a~b if a-b=knQuote:
let 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:
Reflexivity: a-a=0=0n and b-b=0=0n so reflexivity holds.
Symmetry: a~b a-b=kn
Therefore b-a=-kn so symmetry holds.
Transitivity a~b a-b=kn
a-b+b-c=(k+q)n so a-c=(k+q)n so transitivity holds.
a). a-b=2q. This splits into all the odd and even numbers since even-even=even and odd-odd=odd.
is not split. There is one one class- itself.
this implies that a=b. This seems to suggest that equivalence classes cannot exist.
Are these right? c). especially seems a little curious.