Solution for Section 10.3 Question 6

6. Define the relation d on the set of integers Z as follows: for all m and n in Z,  m d n iff.jpg (642 bytes)d | (m - n)  for some integer d.

a) Prove that d is an equivalence relation.

Suppose that m is an integer. Now m - m = 0, and d | 0 since 0 = d � 0. Hence d | (m - m), so m d m. Thus the relation d is reflexive.

Suppose that m and n are integers and that m d n (so d | (m - n) ). Since d | (m - n) we know that  m - n = d � k for some integer k. Thus n - m = d � (-k) and -k is also an integer. Hence d | (n - m) and so n d m. Thus the relation d is symmetric.

Suppose that m, n and p are integers and that m d n and n d p (so d | (m - n) and d | (n - p) ). Since d | (m - n) we know that m - n = d � k for some integer k and since d | (n - p) we know that n - p = d � h for some integer h. Hence m - p = (m - n) + (n - p) = d � k + d � h = d (k + h)  and k + h will also be an integer. Therefore d | (m - p) and so m d p. Thus the relation d is transitive.

Therefore, the relation d is an equivalence relation.

b) Find the equivalence classes for d.

There are d equivalence classes for d:

[0] = [d] = [2d] = ... = { xinred.jpg (595 bytes)Z | x = d � k for some integer k } = { ..., -d, 0, d, 2d, ... }
[1] = [d + 1] = [2d + 1] = ... = { xinred.jpg (595 bytes)Z | x = d � k + 1 for some integer k}  = { ..., -d + 1,  1,  d + 1,  2d + 1, ... }
[2] = [d + 2] = [2d + 2] = ... = { xinred.jpg (595 bytes)Z | x = d � k + 2 for some integer k}  = { ..., -d + 2,  2,  d + 2,  2d + 2, ... }
[3] = [d + 3] = [2d + 3] = ... = { xinred.jpg (595 bytes)Z | x = d � k + 3 for some integer k}  = { ..., -d + 3,  3,  d + 3,  2d + 3, ... }
:
:
[d-2] = [d + (d-2)] = [2d + (d-2)] = ... = { xinred.jpg (595 bytes)Z | x = d � k + (d-2) for some integer k}  = { ..., -d + (d-2),  d-2,  d + (d-2),  2d + (d-2), ... }
[d-1] = [d + (d-1)] = [2d + (d-1)] = ... = { xinred.jpg (595 bytes)Z | x = d � k + (d-1) for some integer k}  = { ..., -d + (d-1),  d-1,  d + (d-1),  2d + (d-1), ... }

Back to Section 10.3