Animated Solution for Mathematics - Sets and Relations: If $R=\{(x,y):x,y\in Z,x^2+3y^2\le 8\}$ is a relation on the set of integers $Z$, then the domain of $R^{-1}$ is:

Unlocking the Relation

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure.

When you see a relation defined by an inequality like $x^2+3y^2\le 8$, your first instinct might be to panic or start plugging in random numbers. But let's pause and look at the elegance hidden within this constraint.

The Inverse Insight

The question asks for the domain of $R^{-1}$. Many students stumble here, trying to invert the inequality itself.

But remember the fundamental definition of an inverse relation: if $(x,y)\in R$, then $(y,x)\in R^{-1}$. The domain of $R^{-1}$ is simply the set of all first coordinates of the inverse relation, which are the second coordinates of the original relation.

In plain English: The domain of $R^{-1}$ is the range of $R$. Our entire objective is now to find all possible integer values of $y$ that can exist in the relation $R$.

The Elliptical Constraint

Look at the inequality: $x^2+3y^2\le 8$. If we were in the realm of real numbers, this would describe the interior of an ellipse.

But we are in the discrete world of integers, $\mathbb{Z}$. This means we are looking for grid points $(x,y)$ that fall inside or on this ellipse. To find the range, we need to isolate $y$. Let's rearrange the inequality:

Here is where the magic happens. We know that for any integer $x$, the square $x^2$ is always non-negative ($x^2\ge 0$).

If $x^2$ is at least $0$, then the expression $8-x^2$ can never be greater than $8$. Therefore, we can establish a strict upper bound for $y$:

The Integer Filter

Now, let's solve for $y^2$:

Since $y$ must be an integer, $y^2$ must also be a perfect square integer. What are the perfect squares less than or equal to $2.66$? Only $0$ and $1$.

This is a massive simplification! It means $y$ can only be $0$, $1$, or $-1$. Any other integer value for $y$ would force $x^2$ to be negative, which is impossible for real integers.

The Final Verification

We have our candidates: $y\in \{-1,0,1\}$. We must verify that for each of these, there exists at least one integer $x$ that satisfies the original inequality.

1. For $y=0$: The inequality becomes $x^2\le 8$. The integers $x$ satisfying this are $\{-2,-1,0,1,2\}$. Since solutions exist, $y=0$ is in the range.

2. For $y=1$: The inequality becomes $x^2+3(1{)}^2\le 8$, which simplifies to $x^2\le 5$. The integers $x$ satisfying this are $\{-2,-1,0,1,2\}$. Since solutions exist, $y=1$ is in the range.

3. For $y=-1$: The inequality becomes $x^2+3(-1{)}^2\le 8$, which again simplifies to $x^2\le 5$. The integers $x$ satisfying this are $\{-2,-1,0,1,2\}$. Since solutions exist, $y=-1$ is in the range.

We have rigorously tested our candidates. The range of $R$ is $\{-1,0,1\}$.

Because the domain of $R^{-1}$ is the range of $R$, our final answer is the set $\{-1,0,1\}$. By using logic to bound the variables, we turned a potentially messy problem into a clear, logical path.