Question

For real numbers a and b, define aRb if b-a+√5 is an irrational number. Then the relation R is:

• A. Symmetric
• B. Reflexive
• C. Transitive
• D. Equivalence

Answer 1

The Correct Option is B

Let's analyze the given relation R for real numbers a and b, where aRb is defined if b-a+√5 is an irrational number. We will evaluate the properties of R: reflexivity, symmetry, and transitivity, to determine its characteristics.

• Reflexive: For a relation to be reflexive, every element must relate to itself. Thus, we verify if aRa is true for all real numbers a. Consider:
  a-a+√5=√5
  Since √5 is an irrational number, aRa holds true for every real number a. Therefore, the relation R is reflexive.
• Symmetric: A relation is symmetric if aRb implies bRa. Assume aRb, implying:
  b-a+√5 is irrational.
  Now, check if bRa holds, i.e., if a-b+√5 is irrational. Note:
  a-b+√5=-(b-a)+√5
  Since we can't generally say if -(b-a)+√5 is irrational (as irrationality isn't necessarily preserved under addition of negatives), R is not symmetric.
• Transitive: A relation is transitive if aRb and bRc imply aRc. Assume:
  • b-a+√5 is irrational
  • c-b+√5 is irrational
  We need to check if c-a+√5 is irrational. As:
  c-a+√5=(c-b)+(b-a)+√5
  The sum of irrational numbers does not inherently preserve irrationality, hence we cannot guarantee c-a+√5 is irrational. Thus, R is not transitive.

Since R is only reflexive, it does not satisfy the conditions to be symmetric or transitive. Therefore, the relation R is Reflexive.