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Sets and Relations question

2023 · 29 Jan · Shift 2 · Q25
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  5. /2023 · 29 Jan · Shift 2 · Q25

Sets and Relations question

2023 · 29 Jan · Shift 2 · Q25

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let R be a relation defined on N\mathbb{N}N as aRba\mathrm{R}baRb if 2a+3b2a+3b2a+3b is a multiple of 5,a,b∈N5,a,b\in \mathbb{N}5,a,b∈N. Then R is
  1. A
    an equivalence relation
  2. B
    non reflexive
  3. C
    symmetric but not transitive
  4. D
    transitive but not symmetric
View written solutionFree

Correct answer: A

  1. Given relation

A relation RRR on N\mathbb{N}N is defined by aRb  ⟺  2a+3b is a multiple of 5.aRb \iff 2a+3b \text{ is a multiple of }5.aRb⟺2a+3b is a multiple of 5. That is, 2a+3b≡0(mod5).2a+3b\equiv 0\pmod{5}.2a+3b≡0(mod5).

We will simplify this condition first.

  1. Rewrite the condition modulo 555

Since 3≡−2(mod5)3\equiv -2 \pmod{5}3≡−2(mod5), we get 2a+3b≡2a−2b=2(a−b)(mod5).2a+3b \equiv 2a-2b = 2(a-b) \pmod{5}.2a+3b≡2a−2b=2(a−b)(mod5). So,

\iff 2(a-b)\equiv 0\pmod{5}.$$ Now $2$ is invertible modulo $5$ (because $\gcd(2,5)=1$), so this is equivalent to $$a-b\equiv 0\pmod{5},$$ that is, $$a\equiv b\pmod{5}.$$ Hence the relation is simply: $$aRb \iff a\equiv b\pmod{5}.$$ --- 3. **Check reflexivity** A relation is reflexive if $aRa$ for every $a\in\mathbb{N}$. Now, $$2a+3a=5a,$$ and $5a$ is always divisible by $5$. So $aRa$ for every $a$. Therefore, $R$ is **reflexive**. So option **B: non reflexive** is false. --- 4. **Check symmetry** A relation is symmetric if $$aRb \implies bRa.$$ Since $$aRb \iff a\equiv b\pmod{5},$$ we immediately have $$b\equiv a\pmod{5},$$ so $$bRa.$$ Therefore, $R$ is **symmetric**. --- 5. **Check transitivity** A relation is transitive if $$aRb \text{ and } bRc \implies aRc.$$ Now, $$aRb \iff a\equiv b\pmod{5},$$ and $$bRc \iff b\equiv c\pmod{5}.$$ From these, $$a\equiv b\pmod{5}, \quad b\equiv c\pmod{5} \implies a\equiv c\pmod{5}.$$ Hence, $$aRc.$$ Therefore, $R$ is **transitive**. --- 6. **Conclusion** Since $R$ is reflexive, symmetric, and transitive, it is an **equivalence relation**. So the correct option is: $$\boxed{\text{A}}$$ --- 7. **Compare with stored answer** Stored correct answer: **A** Our derived answer: **A** They agree.
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