Let a relation R on N \times \mathbbN 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?

Neither (I) nor (II) is correct.

Sets and Relations PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q48 with Solution

View written solutionFree

Correct answer: D

Given relation

On $\mathbb{N}\times \mathbb{N}$ , define

(x_1,y_1)R(x_2,y_2) \iff x_1\le x_2 \;\text{or}\; y_1\le y_2.

We must check the truth of:

(I) $R$ is reflexive but not symmetric.
(II) $R$ is transitive.

Check reflexivity

A relation is reflexive if every element is related to itself.

Take any $(x,y)\in \mathbb N\times \mathbb N$ . Then

(x,y)R(x,y)

means

x\le x \quad \text{or} \quad y\le y.

Both are true, so certainly the relation holds.

Hence, $R$ is reflexive.

Check symmetry

A relation is symmetric if

(a,b)R(c,d) \implies (c,d)R(a,b).

We need only one counterexample to show it is not symmetric.

Take

(1,3),\ (2,1).

Now,

(1,3)R(2,1)

because

1\le 2

is true.

But for the reverse,

(2,1)R(1,3)

would require

2\le 1 \quad \text{or} \quad 1\le 3.

Since $1\le 3$ is true, this example does not fail symmetry. So we need a better counterexample.

Let us choose points so that exactly one direction works.

Take

(2,3),\ (1,2).

Then

(2,3)R(1,2)

requires

2\le 1 \quad \text{or} \quad 3\le 2,

which is false. So not useful.

Try

(1,3),\ (2,2).

Then

(1,3)R(2,2)

because $1\le 2$ is true.

Reverse:

(2,2)R(1,3)

requires

2\le 1 \quad \text{or} \quad 2\le 3.

Again true. Still symmetric for this pair.

To break symmetry, we need

x_1\le x_2 \text{ or } y_1\le y_2

true, but

x_2\le x_1 \text{ or } y_2\le y_1

false.

For the reverse to be false, both must fail:

x_2>x_1 \quad \text{and} \quad y_2>y_1.

But then in the forward direction we have $x_1<x_2$ and $y_1<y_2$ , so forward is true. Good.

Take

(1,1),\ (2,2).

Then

(1,1)R(2,2)

because $1\le 2$ (also $1\le 2$ in second coordinate).

But

(2,2)R(1,1)

requires

2\le 1 \quad \text{or} \quad 2\le 1,

which is false.

Hence $R$ is not symmetric.

So statement (I) is true.

Check transitivity

A relation is transitive if

(a,b)R(c,d) \text{ and } (c,d)R(e,f) \implies (a,b)R(e,f).

We test whether this always holds.

We need a counterexample.

Take

A=(2,1),\quad B=(1,3),\quad C=(3,2).

Now check:

$A R B$ ?
$(2,1)R(1,3) \iff 2\le 1 \text{ or } 1\le 3.$
Since $1\le 3$ is true, $A R B$ holds.
$B R C$ ?
$(1,3)R(3,2) \iff 1\le 3 \text{ or } 3\le 2.$
Since $1\le 3$ is true, $B R C$ holds.
$A R C$ ?
$(2,1)R(3,2) \iff 2\le 3 \text{ or } 1\le 2.$
This is actually true, so this is not a counterexample.

Let us try to force the final relation to fail. For

(x_1,y_1)R(x_3,y_3)

to fail, we need

x_1>x_3 \quad \text{and} \quad y_1>y_3.

Choose:

A=(3,3),\quad B=(1,4),\quad C=(2,2).

Then

$A R B$ :
$(3,3)R(1,4) \iff 3\le 1 \text{ or } 3\le 4.$
True because $3\le 4$ .
$B R C$ :
$(1,4)R(2,2) \iff 1\le 2 \text{ or } 4\le 2.$
True because $1\le 2$ .
$A R C$ :
$(3,3)R(2,2) \iff 3\le 2 \text{ or } 3\le 2,$
which is false.

Thus,

A R B \text{ and } B R C \text{ but } A \not R C.

So $R$ is not transitive.

Hence statement (II) is false.

Conclusion

(I) is true.
(II) is false.

Therefore, the correct option is:

\boxed{\text{D: Only (I) is correct.}}

Comparison with stored answer

Stored correct answer: $\boxed{D}$

My derived answer is also $\boxed{D}$ , so they agree.

