Problem 28 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(R\) be the real line. Consider the following subsets of the plane \(R \times R\). \(S=\\{(x, y): y=x+1\) and \(0

Step-by-Step Solution

Answer

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

Step 1: Define Equivalence Relation

An equivalence relation on a set must satisfy three properties: reflexivity, symmetry, and transitivity. We will check these properties for both subsets \( S \) and \( T \).

Step 2: Check if S is an Equivalence Relation

For reflexivity, in \( S \), there are no elements \((x, x)\) because \( y = x+1 \) and \( y eq x \). Hence, \( S \) is not reflexive. Since \( S \) is not reflexive, it cannot be an equivalence relation. Additionally, it does not satisfy symmetry or transitivity as not all pairs are reflected and chained within the set.

Step 3: Check if T is Reflexive

For reflexivity in \( T \), each \(x\) should relate to itself, which is true as \(x - x = 0\), an integer. Thus, \( T \) is reflexive on \( R \).

Step 4: Check if T is Symmetric

For symmetry in \( T \), if \( (x, y) \in T \), then \( x-y \) is an integer. Thus, \( y-x = -(x-y) \) is also an integer, implying \( (y, x) \in T \). This confirms \( T \) is symmetric.

Step 5: Check if T is Transitive

For transitivity in \( T \), if \( (x, y) \in T \) and \( (y, z) \in T \), then \( x-y \) and \( y-z \) are integers. Thus, \( x-z = (x-y) + (y-z) \) is an integer, which implies \( (x, z) \in T \). Hence, \( T \) is transitive.

Key Concepts

Reflexivity

Reflexivity is one of the key properties that an equivalence relation must have. For a relation to be reflexive, every element must relate to itself. Imagine you have a large mirror. For reflexivity, each object in the room should appear in the mirror exactly where it actually is. Mathematically, in a set with relation \( R \), the reflexive property means that for any element \( x \) belonging to our set, the pair \( (x, x) \) should be within the relation \( R \).

Let's consider our example, the set \( S \) described as \( \{(x, y): y = x + 1 \} \). In this set, there are no reflexive pairs such as \( (x, x) \) because \( y \) is defined as \( x+1 \). Therefore, \( y \) can never equal \( x \), meaning \( S \) is not reflexive.

On the other hand, the set \( T \), defined by \( \{(x, y): x - y \text{ is an integer} \} \), includes pairs where the same element relates to itself, because \( x - x = 0 \), which is an integer. Thus, \( T \) satisfies reflexivity, demonstrating itself as a potential equivalence relation.

Symmetry

Symmetry is another vital property of equivalence relations. It ensures that if one element is related to the other, the reverse is also true. Like a two-way street, if you can go from point A to point B, you should also be able to go from B to A for symmetry.

In a relation \( R \), it is symmetric if whenever \( (x, y) \in R \), \( (y, x) \) is also within \( R \). Let's apply this to our sets. First, for the set \( S \), given it deals with \( y = x + 1 \), there is no guarantee that \( x = y + 1 \) fits within the defined parameters, so it is not symmetric.

For the set \( T \), since \( x - y \) being an integer implies \( y - x = -(x - y) \) is also integer, it satisfies symmetry perfectly. Thus, through this reflection-like property, \( T \) further proves its standing as an equivalence relation.

Transitivity

The final pillar necessary for a relation to qualify as an equivalence relation is transitivity. It employs the principle of chaining. If you know A is connected to B and B is connected to C, transitivity assures you A is also connected to C, forming a consistent link.

For a relation \( R \) to be transitive, whenever pairs \( (x, y) \) and \( (y, z) \) are in \( R \), then \( (x, z) \) must also be in \( R \). In the case of \( S \), when examining two such connections \((x, y)\) and \((y, z)\), the nature of \( y = x + 1 \) does not ensure a combined result that meets the criteria, so \( S \) is not transitive.

Reflecting on set \( T \), the property comes into play well as \( x-y \) and \( y-z \) being integers guarantees \( x-z = (x-y) + (y-z) \) is also an integer. Thus, \( (x, z) \) is in \( T \), confirming \( T \) as transitive and solidifying its status as an equivalence relation on \( R \).