Suppose $F = \mathbb{R}^2$ is a field equipped with the following operations:
Addition: Defined entrywise i.e.
$$(x_1, x_2) + (y_1, y_2) = (x_1 + y_1, x_2 + y_2)$$
Multiplication:
$$(x_1, x_2) \cdot (y_1, y_2) = (x_1y_1 - x_2y_2, x_1y_2 + x_2y_1)$$
Provided with the operations above, and knowing that $F$ is a field, show that no ordered relation makes $F$ into an ordered field.
I'm attempting to solve the problem above. I started with "Assume that < is an ordered relation on $F$ and $(x_1, x_2), (y_1, y_2), (z_1, z_2) \in F$. We see that: " in an attempt to find a contradiction but I haven't been able to find any. I'm not exactly sure how I can proceed by using the axioms though I'm sure it has something to do with multiplication since it isn't defined entrywise. Also, I'm not sure if I can use this but I found the additive identity to be $(0, 0)$ and the multiplicative identity to be $(1, 0)$. Any hints or assistance are much appreciated.