Relation Problem?

Question

Let R be a relation on the set N of natural numbers defined by

R = {(a,b) (element symbol) N x N ׀ a divides b in N }

1.) Is R a partial order on N? Explain.

NOT SURE. I know I need to show using reflexive, asymmetric and transitive somehow.

2.) Is N with the divisibility relation given above a totally ordered set? Explain.

***I think I can answer the second problem. Here is my solution:
2. To be totally ordered every pair of elements a,b in the set is comparable (a <=b or b <= a). The set of natural numbers is not totally ordered since for example, 3 and 5 are not comparble, neither 3 <= 5 nor 5 <= 3 holds.

Answer 1

From mathworld.wolfram:
A relation is a partial order on a set if it has:
1. Reflexivity
2. Antisymmetry
3. Transitivity.

Let's figure them out!
1. Does (a,a) belong in R for all a? Sure it does because a divides a. Therefore we have reflexivity.
2. Does (a,b) E R and (b,a) E R imply a=b? Yes it does. If a divides b and b divides a, then a and b must be the same.
3. Does (a,b) E R and (b,c) E R imply (a,c) E R? Sure does: If a divides b which divides c, you can show that a divides c.

So yes you have a partial order there.

The second part you have right.

Answer 2

1. Why not? if a,b,c are in n then:
a. a divides a (reflexive)
b. If a divides b and b divides a then a=b (asymetric)
c. If a divides b and b divides c then a divides c. (transitive)

2. You gave yourself the correct answer.

Answer 3

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