diff --git a/q2/b.md b/q2/b.md
index 08309f8..1862b1d 100644
--- a/q2/b.md
+++ b/q2/b.md
@@ -1,10 +1,3 @@
-## Question 2
-
-> A relation $\star$ is defined on the set $\mathbb{Q}^{2}$ by
-> 	$\left(x_{1},x_{2}\right)\star\left(y_{1},y_{2}\right)$ if and only if $x_{1}y_{2}=x_{2}y_{1}$.
->
-> For each of the questions below, be sure to provide a proof supporting your answer.
-
 ### Part B
 
 > Is $\star$ symmetric?
diff --git a/q2/c.md b/q2/c.md
index 00d7bdd..6cfd16f 100644
--- a/q2/c.md
+++ b/q2/c.md
@@ -1 +1,7 @@
-WIP
+### Part C
+
+> Is $\star$ anti-symmetric?
+
+If $\star$ is anti-symmetric, then for every distinct pair of $\left(x_{1},x_{2}\right)$ and $\left(y_{1},y_{2}\right)$ in $\Q^2$ such that $\left(x_{1},x_{2}\right)\star\left(y_{1},y_{2}\right)$ and $\left(y_{1},y_{2}\right)\star\left(x_{1},x_{2}\right)$, $x_1=y_1$ and $x_2=y_2$.
+
+Given that $x_1 y_2 = x_2 y_1$ is known to be true, we can rearrange for $y_1$ to find that $y_1=\frac{x_1 y_2}{x_2} (2)$. In a similar manner, rearranging for $y_2$ provides $y_2=\frac{x_2 y_1}{x_1} (3)$. Substituting $(2)$ and $(3)$ into $(1)$, it can be found that $x_2=y_2$ and $x_1 = y_1$. Therefore, the relation is anti-symmetric.
diff --git a/q2/d.md b/q2/d.md
index 00d7bdd..6ebfc47 100644
--- a/q2/d.md
+++ b/q2/d.md
@@ -1 +1,18 @@
-WIP
+## Question 2
+
+> A relation $\star$ is defined on the set $\mathbb{Q}^{2}$ by
+> 	$\left(x_{1},x_{2}\right)\star\left(y_{1},y_{2}\right)$ if and only if $x_{1}y_{2}=x_{2}y_{1}$.
+>
+> For each of the questions below, be sure to provide a proof supporting your answer.
+
+### Part D
+
+> Is $\star$ transitive?
+
+For $\star$ to be transitive, then for any set of 3 elements $(x_1,x_2)$, $(y_1,y_2)$, and $(z_1,z_2)$ where $(x_1,x_2)\star(y_1,y_2)$ and $(y_1,y_2)\star(z_1,z_2)$ are both true, then $(x_1,x_2)\star(z_1,z_2)$ must also be true.
+
+Consider a set of 3 elements $(x_1,x_2)$, $(y_1,y_2)$, and $(z_1,z_2)$ where $(x_1,x_2)\star(y_1,y_2)$ and $(y_1,y_2)\star(z_1,z_2)$ are both true. This means that $x_1y_2=x_2y_1$ and $y_1z_2=y_2z_1$. From this, it can be found that $y_1=\frac{y_2 z_1}{z_2}=\frac{x_1 y_2}{x_2}$. This can be simplified to 
+
+$$\begin{aligned}\frac{z_1}{z_2}&=\frac{x_1}{x_2} \\ x_1&=\frac{z_1x_2}{z_2} \\ x_1z_2&=z_1x_2,\end{aligned}$$
+
+which satisfies the requirement for transitivity. Therefore, the relation $\star$ is transitive.
diff --git a/q2/e.md b/q2/e.md
index 00d7bdd..01c0f3f 100644
--- a/q2/e.md
+++ b/q2/e.md
@@ -1 +1 @@
-WIP
+###