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Consider a relation R defined as aRb if 2+ab>0 where a, b are real numbers. Then, the relation R is

A. reflexive and symmetric
B. symmetric and transitive
C. transitive and reflexive
D. None of these

User Pascamel
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1 Answer

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Final answer:

The relation R is reflexive because 2+a^2 is always positive for any real number a, and it's symmetric because if 2+ab is positive then so is 2+ba. However, it's not transitive as 2+ab and 2+bc being positive does not ensure 2+ac is positive. Option A is the correct answer.

Step-by-step explanation:

You asked how to determine whether the relation R defined as aRb if 2+ab>0 where a, b are real numbers, is reflexive, symmetric, and/or transitive. Let's analyze each property:

  • Reflexive: A relation is reflexive if every element is related to itself. Since for any real number a, the expression 2+a*a (which becomes 2+a^2) is always greater than zero because a^2 is non-negative, the relation R is reflexive.
  • Symmetric: A relation is symmetric if whenever aRb it's also true that bRa. Since if 2+ab > 0 then it is also true that 2+ba > 0, we can see that the relation R is symmetric.
  • Transitive: A relation is transitive if whenever aRb and bRc, then it is also true that aRc. However, we cannot guarantee this for our relation R, because from 2+ab > 0 and 2+bc > 0 it does not necessarily follow that 2+ac > 0.

Given this information, the relation R is reflexive and symmetric, but not transitive. Therefore, the correct answer to your question is Option A: reflexive and symmetric.

User Augusto Hack
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