Prove that $A=B$. Let $A=\{n∈\mathbb Z | 28 \text{ divides }n\},B=\{n∈\mathbb Z|4\text{ divides }n\}\cap\{n∈\mathbb Z|7\text{ divides }n\}$ To prove this is to show that each set is a subset of the other. 
Claim 1: $A ⊆ B$
Let $n ∈ A$. Then there exists some $k ∈ \mathbb Z$ such that $n = 28k$.
 How do I show that $n ∈ B$? So far I have:
$n ∈ B$ such that there exists an integer $m$ such that $n = 4m$ and that there exists some integer $p$ such that $n = 7p$.
Also, Claim 2: $B ⊆ A$
Let $n ∈ B$. Then there exists some $m ∈ Z$ such that $n = 4m$ and
there exists some $p ∈ Z$ such that $n = 7p$. How do I Show that there exists some $k ∈ Z$ such that $n = 28k$.
 A: When $a,b\in\mathbb Z$, $a$ divides $b$ means there is $c\in\mathbb Z$ such that $b=ca$.
Claim $1$:  if $n\in A$, then $n=28k$ ($k\in\mathbb Z$), so $n=4\times7\times k=4(7k)=7(4k)$, so  $n\in B$.
Claim $2$:  if $n\in B,$ then $n=7p=4m$ ($p,m\in\mathbb Z$), so $7|m$ (by Euclid's lemma),
so $m=7s$ so $n=28s$ so $n\in A$.
A: Once you get used to set notation you will see all this means is
Proving $A \subset B$ means proving

If $a \in A$ then $a \in B$, or in other words, that $a \in \{n|4$ divides $n\}$ and $a\in \{n|7$ divides $n\}$ which means proving

If $a$ is a multiple of $28$ then  then $a$ is a multiple of $4$ and $a$ is a multiple of $7 $.


If we put it in those terms I assume it is easy.
There is an integer $k$ so that $a = 28k  = 4*7k$ so....?


Once you prove that you proven that $a \in \{4:n\}$ and $a \in \{7: n\}$ so $a \in \{4:n\}\cap \{7: n\} = B$.

And that proves $A\subset B$.


And Proving $B \subset A$ means proving

If $b \in B$ then $b \in A$ or in other words if $b \in \{n|4$ divides $n\}$ and $b\in \{n|7$ divides $n\}$ then $b\in A$ which means proving

if $b$ is a multiple of $4$ and $b$ is a multiple of $7$ then $b$ is a multiple of $28$.


This is admitted not as easy as the other way around, but it shouldn't be hard.
There is a $k$ so that $b= 4k$ and a $m$ so that $b = 7m$ so $b =4k = 7m$.  Can you finish from there?
Hint: $\gcd(4,7) =1$ and $4$ divides $7m$ but not $7$ and $7$ divides $4k$ but not $4$.


That proves that if $b \in \{4:n\}\cap \{7:n\}$ the $b \in A$.

which proves $B \subset A$>

