Let a relation R on be defined as: if and only if Consider the two statements: (I) R is reflexive but not symmetric. (II) R is transitive. Then which one of the following is true?

Question

Let a relation R on $N \times N$ be defined as:

$(x_{1}, y_{1}) R (x_{2}, y_{2})$ if and only if $x_{1} \leq x_{2} or y_{1} \leq y_{2} .$

Consider the two statements:

(I) R is reflexive but not symmetric.

(II) R is transitive.

Then which one of the following is true? [2024]

Smart Achievers · Accepted Answer

(3)

$(x_{1}, y_{1}) R (x_{2}, y_{2})$

$x_{1} \leq x_{2} or y_{1} \leq y_{2} .$

For reflexive :

$(x_{1}, y_{1}) R (x_{1}, y_{1})$

$\Rightarrow x_{1} \leq x_{1} or y_{1} \leq y_{1} which is true.$

So, R is reflexive.

For symmetric:

$When (x_{1}, y_{1}) R (x_{2}, y_{2})$

$\Rightarrow x_{1} \leq x_{2} or y_{1} \leq y_{2}$

$⇏ x_{2} \leq x_{1} or y_{2} \leq y_{1}$

They may or may not be true.

For example (1, 2) and (3, 4)

$1 \leq 3 and 2 \leq 4 but 3 ≰ 1 and 4 ≰ 2 .$

$∴$ R is not symmetric.

For transitive :

Take pairs as (3, 9), (4, 6), (2, 7)

$(3, 9) R (4, 6) as 4 \geq 3$

$(4, 6) R (2, 7) as 7 \geq 6$

$But (3, 9) R (n o t) (2, 7), as neither 2 \geq 3 nor 7 \geq 9$

So, R is not transitive.

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