Equivalence relation on the Cartesian plane

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Homework Statement


A relation p is defined on R^2 (fancy R, as in Reals) by (a,b)p (c,d) if a+d=b+c
Show that p is an equivalence relation.

b) Consider R^2 to be the Cartesian Plane. Describe p's equivalence classes geometrically. (Consider which points will be in the particular equivalence classes by taking an arbitrary point in the same equivalence class as (x,y). )

I have done part a. In part b I only got as far as drawing the Cartesian axes and a table of values. I'll show you below. I think that I have not made enough of a start for you to be able to give me a clue, but I thought you might be able to point me in the direction of a book that would cover this. I am finding my course really hard because I am studying by distance education and we don't have a textbook. When I browse through the library I am not finding anything that quite fits my course. Direction to an online resource would be particulary good or a text book that you think would be readily available at my university library.

Homework Equations


To be an equivalence relation, p must be reflexive symmetric and transitive. I have shown all that.


The Attempt at a Solution



I drew up a list of values
a b c d
1 2 1 2
4 3 3 2
1 2 2 3
-1 0 1 2


I did about 30 so that I had a really good idea of what was happening.
What I figured out:
Points on line y=x+1 map to (1,2)
Points on line y=x+2 map to (1,3)
Points on line y=x+3 map to (1,4)
etcetera

I discovered that I couldn't draw it on the Cartesian axes. Am I meant to be able to?

I don't even understand if "decribe p's eqivalence classes geometrically" means I am meant to draw or use words.

I have been puzzling over this one for about a month now.
Any clues you can give me to point me in the right direction will be greatly appreciated I assure you.
Many thanks is anticipation.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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Homework Helper
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(a, b) p (c, d) if and only if a+ d= b+ c. Since you are talking about the Cartesian plane, it might help to use (x, y) and (x', y') instead. x+ y= x'+ y' is the same as y- y'= x- x' or (y- y')/(x- x')= 1. That gives exactly the result you talk about. An equivalence class is a straight line with slope 1.


By the way, this same equivalence relation can be used to define the integers given on the positive integers (counting numbers). If x> y and (x, y)p (x', y') then it must be true that x'> y' and, in fact, x'- y'= x- y. So we can think of the equivalence class [(x, y)] as defining "x- y". For x> y, of course, that is just a positive integer but if x= y and (x, y) p (x', y') then x'= y' and we can think of the equivalence class as defining "0". Similarly, if x< y and (x', y')p (x, y), x'< y' and we can think of the equivalence class as defining the negative integer x- y= -(y- x).
 
  • #3
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Thank you very much for your help. Much appreciated!
 

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