Prove that sets don't intersect I am trying to solve a computer algorithm problem that boils down to solving the following. I would appreciate some mathematician assistance on the proof. So here goes:
Having:
Set $S$ - rational numbers that can be represented as $n/d$, where $d$ is such that $\lg_2 d \in \Bbb Z$, and $n \in \Bbb Z$ 
Set $A$ – subset of $S$ where $\lg_4d \in\Bbb Z$ (Ex.: $n/4, n/16, n/64$)
Set $B$ - the rest of numbers from set $S$, that are not in $A$ (Ex.: $n/2, n/8, n/32$)
Set $A'$ - set where every member is $l + m$, where $l\in A, m \in A$
Set $B'$ - set where every member is $j + k$, where $j\in B$, $k\in S$ (either in $A$ or in $B$)
Prove (or disprove) that $A'$ and $B'$ don't intersect.
 A: Answer to the question before edit
That is, when only $n=1$ is considered.
First, $S$ is the sets of numbers $2^n$ with $n\in\Bbb Z$.
$A$ is the subset of numbers $4^n=2^{2n}$ for $n\in\Bbb Z$, hence $B=S\backslash A$ is the set of numbers $2^{2n+1}$ for $n\in \Bbb Z$.
$A'=A+A$, so these are numbers $2^{2p}+2^{2q}$, with $(p,q)\in \Bbb Z^2$. Likewise, numbers in $B'=B+S$, are of the form $2^{2p+1}+2^{q}$.
Numbers in your sets all have a unique binary representation (they have only one or two binary digits, or bits).
Now, a number in $A$ has either two even bits (by even, I mean an even power of $2$ in the binary representation of the number), either one odd bit (if $p=q$).
A number in $B$ has likewise one or two bits. Either $q=2p+1$ and it has one even bit, either $q\neq 2p+1$ and it has one odd bit and another arbitrary (even or odd) bit.
Conclude.

Answer to the edited question
You have to assume that $\gcd(n,d)=1$ when $n\neq 0$ (and $0\in S$, of course, with $n=0$), otherwise $A=S$. Thus, either $n$ is $0$ or odd. Notice also that $S$ is the set of numbers that have a finite binary representation (they are all rationals with only powers of $2$ in denominator). Any such rational has a unique representation $n 2^p$ with odd $n$, when it's not $0$.
Then numbers in $A$ are of the form $n 2^{2p}$ with odd $n$, or $0$. And $B$ are numbers of the form $n 2^{2p+1}$, with odd $n$ (and $0\notin B$ since $B=S\backslash A$).
Then $9$ and $7$ are in $A$ (with $d=1$), hence $16\in A'$.
But $2\in B$ and of course $14\in S$, so $16\in B'$.
Of course there are many other counterexamples. Are $A'$ and $B'$ equal?
Actually, since $0 \in A$, you have also $A \subset A'$. And since $n 2^{2p}+n2^{2p}=n2^{2p+1}$, you have also $B\subset A'$, thus $A'=S$.
On the other hand, for any number in $x\in S$, you have also $x-2 \in S$, ans since $2 \in B$, that means $x=2+(x-2) \in B'$. Hence $S \subset B'$, and $B'=S$.
So $A'=B'=S$.
