This problem involves compositions of relations between sets. Let's break it down step by step.

Given Information:

• Set $A = \{1, 2, 3\}$
• Set $B = \{4, 5\}$
• Set $C = \{a, b, c, d\}$
• Relation $R = \{(1, b), (2, a), (2, b), (2, c), (3, d)\}$, which is a relation from $A$ to $C$
• Relation $S = \{(4, a), (4, d), (5, b), (5, c)\}$, which is a relation from $B$ to $C$

Problem (a):

Find $S^{-1} \circ R$. This is a relation from which set to which other set? Justify your solution.

• $S^{-1}$ is the inverse of relation $S$, which means we swap the pairs in $S$. So, $S^{-1} = \{(a, 4), (d, 4), (b, 5), (c, 5)\}$

• Now, we compute the composition $S^{-1} \circ R$, which means we need to find all pairs $(x, y)$ such that there is some $z$ where $(x, z) \in R$ and $(z, y) \in S^{-1}$.

• From $R$, let's find the connections between $A$ and $B$ through $C$:

  • $(1, b) \in R$ and $(b, 5) \in S^{-1}$, so $(1, 5) \in S^{-1} \circ R$.
  • $(2, a) \in R$ and $(a, 4) \in S^{-1}$, so $(2, 4) \in S^{-1} \circ R$.
  • $(2, b) \in R$ and $(b, 5) \in S^{-1}$, so $(2, 5) \in S^{-1} \circ R$.
  • $(2, c) \in R$ and $(c, 5) \in S^{-1}$, so $(2, 5) \in S^{-1} \circ R$ (again).
  • $(3, d) \in R$ and $(d, 4) \in S^{-1}$, so $(3, 4) \in S^{-1} \circ R$.

• Therefore, the composition $S^{-1} \circ R = \{(1, 5), (2, 4), (2, 5), (3, 4)\}$.

Problem (b):

Find $R^{-1} \circ S$. This is a relation from which set to which other set? Justify your solution.

• $R^{-1}$ is the inverse of relation $R$, which means we swap the pairs in $R$. So, $R^{-1} = \{(b, 1), (a, 2), (b, 2), (c, 2), (d, 3)\}$

• Now, we compute the composition $R^{-1} \circ S$, which means we need to find all pairs $(x, y)$ such that there is some $z$ where $(x, z) \in S$ and $(z, y) \in R^{-1}$.

• From $S$, let's find the connections between $B$ and $A$ through $C$:

  • $(4, a) \in S$ and $(a, 2) \in R^{-1}$, so $(4, 2) \in R^{-1} \circ S$.
  • $(4, d) \in S$ and $(d, 3) \in R^{-1}$, so $(4, 3) \in R^{-1} \circ S$.
  • $(5, b) \in S$ and $(b, 1) \in R^{-1}$, so $(5, 1) \in R^{-1} \circ S$.
  • $(5, b) \in S$ and $(b, 2) \in R^{-1}$, so $(5, 2) \in R^{-1} \circ S$.
  • $(5, c) \in S$ and $(c, 2) \in R^{-1}$, so $(5, 2) \in R^{-1} \circ S$ (again).

• Therefore, the composition $R^{-1} \circ S = \{(4, 2), (4, 3), (5, 1), (5, 2)\}$.

Summary:

• $S^{-1} \circ R$ is a relation from $A$ to $B$.
• $R^{-1} \circ S$ is a relation from $B$ to $A$.

Would you like a further breakdown of any part of the solution?

Here are 5 related questions:

1. How do we define the composition of relations formally?
2. What happens when both relations map between the same sets?
3. Can a composition of relations result in an empty set? If so, when?
4. How does the composition of functions differ from the composition of relations?
5. What is the difference between left and right composition of relations?

Tip: When calculating the composition of relations, be sure to match pairs by the middle element (the common set between the two relations).