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\}$ ?