Prove $\mathbb Q( \sqrt2)$ has only two orderings

Question

I'm having trouble showing that there are only two unique orderings of $\mathbb Q$ restricted to square root of two. I can show that the rationals are ordered, but I can't seem to figure out how to show that $a+b\sqrt2$ and $a-b\sqrt2$ are the only orderings. The only tools I have at my disposal are field axioms and ordered field axioms. Any help would be greatly appreciated.

Answer 1

HINT:

Assume that $\sqrt{2}>0$. We will show for instance that $7 - 5 \sqrt{2} < 0$. Indeed, we have

$$7 - 5 \sqrt{2} = \frac{7^2 - 5^2 \cdot 2}{ 7 + 5 \sqrt{2}}$$

a quotient of a negative rational number by a positive element of the field, hence negative.