Question

Determine whether the relation r on the set of all real numbers as r={(a,b)}:a,b is r and a-b +root 3 is s where s is the set of all the irrational numberes is reflexive,symmetricand transitive

Answer 1

Given:  r={(a,b)}:a,b is r and a-b +root 3

To find: Determine whether the relation is reflexive, symmetric and transitive?

Solution:

• Now we have given r={(a,b)}: a,b is r and a - b +√3

• Now let a belongs R , then

               a - a + √3

               √3 ∈ S

So, (a,a) belongs to R . So, R is reflexive .

• Now let (a,b) belongs to R for any real numbers a & b , then :

               a - b + √3 will be  irrational number

               b - a + √ 3 will also be irrational number

               (b,a) ∈ R

• So hence R is symmetric.

• Now let (a,b) & (b,c) belongs to R , then:

               a-b + √3  is irrational

               b - c + √3 is irrational

• Now adding the two terms, we get:

               a - c + 2√3  is irrational.

• Thus (a,c) belongs to R

• So R is transitive

Answer:

      So the the relation given is reflexive, symmetric and transitive.

Answer 2

Answer:

R is reflexive but neither symmetric nor transitive.

Step-by-step explanation:

For Reflexive:-

As a-a +√3 = √3 belongs to S, which is an irrational number.

Therefore, (a, a) belongs to R for all a belongs to R.

So, R is reflexive.

For Symmetric:-

Let (a, b) belongs to R

=> a - b + √3 is an irrational number.

But, b - a + √3 is not an irrational number.

=> (b, a) doesn't belongs to R.

Example:- Let a = √3 , b = 0

So, R is not symmetric.

For Transitive:-

Let (a, b) belongs to R and (b, c) belongs to R.

=> a - b + √3 is an irrational number and b-c+√3 is an irrational number.

But, a - c + √ 3 is not an irrational number.

=> (a, c) belongs to R.

Example:- Let a= 0 , b= 2√3 and c= √3

So, R is not transitive.