Exercise 1.1 · Q16 (MCQ): $a=b-2,\ b>6$⁠: which pair belongs?

Basics you need

Solution

1. Read the rule and list the pairs of $R$

   $(a,b)\in R$ exactly when

   $$\text{(i) } a = b - 2 \qquad\text{and}\qquad \text{(ii) } b > 6.$$

   Condition (ii) allows $b = 7, 8, 9, \dots$⁠, and then condition (i) fixes $a = b - 2$⁠, which is $5, 6, 7, \dots$⁠, always a natural number. So

   $$\begin{aligned} R &= \{(5,7),\ (6,8),\ (7,9),\ (8,10),\ \dots\}\\ &= \{(b-2,\ b) : b \in \N,\ b \ge 7\}. \end{aligned}$$

   Two features are worth noticing. In every pair the first entry is 2 less than the second, so $a < b$⁠. And the smallest first entry is 5, because the smallest allowed $b$ is 7.

   Writing out a few pairs of $R$ turns an abstract rule into concrete pairs we can compare with the options. It also shows the pattern each option must fit, which makes the wrong options easy to spot.
   "$a = b - 2$" says $a$ is determined by $b$⁠: choose $b$⁠, subtract 2. The condition $b > 6$ then restricts which $b$ may be chosen.

2. Test each option against both conditions

   Option	$(a,b)$	(i) $b - 2$⁠, equal to $a$⁠?	(ii) $b > 6$⁠?	In $R$⁠?
   (A)	$(2,4)$	$4 - 2 = 2$⁠: yes	$4 > 6$⁠: no	no
   (B)	$(3,8)$	$8 - 2 = 6 \ne 3$⁠: no	$8 > 6$⁠: yes	no
   (C)	$(6,8)$	$8 - 2 = 6$⁠: yes	$8 > 6$⁠: yes	yes
   (D)	$(8,7)$	$7 - 2 = 5 \ne 8$⁠: no	$7 > 6$⁠: yes	no

   Only $(6,8)$ passes both tests, so $(6,8)\in R$.

   Each option is a single pair, so checking it is direct: substitute and test both conditions. Recording both tests for every option, even after one has failed, shows exactly which condition is responsible.
   Membership in a set-builder set means every condition holds. One false condition is enough to exclude a pair (Basic 3).

3. Why each wrong option fails

   • (A) $(2,4)$⁠: it does satisfy $a = b - 2$⁠, since $4 - 2 = 2$⁠. But $b = 4$ is not greater than 6, so condition (ii) fails. This option tests whether you check both conditions.
   • (B) $(3,8)$⁠: $b = 8 > 6$ is fine, but $b - 2 = 6$⁠, not 3, so condition (i) fails. The pair of $R$ with $b = 8$ is $(6,8)$⁠, not $(3,8)$⁠.
   • (D) $(8,7)$⁠: $b = 7 > 6$ is fine, but $b - 2 = 5$⁠, not 8, so condition (i) fails. In fact no pair with $a > b$ can ever be in $R$⁠, because $a = b - 2 < b$⁠. This option tests whether you read the order of the pair correctly: the first entry must be the smaller.

   Option (C) $(6,8)$ passes both tests: $8 - 2 = 6$ and $8 > 6$⁠. So the answer is (C).

   Explaining every rejection confirms that exactly one option is correct, and each explanation names the one condition that fails, which is a complete disproof for that option.
   $a = b - 2$ means $a$ is 2 less than $b$⁠, so $a < b$ in every pair of $R$⁠. Swapping the entries of a pair of $R$ never gives another pair of $R$⁠. Basics: what a relation is

Every fact used, and where it comes from