1. **State the problem:** Determine if the set $A = \{ f_{a,b} : a,b \in \mathbb{R} \}$ with the operation of composition $(A, \circ)$ forms a monoid, and whether $(A, \circ)$ is a group. 2. **Recall definitions:** - A **monoid** is a set with an associative binary operation and an identity element. - A **group** is a monoid where every element has an inverse. 3. **Describe the functions:** Each $f_{a,b}: \mathbb{R} \to \mathbb{R}$ is defined by $$f_{a,b}(x) = ax + b$$ 4. **Check closure under composition:** For $f_{a,b}, f_{c,d} \in A$, their composition is $$f_{a,b} \circ f_{c,d} (x) = f_{a,b}(f_{c,d}(x)) = f_{a,b}(cx + d) = a(cx + d) + b = (ac)x + (ad + b)$$ Since $ac, ad + b \in \mathbb{R}$, composition stays in $A$. 5. **Check associativity:** Function composition is associative in general, so $(A, \circ)$ is associative. 6. **Find the identity element:** The identity function $f_{1,0}(x) = 1 \cdot x + 0 = x$ acts as identity: $$f_{1,0} \circ f_{a,b} = f_{a,b} \circ f_{1,0} = f_{a,b}$$ 7. **Check for inverses:** To have an inverse, for $f_{a,b}$ there must be $f_{a',b'}$ such that $$f_{a,b} \circ f_{a',b'} = f_{1,0}$$ Compute: $$f_{a,b} \circ f_{a',b'}(x) = a(a'x + b') + b = (aa')x + (ab' + b)$$ For this to be identity: $$aa' = 1 \quad \text{and} \quad ab' + b = 0$$ So $a \neq 0$ and $$a' = \frac{1}{a}, \quad b' = -\frac{b}{a}$$ 8. **Conclusion:** - If $a = 0$, $f_{a,b}$ is not invertible. - Therefore, $(A, \circ)$ is a monoid but not a group because not all elements have inverses. **Final answer:** - $(A, \circ)$ is a monoid. - $(A, \circ)$ is not a group.