Key Concepts and Formulas

• Equivalence Relation: A relation R on a set A is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Reflexive: $a \mathrm{R} a$ for all $a \in A$.
  • Symmetric: If $a \mathrm{R} b$, then $b \mathrm{R} a$ for all $a, b \in A$.
  • Transitive: If $a \mathrm{R} b$ and $b \mathrm{R} c$, then $a \mathrm{R} c$ for all $a, b, c \in A$.
• Modular Arithmetic: The notation $x \equiv y \pmod m$ means that $x-y$ is a multiple of $m$, or equivalently, $x$ and $y$ have the same remainder when divided by $m$. This is useful for problems involving divisibility.

Step-by-Step Solution

We are given a relation R defined on the set of natural numbers $\mathbb{N}$ as $a \mathrm{R} b$ if $2a+3b$ is a multiple of $5$. We need to determine if R is an equivalence relation. We will check the three properties: reflexivity, symmetry, and transitivity.

Step 1: Checking for Reflexivity

• Goal: To check if $a \mathrm{R} a$ for all $a \in \mathbb{N}$.
• Applying the condition: For $a \mathrm{R} a$ to hold, $2a+3a$ must be a multiple of $5$.
• Calculation: $2a+3a = 5a$
• Explanation: Since $a$ is a natural number, $5a$ is always a multiple of $5$. Thus, $a \mathrm{R} a$ is true for all $a \in \mathbb{N}$.
• Conclusion: The relation R is reflexive.

Step 2: Checking for Symmetry

• Goal: To check if $a \mathrm{R} b \implies b \mathrm{R} a$ for all $a, b \in \mathbb{N}$.
• Assumption: Assume $a \mathrm{R} b$. This means $2a+3b$ is a multiple of $5$. $2a+3b \equiv 0 \pmod 5$
• Derivation: We need to show that $b \mathrm{R} a$, which means $2b+3a$ is a multiple of $5$. We can rewrite $3a+2b$ using modular arithmetic. Notice that $3 \equiv -2 \pmod 5$ and $2 \equiv -3 \pmod 5$. $3a+2b \equiv (-2)a + (-3)b \pmod 5$ $3a+2b \equiv -(2a+3b) \pmod 5$ Since we assumed $2a+3b \equiv 0 \pmod 5$, we substitute this into the expression: $3a+2b \equiv -(0) \pmod 5$ $3a+2b \equiv 0 \pmod 5$
• Explanation: This shows that if $2a+3b$ is a multiple of $5$, then $3a+2b$ is also a multiple of $5$.
• Conclusion: The relation R is symmetric.

Step 3: Checking for Transitivity

• Goal: To check if ($a \mathrm{R} b$ and $b \mathrm{R} c$) $\implies a \mathrm{R} c$ for all $a, b, c \in \mathbb{N}$.
• Assumptions:
  1. $a \mathrm{R} b \implies 2a+3b \equiv 0 \pmod 5$.
  2. $b \mathrm{R} c \implies 2b+3c \equiv 0 \pmod 5$.
• Derivation: We need to show that $a \mathrm{R} c$, which means $2a+3c \equiv 0 \pmod 5$. From the assumptions: From (1): $2a \equiv -3b \pmod 5$. From (2): $3c \equiv -2b \pmod 5$. Now consider the expression $2a+3c$: $2a+3c \equiv (-3b) + (-2b) \pmod 5$ $2a+3c \equiv -5b \pmod 5$ Since $-5b$ is always a multiple of $5$, $-5b \equiv 0 \pmod 5$. Therefore, $2a+3c \equiv 0 \pmod 5$.
• Explanation: This shows that if $a \mathrm{R} b$ and $b \mathrm{R} c$, then $a \mathrm{R} c$.
• Conclusion: The relation R is transitive.

Common Mistakes & Tips

• Thoroughness: Always verify all three properties (reflexive, symmetric, transitive). A relation is an equivalence relation only if all three hold.
• Modular Arithmetic: For divisibility problems, using modular arithmetic can simplify your steps and reduce the chance of algebraic errors.
• Domain Awareness: Ensure your proofs are valid for the entire domain specified (here, $\mathbb{N}$).

Summary

The relation R defined on $\mathbb{N}$ as $a \mathrm{R} b$ if $2a+3b$ is a multiple of $5$ has been shown to be reflexive, symmetric, and transitive. Therefore, R is an equivalence relation.

The final answer is $\boxed{A}$.