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Let $a, b \in \mathbb Z$, and

$$ A = \{x \in \mathbb Z : a|x\} $$ $$ B = \{y \in \mathbb Z : b|y \} $$ Prove: $$b|a \iff A \subseteq B$$

Assuming $A \subseteq B$ , I can understand that there exists and element in sets A and B where $x = y$ and since $ka = x$ and $lb=y$ for some integers $ k $ and $l$, $ka=lb$ and so $$ b = (k/l) a $$ The only problem I am getting now is that I can't prove that $k/l$ is an integer. I have hit a wall.

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    If $n$ divides $m$,then $k/l$ is an integer. SImilarly, if $k/l$ is an integer, then $n$ divides $m$. Use this to finish the proof – 2017-01-17
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    That's what I initially thought, however, if $b = 3$ and $a = 12$ then $k/l = 1/4$ which is not an integer. – 2017-01-17
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    You don't want $k/l$ to be an integer, you want $l/k$ to be an integer, right? That is true, as in the above case it is $4$. – 2017-01-17

2 Answers 2

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Let's be methodical and prove the equivalence as a double implication.

So assume $b|a$. And consider $x\in A$. This means $a|x$. But $b|a$ so we have $a=k\cdot b$ and $x=k'\cdot a$; so we have $x=kk'\cdot b$ and $b|x$ i.e $x\in B$. Therefore $A\subset B$

Now assume $A\subset B$ meaning any element of $A$ is an element of $B$. In particular $a\in A\subset B$ because $a|a$; so $a\in B$ i.e $b|a$.

And we're done.

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    Assuming $A \subseteq B$ we can show that $b|x$ not necessarily $b|a$ – 2017-01-17
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    We have $b|x\,\,\forall x\in A$ but $a\in A$ because $a|a$ so $b|a$ – 2017-01-17
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    @badandboujee1234 what's x in your statement? Note $a|a$ so $a \in A$. If $A \subset B$ then $a \in B$. So $b|a$. No need to bring an arbitrary value of $A$ in when you can bring a specific in. – 2017-01-17
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Assume $A\subset B.$

Notice $a|a $ so $a\in A\subset B $ so $b|a $.

Assume $b|a $

then $a=kb $. If $x \in A $ then $x=ma=mkb $ so $b|x $ so $x\in B $. So $A \subset B$