# Is the sum of two irrational numbers almost always irrational?

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.

• – mrtaurho Jan 6 '19 at 14:45
• I think you mean for your denominator to be the measure of $B(x)$ not of $R\cap B(x)$. – Milo Brandt Jan 6 '19 at 16:46

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 $$1/1$$.
• 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. – AlephNull Jan 6 '19 at 15:03