Algebraic Structures and Groups Questions and Answers

Answer: B

The structure is an Abelian monoid.

Pointwise addition of functions from the non-negative integers to themselves is closed (the sum of two such functions is again one), associative and commutative, because addition of integers is. The zero function \(z(n) = 0\) is an identity: \(f \odot z = f\). So the structure is a commutative (Abelian) monoid.

It is not a group, because inverses do not exist: for \(f(n) = 1\) there is no function g with values in A such that \(1 + g(n) = 0\); that would need \(g(n) = -1\), which is not a non-negative integer. So both group options are false, and since the operation is commutative, the non-Abelian options are false too. That leaves the single correct option B.

Answer: B

The received word is 0111.

The code is every sum (mod 2) of the rows of G. The three rows are 1001, 0101 and 0011, and their combinations give the eight code words:

0000, 1001, 0101, 0011, 1100, 1010, 0110, 1111

Every row has even weight, so every code word has even weight: the last column is a parity bit. A single-bit error changes the weight by one, so a word received after exactly one error must have odd weight.

0111 has weight 3, so it cannot be a code word and must be a corrupted one (it is one bit away from 0011, 0101, 0110 and 1111). 0011, 1111 and 1100 all have even weight and are themselves code words, so receiving them is consistent with no error, not with a one-bit error.

Answer: D

The code can correct 0 errors.

The code words are all sums (mod 2) of the rows of G: 000, 110, 011 and 110 + 011 = 101.

\[ \begin{aligned} d_{min} &= \text{least non-zero weight} = 2 \\ t &= \lfloor (d_{min} - 1)/2 \rfloor = \lfloor 1/2 \rfloor = 0 \end{aligned} \]

With a minimum distance of 2 the code can detect a single error but cannot correct any, because a received word one bit away from two code words is equally close to both. Correcting one error would need a minimum distance of at least 3, so 1, 2 and 3 are all too high.