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\(A = \{0, 1, 2, 3, \ldots\}\) is the set of non-negative integers. Let Ϝ be the set of functions from \(A\) to itself. For any two functions, \(f_{1}, f_{2} \in\) Ϝ, we define

\((f_{1} \odot f_{2})(n) = f_{1}(n) + f_{2}(n)\)

for every number \(n\) in \(A\). Which of the following is/are CORRECT about the mathematical structure (Ϝ, ⨀)?

  • A.(Ϝ, ⨀) is an Abelian group.
  • B.(Ϝ, ⨀) is an Abelian monoid.
  • C.(Ϝ, ⨀) is a non-Abelian group.
  • D.(Ϝ, ⨀) is a non-Abelian monoid.

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.

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