update notations and fix typos

pages/CSE442T/CSE442T_L10.md

@@ -98,7 +98,7 @@ $x_1\equiv x_2\mod N$
 
 So it's one-to-one.
 
-EOP
+QED
 
 Let $y\in \mathbb{Z}_N^*$, letting $x=y^d\mod N$, where $d\equiv e^{-1}\mod \phi(N)$
 
@@ -130,7 +130,7 @@ By RSA assumption
 
 The second equality follows because for any finite $D$ and bijection $f:D\to D$, sampling $y\in D$ directly is equivalent to sampling $x\gets D$, then computing $y=f(x)$.
 
-EOP
+QED
 
 #### Theorem If inverting RSA is hard, then factoring is hard.