GATE CS 2025 Set 1 Q39 — Sets, Relations and Functions

To determine the algebraic structure of $(F, \odot)$, let us analyze the properties of the operation $\odot$ defined on the set of functions $F = \{f \mid f: A \to A\}$, where $A = \{0, 1, 2, 3, \ldots\}$ is the set of non-negative integers.

1. Closure

For any two functions $f_1, f_2 \in F$, and for any $n \in A$: $$(f_1 \odot f_2)(n) = f_1(n) + f_2(n)$$ Since $f_1(n) \ge 0$ and $f_2(n) \ge 0$, their sum $f_1(n) + f_2(n)$ is also a non-negative integer. Thus, $f_1 \odot f_2 \in F$. Closure holds.

2. Associativity

For any $f_1, f_2, f_3 \in F$ and $n \in A$: $$((f_1 \odot f_2) \odot f_3)(n) = (f_1(n) + f_2(n)) + f_3(n)$$ Since addition of non-negative integers is associative: $$(f_1(n) + f_2(n)) + f_3(n) = f_1(n) + (f_2(n) + f_3(n)) = (f_1 \odot (f_2 \odot f_3))(n)$$ Thus, associativity holds.

3. Identity Element

An identity element $e \in F$ must satisfy $f \odot e = f$ for all $f \in F$. This requires: $$f(n) + e(n) = f(n) \implies e(n) = 0 \quad \forall n \in A$$ Since the constant function $e(n) = 0$ maps every element of $A$ to $0 \in A$, we have $e \in F$. Thus, the identity element exists.

4. Commutativity

For any $f_1, f_2 \in F$ and $n \in A$: $$(f_1 \odot f_2)(n) = f_1(n) + f_2(n) = f_2(n) + f_1(n) = (f_2 \odot f_1)(n)$$ Since addition of integers is commutative, the operation $\odot$ is commutative. Hence, the structure is Abelian.

5. Inverse Element

For an inverse $g \in F$ of a function $f \in F$ to exist, we must have: $$f(n) + g(n) = e(n) = 0 \implies g(n) = -f(n)$$ If we choose a function like $f(n) = 1$ for all $n \in A$, its inverse would require $g(n) = -1$. However, $-1 \notin A$ (the codomain contains only non-negative integers). Thus, $g \notin F$, meaning inverses do not exist for all elements in $F$.

Conclusion

Since $(F, \odot)$ satisfies closure, associativity, commutativity, and has an identity element, but lacks inverses, it is an Abelian monoid but not a group.