pure maths- basic proof?

If we have 3 infinite sets- set A is the set of multiples of 2, set B is the set of multiples of 3, set C is the set of multiples of 6. How do I go about proving that the intersection of set A and B is equivalent to set C? Sorry about the way this is written, I'm on my phone in a long car journey but don't know how to go about this proof!

Let x be an element of the set A n B. .

The usual method of tackling something like this is to lay out the proof in both directions:

Show that if x is an element of A n B then x is an element of C

Show that if x is an element of C then x is an element of A n B.

This shows that the 2 sets must be the same.

I'm just not really sure how to... prove stuff. I get that I can say "suppose x is an element of AnB" and then it must therefore be an element in both sets A and B, then I guess I should maybe try and write it in a form where it is therefore the same as C but I really have no idea how and haven't seen enough proofs before to know what steps I can and can't make D=

If we have 3 infinite sets- set A is the set of multiples of 2, set B is the set of multiples of 3, set C is the set of multiples of 6. How do I go about proving that the intersection of set A and B is equivalent to set C? Sorry about the way this is written, I'm on my phone in a long car journey but don't know how to go about this proof!

Well, if a number x is a multiple of 6 then we can write x = 6m, where m is an integer.
If a number is a multiple of 2 we can write x = 2n where n is an integer.
If a number is a multiple of 3 we can write x = 3p where p is an integer.

Can you see how to deduce the 2nd and 3rd statements from the first one?

Ignore this post until you've solved the problem, and also ignore it if you've never worked with ideals of rings. It won't do as an answer to the question, probably, because it uses enough theory that it's probably cheating. It's just for background if you've covered the relevant stuff, and it's useless if you haven't.

The question asks you to show that the intersection of the ideals $(3)$ and $(2)$ is the ideal $(6)$ in $\mathbb{Z}$. Certainly it is an ideal, because the intersection of ideals is. Certainly it contains $(6)$, because it lies in $(3)$ and $(2)$. It does not contain either $3$ or $2$, and it clearly doesn't contain 1, 4 or 5. Since $\mathbb{Z}$ is a principal ideal domain, in which all ideals are generated by their least positive element, it must be the ideal $(6)$, as required.

Youre right, I haven't seen that before I had a go at the proof though- I'm not really sure if I did it properly, but it looks okay to me- mind taking a look?