1. Check that this relation is an equivalence relation
2. Find the two equivalence classes, and name them appropriately.
3. How would you add these equivalence classes, if at all ?

reflexive:

x - x is equal to zero, which is divisible by two. Hence, every element is related to itself.

symmetry:

if x ~ y, then y - x is divisible by 2. But then - (y - x) = x - y is divisible by two. Hence, y ~ x

transitivity:

• if x ~ y then y - x = 2n for some integer n
• if y ~ z then z - y = 2m for some integer m

But then z - x = (2m + y) - (y - 2n) = 2 (m + n), so that z - x is divisible by 2. In other words, x and z are related.

Two elements are in the same equivalence class if and only if they are related. If x and y are in the same class, then y - x = 2m for some integer n.

• If y was even, then y = 2m for some integer m, and x = 2m - 2n must also be even.
• If y was odd, then y = 2m + 1 for some integer m, and x = 2m + 1 - 2n must also be odd

• E = even numbers: E = [(2)] contains all even numbers
• O = odd numbers: O = [(3)] contains all odd numbers.

Define [x] + [y] = [x + y]. We need to show that this is well-defined, i.e. independent of the particular representative of the equivalence classes of [x] and [y]. Take x ~ x' and y ~ y'.

• Then x' - x = 2m and y' - y = 2n for some integers n and m.
• Then (x' + y') - (x + y) = x' - x + y' - y = 2m + 2n = 2 (m + n)

• [x] + [y] = [x + y] = [x' + y'] = [x'] + [y']

A better method would be the following: Define two classes 0 and 1 by saying:

• all numbers divisible by 2 with no remainder are in class 0
• all numbers divisible by 2 with remainder 1 are in class 1