If R = { (x, y) | x, y in Z, x^2 + y^2 le 4 } | vedclass

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(C) The relation is defined as $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$.To find the domain,we identify all possible values of $x \in Z$ such t

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$\{-2, -1, 0, 1, 2\}$

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(C) The relation is defined as $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$.To find the domain,we identify all possible values of $x \in Z$ such that there exists at least one $y \in Z$ satisfying $x^2 + y^2 \le 4$.If $x = 0$,then $y^2 \le 4 \implies y \in \{-2, -1, 0, 1, 2\}$.If $x = 1$,then $1^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.If $x = -1$,then $(-1)^2 + y^2 \le 4 \implies y^2 \le 3 \implies y \in \{-1, 0, 1\}$.If $x = 2$,then $2^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.If $x = -2$,then $(-2)^2 + y^2 \le 4 \implies 4 + y^2 \le 4 \implies y^2 \le 0 \implies y = 0$.If $|x| > 2$,then $x^2 > 4$,so $x^2 + y^2 > 4$ for any $y \in Z$.Thus,the set of all possible values of $x$ is $\{-2, -1, 0, 1, 2\}$.Therefore,the domain of $R$ is $\{-2, -1, 0, 1, 2\}$.

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is

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