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!