If $A \subseteq B'$, then $A \cap B = \emptyset$
subset def... $[x: x \in A \rightarrow x \notin B]$ since nothing belongs in B... then A is a subset of a B' which means that while $ x \in A$, $x \notin B$.. so $A \cap B$ would mean that there are elements inside $A$ and $B$, but since $B$ for the subset isn't there...as in x doesn't belong to B in the subset... then there's an emptyset for $A \cap B$... ?!
Let $s$ be a complex number with a strictly positive real part ($Re(s)>0$).
Let $f(s)=\sum_{a^2+b^2>0} \dfrac{(-1)^{a^2+b^2}}{(a^2+b^2)^s}$
where the sum runs over all positive integers $a,b$ such that $a^2+b^2>0$ and allowing multiplicity (as e.g. $x^2+y^2=q^2+s^2$ ).
Notice $a^2+b^2$ is a n...
