A relation R is defined on the set Z of integers as follows- x- y - R- x2-y2- 25 Express R and R-1 as the sets of ordered pairs and hence find their respective domains-

y= ±√(25 x^2) where both x and y are integers.Cleraly for x= 0,± 3, ± 4, ± 5 , y will be integers as y = ± 5, ± 4, ± 3, 0 ∴ R = (0, ± 5), (± ...

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Standard XII

Mathematics

Domain

A relation R ...

Question

A relation R is defined on the set Z of integers as follows: (x, y) $\in R^{'} x^{2} + y^{2} = 25$
Express R and $R^{- 1}$ as the sets of ordered pairs and hence find their respective domains.

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Solution

$y = \pm \sqrt{25 - x^{2}}$
where both x and y are integers.
Cleraly for $x = 0, \pm 3, \pm 4, \pm 5$ , y will be integers as
$y = \pm 5, \pm 4, \pm 3, 0$
$∴ R = (0, \pm 5), (\pm 3, \pm 4), (\pm 4, \pm 3), (\pm 5, 0)$
$R^{- 1} = (\pm 5, 0), (\pm 4, \pm 3), (\pm 3, \pm 4), (0, \pm 5)$
Each R and $R^{- 1}$ consists of 2+ 4 + 4 +2 = 12 ordered pairs.
Domain of R ={ $0, \pm 3, \pm 4, \pm 5$ } = domain of $R^{- 1}$ .

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A relation R is defined on the set Z of integers as follows: (x, y)∈R′x2+y2=25 Express R and R−1 as the sets of ordered pairs and hence find their respective domains.

y=±√25−x2where both x and y are integers.Cleraly for x=0,±3,±4,±5, y will be integers asy=±5,±4,±3,0∴R=(0,±5),(±3,±4),(±4,±3),(±5,0)R−1=(±5,0),(±4,±3),(±3,±4),(0,±5)Each R and R−1 consists of 2+ 4 + 4 +2 = 12 ordered pairs.Domain of R ={0,±3,±4,±5} = domain of R−1.

A relation R is defined on the set Z of integers as follows: (x, y) $\in R^{'} x^{2} + y^{2} = 25$
Express R and $R^{- 1}$ as the sets of ordered pairs and hence find their respective domains.