Yes.

1)Reflexcise. Note that a|a is a true statement for any "a" in N.

2)Anti-symmettric. Note if a|b and b|a then a<= b and b<= a. Thus, a=b.

3)Transitive. Yes, if a|b and b|c then a|c. I leave that to thee to prove.

Thus, "|" is an ordering relation on this set.

By definition, "total order" is that all elements are comparable, i.e. the full set is a chain.Is N with the divisibility relation given above a totally ordered set? Explain.

But that is note the case,

(a,b) = (3,5) in R.

But 3|5 is not true.

Thus, it is not a totally ordered set.