Homework Help: Linear Order on RxR. ?

1. Jan 31, 2010

sutupidmath

1. The problem statement, all variables and given/known data
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!

2. Feb 1, 2010

Anyone?

3. Feb 1, 2010

Hurkyl

Staff Emeritus
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".

4. Feb 1, 2010

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?

Last edited: Feb 1, 2010
5. Feb 1, 2010

...??

6. Feb 2, 2010

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

7. Feb 3, 2010