Checking/Verifying Proof for all $a,\,b\in N$, $a|b$ if and only if $bZ\subseteq aZ$

Question

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!

Answer 1

1. 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.

2. 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 $x\in\Bbb{Z}$.

3. 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 $p\in a\Bbb{Z}$.

4. 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.

5. 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 $a\mid b$.

Answer 2

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......