Question

Among the relations 
$S=\left\{(a, b): a, b \in R -\{0\}, 2+\frac{a}{b}>\right\}$ 
and $T=\left\{(a, b): a, b \in R , a^2-b^2 \in Z\right\}$,

• A. $S$ is transitive but $T$ is not
• B. both $S$ and $T$ are symmetric
• C. neither $S$ nor $T$ is transitive
• D. $T$ is symmetric but $S$ is not

Hint:

When analyzing relations for symmetry, check if \( (a, b) \) implies \( (b, a) \) for the relation to be symmetric.

Answer 1

The Correct Option is D

To solve this question, we must determine the properties of the given relations \(S\) and \(T\).

Analysis of Relation \(S\):

The relation \(S\) is defined as: \(S=\left\{(a, b): a, b \in R -\{0\}, 2+\frac{a}{b} > 0\right\}\).

• Symmetric Property: For \(S\) to be symmetric, if \((a, b) \in S\), then \((b, a)\) must also be in \(S\). For the relation \(2+\frac{a}{b}>0\), if it holds true, the reverse condition \(2+\frac{b}{a} > 0\) may not necessarily hold true as well. For example, let \(a = 1\) and \(b = -2\). \(2+\frac{1}{-2} = 1.5 > 0\), but \(2+\frac{-2}{1} = 0\) is not greater than 0. Hence, \(S\) is not symmetric.
• Transitive Property: For \(S\) to be transitive, if \((a,b) \in S\) and \((b,c) \in S\), then \((a,c)\) must be in \(S\). Since the expression involves non-linearity due to reciprocal terms, transitivity is not guaranteed in all cases depending on individual values of \(a\), \(b\) and \(c\).

Analysis of Relation \(T\):

The relation \(T\) is defined as: \(T=\left\{(a, b): a, b \in R , a^2-b^2 \in Z\right\}\).

• Symmetric Property: If \((a, b) \in T\), then \(a^2-b^2 \in Z\). This implies \((b, a) \in T\) as well, since reversing the terms gives \(b^2 - a^2 = -(a^2 - b^2) \in Z\). Hence, \(T\) is symmetric.
• Transitive Property: For \(T\) to be transitive, if \((a,b) \in T\) and \((b,c) \in T\), then \((a,c)\) must be in \(T\). However, continuity of integer values may not hold, thus transitivity isn't always ensured.

Based on the above analysis, the correct option that describes the properties of relations \(S\) and \(T\) is: $T$ is symmetric but $S$ is not.