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**Haven** This problem is causing me some trouble, and I fear that I am overthinking it.

Prove that a binary $\displaystyle [n,M]$-code C of distance $\displaystyle d=2t+1$ exists if and only if a binary $\displaystyle [n+1,M]$-code C' of distance $\displaystyle d'=2t+2$ exists.

Skip this if you understand the notation above

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For those unfamiliar with the notation of coding theory. A code is merely a subset of M vectors in the vector space A^n, where A defines the alphabet of the code. In this problem the alphabet is $\displaystyle A=\left{0,1\right}$. So, a binary $\displaystyle [n,M]$-code is a set of M n-tuples such that, each entry is either 0 or 1

The Hamming distance of two vectors, is the number of entries where the two vectors differ. So taking all possible Hamming distances of the vectors in the code, the distance of an [n,M]-code C, is the minimum of these distances. So, if a code C has distance d, then there exist a pair of vectors $\displaystyle c_i,c_j \in C$ such that $\displaystyle d(c_i,c_j)=d$, and for all other pairs $\displaystyle a,b \in C, d(c_i,c_j)>d$

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What i am attempting to do for the direction from left to right, is to assume the existence of a code with distance d=2t+1 and then show that I can construct a new code C' such that the distance is 2t+2. So, I do the natural thing and append all the old vectors with new vectors. However no matter what I try, I cannot get a construction to work out.

Any help would be greatly appreciated.