modified ds 8 for pandoc compatibility

Author: hi-im-buggy

_posts/2020-10-08-ds-lecture-8.md

@@ -28,7 +28,7 @@ Given an equivalence relation $R$ on a set $A$, the *equivalence class* of an ar
 The set of all equivalence classes of $A$ with respect to $R$ form a *partition* of $A$. A partition of a set $A$ is a collection of disjoint subsets $A_{i}$ (called *parts*) of $A$ whose union is $A$, *i.e.*, mutually exclusive and exhaustive.
 
 Also, given a partition $P = \{A_{1}, A_{2},...,A_{k}\}$ of $A$, an equivalence relation can be defined on A as 
-<div>$$R = \{(x,y) \in A \times A | x,y \in  $A_{i}\}$$</div>
+<div>$$R = \{(x,y) \in A \times A | x,y \in  A_{i}\}$$</div>
 
 Thus, defining an equivalence relation on a set is precisely the same as partitioning it, since the equivalence classes form a partition, and any partition gives rise to an equivalence relation.
 
@@ -65,11 +65,11 @@ Since $k$ ranges from 0 to $n$, we obtain the following recursion:
 
 <div>
 $$
-\begin{align}
+\begin{aligned}
 B_{n+1} & = \sum_{k = 0}^{n} { {n} \choose {k} }{B_{n-k} } \\
 		& = \sum_{k=0}^{n} { {n} \choose {n-k} }{B_{k} } \\
 		& = \sum_{k=0}^{n} { {n} \choose {k} }{B_{k} }
-\end{align}
+\end{aligned}
 $$
 </div>