Let's break down the problem step by step for the given relation $R$ defined on $\mathbb{N}$ (the set of natural numbers) by the equation $x + 3y = 12$.

(a) Write $R$ as a set of ordered pairs.

The relation $R$ consists of ordered pairs $(x, y)$ such that $x + 3y = 12$ holds. Since $x$ and $y$ are natural numbers ($\mathbb{N}$), we need to find all such pairs that satisfy this equation.

Let's solve the equation $x + 3y = 12$ for natural number values of $y$:

• For $y = 0$: $x + 3(0) = 12 \Rightarrow x = 12$.
• For $y = 1$: $x + 3(1) = 12 \Rightarrow x = 9$.
• For $y = 2$: $x + 3(2) = 12 \Rightarrow x = 6$.
• For $y = 3$: $x + 3(3) = 12 \Rightarrow x = 3$.
• For $y = 4$: $x + 3(4) = 12 \Rightarrow x = 0$.

Hence, the relation $R$ as a set of ordered pairs is:

$R = \{ (12, 0), (9, 1), (6, 2), (3, 3), (0, 4) \}.$

(b) Find the domain and range of $R$.

• Domain: The domain of $R$ is the set of all possible values of $x$ that appear in the ordered pairs. From the set of ordered pairs $R$, the domain is: $\text{Domain of } R = \{ 12, 9, 6, 3, 0 \}.$

• Range: The range of $R$ is the set of all possible values of $y$ that appear in the ordered pairs. From the set of ordered pairs $R$, the range is: $\text{Range of } R = \{ 0, 1, 2, 3, 4 \}.$

(c) Find $R^{-1}$.

The inverse of $R$, denoted $R^{-1}$, is the set of ordered pairs obtained by swapping the elements of each pair in $R$. So:

$R^{-1} = \{ (y, x) | (x, y) \in R \}.$

Thus, by swapping the elements of each pair in $R$:

$R^{-1} = \{ (0, 12), (1, 9), (2, 6), (3, 3), (4, 0) \}.$

(d) Find the composition relation $R \circ R$.

The composition $R \circ R$ means we are finding all pairs $(x, z)$ such that there exists a $y$ where $(x, y) \in R$ and $(y, z) \in R$.

We will go through the ordered pairs in $R$:

1. $(12, 0)$: There is no pair in $R$ where the first element is 0, so no pair for $x = 12$.
2. $(9, 1)$: There is no pair in $R$ where the first element is 1, so no pair for $x = 9$.
3. $(6, 2)$: There is no pair in $R$ where the first element is 2, so no pair for $x = 6$.
4. $(3, 3)$: The pair $(3, 3)$ exists, so we have $(3, 3) \in R \circ R$.
5. $(0, 4)$: There is no pair in $R$ where the first element is 4, so no pair for $x = 0$.

Thus, the composition relation $R \circ R$ is:

$R \circ R = \{ (3, 3) \}.$

Summary of Answers:

(a) $R = \{ (12, 0), (9, 1), (6, 2), (3, 3), (0, 4) \}$.

(b) Domain of $R = \{ 12, 9, 6, 3, 0 \}$, Range of $R = \{ 0, 1, 2, 3, 4 \}$.

(c) $R^{-1} = \{ (0, 12), (1, 9), (2, 6), (3, 3), (4, 0) \}$.

(d) $R \circ R = \{ (3, 3) \}$.

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Do you have any questions or would you like more details?

Here are 5 related questions to consider:

1. How does the domain and range of a relation relate to the inverse of that relation?
2. What is the significance of the composition of two relations?
3. Can you provide an example where $R \circ R$ includes more than one pair?
4. How would the relation change if the equation was $x + 2y = 10$?
5. What would happen if the relation was defined over integers instead of natural numbers?

Tip: Inverse relations swap the roles of inputs and outputs. Always check how this affects both domain and range!