Ordered Fields

Let $D$ be an ordered integral domain and $a \in D$. Prove that

Prove Proposition 1.15.

Proposition 1.15: The axioms for multiplication imply the following statements. (a) If $x \neq 0$ and $x y = x z$ then $y = z$. (b) If $x \neq 0$ and $x y = x$ then $y = 1$. (c) If $x \neq 0$ and $x y = 1$ then $y = 1 / x$. (d) If $x \neq 0$ then $1 /(1 / x) = x$.

The axioms for multiplication in a field $F$ (Definition 1.12) are: (M1) If $x \in F$ and $y \in F$, then their product $x y$ is in $F$. (M2) Multiplication is commutative: $x y = y x$ for all $x, y \in F$. (M3) Multiplication is associative: $(x y) z = x(y z)$ for all $x, y, z \in F$. (M4) $F$ contains an element $1 \neq 0$ such that $1 x = x$ for every $x \in F$. (M5) If $x \in F$ and $x \neq 0$ then there exists an element $1 / x \in F$ such that $x \cdot (1 / x) = 1$.

Prove that no order can be defined in the complex field that turns it into an ordered field. Hint: -1 is a square.