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
Answers
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:
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.