Question

Let Q be the set of rational numbers and R be a relation on Q defined by

R = { (x,y) : x , y ∈ Q , x2

+ y2

= 5 } Check whether each of the above relation is reflexive ,

symmetric and transitive​

Answer 1

Answer:

The relation R is defined as

R={(x,y):1+xy>0}

The relation R is reflexiva if (x,x) is an element of R. For that (x,x) to be in R 1+(x)(x) must be positive.

1+x²>0 irrespective of any value of x.

So, R is reflexive.

As (x,x) is an element of R for any x.

Similarly, (x,y) is an element of R if 1+xy>0

Since it is a product it satisfies the commutative law, so 1+yx>0

Since, 1+yx>0, then (y,x) is the element of R.

So, if (x,y) is an element of R then (y,x) is also the element of R.

So, R is symmetric.

Consider, x,y,z are any three elements of set Q,

Then say, (x,y) and (y,z) are the elements of R

1+xy>0 and 1+yz>0

So, 1+xy+1+yz>0

2+xy+yz>0

If (x,z) is the elements of R then 1+xz>0 but the above simplified term is different, so (x,z) is not the element of R.

So, this is not transitive relation.

So, the relation R is both reflexive and symmetric but not transitive.

Step-by-step explanation:

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Answer 2

SOLUTION

TO DETERMINE

Let Q be the set of rational numbers and R be a relation on Q defined by

R = { (x,y) : x , y ∈ Q , x² + y² = 5 }

Check whether each of the above relation is reflexive , symmetric and transitive

EVALUATION

Here the given relation is defined as

R = { (x,y) : x , y ∈ Q , x² + y² = 5 }

CHECKING FOR REFLEXIVE

1 ∈ Q

But 1² + 1² = 2 ≠ 5

So (1, 1) ∉ R

So R is not Reflexive

CHECKING FOR SYMMETRIC

Let x , y ∈ Q and (x, y) ∈ R

⇒ x² + y² = 5

⇒y² + x² = 5

⇒(y, x) ∈ R

Thus (x, y) ∈ R implies (y, x) ∈ R

So R is symmetric

CHECKING FOR TRANSITIVE

1 , 2 ∈ Q

Now 1² + 2² = 5 and 2² + 1² = 5

So (1, 2) ∈ R and (2, 1) ∈ R

But (2,1) ∉ R

Thus (1, 2) ∈ R and (2, 1) ∈ R but (2,1) ∉ R

R is not transitive

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