a~b if a-b=knlet n be a fixed non-negative integer, and define a relation ~ on the set of integers by setting a~b iff a-b is an integer multiple of n. Show that ~ is an equivalence relation and describe the corresponding equivalence classes when:

a). n=2

b). n=1

c). n=0

Reflexivity: a-a=0=0n and b-b=0=0n so reflexivity holds.

Symmetry: a~b $\displaystyle \Rightarrow$ a-b=kn

Therefore b-a=-kn so symmetry holds.

Transitivity a~b $\displaystyle \Rightarrow$ a-b=kn

b~c $\displaystyle \Rightarrow$ b-c=qn

a-b+b-c=(k+q)n so a-c=(k+q)n so transitivity holds.

a). a-b=2q. This splits $\displaystyle \mathbb{Z}$ into all the odd and even numbers since even-even=even and odd-odd=odd.

b). a-b=q

$\displaystyle \mathbb{Z}$ is not split. There is one one class- $\displaystyle \mathbb{Z}$ itself.

c). a-b=0

this implies that a=b. This seems to suggest that equivalence classes cannot exist.

Are these right? c). especially seems a little curious.