Proving Comparability of a Relation on the Plane (RxR): A Dilemma

[sutupidmath]

Homework Statement

Define a relation on the plane by setting:

$$(x_o,y_o)<(x_1,y_1) \mbox{ if either } y_0-x_0^2<y_1-x_1^2, \mbox{ or } y_o-x_o^2=y_1-x_1^2 \mbox{ and } x_0<x_1.$$

I have easily showed that Nonreflexivity and Transitivity hold. The only dilemma i am facing is to show that Comparability holds as well.
That is to show that for any two elements
$$(x_0,y_0),(x_1,y_1)\in R^2$$
such that :
$$(x_0,y_0)\not=(x_1,y_1)$$

then either
$$(x_0,y_0)<(x_1,y_1) \mbox{ or } (x_1,y_1)<(x_0,y_0)$$

To be more specific, is proof by cases the only way to go about it, or is there any way around it? There seem to be too many cases, and i don't really want to pursue this route. I am also thinking about proving its contrapositive, but still, it looks like i would have to treat a few cases separately.

Any suggestions would be appreciated!

 

[sutupidmath]

Anyone?

 

[Hurkyl]

Too many cases?

How many do you think there are? And how involved is working through each one?

I haven't worked the problem, but my instinct says there shouldn't be much more than 4, and each one very quick. So nowhere near "too many".

 

[sutupidmath]

Ok then, let:

$$(x_o,y_o),(x_1,y_1)\in R^2, \mbox{ such that } (x_o,y_o)\not=(x_1,y_1).$$

$$\mbox{ Then } x_o\not=x_1 \mbox { or } y_o\not=y_1.$$

Without loss of generality(?) assume that

$$x_o<y_o \mbox{ or } y_o<y_1.----(1)$$

Then we would have to consider the following cases:

$$(i) x_o>0,x_1>0 \mbox{ or } y_o>0,y_1>0 .$$

$$(ii) x_0<0,x_1<0 \mbox{ or } y_o<0,y_1<0.$$

$$(iii) x_o<0,x_1>0 \mbox{ or } y_o<0,y_1>0.$$

$$(iv) x_o>0, x_1<0 \mbox{ or } y_o>0,y_1<0.$$

First, can i make the assumption on (1), and second are these all the cases or?

 

[sutupidmath]

...??

 

[HallsofIvy]

It looks pretty straight forward to me. Assume that $(x_0,y_0)\ne (x_1,y_1)$ and $(x_0, y_0) is not < (x1, y1)$. Then it is NOT the case that $y_0- x_0^2< y_1- x_1^2$ so either $y_0- x_0^2> y_1- x_1^2$ or $y_0- x_0^2= y_1- x_1^2$.
If the former, then $y_1- x_1^2< y_0- x_0^2$ so $(x_0, y_0)< (x_1, y_1)$.

If the latter, then $y_0- x_0^2= y_1- x_1^2$ but $x_0$ is not less than $x_1$.

If $x_0= x_1$ then $x_0^2= x_1^2$ so from $y_0- x_0^2= y_1- x_1^2$ we get $y_0= x_0$ which contradicts $(x_0, y_0)\ne (x_1, y_1)$.

If $x_1< x0$ then that, together with $y_1- x_1^2= y_0- x_0^2$ gives $(x_1,y_1)< (x_0,y_0)$.

You still need to show that only one of those three can hold but that should be fairly easy.


We want to prove that, in that case, $(x_1, y_1)< (x_0, y_0)$. Then

 

[sutupidmath]

Thanks! I appreciate your help!