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Let $a,b,q,r \in \mathbb{Z}$ so that $a=qb+r$. Consider:

$A=\{n \in \mathbb{N}: n|a\}$, $B=\{n \in \mathbb{N}: n|b\}$, $C=\{n \in \mathbb{N}: n|r\}$.

Show that "$A\subseteq B\cap C$" is false.

I'm trying to understand why is this statement false.

So, for $B\cap C$ , if $n|b$ and $n|r$ then $n|bx+ry$. (Proposition)

We have $a=qb+r$, so $n$ should also divide $a$.(Just assume $x=q$ and $y=1$)

If you have $B\cap C$ you also get $A$, so shouln't $A\subseteq B\cap C$ be true?

Note: $p|q$ represents $p$ divides $q$

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You proved that $n\in B\cap C\Rightarrow n\in A$, but that only means $B\cap C\subseteq A$. However, consider, for example $3=2\cdot1+1$. Then $n=3|3$, so $n\in A$, but $3$ doesn't divide either $b=2$ or $r=1$.

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    Oh I see my problem Thanks. $A\subseteq B\cap C$ would only be true if I start from $A$ and get $B\cap C$. I prooved the other way around. – 2017-02-20