Problem 31 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(R\) be the set of all real numbers. Statement 1: \(A=\\{(x, y) \in R \times R: y-x\) is an integer \(\\}\) is an equivalence relation on \(R\). Statement \(2: B=\\{(x, y) \in R \times R: x=\alpha y\) for some rational number \(\alpha\\}\) is an equivalence relation on \(R\). (A) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (B) Statement 1 is true, Statement 2 is false. (C) Statement 1 is false, Statement 2 is true. (D) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1

Step-by-Step Solution

Answer

(A) Both statements are true; Statement 2 is not an explanation for Statement 1.

Step 1: Define Equivalence Relation

An equivalence relation is a relation that satisfies three properties: reflexivity, symmetry, and transitivity. We'll check each for both statements.

Step 2: Analyze Statement 1: Reflexivity

For reflexivity, for every \((x, x)\), y-x = x-x = 0, which is an integer. So, reflexivity holds for set \(A\).

Step 3: Analyze Statement 1: Symmetry

For symmetry, if \((x, y) \in A\), then \(y-x\) is an integer. This means \(x-y\) is also an integer (negative of an integer is an integer), so \((y, x) \in A\). Hence, symmetry holds for \(A\).

Step 4: Analyze Statement 1: Transitivity

For transitivity, suppose \((x, y) \in A\) and \((y, z) \in A\), meaning \(y-x\) and \(z-y\) are integers. Therefore, \((z-x) = (z-y) + (y-x)\) is also an integer. Thus, \((x, z) \in A\). Transitivity holds for \(A\).

Step 5: Conclude Statement 1

Since \(A\) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation. Statement 1 is true.

Step 6: Analyze Statement 2: Reflexivity

For reflexivity, \((x, x) \in B\) if \(x = \alpha x\) for some rational number \(\alpha\). This is true for \(\alpha = 1\), which is rational, so reflexivity holds for set \(B\).

Step 7: Analyze Statement 2: Symmetry

For symmetry, if \((x, y) \in B\), then \(x = \alpha y\) for some rational \(\alpha\). This implies \(y = \frac{1}{\alpha} x\) and since \(\frac{1}{\alpha}\) is rational when \(\alpha\) is rational, \((y, x) \in B\). Symmetry holds.

Step 8: Analyze Statement 2: Transitivity

For transitivity, suppose \((x, y) \in B\) and \((y, z) \in B\). There exist rational numbers \(\alpha\) and \(\beta\) such that \(x = \alpha y\) and \(y = \beta z\). Then, \(x = \alpha (\beta z) = (\alpha \beta) z\). Since the product of two rational numbers is rational, \((x, z) \in B\). Transitivity holds for \(B\).

Step 9: Conclude Statement 2

Since \(B\) satisfies reflexivity, symmetry, and transitivity, it is an equivalence relation. Statement 2 is true.

Step 10: Evaluate the Relationship between Statements

Statement 2 describes a different type of relation compared to Statement 1 and does not explain Statement 1. The sets \(A\) and \(B\) are based on different conditions: an integer difference and a rational multiple, respectively.

Step 11: Determine the Correct Option

Both statements are true, but Statement 2 is not an explanation of Statement 1. Thus, the correct option is (A).

Key Concepts

Reflexivity

Reflexivity is one of the three essential properties that define an equivalence relation. It applies to every element within a set. In simpler terms, a relation is reflexive if every element is related to itself. Let's take a closer look. For Example, considering set \(A\): If we have a pair \((x, x)\), then \(y-x = x-x = 0\), which is indeed an integer. Therefore, for set \(A\), reflexivity holds true because any real number minus itself is zero, an integer.Now, looking at set \(B\): Reflexivity here implies \((x, x)\) would mean \(x = \alpha x\) for some rational number \(\alpha\). This holds true for \(\alpha = 1\), as multiplying any number by 1 leaves it unchanged. Thus, every number is related to itself under this rule, maintaining reflexivity.

Symmetry

Symmetry is another vital property of an equivalence relation. It means if one element is related to another, then the second is also related to the first. This bidirectional relationship forms the core of symmetry.For set \(A\), symmetry can be seen as: if \((x, y) \in A\), then \(y-x\) is an integer. By reversing this, \(x-y = -(y-x)\) is still an integer. This means if \((x, y)\) is in set \(A\), so is \((y, x)\). Hence, set \(A\) satisfies the symmetry condition.Considering set \(B\), if we have \((x, y)\) such that \(x = \alpha y\), symmetry implies \(y = \frac{1}{\alpha} x\). Since the inverse of a rational number is also rational, \(\frac{1}{\alpha}\) remains rational. Thus, \((y, x)\) is a part of set \(B\), showing the set is symmetric.

Transitivity

Transitivity is the third pillar of an equivalence relation. It's the idea that if an element is related to two others, those two are also related. In other words, it's a chain of relations. For set \(A\), if \((x, y)\) and \((y, z)\) are both in \(A\), then \(y - x\) and \(z - y\) are integers. By the property of integers, \(z - x = (z - y) + (y - x)\) must also be an integer, bringing \((x, z)\) into set \(A\). Thus, transitivity is upheld.In set \(B\), transitivity is determined as follows: if \(x = \alpha y\) and \(y = \beta z\) for rational numbers \(\alpha\) and \(\beta\), then \(x = (\alpha\beta) z\). Since multiplying rational numbers yields a rational product, \(x = \alpha z\) for some rational \(\alpha\), making \((x, z)\) part of set \(B\). Consequently, set \(B\) obeys the transitive nature of equivalence relations.