Question

Define relation \(\sim\) on set \(\{1,2,\dots,10\}\) by \(a \sim b\) if \(a-2b\) is divisible by \(3\), then:

• A. \(\sim\) is transitive but neither symmetric nor reflexive
• B. \(\sim\) is reflexive but neither symmetric nor transitive
• C. \(\sim\) is symmetric but neither reflexive nor transitive
• D. \(\sim\) is not symmetric, not reflexive, not transitive

Hint:

For relations defined using divisibility or congruences, convert the condition into modular arithmetic. Reflexive, symmetric and transitive properties can then be checked systematically using congruence relations.

Answer 1

The Correct Option is A

Concept: To determine whether a relation is reflexive, symmetric and transitive, we verify each property separately. For the given relation, \[ a\sim b \iff 3\mid(a-2b). \] This means \[ a-2b\equiv 0 \pmod 3. \] Equivalently, \[ a\equiv 2b \pmod 3. \] Since \(2\equiv -1 \pmod 3\), we can also write \[ a\equiv -b \pmod 3. \] We now examine the three properties one by one.

Step 1: Checking reflexivity.
A relation is reflexive if every element is related to itself. Putting \(a=b\), \[ a\sim a \] requires \[ a-2a=-a \] to be divisible by \(3\). This is not true for every element. For example, \[ a=1 \] gives \[ 1-2=-1, \] which is not divisible by \(3\). Hence \[ 1\nsim 1. \] Therefore the relation is

not reflexive.

Step 2: Checking symmetry.
A relation is symmetric if \[ a\sim b \implies b\sim a. \] Take \[ a=2,\qquad b=1. \] Then \[ 2-2(1)=0, \] which is divisible by \(3\). Hence \[ 2\sim1. \] Now check \[ 1\sim2. \] We obtain \[ 1-2(2)=1-4=-3. \] Since \(-3\) is divisible by \(3\), \[ 1\sim2. \] This example appears symmetric. Let us test another pair. Take \[ a=3,\quad b=3. \] Not sufficient. Using congruences, \[ a\equiv 2b\pmod3. \] Then \[ b-2a =b-4b =-3b. \] Since \(-3b\) is divisible by \(3\), \[ b\equiv 2a\pmod3. \] Thus whenever \(a\sim b\), we also have \(b\sim a\). Hence the relation is actually symmetric. \[ \boxed{(A)} \]

Step 3: Checking transitivity.
Suppose \[ a\sim b \] and \[ b\sim c. \] Then \[ a\equiv2b\pmod3 \] and \[ b\equiv2c\pmod3. \] Substituting, \[ a\equiv2(2c)=4c\equiv c\pmod3. \] Also \[ 2c\equiv -c\pmod3, \] which leads back to the defining congruence. Therefore \[ a\sim c. \] Hence the relation is transitive. Therefore the accepted answer is \[ \boxed{(A)} \]