Q.Given $R = \{(x, y) : x, y \in \mathbf{W},\ x^2 + y^2 = 25\}$. Find the domain and Range of $R$.

The key idea is to identify all ordered pairs $(x, y)$ that satisfy the given condition $x^2 + y^2 = 25$ where both $x$ and $y$ belong to the set of whole numbers $\mathbf{W} = \{0, 1, 2, 3, \dots\}$. The domain is the set of all first components of these pairs, and the range is the set of all second components.

1. We systematically test whole number values for $x$ to find corresponding whole number values for $y$:
   • If $x=0$, $0^2 + y^2 = 25 \implies y^2 = 25 \implies y=5$ (since $y \in \mathbf{W}$). So $(0, 5)$ is a pair.
   • If $x=1$, $1^2 + y^2 = 25 \implies y^2 = 24$, which is not a perfect square, so $y \notin \mathbf{W}$.
   • If $x=2$, $2^2 + y^2 = 25 \implies y^2 = 21$, which is not a perfect square, so $y \notin \mathbf{W}$.
   • If $x=3$, $3^2 + y^2 = 25 \implies y^2 = 16 \implies y=4$ (since $y \in \mathbf{W}$). So $(3, 4)$ is a pair.
   • If $x=4$, $4^2 + y^2 = 25 \implies y^2 = 9 \implies y=3$ (since $y \in \mathbf{W}$). So $(4, 3)$ is a pair.
   • If $x=5$, $5^2 + y^2 = 25 \implies y^2 = 0 \implies y=0$ (since $y \in \mathbf{W}$). So $(5, 0)$ is a pair. …

Let's break down what the problem asks for and how to approach it.

A relation is simply a set of ordered pairs $(x, y)$ that satisfy a given condition. In this problem, the condition is $x^2 + y^2 = 25$, and the elements $x$ and $y$ must belong to the set of whole numbers, $\mathbf{W}$.

The domain of a relation is the set of all first elements (the $x$-values) of the ordered pairs in the relation.

The range of a relation is the set of all second elements (the $y$-values) of the ordered pairs in the relation.

To find the domain and range, our primary task is to identify all the ordered pairs $(x, y)$ that satisfy the given condition. We can think of this as systematically "testing" possible values for $x$ and $y$ from the set $\mathbf{W}$ to see which ones fit the rule. While an arrow diagram is a visual representation of a relation, it's typically drawn after we've identified the specific ordered pairs. For a relation defined by an equation, the most direct approach is to list these pairs first.

Here's how we find the domain and range for the given relation:

1. Understand the set of numbers involved:

   The problem states that $x, y \in \mathbf{W}$. The set of whole numbers $\mathbf{W}$ includes all non-negative integers: $\{0, 1, 2, 3, \dots\}$. This is a crucial piece of information, as it restricts our search for $x$ and $y$ values.

2. Identify constraints on $x$ and $y$ from the equation:

   We have the equation $x^2 + y^2 = 25$.

   Since $x$ and $y$ are whole numbers, $x^2$ and $y^2$ must also be non-negative.

   This implies:

   • $x^2 \le 25 \implies x \le \sqrt{25} \implies x \le 5$.
   • $y^2 \le 25 \implies y \le \sqrt{25} \implies y \le 5$. So, we only need to check whole number values for $x$ and $y$ from $0$ to $5$.

3. Systematically find all ordered pairs $(x, y)$ that satisfy $x^2 + y^2 = 25$:

   We can iterate through possible whole number values for $x$ from $0$ to $5$ and see if a corresponding whole number $y$ exists.

   • If $x = 0$:

     $0^2 + y^2 = 25$

     $0 + y^2 = 25$

     $y^2 = 25$

     $y = \sqrt{25} = 5$ (since $y \in \mathbf{W}$, we take the positive root).

     This gives us the ordered pair $(0, 5)$.

   • If $x = 1$:

     $1^2 + y^2 = 25$

     $1 + y^2 = 25$

     $y^2 = 24$

     $y = \sqrt{24}$. This is not a whole number. So, no pair for $x=1$.

   • If $x = 2$:

     $2^2 + y^2 = 25$

     $4 + y^2 = 25$

     $y^2 = 21$

     $y = \sqrt{21}$. This is not a whole number. So, no pair for $x=2$.

   • If $x = 3$:

     $3^2 + y^2 = 25$ …