remove question from Q2.d

q2/a.md

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+## 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 A
+
+> Is $\star$ reflexive?
+
+A relation on $\mathbb{Q}^2$ is reflexive if $a\star a$ for all $a\in\mathbb{Q}^2$. To prove reflexivity, assume that $x_{1}x_{2}=y_{2}y_{1}$ is true (let this be $(1)$).
+
+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, $x_{1}x_{2}=y_{2}y_{1}=x_1 y_2 = x_2 y_1$. Since every $(x_1,x_2)$ satisfies the requirement that $x_{1}x_{2}=y_{2}y_{1}$, the relation is reflexive.
+

q2/b.md

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+### Part B
+
+> Is $\star$ symmetric?
+
+For $\star$ to be symmetric, every set of elements $(x_1,x_2)\in\Q^2,(y_1,y_2)\in\Q^2$ which satisfies $(x_1,x_2)\star(y_1,y_2)$ must also satisfy $(y_1,y_2)\star(x_1,x_2)$.
+
+Assume $(y_1,y_2)\star(x_1,x_2)$ is true. This means that $y_1 x_2 =y_2 x_1$, which is the same as as $x_{1}y_{2}=x_{2}y_{1}$ (i.e. the original relation). Therefore the requirement is satisfied, and $\star$ is symmetric.
+

q2/c.md

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+### 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.

q2/d.md

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+### 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.

q2/e.md

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+### Part E
+
+> Is $\star$ an equivalence relation, a partial order, both, or neither?
+
+A relation is an equivalence relation if it is reflexive, symmetric and transitive. It was shown in parts A, B, and D that $\star$ meets these criteria.
+
+A relation is a partial order if it is reflexive, anti-symmetric, and transitive. It was shown in parts A, B, and C that $\star$ meets these criteria.
+
+Therefore, $\star$ is both an equivalence relation and a partial order.