Show $(N, ≤)$ is a partial order with least upper bounds (lubs) and greatest lower bounds (glbs) of all pairs.

Question

does anyone have any solution or a good hint?

Let $(\mathbb{N}, ≤)$ be the set of natural numbers with the relation $m ≤ n$, meaning $m$ divides $n$. Show $(\mathbb{N}, ≤)$ is a partial order with least upper bounds (lubs) and greatest lower bounds (glbs) of all pairs. What are these lubs and glbs in more traditional terms? If $\mathbb{N}$ is replaced by $\mathbb{Z}$, does the divides relation still yield a partial order?

Ok, this is what I got:

I have the set of all divisors for n, $(N,≤)=(m_1,…,m_k│m_i divides n)$,where $i=1,2,…,k$. In order for $(N,≤)$ to be of partial order it need to be: reflexive (each element is comparable to itself), antisymmetric (no two different elements precede each other and transitive (the start of a chain of precedence relations must precede the end of the chain).

Reflexive: $m_i$ divides $m_i$

Antisymmetric:if $m_i |m_j$,where $i≠j$,then $m_j∤m_i$ since $m_j$ must be greater than $m_i$.

Transitive:if $m_i|m_j$ and $m_j|m_k$, then $m_i|m_j*q=m_k$ for some $q∈N$.

Here is where it gets tricky. I'm reasoning like this:

Take any pair say m_i and m_j, then I have the union of the set of all divisors for m_i and the set of all divisors for m_j. The union is not empty, because it contains 1, and every non-empty subset of the natural number has a least member. Also, any common divisor c satisfies c≤m_i and c≤m_j, so the set has a greatest member. The greatest member has to be the least upper bound since the union of divisors is finite. 1 must always be a lower bound of the union since 1 is less than or equal to all elements in the union. It must also be the greatest lower bound since 1 is the greatest lower bound for the natural numbers (and the union is a subset, containing 1, of the natural numbers).

I'm sure this is wrong, help please.

Answer 1

I'll continue the good example, given by @bof. Let's consider the pair $\{6, 10\}$.

1. GLB. All the divisors of $6$ are $\{1,2,3,6\}$. All the divisors of $10$ are $\{1,2,5,10\}$. Intersection (by the way, you wrote union in your question) of these two sets will give you a set $\{1,2\}$ of all the lower bounds. Then you can choose $2$ as the greatest lower bound, because $1\le2$ (but not because $1$ is less than $2$ in arithmetical sense). Note now that $GCD(6,10)=2$.
2. LUB. Number of all the upper bounds for the pair $\{6,10\}$ is infinite, but you can choose a most evident one - $60$, because it's a product of $6$ and $10$. However it won't be the least one. You need to represent both $6$ and $10$ as products of prime factors. So, $6=2\cdot3$ and $10=2\cdot5$. The least upper bound, represented as product of prime factors, must contain all the prime factors of $6$ and $10$ - so we get $30=2\cdot3\cdot5$. Note now that $LCM(6,10)=30$.

The picture below shows a piece of the lattice $(N,\le)$, where partial order is induced by divisibility:

And now you need to generalize all this for any pair of natural numbers - good luck!

As for your second question about the $\mathbb{Z}$ - please think about antisymmetry axiom. If $(-n)$ divides $n$ and $n$ divides $(-n)$, then what?