how to give elements of $A × B$ and $A ∩ B$ $A$ = {$3n+1: n∈Z$} and $B$ = {$12m+7: m∈Z$}
I have proven that $A⊄B$ by showing that $3n+1$ $⇒$ $3n+1+7-7$ ⇒ $3n-6+7$ ⇒ $3(n-2)+7$ ⇒ $4((3(n-2))+7)/4$ ⇒ $12(n-2)(1/4) +7$ ⇒ $12m-7$, where $m = (n-2)(1/4)$, and this proves that $A⊄B$, since $m$ will not always give an integer value.
I have also proved that $B⊆A$ by showing that $12m+7$ ⇒ $12m+7-6+6$ ⇒ $12m+6+1$ ⇒ $3(4m+2)+1$ ⇒ $3n+1$, where $n = (4m+2)$.
If my proofs for the above statements can be improved, let me know.
Now, I am trying to work on giving two concrete elements of $A × B$ and giving two concrete elements of $A ∩ B$.
$A × B$ is defined as $:=$ {$(a,b): a∈A$ and $b∈B$}
$A ∩ B$ is defined as $=$ {$x: (x∈A)$ and $(x∈B)$}
I am not exactly sure how to give concrete elements on $A × B$ and $A ∩ B$ or what it means to give concrete elements. If someone can explain this to me, I would really appreciate it.
Could I say that a concrete element of $A×B$ is $(1,6)$ and $(3, 10)$? I say that $b=6$ and $b=10$ because $(n−2)(1/4)$ will give an integer with those values of $b$? I am not sure what makes an element "concrete" and if I am understanding this correctly.
Could I say that a concrete element of $A∩B$ is $x=10$ or $x=6$ or $x=0$ or $x=14$. I am basically giving out numbers that $(n−2)(1/4)$ will yield an integer value for, since $B$ is contained in $A$, so all values of $B$ will work for $A$ but not the other way around?
Thanks for any answers given!
 A: $A\times B$ is the set of all ordered pairs or couples $(x,y)$ satisfying these 2 conditons :
(1) x belongs to $A$
(2) y belongs to $B$.
So , simply find an $x$ that meets the conditions to belong to $A$ and a $y$ that meets the conditions to belong to $B$, and "build" an ordered pair ( properly ordered of course!) with them; this ordered pair will automatcally be an element of $A\times B$.
In order to find  a member of $A\cap B$, just find an object (here , a number) that meets both the conditions to belong to $A$ and to belong to $B$ .
It could happen that no number meets these 2 conditions. In that case $ A\cap B$ would be identical to $\emptyset$ ( the empty set).
Now, in order to know whether it is the case, you could suppose you have found a number X that belongs to both sets.
You'd have $X = 3n+1$ and $X = 12m+7$.
This implies : $3n+1 = 12m+7$.
Some algebraic  manipulations will give you a certain relation between $n$ and $m$ that must occur in order $X$ to belong both to $A$ and to $B$.
Note : this relation must be compatible with the fact that $n$ and $m$ are integers.
A: Giving a concrete element of $A \times B$ amounts to finding a concrete element $a$ of $A$, and a concrete element $b$ of $B$, and then writing $(a, b)$.
Similarly, giving a concrete element of $A \cap B$ amounts to finding a concrete element $a$ of $A$ which also happens to be an element of $B$.
Also, you are unfortunately mistaken that $A \subseteq B$. We have $1 \in A$ but $1 \notin B$. Perhaps you were trying to prove the other direction?
