# Equivalence Relation Help!

• July 7th 2008, 09:08 AM
Vedicmaths
Equivalence Relation Help!
I need help in equivalence relationship...

Question> For points (x1, y1) and (x2, y2) in a plane with rectangular
coordinate system, let (x1, y1) ~ (x2, y2) mean that either x1=x2 or y1=y2.
i) Explain why ~ is not an equivalence relation on the set of points in the
plane.
ii) And describe the equivalence classes geometrically..And give a complete
set of equivalence class representatives.

Thank you!
• July 7th 2008, 09:25 AM
Plato
Quote:

Originally Posted by Vedicmaths
I need help in equivalence relationship...

Question> For points (x1, y1) and (x2, y2) in a plane with rectangular
coordinate system, let (x1, y1) ~ (x2, y2) mean that either x1=x2 or y1=y2.
i) Explain why ~ is not an equivalence relation on the set of points in the
plane.
ii) And describe the equivalence classes geometrically..And give a complete
set of equivalence class representatives.

$\left( {1,2} \right) \sim \left( {1, - 2} \right)\,\& \,\left( {1, - 2} \right) \sim \left( { - 1, - 2} \right)\,but\,\left( {1,2} \right)\mathop \sim \limits^? \left( { - 1, - 2} \right)$

For ii) it is not equivalence relation. So I don't understand the point.
• July 7th 2008, 12:24 PM
topsquark
Quote:

Originally Posted by Vedicmaths
I need help in equivalence relationship...

Question> For points (x1, y1) and (x2, y2) in a plane with rectangular
coordinate system, let (x1, y1) ~ (x2, y2) mean that either x1=x2 or y1=y2.
i) Explain why ~ is not an equivalence relation on the set of points in the
plane.
ii) And describe the equivalence classes geometrically..And give a complete
set of equivalence class representatives.

Thank you!

Quote:

Originally Posted by Plato
$\left( {1,2} \right) \sim \left( {1, - 2} \right)\,\& \,\left( {1, - 2} \right) \sim \left( { - 1, - 2} \right)\,but\,\left( {1,2} \right)\mathop \sim \limits^? \left( { - 1, - 2} \right)$

Perhaps the "or" condition is supposed to be "at least one of $x_1 = x_2$ or $y_1 = y_2$? But then the first part is faulty because that is an equivalence relation!

-Dan
• July 7th 2008, 01:09 PM
Plato
Quote:

Originally Posted by topsquark
Perhaps the "or" condition is supposed to be "at least one of $x_1 = x_2$ or $y_1 = y_2$?

No. At least one is equivalent to or.
In fact, the counter example to transitivity works for that.

There is a classical equivalence relation similar to that one.
$\left( {a,b} \right) \sim \left( {x,y} \right)\,iff\,\left| a \right| = \left| x \right| \wedge \left| b \right| = \left| y \right|$.
Note that that is and.
• July 7th 2008, 03:03 PM
Vedicmaths
Yes Sir! you are right, I forgot to mention a point in the (ii) one..
Let (x1, y1) ~ (x2, y2) means that y1=y2. I mistakenly mentioned either and or.
ii) Describe the equivalence classes geometrically..And give a complete
set of equivalence class representatives.
• July 7th 2008, 03:13 PM
Plato
Quote:

Originally Posted by Vedicmaths
Let (x1, y1) ~ (x2, y2) means that y1=y2.
ii) Describe the equivalence classes geometrically. And give a complete
set of equivalence class representatives.

Well that is much better. Clearly that is an equivalence relation.
We can describe an equivalence class as “the set of pairs having the same second term”.
Geometrically that is the set of all horizontal lines. Each class is one of those lines.