More from Sets and Relations

View written solutionFree

Correct answer: D

Given relation

On $\mathbb{N}\times \mathbb{N}$ , define

(x_1,y_1)R(x_2,y_2) \iff x_1\le x_2 \;\text{or}\; y_1\le y_2.

We must check the truth of:

(I) $R$ is reflexive but not symmetric.
(II) $R$ is transitive.

Check reflexivity

A relation is reflexive if every element is related to itself.

Take any $(x,y)\in \mathbb N\times \mathbb N$ . Then

(x,y)R(x,y)

means

x\le x \quad \text{or} \quad y\le y.

Both are true, so certainly the relation holds.

Hence, $R$ is reflexive.

Check symmetry

A relation is symmetric if

(a,b)R(c,d) \implies (c,d)R(a,b).

We need only one counterexample to show it is not symmetric.

Take

(1,3),\ (2,1).

Now,

(1,3)R(2,1)

because

1\le 2

is true.

But for the reverse,

(2,1)R(1,3)

would require

2\le 1 \quad \text{or} \quad 1\le 3.

Since $1\le 3$ is true, this example does not fail symmetry. So we need a better counterexample.

Let us choose points so that exactly one direction works.

Take

(2,3),\ (1,2).

Then

(2,3)R(1,2)

requires

2\le 1 \quad \text{or} \quad 3\le 2,

which is false. So not useful.

Try

(1,3),\ (2,2).

Then

(1,3)R(2,2)

because $1\le 2$ is true.

Reverse:

(2,2)R(1,3)

requires

2\le 1 \quad \text{or} \quad 2\le 3.

Again true. Still symmetric for this pair.

To break symmetry, we need

x_1\le x_2 \text{ or } y_1\le y_2

true, but

x_2\le x_1 \text{ or } y_2\le y_1

false.

For the reverse to be false, both must fail:

x_2>x_1 \quad \text{and} \quad y_2>y_1.

But then in the forward direction we have $x_1<x_2$ and $y_1<y_2$ , so forward is true. Good.

Take

(1,1),\ (2,2).

Then

(1,1)R(2,2)

because $1\le 2$ (also $1\le 2$ in second coordinate).

But

(2,2)R(1,1)

requires

2\le 1 \quad \text{or} \quad 2\le 1,

which is false.

Hence $R$ is not symmetric.

So statement (I) is true.

Check transitivity

A relation is transitive if

(a,b)R(c,d) \text{ and } (c,d)R(e,f) \implies (a,b)R(e,f).

We test whether this always holds.

We need a counterexample.

Take

A=(2,1),\quad B=(1,3),\quad C=(3,2).

Now check:

$A R B$ ?
$(2,1)R(1,3) \iff 2\le 1 \text{ or } 1\le 3.$
Since $1\le 3$ is true, $A R B$ holds.
$B R C$ ?
$(1,3)R(3,2) \iff 1\le 3 \text{ or } 3\le 2.$
Since $1\le 3$ is true, $B R C$ holds.
$A R C$ ?
$(2,1)R(3,2) \iff 2\le 3 \text{ or } 1\le 2.$
This is actually true, so this is not a counterexample.

Let us try to force the final relation to fail. For

(x_1,y_1)R(x_3,y_3)

to fail, we need

x_1>x_3 \quad \text{and} \quad y_1>y_3.

Choose:

A=(3,3),\quad B=(1,4),\quad C=(2,2).

Then

$A R B$ :
$(3,3)R(1,4) \iff 3\le 1 \text{ or } 3\le 4.$
True because $3\le 4$ .
$B R C$ :
$(1,4)R(2,2) \iff 1\le 2 \text{ or } 4\le 2.$
True because $1\le 2$ .
$A R C$ :
$(3,3)R(2,2) \iff 3\le 2 \text{ or } 3\le 2,$
which is false.

Thus,

A R B \text{ and } B R C \text{ but } A \not R C.

So $R$ is not transitive.

Hence statement (II) is false.

Conclusion

(I) is true.
(II) is false.

Therefore, the correct option is:

\boxed{\text{D: Only (I) is correct.}}

Comparison with stored answer

Stored correct answer: $\boxed{D}$

My derived answer is also $\boxed{D}$ , so they agree.

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