Question:
If R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4} is a relation on Z, then the domain of R is
(a) [0, 1, 2]
(b) [0, −1, −2]
(c) {−2, −1, 0, 1, 2]
(d) None of these
Solution:
(c) {−2, −1, 0, 1, 2}
R = {(x, y) : x, y ∈ Z, x2 + y2 ≤ 4}
We know that,
$(-2)^{2}+0^{2} \leq 4$
$\Rightarrow(2)^{2}+0^{2} \leq 4$
$\Rightarrow(-1)^{2}+0^{2} \leq 4$
$\Rightarrow(-1)^{2}+0^{2} \leq 4$
$\Rightarrow(1)^{2}+0^{2} \leq 4$
$\Rightarrow 0^{2}+0^{2} \leq 4$
$\Rightarrow(1)^{2}+(1)^{2} \leq 4$
$\Rightarrow(-1)^{2}+(-1)^{2} \leq 4$
Hence, domain $(\mathrm{R})=\{-2,-1,0,1,2,$,
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