit overkill - if you can prove that a single line has measure zero (not so hard by explicit construction of covers by open balls), then this is a countable union of lines, so has measure zero. $\endgroup$ – Milo Brandt Jan 6 '19 at 16:49
  • $\begingroup$ I guess the answer is YES? (notice that the question was actually answerable by a simple Yes/No... I read this answer and I think it says Yes but I am sincerely not sure) $\endgroup$ – Stijn de Witt Jan 6 '19 at 23:00

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