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.