Q.Let R be the relation in the set N given by $R = \{(a, b) : a = b - 2, b > 6\}$. Choose the correct answer. (A) $(2, 4) \in R$ (B) $(3, 8) \in R$ (C) $(6, 8) \in R$ (D) $(8, 7) \in R$

The key idea is that a relation is defined by a specific condition — here, each ordered pair $(a, b)$ must satisfy $a = b - 2$ and $b > 6$.

Step 1: Check condition $b > 6$ for each option.

• (A) $b = 4$ → not $> 6$, so reject.
• (B) $b = 8$ → $8 > 6$ holds.
• (C) $b = 8$ → $8 > 6$ holds.
• (D) $b = 7$ → $7 > 6$ holds.

Step 2: Now check $a = b - 2$ for the remaining options.

• (B) $a = 3$, $b - 2 = 6$ → $3 \neq 6$, reject. …

We need to understand what the relation $R$ actually means before we check any of the given pairs. The definition is $R = \{(a, b) : a = b - 2, b > 6\}$, where $a$ and $b$ are natural numbers ($\mathbb{N}$). This is not a vague "related if" condition — it's a precise rule: for a pair $(a, b)$ to belong to $R$, two things must be true simultaneously.

First, the second element $b$ must be strictly greater than 6. Second, the first element $a$ must equal $b - 2$. That's it. There is no other condition. So if we take any natural number $b$ that is 7 or more, then $a = b - 2$ is automatically determined, and that pair is in $R$. For example, $(5, 7)$ is in $R$ because $7 > 6$ and $5 = 7 - 2$. Similarly, $(6, 8)$ would be in $R$ because $8 > 6$ and $6 = 8 - 2$.

Now let's test each option one by one.

1. Option (A): $(2, 4)$

   Here $b = 4$. The condition $b > 6$ fails because $4$ is not greater than $6$. So this pair cannot be in $R$ regardless of the $a$ value.

   Result: Not in $R$.

2. Option (B): $(3, 8)$

   Here $b = 8$, which is greater than 6 — so the first condition is satisfied. Now check $a = b - 2$: $8 - 2 = 6$, but the given $a$ is $3$. Since $3 \neq 6$, the second condition fails.

   Result: Not in $R$.

3. Option (C): $(6, 8)$

   $b = 8 > 6$ — good. Now $b - 2 = 6$, and the given $a$ is exactly $6$. Both conditions hold.

   Result: In $R$.

4. Option (D): $(8, 7)$ …

Method: Checking whether a pair belongs to a defined relation

When a relation is defined by a condition on $(a,b)$, a pair belongs to it only if it satisfies EVERY part of the condition at once.

Steps

Step 1: Split the definition into separate conditions

Here $R=\{(a,b): a=b-2,\ b>6\}$ carries two conditions joined by "and": (i) $a=b-2$ and (ii) $b>6$.

Step 2: Apply the cheapest filter first …

Common Mistakes

Mistake 1: Reading the two conditions as "or" instead of "and".

Why it's wrong: a pair is in $R$ only if $a=b-2$ AND $b>6$ both hold; satisfying just one is not enough. Correct approach: require both — e.g. $(3,8)$ has $b>6$ but $3\neq 8-2$, so it is rejected.

Mistake 2: Swapping the roles of $a$ and $b$. …