First slide

Relations XII

Question

For real numbers x and y, we write $xRy \Leftrightarrow x - y + \sqrt{2}$ is an irrational number. Then, the relation R is

Difficult

A

reflexive
B

Not symmetric
C

transitive
D

Not transitive

Solution

$xRy \Leftrightarrow x - y + \sqrt{2}$ is an irrational number.
Let $(x, x) \in R$
Then, $x - x + \sqrt{2} = \sqrt{2}$ which is an irrational number
$∴ xR \times, \forall x \in R$
R is an reflexive relation.
R is not symmetric since $(\sqrt{2}, 1) \in R \sin c e \sqrt{2} - 1 + \sqrt{2} i s i r r a t i o n a l b u t (1, \sqrt{2}) \notin R \sin c e 1 - \sqrt{2} + \sqrt{2} = 1, w h i c h i s r a t i o n a l$
R is not transitive because $(1, \sqrt{3}) \in R, since (1 - \sqrt{3} + \sqrt{2})$ is an irrational number
and $(\sqrt{3}, \sqrt{2}) \in R, since (\sqrt{3} - \sqrt{2} + \sqrt{2})$ = $\sqrt{3}$ is an irrational number.
but $(1, \sqrt{2}) \notin R sinc e (1 - \sqrt{2} + \sqrt{2}) = 1$ is a rational number.
$∴$ R is not transitive relation.

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