Question

Let S be the set of all real numbers. Then the relation R = {(a, b): 1 + ab $\ge$ 0} on S is
A. Reflexive and symmetric but not transitive
B. Reflexive and transitive but not symmetric
C. Symmetric and transitive but not reflexive
D. An equivalence relation
E. None of the above

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step by step answer:
It is given that aRb is defined such that R = {(a, b): 1 + ab $\ge$ 0} . This means that for any two real numbers a and b, 1 + ab $\ge$ 0. We will check if it is reflexive, symmetric or transitive or not.
For a reflexive relation, aRa should exist. That is, 1 + ab $\ge$ 0. So we can write that-
1 + a.a$\ge$ 0
1 + $a^2\ge$ 0
We know that the square of any real number is greater than or equal to zero. So, 1 + $a^2$ should always be greater than zero.
The given relation is reflexive.
For a symmetric relation, both aRb and bRa should exist. So we can proceed as-
Let aRb exist, that is, 1 + ab $\ge$ 0
We know that ab = ba, hence-
1 + ba $\ge$ 0
This implies that bRa also exists. Hence, the given relation is symmetric.
There is no direct method to prove if the relation is transitive or not. So we will use an example to check if it satisfies the relation or not. Let us assume that-
a = 1, b = 0 and c = -1
So we can write that-
1 + ab = 1 + 1(0) = 1 $\ge$ 0
Hence, aRb exists. Also,
1 + bc = 1 + 0(-1) = 1 $\ge$ 0
Hence, bRc also exists. We will now check if aRc exists or not.
1 + ac = 1 + 1(-1) = 0
Hence we can conclude with this example that the relation is transitive.

So, the correct answer is “Option D”.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.