Let $A = \{0, 1\}^8$. Define the relation $R$ on $A$ as
$R = \{ (u, v) \in A \times A | \text{u and v have the same number of entries equal to 0}\}$
How can I show that $R$ is an equivalence relation on $A$? I know that $R$ must be reflexive, symmetric, and transitive, but I'm a bit stuck on the actual proof.
For reflexivity, I believe the following work I have is true. $(u, u) \in A$ since $u$ and $u$ obviously have the same number of entries equal to 0. Is this sufficient to prove reflexivity?
I also think I've worked out the symmetry part. For $(u, v) \in R$, since $u$ and $v$ have the same number of entries equal to 0, then naturally $v$ and $u$ are the same case, and so $(v, u) \in R$. Correct?
I have no idea about the transitivity part yet.
The first two parts are fine as they are (except perhaps the formulation "are the same case"). For the transitivity: If $u$ and $v$ have the same number of zeros and $v$ and $w$ have the same number of zeros, what can you say about $u$ and $w$?
A general piece of advice: Don't be daunted by formal expressions that you're not quite familiar with. Things often turn out to be a lot more common-sensical than they may seem when you first encounter them. In the present case, transitivity is a very natural concept, which you can get a feel for if you see it as something that you can grok rather than a forest of variables and quantifiers.