Exercise 3.6

Question

ewcommand\braces[1]{\left\{ #1 ight\}}

ewcommand\pptp[1]{(x_{#1}, y_{#1})}

ewcommand\pptt[1]{x_{#1} 	imes y_{#1}}

ewcommand\peq[1]{y_{#1} - x_{#1}^2}

Define a relation on the plane by setting
\begin{gather*}
  \pptp{0} < \pptp{1}
\end{gather*}
if either $y_0 - x_0^2 < y_1 - x_1^2$ or $y_0 - x_0^2 = y_1 - x_1^2$ and $x_0 < x_1$.
Show that this is an order relation on the plane, and describe it geometrically.

Dan Whitman · Accepted Answer

First we show that $<$ is an order relation on the plane.
\begin{proof}
  As clearly $<$ is a relation on the plane, we need only show that it has the three required properties of an order relation:

\begin{itemize}
  \item (Comparability) Consider distinct $\pptp{0}$ and $\pptp{1}$ in the plane so that either $x_0 
eq x_1$ or $y_0 
eq y_1$.
    Obviously if $\peq{0} < \peq{1}$ (or $\peq{1} < \peq{0}$) then of course $\pptp{0} < \pptp{1}$ (or $\pptp{1} < \pptp{0}$) so we are done.
    So assume that $\peq{0} = \peq{1}$.
    It it were the case that $x_0 = x_1$ it would have to be that $y_0 
eq y_1$, but we would have
    \begin{gather*}
      \peq{0} = \peq{1} = y_1 - x_0^2 \
      y_0 = y_1 \,,
    \end{gather*}
    which is a contradiction.
    So it must be that $x_0 
eq x_1$.
    So either $x_0 < x_1$ and so $\pptp{0} < \pptp{0}$ or $x_1 < x_0$ and so $\pptp{1} < \pptp{0}$.
    This shows that $<$ is comparable in the plane.

\item (Nonreflexivity) Consider $\pptp{}$ in the plane so that obviously $\peq{} = \peq{}$.
    As we also have that $x = x$, it is not the case that $x < x$ so that it is not true that $\pptp{} < \pptp{}$.

\item (Transitivity) Suppose that $\pptp{0} < \pptp{1}$ and $\pptp{1} < \pptp{2}$.
    We then have the following:
ewline

Case: $\peq{0} < \peq{1}$.
    \begin{itemize}
    \item Case:  $\peq{1} < \peq{2}$.
      Then of course $\peq{0} < \peq{1} < \peq{2}$ so that $\pptp{0} < \pptp{2}$.

\item Case: $\peq{1} = \peq{2}$ and $x_1 < x_2$.
      Then we have $\peq{0} < \peq{1} = \peq{2}$ so that again $\pptp{0} < \pptp{2}$.
    \end{itemize}

Case: $\peq{0} = \peq{1}$ and $x_0 < x_1$.
    \begin{itemize}
    \item Case: $\peq{1} < \peq{2}$.
      Then $\peq{0} = \peq{1} < \peq{2}$ so that $\pptp{0} < \pptp{2}$.

\item Case: $\peq{1} = \peq{2}$ and $x_1 < x_2$.
      Then $\peq{0} = \peq{1} = \peq{2}$ and $x_0 < x_1 < x_2$ so that $\pptp{0} < \pptp{2}$.
    \end{itemize}

Thus in all cases $\pptp{0} < \pptp{2}$, which shows that $<$ is transitive in the plane.
  \end{itemize}
\end{proof}

Geometrically, we refer back to \href{./exercise_3-1}{Exercise 1} and consider a parabola $y = x^2$ shifted up or down on the $y$-axis be some amount $c$.
Then $y = x^2 + c$ so that $\peq{} = c$, and hence every $\pptp{}$ point on the parabola has the same value for $\peq{}$, namely $c$.
Therefore if two distinct points $\pptp{0}$ and $\pptp{1}$ lie on the different parabolas then $\peq{0}$ and $\peq{1}$ will have different values, say $c$ and $d$, respectively.
Then clearly if $\pptp{1}$ is on a higher parabola on the $y$-axis then $c < d$ so that $\peq{0} = c < d = \peq{1}$ so that $\pptp{0} < \pptp{1}$ in our order.
If the points lie on the same parabola then $\peq{0} = \peq{1}$ and whichever points is further to the right will be larger in our order since then, for example, $x_0 < x_1$ so that $\pptp{0} < \pptp{1}$.

Exercise 3.6

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Exercise 3.6

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