Exercise 3.8

Proof. We show that this satisfies the three properties of an order:

• (Comparability) Suppose that $x$ and $y$ are distinct real numbers. If $x^2<y^2$ (or $y^2<x^2$) then of course $\mathit{xCy}$ (or $\mathit{yCx}$) so we are done. So assume that $x^2=y^2$. Since we know that $x\ne y$, it has to be that $y=-x$ so that still $x^2=y^2$. This also implies that $x,y\ne 0$ since otherwise we would have $0=y=-x=0=x$. If $x>0$ then we have $y=-x<0<x$. If $x<0$ then we have $x<0<-x=y$. Hence either way $x^2=y^2$ but $x<y$ (or $y<x$) so that $\mathit{xC}$ (or $\mathit{yCx}$). This shows that $x$ and $y$ are comparable in $C$.
• (Nonreflexivity) Suppose that $x\in ℝ$ so that of course $x^2=x^2$. However clearly it is not the case that $x<x$ so that it cannot be that $\mathit{xCx}$ in this relation.

• (Transitivity) Suppose that $\mathit{xCy}$ and $\mathit{yCz}$. We then have the following cases:

  Case: $x^2<y^2$.

  • Case: $y^2<z^2$. Then clearly $x^2<y^2<z^2$ so that $\mathit{xCz}$.
  • Case: $y^2=z^2$ and $y<z$. Then $x^2<y^2=z^2$ so that again $\mathit{xCz}$.

  Case: $x^2=y^2$ and $x<y$.

  • Case: $y^2<z^2$. Then we have $x^2=y^2<z^2$ so that $\mathit{xCz}$.
  • Case: $y^2=z^2$ and $y<z$. Then $x^2=y^{\text{=}}{z}^2$ and $x<y<z$ so that again $\mathit{xCz}$.

  Hence in all cases $\mathit{xCz}$ so that $C$ is also transitive.

□