My question is one aspect of the question below.

I have already shown that the relation is reflexive, symmetric, and transitive. What is meant by, and how do I describe the corresponding partition on ?

a ~ b if a-b is divisible by 3.

Thank you.

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- May 16th 2011, 04:17 PMdwsmithEquiv Class
My question is one aspect of the question below.

I have already shown that the relation is reflexive, symmetric, and transitive. What is meant by, and how do I describe the corresponding partition on ?

a ~ b if a-b is divisible by 3.

Thank you. - May 16th 2011, 06:25 PMTKHunny
How did you demonstrate te properties without the ability to describe the relation?

Maybe 3|(a-b) or

Not too sure what you're looking for. - May 16th 2011, 06:28 PMdwsmith
- May 16th 2011, 08:48 PMDeveno
well, for a given integer, let's call it k, let us find out what [k], the equivalence class of k under ~ is.

we have that if a is in [k], then k - a = 3m, for some integer m (k-a is divisible by 3).

so the possible values for a, are k, k-3, k+3, k-6, k+6, k-9, k+9, etc. if we choose the smallest positive integer of [k], to represent [k],

we see that [k] = [0], [1], or [2] (depending on whether a is of the form 3n, 3n+1, or 3n+2).

these 3 equivalence classes are disjoint, they share no common members, and Z = [0] U [1] U [2].

as we saw above, [0] = 0,-3,3,-6,6,-9,9, etc. this is the same set as 3Z = {3k : k in Z}.

similarly, [1] = 1,-2,4,-5,7,-8,10, etc. which is the same set as 1+3Z = {3k+1 : k in Z},

and [2] = 2,-1,5,-4,8,-7,11, etc. which is the same set as 2+3Z = {3k+2 : k in Z}.

this is typical of any equivalence relation, the equivalence classes form a partition of the set the equivalence is defined on. - May 17th 2011, 05:43 AMTKHunny
That's fine. It's not always clear what notation is wanted. Getting used to your text or professor is a good idea. I was defining the "0, 1, and 2", and suggesting how to determine the class. The notation you presented actually names the classes and includes all values in the class. It should bother you for 3-5 milliseconds how we used to have a set of all integers and now we have three sets of all integers but we don't have any more integers than we had before.