Understanding the proof of unique natural divisor Let $m,n \in \mathbb{Z} $. There exists a unique natural number $d \in \mathbb{N}$ such that $\text{div} (m) \cap \text{div} (n)= \text{div} (d)$. 
Now I know this is just saying that between any two numbers, there is a natural number that divides both. It is a fact learned in primary school and is obvious. Apologies in advance if this proof is super easy and I am just a slow learner. 
I will put  asterisks **  before any line of the proof I am struggling with.
Proof:
The uniqueness follows from the fact that 
$$\text{div}(d_1)=\text{div}(d_2)\;\text{if and only if}\; d_1=d_2$$
assuming that $d_1,d_2 \in \mathbb{N}$. When proving the existence of $d$ we may assume $m,n \in \mathbb{N}$, since $\text{div}(x)=\text{div}(-x)$ for $x \in \mathbb{Z}$. We proceed using induction on $\min(m,n)$, where $\min(m,n)=m$ if $m \le n$ and $\min(m,n)=n$ if $m \gt n$. If $\min(m,n)=0$ we may assume $n=0$.   


*

*** Therefore 
$$\text{div}(m) \cap \text{div}(n)=\text{div}(m)$$
This is where I got confused, if $m \gt n$ how can this be true?

*** This settles the intial step 
$$\min(m,n)=0$$
of the induction. So how does this line tie into what we were trying to prove?

*** Now assume that we have proved 
$$\text{div}(m) \cap \text{div}(n)=\text{div}(d)$$ for every $m,n \in \mathbb{N}$ with $\min(m,n) \lt N$, where $N \gt 0$. 
How can we assume the very thing we are trying to prove? Suppose for the induction step that we are given $m,n \in \mathbb{N}$ with $\min(m,n)=N$. Then we may write $m=qn+r$, where $0 \le r \lt n$.

*** But 
$$\text{div}(m) \cap \text{div}(n)= \text{div}(m-qn) \cap \text{div}(n)= \text{div}(r) \cap \text{div}(n)$$ 
Since a number divides $m$ and $n$ if and only if it divides $m-qn$ and $n$ (this line I understand). 

*** By induction we know that 
$$\text{div}(r) \cap \text{div}(n)=\text{div}(d)$$
for some $d \in \mathbb{N}$, since $\min(r,n)= r \lt n=N $. And this completes the proof. 
Where I have put an asterisk is either I don't understand how they came to that conclusion, or I don't understand how it ties into what we are trying to prove. Sorry, sometimes it is hard for me to explain exactly what I don't understand. 
 A: I don't like that proof at all.  It's too hard to read.  So I wrote a new proof.
A nonempty subset $I$ of $\mathbb{Z}$ is called an ideal if:

(i) If $x,y \in I$, so is $x+y$.
(ii) If $x \in I$, and $y$ is any integer, then $xy \in \mathbb{Z}$.

Here are some examples:

(I): $\mathbb{Z}$ is an ideal.
(II): If $n$ is any integer, then $n\mathbb{Z}$, by definition the set of all integers which are divisible by $n$, is an ideal (check that this satisfies the definition).

Actually, these are all the ideals of $\mathbb{Z}$.
Lemma: If $I$ is an ideal, then there exists a positive integer $n$ such that $I = n\mathbb{Z}$.  If $m$ is another integer, and $n\mathbb{Z} = m\mathbb{Z}$, then $n = \pm m$.
Proof: If $I$ just consists of the number $0$, then we are done, because then $ I = 0\mathbb{Z}$, which is an ideal.  So assume $I$ has a nonzero element $x$.
Then $I$ has a positive element.  If $x$ is positive, great.  If $x$ is negative, then $(-1) \cdot x = -x$ is a positive member of $I$, by property (ii) of the definition of an ideal.
So $I$ has positive elements.  Let $n$ be the smallest positive member of $I$ (remember that every nonempty subset of $\mathbb{N}$ has a smallest element).  I claim that $$I = n\mathbb{Z}$$ If $a \in n\mathbb{Z}$, then $a = nk$ for some integer $k$.  But since $n \in I$, so is $nk$ by property (ii).  This shows that $n\mathbb{Z} \subseteq I$.
Now let $x \in I$.  Then there exist integers $q, r$ such that $$x = qn + r$$ with $0\leq r \leq n-1$.  Now $x$ and $qn$ are elements of $I$.  Hence $(-1)qn = -qn$ is an element of $I$.  Hence $x + (-1)qn = r$ is an element of $I$ by property (i).  But by the way we defined $n$, the set $I$ does not contain any positive elements which are smaller than $n$. So we must have $r = 0$.  This shows that $x = qn$ lies in $n\mathbb{Z}$.  Thus $I = n\mathbb{Z}$.
If $m\mathbb{Z} = n\mathbb{Z}$, then since $m \in m\mathbb{Z}$, it is also an element of $n\mathbb{Z}$, hence $m = kn$ for some integer $k$.  Therefore, $|m| = |kn| \geq |n|$.  By the same reasoning, $|n| \geq |m|$.  Therefore, $|n| = |m|$, which means that $n = \pm m$.  $\blacksquare$
Proposition: Let $n,m$ be integers.  There exists a unique natural number $d$ whose divisors are exactly those numbers which are divisors of both $m$ and $n$.
Proof: Let $n\mathbb{Z} + m\mathbb{Z}$ denote the set of integers which can be written in the form $x+y$, where $x \in n\mathbb{Z}$ and $y \in m\mathbb{Z}$.  You can check that $n\mathbb{Z} + m\mathbb{Z}$ is an ideal.  So by the lemma, we know that there exists a positive integer $d$ such that $$n\mathbb{Z} + m\mathbb{Z} = d \mathbb{Z}$$
We claim that $d$ does what you want.  First, note that $n\mathbb{Z} + m\mathbb{Z}$ contains $n\mathbb{Z}$ as a subset.  Since $n \in n\mathbb{Z}$, we then see that $n$ is divisible by $d$.  Similarly, $m$ is divisible by $d$.  So any divisor of $d$ must also divide both $n$ and $m$.
We now just have to show that if $x$ divides both $n$ and $m$, then $x$ divides $d$.  Since $d$ is an element of $n\mathbb{Z} + m\mathbb{Z}$, there exist integers $a, b$ such that $$d = an + bm$$  Since $x$ divides $n$ and $m$, we can write $n = kx$ and $m = lx$ for some integers $k, l$.  But then $$d = akx + blx = (ak + bl)x$$ so $x$ divides $d$.
The last thing to establish is the uniqueness of $d$.  If $d'$ is any other number with the same properties, then, since $d'$ is a divisor of itself, it must be a divisor of both $m$ and $n$.  But $d$ divides all numbers with those properties.  So $d$ divides $d'$.  But reversing the roles of $d$ and $d'$, we see that $d'$ divides $d$.  The only way two integers $d$ and $d'$ can divide each other is if $d = \pm d'$.  $\blacksquare$
A: The language and choice of variables is confusing. 
Let me rewrite it.
Prove that for $m, n \in \mathbb N+\{0\}$ $m \ne n$, there exist a $d$ such that $div(m) \cap div(n) = div(d)$. 
Since either $m > n$ or $n > m$ we might as well assume $m > n$ (otherwise we'd just relabel and get the same result.
Initial step:
$n = 0$
Then every number divides $0$ so $div(0) = \mathbb Z$.
So $div(m) \cap div(0) = div(m) \cap \mathbb Z = div(m)$.
Induction step:
Assume we have shown it is true for all $n \le k-1$.  (We have proven it is true for $k = 1$ as we have proven it is true for $n = 0$).  We want to prove it must be true for $n = k$.
Let $m = q*k + r$ with $0 \le r < 0$.
Then 
$div(m) \cap div(k) = div(m - kq) \cap div(k) = div(r) \cap div(k)$
Since $r < k$ and we have assumed $div(m) \cap div(n)= div(d)$ for some $d$ if $m>n$ and $n \le k-1$, we can conclude $div(r) \cap div(k) = div(d)$ for some $d$.
Thus we have shown this is not only true for $n \le k-1$ it is true for $n = k$ too.

