I have the following error correcting code: $$\begin{array}{c|c} 00&00000\\ 01&01101\\ 10&10110\\ 11&11011 \end{array}$$
I should show, that every error at 2 positions cannot be (unique) corrected.
So the code 00111 can correlate with 01 or with 10.
The obvious proof, is to check every possible error, but I guess, there have to be something nicer.
I know, that every $a\in \{0,1\}^5$ has a hamming-distance $<=1$ to one of the code words, except 10001 and 01010. Can I somehow use this fact?
EDIT: Today our tutor showed us a really short and simple solution.
If you look at the matrix of the error-free codes, $$\begin{array}{ccccc} 0&0&0&0&0\\ 0&1&1&0&1\\ 1&0&1&1&0\\ 1&1&0&1&1 \end{array}$$ you can notice that there are exactly two 1s and two 0s in each column. If we take one of the 4 codes and add one error, the hamming-distance to two of the other tree codes will be decrease by 1, and the hamming-distance to the third word will be increase by 1. If we repeat this procedure, the same thing will happen. Because there are only three other code-words, the hamming-distance to one of the three will be decrease by 2 (pigeonhole principle). The max. hamming-distance beweet two of the code-words is $\le 4$, so there exists a code-word with hamming-distance $\le 4-2 = 2$.
In $\{0,1\}^5$, there are $4$ code words without error. Each error-free code word is adjacent to $5$ code words with exactly one error. By observation of the $\binom{4}{2}=6$ pairs of error-free code words, no code word with one error is adjacent to two error-free code words, since the pairwise differences of the error-free code words have at least three 1's. At this point we have counted that the error-free code words and the error-1 code words account for $24$ codes, leaving $8$ code words which have error $\ge2$.
If we could identify these $8$ code words and demonstrate that each is distance 2 from two error-free code words, I believe we have established what you are trying to establish. Comments have identified two of the offending code words; can you find four more?
Again by observation, 00000 and 11011 are 4 units apart. So there are $\binom{4}{2}$ code words that are halfway between them: 11000, 10010, 10001, 01010, 01001, and 00011. By observation of comparison with the other two error-free code words, some of these are in fact error-2 code words: 11000, 10001, 01010, and 00011. We only have four error $\ge2$ code wordsleft to find and verify that they cannot be corrected uniquely. Now look again for a pair of error-free code wordsthat are distance 4 from each other...
01101 and 10110 are four units apart. Midway between them are 10101, 11111, 11100, 00111, 00100, and 01110. Of these, comparison with the error-free code wordsconfirms that 10101, 11100, 00111, and 01110 are in fact error-2.
In summary, we have identified all of the error-2 code wordsas lying halfway between two error-free code words, 2 units away from them making all of these 8 code wordsnot have unique corrections:
11000, 10001, 01010, and 00011 are between 00000 and 11011,
10101, 11100, 00111, and 01110 are between 01101 and 10110.
(We've also ruled out code words with error greater than 2.)