Define a relation | on the set of natural numbers by aRb if a|b So for the division relation (or divides relation, depending on how one says it), I have to show the following:
a. Prove that | is a partial order on the set of Natural Numbers.
b. Prove that | has no maximum element.
c. Prove that | has a minimum element.
d. Consider the relation | on the set ℕ \ {1}. Prove that there are inﬁnitely many minimal
elements that are not minimums.
For a, I can prove everything on it except for antisymmetry. For reflexive, I have shown that it a divides itself, and for transitivity, I have shown that for all a,b, and c in the set of Natural numbers, then if a|b, and b|c, then we need only apply properties of division to show that c is a multiple of b which is a multiple of a, and therefore, a divides c.
For part b, could I prove there is no maximum by proving that the set of all natural numbers is infinite, therefore there cannot be a maximum in the relation?
For part c, I am a little lost on how to prove such a thing because if it is the set of all natural numbers, which would start at 1, wouldn't 1 be the minimum element?
The same question I have for c goes for part d.
Any help would be appreciated. Thank you kindly.
 A: Antisymmetry for (a): Suppose $a|b$ and $b|a.$ Then there are positive integers $s,t$ such that both $as=b$ and $bt=a.$ Substitution then gives $ast=a,$ so that $st=1.$ But since $s,t \ge 1$ this implies $s=t=1$ which then gives $a=b.$
For part (b) all you need to note is that, if $m$ is assumed to be maximal, it should be that every positive integer $x$ is a divisor of $m,$ i.e $x|m$ for all positive $x.$ But this cannot be true for example if $x=2m,$ since then there is $k$ for which $(2m)k=m,$ giving $2k=1$ which is not possible for a positive integer $k.$
In part (c) you are correct in that $1$ is a divisor of every positive integer $x$ since one only needs some $k$ with $1 \cdot k=x,$ and taking $k=x$ works.
For part (d) one is throwing $1$ out of the positive integers, and then you can show that any prime $p$ is a minimal element, in the sense that no other element divides it (in the set of naturals other than $1$). However none of these primes $p$ is a minimum because there are other positive non-1 integers which any specific prime $p$ does not divide.
