Proof of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field

Throughout, numbered claims are those of Elementary Order Arithmetic in an Ordered Field, while parts 1, 2 and 3 are those of the statement. The order $\le$ of an ordered field is a total order, so it is reflexive, antisymmetric, and any two elements of $F$ are comparable; and $\gamma\cdot 0=0$ for every $\gamma$ in the underlying field.

Part 1, forward direction. Suppose $\alpha<\beta$. From $0\le\alpha$ and $\alpha<\beta$, claim 2 gives $0<\beta$, so claim 10 with multiplier $\beta$ gives $\beta\alpha<\beta\beta=\beta^{2}$. Also $\alpha\alpha\le\beta\alpha$: since $0\le\alpha$, either $\alpha=0$, in which case both products are $0$, or $0<\alpha$, in which case claim 10 with multiplier $\alpha$ gives $\alpha\alpha<\alpha\beta=\beta\alpha$. Hence $\alpha^{2}=\alpha\alpha\le\beta\alpha$ and $\beta\alpha<\beta^{2}$, so $\alpha^{2}<\beta^{2}$ by claim 2.

Part 1, converse. Suppose $\alpha^{2}<\beta^{2}$, so in particular $\alpha^{2}\ne\beta^{2}$. If $\alpha=\beta$ then $\alpha^{2}=\beta^{2}$, which is excluded. If $\beta<\alpha$, then the forward direction applied with $\beta$ in place of $\alpha$ and $\alpha$ in place of $\beta$, which is legitimate because $0\le\beta$ and $0\le\alpha$, gives $\beta^{2}<\alpha^{2}$; combined with $\alpha^{2}<\beta^{2}$, claim 2 gives $\alpha^{2}<\alpha^{2}$ and hence $\alpha^{2}\ne\alpha^{2}$, which is impossible. By comparability, $\alpha\le\beta$ or $\beta\le\alpha$; in the second case either $\beta=\alpha$ or $\beta<\alpha$, and both have been excluded. Hence $\alpha\le\beta$, and $\alpha\ne\beta$, that is, $\alpha<\beta$.

Part 3. If $\alpha=\beta$ then $\alpha^{2}=\alpha\alpha=\beta\beta=\beta^{2}$. Conversely, suppose $\alpha^{2}=\beta^{2}$ and, for a contradiction, that $\alpha\ne\beta$. By comparability, $\alpha\le\beta$ or $\beta\le\alpha$, so $\alpha<\beta$ or $\beta<\alpha$. In the first case part 1 gives $\alpha^{2}<\beta^{2}$ and in the second it gives $\beta^{2}<\alpha^{2}$; each contradicts $\alpha^{2}=\beta^{2}$, since a strict inequality between two elements forces them to be distinct.

Part 2. Suppose $\alpha\le\beta$. Either $\alpha=\beta$, and then $\alpha^{2}=\beta^{2}$ by part 3, or $\alpha\ne\beta$, and then $\alpha<\beta$, so $\alpha^{2}<\beta^{2}$ by part 1; in both cases $\alpha^{2}\le\beta^{2}$. Conversely suppose $\alpha^{2}\le\beta^{2}$. If $\alpha^{2}=\beta^{2}$ then $\alpha=\beta$ by part 3, so $\alpha\le\beta$ by reflexivity of $\le$. Otherwise $\alpha^{2}<\beta^{2}$, so $\alpha<\beta$ by part 1 and in particular $\alpha\le\beta$.