Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see [here](https://math.stackexchange.com/q/381690/462). 



> 
> What would be an explicit example of a number $r$ with this property?
> 
> 
> 


Short of an explicit example, are there any references addressing this question? 


A natural approach would be to see that all irrationals in $C$ are transcendental, so it would suffice to take $r=\sqrt2$. But this is open, see [here](https://mathoverflow.net/q/114758/6085).




---


Many thanks for the answers. (It would be interesting to know whether $\sqrt2$ works.) 



A translation of the Cantor set contained in the irrationals
One way to obtain explicit examples, which combines the ideas of (weak forms of) randomness and base 3 expansions, is to use the fact that normality in a given base is preserved under rational addition, which was proved by D. D. Wall in his 1949 Berkeley PhD Dissertation. (I'm relying on D. Doty, J. H. Lutz, and S.Nandakumar [*Finite-state dimension and real arithmetic*, Information and Computation 205(11):1640-1651, 2007] for this reference.) Here a number $r$ is *normal in base $b$* if for any finite nonempty string $\sigma$ drawn from the alphabet $\{0,...,b-1\}$, the limiting frequency of the appearances of $\sigma$ as a substring of the base $b$ expansion of $r$ is $b^{-|\sigma|}$. Since elements of $C$ are not normal in base $3$, any number $r$ that *is* normal in base $3$ has the desired property. Examples of such numbers can be found at <http://en.wikipedia.org/wiki/Normal_number>, for instance.


In fact, normality is overkill. Let $r$ be disjunctive in base $3$, i.e., every finite ternary string appears as a substring of the ternary expansion of $r$ (which is both a comeager and a conull property). We need to show is that if $q$ is a positive rational then $r+q \notin C$. (Here addition is mod $1$.) If the ternary expansion of $q$ has infinitely many $1$'s then the fact that the ternary expansion of $r$ contains $0^n$ for all $n$ means that $q+r \notin C$. Otherwise, the fact that this expansion contains $0^m10^n$ for all $m,n$ does the trick.


