Key Concepts and Formulas

1. Relation: A relation $R$ on a set $Z$ is a subset of $Z \times Z$.
2. Domain of a Relation: For a relation $R$, the domain, $\text{Dom}(R)$, is the set of all first elements of the ordered pairs in $R$.
3. Range of a Relation: For a relation $R$, the range, $\text{Rng}(R)$, is the set of all second elements of the ordered pairs in $R$.
4. Inverse Relation: If $R = \{(x, y) : (x, y) \in R\}$, then $R^{-1} = \{(y, x) : (x, y) \in R\}$.
5. Property of Inverse Relation: The domain of $R^{-1}$ is the range of $R$, i.e., $\text{Dom}(R^{-1}) = \text{Rng}(R)$.

Step-by-Step Solution

Step 1: Understand the Given Relation and the Goal The relation is given by $R = \{(x, y) : x, y \in Z, x^2 + 3y^2 \le 8\}$. We are asked to find the domain of the inverse relation $R^{-1}$. According to the property of inverse relations, $\text{Dom}(R^{-1}) = \text{Rng}(R)$. Therefore, our goal is to find all possible integer values of $y$ for which there exists at least one integer $x$ satisfying the inequality $x^2 + 3y^2 \le 8$.

Step 2: Systematically Find Possible Integer Values for $y$ Since $x, y$ are integers and $x^2 \ge 0$, the term $3y^2$ must be less than or equal to 8. Let's examine possible integer values for $y$:

• If $y = 0$: The inequality becomes $x^2 + 3(0)^2 \le 8$, which simplifies to $x^2 \le 8$. The integer values of $x$ satisfying $x^2 \le 8$ are $x \in \{-2, -1, 0, 1, 2\}$. Since we found integer values for $x$ when $y=0$, $y=0$ is a possible value in the range of $R$.

• If $y = 1$: The inequality becomes $x^2 + 3(1)^2 \le 8$, which simplifies to $x^2 + 3 \le 8$, so $x^2 \le 5$. The integer values of $x$ satisfying $x^2 \le 5$ are $x \in \{-2, -1, 0, 1, 2\}$. Since we found integer values for $x$ when $y=1$, $y=1$ is a possible value in the range of $R$.

• If $y = -1$: The inequality becomes $x^2 + 3(-1)^2 \le 8$, which simplifies to $x^2 + 3 \le 8$, so $x^2 \le 5$. The integer values of $x$ satisfying $x^2 \le 5$ are $x \in \{-2, -1, 0, 1, 2\}$. Since we found integer values for $x$ when $y=-1$, $y=-1$ is a possible value in the range of $R$.

• If $y = 2$: The inequality becomes $x^2 + 3(2)^2 \le 8$, which simplifies to $x^2 + 12 \le 8$, so $x^2 \le -4$. There are no real numbers $x$ (and hence no integers) that satisfy $x^2 \le -4$. Thus, $y=2$ is not a possible value in the range of $R$.

• If $y = -2$: The inequality becomes $x^2 + 3(-2)^2 \le 8$, which simplifies to $x^2 + 12 \le 8$, so $x^2 \le -4$. Again, there are no real numbers $x$ that satisfy $x^2 \le -4$. Thus, $y=-2$ is not a possible value in the range of $R$.

For any integer $y$ such that $|y| \ge 2$, $3y^2 \ge 3(2^2) = 12$. This will always lead to $x^2 \le 8 - 3y^2 < 0$, which has no real solutions for $x$. Therefore, we have considered all possible integer values for $y$.

Step 3: Determine the Range of $R$ The possible integer values for $y$ that allow for integer values of $x$ to satisfy the inequality are $\{-1, 0, 1\}$. This set constitutes the range of the relation $R$. So, $\text{Rng}(R) = \{-1, 0, 1\}$.

Step 4: Determine the Domain of $R^{-1}$ Using the property $\text{Dom}(R^{-1}) = \text{Rng}(R)$, we find that the domain of $R^{-1}$ is $\{-1, 0, 1\}$.

Common Mistakes & Tips

• Confusing Domain and Range: Remember that the domain of $R^{-1}$ is the range of $R$, not the domain of $R$.
• Forgetting Integer Constraint: Ensure that both $x$ and $y$ are integers. The inequality $x^2 \le 8$ has many non-integer solutions, but only integers are relevant here.
• Inefficient Search: Systematically checking values for $y$ (the variable with the larger coefficient) first is more efficient as it restricts the possibilities more quickly.

Summary To find the domain of the inverse relation $R^{-1}$, we need to determine the range of the original relation $R$. The relation $R$ is defined by $x^2 + 3y^2 \le 8$ for integers $x$ and $y$. We systematically tested integer values for $y$ to see if corresponding integer values for $x$ exist. We found that $y$ can only be $-1, 0,$ or $1$ for integer $x$ to satisfy the inequality. Therefore, the range of $R$ is $\{-1, 0, 1\}$, which is also the domain of $R^{-1}$.

The final answer is $\boxed{\text{{-1, 0, 1}}}$.