Update Modular.mdx

content/4_Gold/Modular.mdx

@@ -286,8 +286,21 @@ assert 2 * x % MOD == 1
 
 We can also find modular inverses through Euclidean Division.
 Given the prime modulus $m > a$ we have:
-$m = k \cdot a + r$, where k = $floor(\frac{m}{a})$	and $r = m \mod a$. Then: $0 = k \cdot a + r \mod m \iff r = -k \cdot a \mod m
-\iff r \cdot a^{-1} = -k \mod m \iff a^{-1} = -k \cdot r^{-1} \mod m$.
+
+$$
+m = k \cdot a + r
+$$
+
+where $k = \lfloor \frac{m}{a} \rfloor$ and $r = m \mod a$. Then:
+
+$$
+\begin{align*}
+         & 0 = k \cdot a + r \mod m        \\
+  \iff{} & r = -k \cdot a \mod m           \\
+  \iff{} & r \cdot a^{-1} = -k \mod m      \\
+  \iff{} & a^{-1} = -k \cdot r^{-1} \mod m
+\end{align*}
+$$
 
 Here is a short recursive implementation of the above formula: