For real numbers x and y , define x R y if and only if x - y + 2 is an irrational number. Then the relation R is:

Question

For real numbers $x$ and $y$ , define $x R y$ if and only if $x - y + \sqrt{2}$ is an irrational number. Then the relation $R$ is:

Smart Achievers · Accepted Answer

(1)

Ans. reflexive

Explanation:

$R_{x} = x - x + \sqrt{2} = \sqrt{2}$ is an irrational number.

Thus, $R$ is reflexive.

Also, $(\sqrt{2}, 1) \in R$ as $\sqrt{2} - 1 + \sqrt{2} = 2 \sqrt{2} - 1$ is an irrational number, but $(1, \sqrt{2}) \notin R$ as $1 - \sqrt{2} + \sqrt{2} = 1$ is a rational number. So $R$ is not symmetric. Since, ${}_{1}R_{2}$ and ${}^{\sqrt{2}}R_{\sqrt{2}}$ but 1 is not related to $\sqrt{2}$ . So, $R$ is not transitive.

Solution

Q.
For real numbers $x$ and $y$ , define $x R y$ if and only if $x - y + \sqrt{2}$ is an irrational number. Then the relation $R$ is:

1 reflexive

2 symmetric

3 transitive

4 equivalence

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