Problem 123 - A Complete Resource Book in Mathematics for JEE Main

Question

Consider the following relations: \(R=\\{(x, y) \mid x, y\) are real numbers and \(x=w y\) for some rational number \(w\\} ;\) \(S=\left\\{\begin{array}{l}\left(\frac{m}{p}, \frac{p}{q}\right) m, n, p \text { and } q \in \mathbf{Z} \\ \text { such that } n, q \neq 0 \text { and } q m=p n\end{array}\right\\}\) Then [2010] (A) neither \(R\) nor \(S\) is an equivalence relation (B) \(S\) is an equivalence relation but \(R\) is not an equivalence relation (C) \(R\) and \(S\) both are equivalence relations (D) \(R\) is an equivalence relation but \(S\) is not an equivalence relation

Step-by-Step Solution

Answer

(D) R is an equivalence relation but S is not an equivalence relation.

Step 1: Assess Equivalence Relation Criteria

An equivalence relation on a set should be reflexive, symmetric, and transitive. Let's determine if either of the given relations R and S satisfies these criteria.

Step 2: Analyzing Reflexivity of R

For R to be reflexive, every real number x must relate to itself, i.e., \((x, x) \in R\). This implies that x should equal wx for some rational w. Choosing w = 1 (which is rational) satisfies this condition, so R is reflexive.

Step 3: Analyzing Symmetry of R

For R to be symmetric, if \((x, y) \in R\), then \((y, x) \in R\) should also hold. If x = wy for some rational w, then y = x/w, and x/w is rational as long as w is not zero (which it cannot be in this context). Thus, R is symmetric.

Step 4: Analyzing Transitivity of R

For R to be transitive, if \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). Suppose x = wy and y = vz for some rational w and v. Thus, x = (wv)z. Since w and v are rational, their product wv is also rational, making \((x, z) \in R\). Hence, R is transitive.

Step 5: Conclusion on R's Equivalence

R is reflexive, symmetric, and transitive, so R is an equivalence relation.

Step 6: Analyzing Relation S

For S, we have pairs \((\frac{m}{p}, \frac{p}{q})\) with the condition qm = pn, for integers m, n, p, q and nonzero n, q.

Step 7: Check Reflexivity of S

For S to be reflexive, \((\frac{m}{p}, \frac{m}{p})\) should satisfy qm = pm, which holds trivially, so S is reflexive.

Step 8: Check Symmetry of S

For S to be symmetric, if \((\frac{m}{p}, \frac{p}{q}) \in S\), then \((\frac{p}{q}, \frac{m}{p}) \in S\). The condition qm = pn implies pq = mp, which is symmetric in terms of the condition, thus S is symmetric.

Step 9: Check Transitivity of S

For S to be transitive, if \((\frac{m}{p}, \frac{p}{q}) \in S\) and \((\frac{a}{b}, \frac{p}{q}) \in S\), then \((\frac{m}{p}, \frac{a}{b}) \in S\) should hold. However, the given conditions do not easily lead to an equivalent form for this requirement, indicating S might not be transitive.

Step 10: Conclusion on S's Equivalence

Since S may fail the transitivity criterion, it's not guaranteed to be an equivalence relation.

Key Concepts

Reflexive Relation

A reflexive relation is one where every element is related to itself. This is the foundational step in determining equivalence relations. To check if a relation is reflexive, we need to see if every element in the set under consideration can be paired with itself.

For example, consider the relation \(R\) defined as \((x, y)\) such that \(x = w y\) for some rational \(w\). For \(R\) to be reflexive, we must have \( (x, x) \) for all real numbers \(x\). By choosing \(w = 1\), which is rational, we satisfy the condition \(x = 1 \times x\), making \( (x, x) \) a pair in \(R\). Thus, \(R\) demonstrates reflexivity because each number can relate to itself when multiplied by one.

Similarly, the expression \(S\) uses mathematical fractions: \((\frac{m}{p}, \frac{m}{p})\). Here, the condition that needs checking is if \(q m = p m\) holds true, which it does because both sides are equivalent by default. Thus, everything within \(S\) also relates to itself, demonstrating reflexivity.

Symmetric Relation

A symmetric relation is defined by the ability for any two elements \((a, b)\) in a relationship to be flipped, meaning \((b, a)\) is also in the relationship. If you can reverse the pairs and maintain the relationship, the relation is symmetric.

In relation \(R\) again, consider two real numbers \(x\) and \(y\) such that \((x, y)\) satisfies \(x = w y\) for rational \(w\). For symmetry, if \((x, y)\) is true, then \(y = x / w\) should also be valid, where \(x / w\) remains a rational number. This essentially means if we flip the pair, it still meets the condition set by \(R\), proving symmetry.

Now, in relation \(S\), suppose \((\frac{m}{p}, \frac{p}{q})\) is within \(S\). By the fact \(q m = p n\), symmetry translates to checking whether \((\frac{p}{q}, \frac{m}{p})\) maintains the same condition. Since \(q m = p n\) inherently maintains \(pq = mp\) when flipped, \(S\) also displays symmetry.

Transitive Relation

Transitivity is a bit more complex than reflexivity and symmetry but equally important. It states that if a relationship holds from \(a\) to \(b\) and from \(b\) to \(c\), then it should also hold directly from \(a\) to \(c\). This means bridging the gap over intermediary steps must still obey the relation's condition.

In \(R\), if you have \((x, y)\) and \((y, z)\) in the relation, the condition \(x = w y\) and \(y = v z\) suggests \(x = (w v) z\). Both \(w\) and \(v\) being rational ensures \((w v)\) is rational, therefore \((x, z)\) naturally fits within \(R\), displaying transitivity.

However, the scenario is a tad different with \(S\). Imagine needing to move from \((\frac{m}{p}, \frac{p}{q})\) to \((\frac{a}{b}, \frac{p}{q})\), where both pairs should logically imply \((\frac{m}{p}, \frac{a}{b})\). The conditions don't easily combine into an equivalent expression, making it hard for the transitive nature to evident in \(S\). Hence, transitivity might not be clearly visible, causing a potential failure in being an equivalence relation.