So by induction we know it is true for $n = 0$ (because $div(m) \cap div(0) = div(m)$).
Therefore it must be true for $n = 0+ 1= 1$ (because $m = q*1 + 0$ and $div(m) \cap div(1) = div(m - q*1) \cap div(1) = div(0) \cap div(1) = div(1)$)
Therefore it must be true for $n = 1+ 1 = 2$ (because $m = q*2 + r$ $r < 2$ and $div(m) \cap div(2) = div(m - 2q) \cap div(2) = div(r) \cap div(2)$.  div(r) = either $div(1)= \{1\}$ or $div(0)$ so the intersection equals either $div(2)$ or $div(1)$.)
Therefore it must be true fore $n = 2+1 = 3$ (because $m = 3q + r$ and $div(m) \cup div(3) = div(m - 3q) \cup div(3) = div(r) \cup div(3)$ and we've proven that for $r < 3$ that $div(r) \cup div(m) = div(d)$ for some $d$.)
Etc.  repeat forever.
A: I am short of time, but I try to explain the first two questions.
An example for $\text{div}()$: 
$$\text{div}(6)=\{-6,-3,-2,-1,1,2,3,6\}$$
and
$$\text{div}(-6)=\{-6,-3,-2,-1,1,2,3,6\}$$
So $\text{div}$ is a subset of $\mathbb {Z}$ that contains negative and poisitive numbers. it does not contain $0$ except for 
$$\text{div}(0)=\mathbb{Z}$$
because 
$$\forall n \in \mathbb{Z}: \, n\cdot 0=0$$
 and so 
$$\forall n \in \mathbb{Z}: \, n \mid 0  $$


*

*we assume $n=0$. Therefore $\text{div}(n)=\mathbb{Z}$ and so 
$$\text{div}(m) \cap \text{div}(n) = \text{div}(m) \cap \mathbb{Z} =\text{div}(m) $$

*We have
$$\mathbb{N}=\{0,1,2,3,\cdots\}$$
We want to proof 
$$\forall m \in \mathbb{N}, \forall n \in \mathbb{N}:\,\exists d \in \mathbb{N}: \; \text{div} (m) \cap \text{div} (n)= \text{div} (d)  \tag{1}$$
by induction on 
$\min(m,n)$
so we formulate $(1)$ as
$$\forall N \in \mathbb{N}: \, \left(m \in \mathbb{N} \land  n \in \mathbb{N}\land \min(m,n)\le N \implies \; \exists d \in \mathbb{N}: \; \text{div} (m) \cap \text{div} (n)= \text{div} (d) \right) \tag{2}$$
If  $m,n \in \mathbb{N}$ then
$$\min_{m,n \in \mathbb{N}}(m,n)= \min(\min_{m \in \mathbb{N}}(m),\min_{n \in \mathbb{N}}(n)) = \min(0,0) = 0$$
So the smallest $\min(m,n)$ we encounter is $0$. Therefore we have to check $(2)$ for $N=0$, which is done in paragraph 1.


The remaining part of the proof is to show that

If $(2)$ is valid for an arbitrary $N=k \in \mathbb{N}$ then $(2)$ is valid for $N=k+1$

