Given the sets $A = \{1, 2, 3, 4, 5, 6, 7\}$ and $B = \{1, 2, 3, 4, 5\}$, and the relations $R_1$ and $R_2$:

• $R_1 = \{(a, b) \mid a > b\}$
• $R_2 = \{(a, b) \mid a \geq b\}$

The problem asks to find the compositions $R_2 \circ R_1$ and $R_1 \circ R_2$.

1. Composition $R_2 \circ R_1$

The composition $R_2 \circ R_1$ is defined as:

$R_2 \circ R_1 = \{(a, c) \mid \exists b \text{ such that } (a, b) \in R_1 \text{ and } (b, c) \in R_2\}$

This means for every pair $(a, c)$ in the composition, there must be an intermediate element $b$ such that $a > b$ (from $R_1$) and $b \geq c$ (from $R_2$).

2. Composition $R_1 \circ R_2$

Similarly, the composition $R_1 \circ R_2$ is defined as:

$R_1 \circ R_2 = \{(a, c) \mid \exists b \text{ such that } (a, b) \in R_2 \text{ and } (b, c) \in R_1\}$

This means for every pair $(a, c)$ in the composition, there must be an intermediate element $b$ such that $a \geq b$ (from $R_2$) and $b > c$ (from $R_1$).

Solution Strategy

To solve this problem, we need to evaluate each composition by checking the conditions on the elements of the sets $A$ and $B$. Specifically:

1. For $R_2 \circ R_1$:

   • Find all pairs $(a, c)$ where $a > b$ and $b \geq c$.

2. For $R_1 \circ R_2$:

   • Find all pairs $(a, c)$ where $a \geq b$ and $b > c$.

I'll calculate both compositions now.The compositions $R_2 \circ R_1$ and $R_1 \circ R_2$ yield the following sets of pairs:

1. Composition $R_2 \circ R_1$:

$R_2 \circ R_1 = \{(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5)\}$

2. Composition $R_1 \circ R_2$:

$R_1 \circ R_2 = \{(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4), (6, 1), (6, 2), (6, 3), (6, 4), (7, 1), (7, 2), (7, 3), (7, 4)\}$

Observations:

• $R_2 \circ R_1$ contains more pairs, specifically those involving the largest elements in $A$, where the final element $c$ can be equal to the intermediate element $b$.
• $R_1 \circ R_2$ is missing the pairs $(6, 5)$ and $(7, 5)$ because there is no intermediate element $b$ where $b > 5$.

If you'd like further details or have questions, feel free to ask!

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Here are 5 related questions for deeper exploration:

1. How would the compositions change if $A$ and $B$ had different elements?
2. What happens if we reverse the conditions for $R_1$ and $R_2$?
3. Can you generalize a formula to compute the compositions for any sets $A$ and $B$?
4. How does the concept of composition relate to functions in mathematics?
5. What is the effect of restricting the relations to specific subsets of $A$ and $B$?

Tip: Composition of relations can be seen as chaining conditions, where the output of one relation serves as the input to another. This is conceptually similar to function composition.