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: 
This is Proposition 9.2 on p. 212 of *Elements of Algebraic Coding Theory* by L. R. Vermani.

> 
> **Definiton 9.2**
> 
> 
> We have shown that linear $[n, 1, n]$, $[n, n- 1, 2]$ and $[n, n, 1]$ codes exist
>  over any finite field $F$ and these are MDS codes. These are called trivial MDS
>  codes.
> 
> 
> **Proposition 9.2**
> 
> 
> The only binary MDS codes are the trivial codes.
> 
> 
> ***Proof***
> 
> 
> Let $C$ be a binary $[n, k, d]$ MDS code. If $k = 1$, then $C$ is a trivial MDS code and
>  so we may suppose that $k > 1$. Let $G$ be a generator matrix of $C$ with the first
>  $k$ columns of $G$ forming the identity matrix. If $n > k + 1$, then $C$ has a column,
>  say $j$th, of weight less than $k$ and greater than $1$. Suppose that the $i$th entry
>  of this column is $0$. Then the first $k$ columns of $G$ except the $i$th together
>  with the $j$th column are linearly dependent. This proves that $C$ cannot be an
>  MDS code. Hence
> 
> 
> $$k \le n \le k + 1$$
> 
> 
> and $C$ is a trivial MDS code.
> 
> 
>
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>