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A binary relation $R$ on $\mathbb{N} \times \mathbb{N}$ is defined as follows: $(a, b) R(c, d)$ if $a \leq c$ or $b \leq d$. Consider the following propositions:

  • $P:$ $R$ is reflexive.
  • $Q:$ $R$ is transitive.

Which one of the following statements is TRUE?

  1. Both $P$ and $Q$ are true.
  2. $P$ is true and $Q$ is false.
  3. $P$ is false and $Q$ is true.
  4. Both $P$ and $Q$ are false.
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Best answer
91 votes
91 votes

$(B)$ Reflexive, but not transitive.

it is "$a \leq c$ OR $b \leq d$",

NOT
"$a \leq c$ AND $b \leq d$"

$(2,5) R (6, 3), \quad (6,3)R (1, 4),$ but $(2,5) \not R (1, 4)$

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25 votes
25 votes

Given relation  (a,b)R(c,d) if a≤c or b≤d is satifies ,

            1.Reflexive property :YES ,bcoz (a,b)R(a,b)

            2.Transitive property :NO  ,counter example  (4,2)R(1,4) and (1,4)R(3,1) but not (4,2)R(3,1).

            3.Symmetric property:NO ,counter eg (1,2)R(3,4) but not (3,4)R(1,2)

Correct ans is,(B) P is true and Q is false.

9 votes
9 votes

It is obvious that it will be it will be reflexive by definition bcz if a relation "<=" then such relation must be reflexive and when we read about the definition about Transititve we found that "A relation "< || > || subset || superset || \" always be transitive so P true and Q is false.

Above is by Theory.

Now in transitive relation it is always given that if a<b,b<c then a<c but there is no such condition is reflecting in above.

4 votes
4 votes
$(a,b)R(a,b)$ is always true , since $a<=a \ and\  b <= b$ is always true , thus the relation is reflexive for sure.

According to the definition of transitivity :-

$If \ (a,b)R(c,d) \ and \ (c,d)R(e,f) \ then (a,b)R(e,f) \ must \ also \ hold \ true.$

Here , the statement is in the form of an implication ,

If A then B.

Thus ,

$(\ (a,b)R(c,d) \ and \ (c,d)R(e,f) )\ => (a,b)R(e,f) $

If the RHS can be proved to be false and LHS can be proved to be true at the same point of time we can surely say that the statement is invalid and thus the relation in not transitive.

Let RHS is false.

Thus , $((a,b),(e,f)) \notin R => a>e \ and \ b>f.$

Let's reverse engineer this .

Let $a = 4 , b=5 ,e=3 \ and \ f=3$.

It can be easily shown that:-

$(4,5)R(3,6) \ holds \ and \ (3,6)R(3,3) \ holds \ but \ (4,5)R(3,3) \ doesn't \ hold. $

Thus we can show that the LHS of the implication is true , and RHS is false at the same point of time , thus the statement is invalid.

Thus , the relation is not following the property of transitivity.
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