The Cantor Set and Triadic Expansion let $K$ be the Cantor set. 
I say that a number $x$ in $[0,1]$ is triadic if 
$x=\frac{m}{3^n}$ for some nonnegative integers $m, n$.
Let $z$ be a triadic number in $[0,1]$. Do there exist two triadic numbers 
$x, y$ in K such that $z= x + y$?
I am pretty sure that the answer is yes, but I could find a proof.
Thank you very much in advance for your help.
 A: It is a standard fact that the sum of two cantor sets is the interval $[0,2]$ (googleize it) see for instance
http://www.cut-the-knot.org/do_you_know/cantor5.shtml
Therefore, any number $z\in[0,2]$ writes as $x+y$ with $x,y\in K$.
Now, $z$ is triadic if and only if its develop in base $3$ is finite: $z=0,a_1\dots a_k$. (or of the type $0,a_1\dots a_k2222222\dots$)
A number is in $K$ if and only if it develops without $1$'s.
Suppose $z=x+y$ with $x,y\in K$ not triadic. Then the develop of both $x$ and $y$ must be infinite and their tails must sum to a triadic number. More precisely, we can write
$x=0,b_1\dots b_n x_{n+1}x_{n+2}\dots$
$y=0,c_1\dots c_n y_{n+1}y_{n+2}\dots$
with $0,x_{n+1}x_{n+2}\dots + 0,y_{n+1}y_{n+2}\dots$ is $1$ or $2$.
Moreover, we can choose $n$ so that $b_n=0$ (otherwise the tail of $x$ is $0,\dots02222222\dots$ and $x$ is triadic whence $y$ is and we are done)
Now $z=x+y=(x+0,0\dots0 y_{n+1}y_{n+2}\dots) + 0,c_1\dots c_n00000\dots$
Set $X=x+0,0\dots0 y_{n+1}y_{n+2}\dots$ and $Y= 0,c_1\dots c_n00000\dots$
Since $b_n=0$ we have that $X$ is triadic with all digit $0$ or $2$ except the last one (the $n^{th}$) that could be $1$.
But $0,\dots01000000\dots=0,0\dots002222222222222$ therefore $X$ is in the cantor set. $Y$ is clearly triadic and in the cantor set.
By contsruction $z=X+Y$
