Prove the following is a subring Let $R=\{m+n\sqrt{2} \mid m,n\in \mathbb{Z}\}$. Prove that $R$ is a subring of the real numbers. 
I just want to know how to get started really. My professor has used the same example for the past two months on everything we've done and hasn't really showed us any other examples on proving whether something is a subring or not. 
 A: Have you been given the definition of a ring? If not then there is a list of properties that must hold ont he wiki page: https://en.wikipedia.org/wiki/Ring_%28mathematics%29#Definition_and_illustration/wiki/Ring_%28mathematics%29
For example, is there an additive identity.
Well, 0 would seem like an obvious choice.
Is it true that $0 +(a+b\sqrt 2 ) = a+b\sqrt 2$?
It is, since $0 +(a+b\sqrt 2 ) = (0+a) + (0+b\sqrt 2) = a+b\sqrt 2$.
You need to check all the parts of the definition in similar manner.
Feel free to ask in the comments if any parts of the definition don't make sense.
Edit: As pointed out in the comments, you do not need to prove that 0 is the identity since you are proving it is a subring, rather than just proving it is a ring. It should still serve as an example for proving the other parts of the definition are satisfied though.
A: How to get started: Prove that it is closed under subtraction and multiplication (maybe and contains the $1$ of $\mathbb{R}$), e.g. why is $(m+n\sqrt{2})+(k+\ell\sqrt{2})$ again in $R$?
A: You want to prove that $R$ is a subring of the real numbers. First note that this just means that you want to show that $R$ is subset and that $R$ itself is a ring. That $R$ is a subset (containing $0$ and $1$) is clear, so you "just" want to prove that that the set $R$ satisfies the axioms of a ring. Now you get most of these axioms for free. For example: $R$ being a subset of the real numbers means that it already satisfies the other properties of a ring. So you for example will have $a(b + c) = ab + ac$ simply because this is true in the real numbers. So the only things left for you to prove are:
(1) First you would need to prove that the sum of two elements in $R$ is again in $R$. Say $m_1 + n_1\sqrt{2}, m_2 + n_2\sqrt{2}\in R$. Then
$$
m_1 + n_1\sqrt{2} + m_2 + n_2\sqrt{2} = (m_1 + m_2) + (n_1 + n_2)\sqrt{2}
$$
and this is also in $R$.
(2) Second need to prove that the product of two elements is again in $R$. So 
$$
(m_1 + n_1\sqrt{2})(m_2 + n_2\sqrt{2}) = \dots
$$
(you can probably figure this out.)
A: Hint $\ $ By the subring test it suffices to show that $\,R\,$ is closed under subtraction and multiplication, which are straightforward verifications.
A: This are the important conditions for a set to be a subring of a main ring.
Let s be a subring.
Then show that zero is in s.
Then show that for all a,b in s a-b is in s.
Finally for all a,b in s then ab is in s.
