Checking/Verifying Proof for all $a,\,b\in N$, $a|b$ if and only if $bZ\subseteq aZ$ Prove that for all $a,\,b\in\Bbb{N}$, $a\mid b$ if and only if $b\Bbb{Z}\subseteq a\Bbb{Z}$.
Answer: = Since $a|b$ $\implies b=ax$ for some $x \in\Bbb{Z} \implies b=ax\in a\Bbb{Z}$
Therefore, $b\in b\Bbb{Z} \implies b\in a\Bbb{Z}$ thus $b\Bbb{Z} \subseteq a\Bbb{Z}$.
Conversely,
Suppose $b\Bbb{Z} \subseteq a\Bbb{Z}$ then in order to show $a|b$ let $x \in b\Bbb{Z} \implies x=b\cdot y$ for some $y\in \Bbb{Z}$ ; at $y=1$ you have $x=b\in a\Bbb{Z} \implies a\cdot k$ for some $k\in \Bbb{Z} \implies a|b.$ 
So because $b\in a\Bbb{Z}$, there exists $x\in \Bbb{Z}$ such that $b=ax$ thus meaning that $a|b$.
Is my answer all correct? Thanks!
 A: *

*The opening statements
$$a\Bbb{Z}=\{a\Bbb{Z};\ \text{ for any }\ x\in\Bbb{Z}\}
 \qquad\text{ and }\qquad
 b\Bbb{Z}=\{b\Bbb{Z};\ \text{ for any }\ y\in\Bbb{Z}\},$$
do not make any sense.

*You write 

However, as $a$ divides $b$ 
$\implies$ $b=ax$ for any $x\in\Bbb{Z}$.

This is false; it should be $b=ax$ for some $x\in\Bbb{Z}$.

*Next you write

$\implies$ $p\in a\Bbb{Z}$ therefore $x\in\Bbb{Z}$, $y\in\Bbb{Z}$ $\implies$ $xy\in\Bbb{Z}$.

This the wrong way around; because $xy\in\Bbb{Z}$ you have $p\in a\Bbb{Z}$.

*You write

Now take $b\Bbb{Z}\subseteq a\Bbb{Z}$. As $b\Bbb{Z}\subseteq a\Bbb{Z}$ then for any $q\in b\Bbb{Z}$ then $q\in a\Bbb{Z}$ therefore $b\Bbb{Z}\subseteq a\Bbb{Z}$.

This is a pointless tautology.

*The remainder of the proof makes no sense. In stead, argue that because $b\in a\Bbb{Z}$, there exists $x\in\Bbb{Z}$ such that $b=ax$. This means precisely that $a\mid b$.
A: In the first part:
⟹p∈aZ therefore x∈Z,y∈Z⟹xy∈Z
is backwards.  You don't conclude $x \in \mathbb Z$ and $y \in \mathbb Z$ from $p \in a\mathbb Z$.
You conclude that because $p = a(xy)$ and because $xy$ is an integer (because $x$ and $y$ are both integers so $xy$ is too) that $p\in a\mathbb Z$.
But otherwise 1) is good.
2) however is a complete mess.
"Now suppose bZ⊆aZ. As bZ⊆aZ then for any q∈bZ then q∈aZ therefore bZ⊆aZ."
This is completely circular and unnecessary.  This is like saying: Suppose Henry is a carnivore.  As Henry is a carnivore then Henry eats meat and therefore Henry is a carnivore.
Instead just state: "Suppose $a\mathbb Z \subset b\mathbb Z$. Let $q \in a\mathbb Z$.  therefore $q \in b\mathbb Z$."
You say: ⟹q=ax for some x∈Z. But aq∈bZ
THis is weird. Why are you working with $aq$ instead of just $q$.  Yes $aq \in a\mathbb Z \subset b\mathbb Z$ but $aq$ is not relevent.  Or was that a typo and you meant but $ax \in b\mathbb Z$?
⟹q=by for any y∈Z.
I think you meant that $q = by$ for SOME integer $y$.  Obviously $q =by$ for ANY $y\in \mathbb Z$ is an absurd and false statement.  That would mean $q = b*1=b$  and that $q = b*513= 513b$ so $q = b = 513b$.  That's only true if $b=q = 0$.
But maybe you meant to say:  $q =by$ for some $y\in \mathbb Z$.
"Therefore, q=by=ax and b=a(x∣y) for all y≠0"
Is $(x|y)$ supposed to be $\frac xy$?  It is true that $b = a \frac xy$ but we have utterly no reason to assume $\frac xy$ is an integer.
If you meant $x$ divides $b$ then this statement is "$b$ equals $a$ times $x$ divides $y$" which isn't even in English.  
You need work.  Notice if $b = 5$ and $a = 4$ and you have $40 = 5*8 = 4*10$.  That does not mean that $4|5$.
You need to work with the idea that for ANY $bx \in b\mathbb Z$ there is some $ay\in b\mathbb Z$ so that $bx = ay$.  So for every $x$ there is *some $y$ so that $bx = ay$.  You need to prove that means $a|b$.  
You need to do more than $a = b\frac xy$.  You need to use that $x$ may be any integer.
Hint:  $b= b*1 \in b\mathbb Z\subset a\mathbb Z$ so......
