Q.If A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 }, C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }; find

Set difference $X-Y$ = elements of $X$ that are NOT in $Y$. Using $A=\{3,6,9,12,15,18,21\}$, $B=\{4,8,12,16,20\}$, $C=\{2,4,6,8,10,12,14,16\}$, $D=\{5,10,15,20\}$:

1. $A-B$: remove $12$ -> $\{3,6,9,15,18,21\}$
2. $A-C$: remove $6,12$ -> $\{3,9,15,18,21\}$
3. $A-D$: remove $15$ -> $\{3,6,9,12,18,21\}$
4. $B-A$: remove $12$ -> $\{4,8,16,20\}$
5. $C-A$: remove $6,12$ -> $\{2,4,8,10,14,16\}$
6. $D-A$: remove $15$ -> $\{5,10,20\}$
7. $B-C$: remove $4,8,12,16$ (all in $C$); $20\notin C$, so it stays -> $\{20\}$
8. $B-D$: remove $20$ -> $\{4,8,12,16\}$
9. $C-B$: remove $4,8,12,16$ -> $\{2,6,10,14\}$
10. $D-B$: remove $20$ -> $\{5,10,15\}$ (xi) $C-D$: remove $10$ -> $\{2,4,6,8,12,14,16\}$ …

Set difference is not symmetric -- $A-B$ is generally not the same as $B-A$. For each pair, take the first set, go through its elements, and keep only those NOT present in the second set.

Given: $A=\{3,6,9,12,15,18,21\}$, $B=\{4,8,12,16,20\}$, $C=\{2,4,6,8,10,12,14,16\}$, $D=\{5,10,15,20\}$.

(i) $A-B$: Only $12$ from $A$ is also in $B$. Remove it: $\{3,6,9,15,18,21\}$

(ii) $A-C$: $6$ and $12$ from $A$ are in $C$. Remove them: $\{3,9,15,18,21\}$

(iii) $A-D$: Only $15$ from $A$ is in $D$. Remove it: $\{3,6,9,12,18,21\}$

(iv) $B-A$: Only $12$ from $B$ is in $A$. Remove it: $\{4,8,16,20\}$

(v) $C-A$: $6$ and $12$ from $C$ are in $A$. Remove them: $\{2,4,8,10,14,16\}$

(vi) $D-A$: Only $15$ from $D$ is in $A$. Remove it: $\{5,10,20\}$

(vii) $B-C$: From $B=\{4,8,12,16,20\}$, the elements $4,8,12,16$ are all in $C$, but $20$ is NOT in $C$ (C's largest element is 16). Remove only $4,8,12,16$: $\{20\}$

(viii) $B-D$: Only $20$ from $B$ is in $D$. Remove it: $\{4,8,12,16\}$

(ix) $C-B$: $4,8,12,16$ from $C$ are in $B$. Remove them: $\{2,6,10,14\}$

(x) $D-B$: Only $20$ from $D$ is in $B$. Remove it: $\{5,10,15\}$ …

Set Difference Method

Method Name: Set Difference (Relative Complement) — Finding elements present in one set but not in another.

Concept First (Why)

The difference $A - B$ (also written $A \setminus B$) means:

Take all elements of A, and remove any that also appear in B.

So $A - B = \{ x \mid x \in A \text{ and } x \notin B \}$.

Steps for Any Set Difference

1. List all elements of the first set.
2. Cross out any element that also appears in the second set.
3. Write the remaining elements as the answer.

