In Introduction to Algorithms (CLRS), Exercise 33-1-1, we are asked to prove that if $p_1 \times p_2$ is positive then $p_1$ is clockwise from $p_2$ and if it's negative, then $p_1$ is counter-clockwise from $p_2$.
I found the following proof: https://sites.math.rutgers.edu/~ajl213/CLRS/Ch33.pdf
The proof above says to consider the angles both vectors make with the x-axis. We know the angle is $\arctan y/x$. We also know the cross product $p_1 \times p_2 = x_1y_2 - y_1x_2$. If this product is greater than zero, then this means that $y_2/x_2 > y_1/x_1$. Since $\arctan$ is monotone then the angle $p_2$ makes with the x-axis is larger. which means you need to move clockwise direction to get from $p_2$ to $p_1$
BUT the question is asking to prove that $p_1$ is clockwise relative $p_2$. What am I missing? is the above proof wrong?