Question: I am really stuck with this problem, and I cannot come out with a solution.

I know that to prove a relation is an equivalence relation we have to prove that it's reflexive, symmetric and transitive, but in this case I don't know how, maybe because I don't much experience (but this exercise should be easy) 

This is the relation:

$S=\{(x,y) \in \mathbb{R}\times \mathbb{R}|x - y \in \mathbb{Q} \}$

Answer: 
Let $G$ be the generator matrix of the code in standard form $[I\_k|A]$ where $A$ is a $k \times (n - k)$ matrix. Since the minimum weight of the code is $d = n - k + 1$, all entries of $A$ are non-zero, and hence $A$ is the all one matrix. If $A$ has only one row, then $k = 1$ and the parameters of the code are $[n, 1, n]$. So, say that $A$ has at least two rows. The sum of first two rows is a vector of weight $2$ which is a codeword. Therefore, $d = n - k + 1 \leq 2$, which shows that $n = k$ or $n = k + 1$. For both cases we have a trivial code, of parameters $[n, n, 1]$ and $[n, n-1, 2]$, respectively.
The only binary codes that are MDS are the trivial (n, n, 1) universe codes, the (n, n − 1, 2) single-parity-check (SPC) codes, and the (n, 1, n) repetition codes.
<http://web.stanford.edu/class/ee392d/Chap8.pdf>