# if $B$ is a boolean algebra and $a\neq b$ in $B$ there exist an ultrafilter containing $a$ but not $b$.

Suppose $$a$$ and $$b$$ are distinct elements in a boolean algebra $$B$$. I am trying to show there is an ultrafilter on $$B$$ containing $$a$$ but not $$b$$. Does this follow from the fact (is this a fact?) that $$a = \wedge \{F:F \text{ is an ultrafilter on B and } a\in F\}$$?