# partition for each equivalence relation

• Mar 23rd 2009, 12:55 PM
chicken
partition for each equivalence relation
hello, i am trying to figure out what i need to do for this problem

for (x,y) is an element of reals, xRy iff x-y is an element of integers

i am supposed to describe the partition for the equivalence relation. im not exactly sure how to do this. i have a similar example, which i don't really understand either...this example is

for n,m is an element of integers, nRm iff n+m is even
now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.

i don't really understand why this would be so, because odd + even= odd

• Mar 23rd 2009, 01:25 PM
Plato
Quote:

Originally Posted by chicken
for (x,y) is an element of reals, xRy iff x-y is an element of integers. I am supposed to describe the partition for the equivalence relation.

The cells in the partition determined by an equivalence relation are simply the equivalence classes.
For this equivalence relation, $\displaystyle \mathcal{R} = \left\{ {\left( {x,y} \right):x \in \Re ,y \in \Re \;\& \,\left( {x - y} \right) \in \mathbb{Z}} \right\}$, the classes are defined using the floor function.
$\displaystyle x/\mathcal{R} = \left[ x \right]_\mathbb{R} = \left\{ {(x - \left\lfloor x \right\rfloor ) + n:n \in \mathbb{Z}} \right\}$
• Mar 23rd 2009, 01:32 PM
chicken
Quote:

Originally Posted by Plato
$\displaystyle x/\mathcal{R} = \left[ x \right]_\mathbb{R} = \left\{ {(x - \left\lfloor x \right\rfloor ) + n:n \in \mathbb{Z}} \right\}$

i don't really understand what is meant by this. im not familiar with the floor function
• Mar 23rd 2009, 01:34 PM
chicken
i think it would help if i understood the example problem, but im not quite getting it.

ex. for n,m is an element of integers, nRm iff n+m is even
now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.
• Mar 23rd 2009, 01:57 PM
Plato
Quote:

Originally Posted by chicken
ex. for n,m is an element of integers, nRm iff n+m is even now this answer would be {X,Y} where X is the set of even intergers and Y is the set of odd integers.

The sum of two even integers is even.
The sum of two odd integers is even.
So even integers are related & odd integers are related.
There are just two equivalance classes.
• Mar 23rd 2009, 07:18 PM
chicken
for x,y as elements of reals xRy iff x-y is an element of integers

so would the answer to this one be all integers?