Why can't we have a codeword of weight $d-1$?

Question

There are no two words of a code $C$ that have weight $\leq \lfloor \frac{d-1}{2} \rfloor$.

Proof:

$e_1, e_2 \in x+C$ with $||e_1||, ||e_2|| \leq \lfloor \frac{d-1}{2} \rfloor$

then $||e_1-e_2|| \leq ||e_1||+||e_2||\leq 2 \lfloor \frac{d-1}{2}\rfloor \leq d-1$, contradiction.

Could you explain to me why this is a contradiction? The minimum weight of an element is $\lfloor \frac{d-1}{2}\rfloor$ and $d-1 \geq \lfloor \frac{d-1}{2}\rfloor$. So why can't we have a codeword of weight $d-1$ ?

EDIT: If we calculate $x+C$ we will not get the same words as these of the code, but we will know that the word with weight $\leq \lfloor \frac{d-1}{2}\rfloor$ is the error since $C$ cannot have words with weight $\leq \lfloor \frac{d-1}{2}\rfloor$ and so $x+C$ should also not have? Or isn't it like that?

Answer 1

By definition, $d$ is the smallest distance between two-distinct-words.

That is, $d(C)=d=\min\{d(e_1,e_2)|e_1,e_2\in C,c\neq d\}$

If $||e_1-e_2||\lt d-1$ you have two words that are closer to each other than the minimum distance-a contradiction.

Equivalently, the weight of a word represents the number of its non-zero characters. If we had a word with weight less than $d$, then it's distance from the all-zero word would not be the minimum.(Bare in mind that every linear code contains an all-zero word).