Solutions

(i) $A - B$

$A = \{3, 6, 9, 12, 15, 18, 21\}$, $B = \{4, 8, 12, 16, 20\}$

• Common element: 12 (present in both)
• Remove 12 from A

Answer: $\{3, 6, 9, 15, 18, 21\}$

(ii) $A - C$

$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$

• Common elements: 6, 12
• Remove these from A

Answer: $\{3, 9, 15, 18, 21\}$

(iii) $A - D$

$D = \{5, 10, 15, 20\}$

• Common element: 15
• Remove 15 from A

Answer: $\{3, 6, 9, 12, 18, 21\}$

(iv) $B - A$

$B = \{4, 8, 12, 16, 20\}$, $A = \{3, 6, 9, 12, 15, 18, 21\}$

• Common element: 12
• Remove 12 from B

Answer: $\{4, 8, 16, 20\}$

(v) $C - A$

$C = \{2, 4, 6, 8, 10, 12, 14, 16\}$, $A = \{3, 6, 9, 12, 15, 18, 21\}$

• Common elements: 6, 12
• Remove these from C

Answer: $\{2, 4, 8, 10, 14, 16\}$

(vi) $D - A$

$D = \{5, 10, 15, 20\}$, $A = \{3, 6, 9, 12, 15, 18, 21\}$

• Common element: 15
• Remove 15 from D

Answer: $\{5, 10, 20\}$

(vii) $B - C$

$B = \{4, 8, 12, 16, 20\}$, $C = \{2, 4, 6, 8, 10, 12, 14, 16\}$

• Common elements: 4, 8, 12, 16
• Remove these from B

Answer: $\{20\}$

(viii) $B - D$

$D = \{5, 10, 15, 20\}$

• Common element: 20
• Remove 20 from B …

Here is a concept-first breakdown of the common mistakes students make with Set Difference (also called Set Membership subtraction), using your given sets.

Core Concept (The "Why")

The difference $A - B$ (also written $A \setminus B$) means: Start with all the elements in set A, then remove any element that also appears in set B. You are left with only the elements that are exclusively in A.

Common Mistake #1: Confusing the Order (A – B vs B – A)

• The Error: Students think $A - B$ is the same as $B - A$. They swap the "starting set" and the "removing set."
• Example from your question:
  • $A - B$: Start with A = {3, 6, 9, 12, 15, 18, 21}. Remove any number that is also in B (12 is in both). Result: {3, 6, 9, 15, 18, 21}.
  • $B - A$: Start with B = {4, 8, 12, 16, 20}. Remove any number that is also in A (12 is in both). Result: {4, 8, 16, 20}.
  • Notice: The results are completely different sets.
• How to Avoid: Read the operation aloud. Say to yourself: "A minus B means 'take A, throw away anything that belongs to B'." The first letter is your "home base."

Common Mistake #2: Forgetting to Remove All Common Elements

• The Error: A student correctly identifies that 12 is common between A and B, but they forget to check if other numbers are also common. They only remove one element.
• Example: For $C - D$:
  • C = {2, 4, 6, 8, 10, 12, 14, 16}
  • D = {5, 10, 15, 20}
  • A hasty student might only see "10" is common and write {2, 4, 6, 8, 12, 14, 16}. But they missed that 20 is not in C, so it's fine. However, the real danger is when multiple overlaps exist.
• How to Avoid: Use a systematic check. For every element in the first set, ask: "Is this number present in the second set?" If yes, cross it out. Do not assume only one match exists.

Common Mistake #3: Including Elements from the Second Set

• The Error: Students treat $A - B$ like a union or a subtraction of numbers. They accidentally include elements from B that are not in A.
• Example: For $A - D$:
  • A = {3, 6, 9, 12, 15, 18, 21}
  • D = {5, 10, 15, 20}
  • A student might incorrectly write {3, 6, 9, 12, 18, 21, 5, 10, 20} (keeping 15 removed, but adding 5, 10, 20 from D).
• How to Avoid: Remember the definition. You are only allowed to keep elements that were originally in the first set. The second set is just a "filter" to remove things; you never add anything from it.

Common Mistake #4: Misidentifying the Result as "Empty Set" When It Isn't

• The Error: Students see a common element and panic, thinking the whole set disappears.
• Example: For $B - C$:
  • B = {4, 8, 12, 16, 20}
  • C = {2, 4, 6, 8, 10, 12, 14, 16}
  • Common elements: 4, 8, 12, 16. Remove them from B. Result: {20}. Many students write $\emptyset$ (empty set) because they think "everything is common." …