Is a sum of equidistributed sequences $\sum \{n \alpha\}$ equidistributed? We have two equidistributed sequences {n a} and {n b} (mod 1), in which a and b are irrational.
1) Is it true that the sum {na} + {nb} is equidistributed?
and 
2) Is it true that {na} + {nb} = {n(a+b)} ? 
For (1), n*a and n*b are both polynomials satisfying Weyl's criteria, and so the sum of these is I think also a polynomial satisfying the criteria (but this does not necessarily mean {na} + {nb} is equidistributed). For (2), well, I just tried a few calculations, so I am not sanguine about it but it seems true. Note: I do see that a + b might not be irrational, and in that case we would not satisfy Weyl's criteria in (2). 
Thanks for any insights. 

(1): Not necessarily. 


*

*We could take $a$ to be any irrational number in $(0,1)$ and $b=1-a$.  In this case, for all $n>0$, $na+nb=n$ and $\{na\}+\{nb\}=1$.  

*If 1, $a$ and $b$ are linearly independent over $\Bbb Q$, then by the Weyl equidistribution criterion, the sequence of points $(\{na\},\{nb\})$ will be equidistributed in the unit square.  This means that $\{na\}+\{nb\}$ will not be equidistributed in the interval $[0,2]$ but will have a triangular probability density function, with maximum density at 1.
On the other hand, if $a=b$ and $a$ is irrational, then $\{na\}+\{nb\}=2\{na\}$ will be equidistributed in $[0,2]$, because $\{na\}$ is equidistributed in $[0,1]$.  Also,
it is possible for $\{na\}+\{nb\}$ to be equidistributed in a subinterval of $[0,2]$.  For example, if $a=\sqrt{2}$, $b=-1/\sqrt{2}$, set $z:=\{nb\}$; then $na=-2nb$, so 
$$\{na\}+\{nb\}=\{-2z\}+z=\left\{
\begin{array}{l}
1-z, \qquad z\in(0,\frac12],\\
2-z, \qquad z\in(\frac12,1].
\end{array}
\right.$$
Since $z$ is equidistributed in $[0,1]$, it follows that $\{na\}+\{nb\}$ is equidistributed in $[\frac12,\frac32]$.
(2): Both $\{\{na\}+\{nb\}\}$ and $\{n(a+b)\}$ are integer translates of $na+nb$ which are in $[0,1)$, so $\{\{na\}+\{nb\}\}=\{n(a+b)\}$.  However, $\{na\}+\{nb\}$ and $\{n(a+b)\}$ may differ by 1.  For example, if you take $a=1/\sqrt{2}$, $b=1/\sqrt{2}$ and $n=1$, then $\{na\}+\{nb\}=2/\sqrt{2}=\sqrt{2}$, but $\{n(a+b)\}=\{\sqrt{2}\}=\sqrt{2}-1$.
