I need to prove the following statement :
$$\{x \in \mathbb{Z}:2∣ x\} \cap \{x \in \mathbb{Z}:9∣x \}\subset \{x\in \mathbb{Z}:6∣x\}$$
Solution :
Verbally this means, that every whole number which is divisible by both 2 and 9, is divisible by 6 as well. So this intersection forms a set, which has an element x. Lets assume that x is divisible by both 2 and 9.
$$x = 2 ⋅9 ⋅ k , k∈Z$$
$$\frac{x}{2 ⋅ 9}=k$$
$$\frac{x}{18}=k$$
In other words, if x is divisible by 2 and 9, it's the same as saying it's divisible by 18. Let's now show that x is divisible by 6 as well.
$$\frac{3⋅x}{3⋅6}=3k$$
$$\frac{x}{6}=3k$$
This shows, that every element x of our intersection is also an element of the rightwise set, and this intersection is, by definition, a subset of that set.
Let $$x\in LHS$$
$x=2k_1$ and $x=9k_2$
Since $2$and $9$ are relatively prime we have $x=18 k_3$
Thus $$x=6(3k_3)=6k_4$$
Thus $$x\in RHS. $$