Coding Theory - triangle inequality a different proof

[xenonforlife]

Prove that the Hamming distance on Fqn satisfies


d(x, z) ≤ d(x, y) + d(y, z)


for all x, y, z ∈ Fqn. You may assume without loss of generality that x and z
are of the form


x = (a1, . . . , ak, b1, . . . , br)
z = (a1, . . . , ak, c1, . . . , cr)


with bi != ci for all 1 ≤ i ≤ r.


I know the normal method to prove this one which uses common sense, However I am not being able to proceed in this special scenario. Please help!

 

[SlipEternal]

Consider the Hamming distance on only one character at a time.
Let $\displaystyle d_i(x,y) = d(x_i,y_i) = \left\{\begin{array}{l}0\text{ if }x_i = y_i \\ 1\text{ if }x_i \ne y_i\end{array}\right.$

Now, consider the five cases:
Case 1: $\displaystyle x_i = y_i = z_i$

$\displaystyle d_i(x,z) = 0 \le 0 = 0 + 0 = d_i(x,y) + d_i(y,z)$​

Case 2: $\displaystyle x_i = y_i \ne z_i$

$\displaystyle d_i(x,z) = 1 \le 1 = 0 + 1 = d_i(x,y) + d_i(y,z)$​

Case 3: $\displaystyle x_i = z_i \ne y_i$

$\displaystyle d_i(x,z) = 0 \le 2 = 1 + 1 = d_i(x,y) + d_i(y,z)$​

Case 4: $\displaystyle x_i \ne y_i = z_i$

$\displaystyle d_i(x,z) = 1 \le 1 = 1 + 0 = d_i(x,y) + d_i(y,z)$​

Case 5: $\displaystyle x_i \ne y_i \ne z_i$

$\displaystyle d_i(x,z) = 1 \le 2 = 1 + 1 = d_i(x,y) + d_i(y,z)$​

Since it is always true that $\displaystyle d_i(x,z) \le d_i(x,y) + d_i(y,z)$, we know that $\displaystyle d(x,z) = \sum_{i=1}^r{d_i(x,z)} \le \sum_{i=1}^r{(d_i(x,y) + d_i(y,z))} = d(x,y) + d(y,z)$