Q.Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find

The complement of a set $X$ (written $X'$) is everything in the universal set $U$ that is NOT in $X$.

Complements of single sets: $A' = U-A = \{5,6,7,8,9\}$; $B' = U-B = \{1,3,5,7,9\}$.

Complements of unions: $A\cup C = \{1,2,3,4,5,6\}$, so $(A\cup C)' = \{7,8,9\}$. $A\cup B = \{1,2,3,4,6,8\}$, so $(A\cup B)' = \{5,7,9\}$.

Double complement and set difference: $(A')' = A = \{1,2,3,4\}$. $B-C = \{2,8\}$, so $(B-C)' = \{1,3,4,5,6,7,9\}$.

Given $U=\{1,2,3,4,5,6,7,8,9\}$, $A=\{1,2,3,4\}$, $B=\{2,4,6,8\}$, $C=\{3,4,5,6\}$.

A complement $X'$ contains every element of $U$ that is NOT in $X$.

(i) $A'$: Elements of $U$ missing from $A$: $5,6,7,8,9$. So $A'=\{5,6,7,8,9\}$.

(ii) $B'$: Elements of $U$ missing from $B$: $1,3,5,7,9$. So $B'=\{1,3,5,7,9\}$.

(iii) $(A\cup C)'$: $A\cup C = \{1,2,3,4,5,6\}$. Elements of $U$ missing from this: $7,8,9$. So $(A\cup C)'=\{7,8,9\}$.

(iv) $(A\cup B)'$: $A\cup B = \{1,2,3,4,6,8\}$. Elements of $U$ missing from this: $5,7,9$. So $(A\cup B)'=\{5,7,9\}$.

(v) $(A')'$: $A'=\{5,6,7,8,9\}$; its complement is everything in $U$ not in $A'$, which is $1,2,3,4$ -- exactly $A$ again. So $(A')'=A=\{1,2,3,4\}$.

(vi) $(B-C)'$: First, $B-C$: from $B=\{2,4,6,8\}$, remove elements also in $C=\{3,4,5,6\}$ -- that removes $4$ and $6$, leaving $B-C=\{2,8\}$. Its complement is everything in $U$ except $2$ and $8$: $\{1,3,4,5,6,7,9\}$.

Set Membership: Complement of Sets

Method: Direct Set Listing Method — Find the complement by listing all elements of the universal set that are not in the given set.

Steps

1. Identify the universal set $U$ and the given set.
2. List all elements of $U$.
3. Remove all elements that belong to the given set.
4. The remaining elements form the complement.

Solutions

(i) $A'$

$U = \{1,2,3,4,5,6,7,8,9\}$, $A = \{1,2,3,4\}$

Remove $\{1,2,3,4\}$ from $U$.

$A' = \{5,6,7,8,9\}$

(ii) $B'$

$B = \{2,4,6,8\}$

Remove $\{2,4,6,8\}$ from $U$.

$B' = \{1,3,5,7,9\}$

(iii) $(A \cup C)'$

First find $A \cup C$:

$A = \{1,2,3,4\}$, $C = \{3,4,5,6\}$

$A \cup C = \{1,2,3,4,5,6\}$

Now complement: remove $\{1,2,3,4,5,6\}$ from $U$.

$(A \cup C)' = \{7,8,9\}$

(iv) $(A \cup B)'$

First find $A \cup B$:

$A = \{1,2,3,4\}$, $B = \{2,4,6,8\}$

$A \cup B = \{1,2,3,4,6,8\}$

Complement: remove these from $U$.

$(A \cup B)' = \{5,7,9\}$

(v) $(A')'$

We already have $A' = \{5,6,7,8,9\}$.

Complement of $A'$ means: remove $\{5,6,7,8,9\}$ from $U$.

$(A')' = \{1,2,3,4\} = A$

Key insight: The complement of a complement returns the original set.

(vi) $(B - C)'$

First find $B - C$ (elements in $B$ but not in $C$):

$B = \{2,4,6,8\}$, $C = \{3,4,5,6\}$

$B - C = \{2,8\}$ (since $4$ and $6$ are in $C$)

Now complement: remove $\{2,8\}$ from $U$.

$(B - C)' = \{1,3,4,5,6,7,9\}$

Quick Reference

Common Mistakes in Set Membership & Complement Problems

The Question at a Glance

We have:

• $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$
• $A = \{1, 2, 3, 4\}$
• $B = \{2, 4, 6, 8\}$
• $C = \{3, 4, 5, 6\}$

We need complements ($'$ means complement with respect to $U$).

Mistake #1: Forgetting the Universal Set

The error: Students write $A'$ as "everything except A" without checking what $U$ actually contains. They might include 10, 11, etc., or forget that $U$ is limited to $\{1,2,3,4,5,6,7,8,9\}$.

How to avoid: Always draw or list $U$ first. The complement is only those elements in $U$ that are not in the given set.

✓ Correct: $A' = U - A = \{5, 6, 7, 8, 9\}$

Mistake #2: Confusing Complement with "Opposite" or "Not in Set"

The error: For $(A \cup C)'$, students might first find $A \cup C$ correctly, but then write the complement as if it's "everything not in A and not in C" — which is actually $A' \cap C'$, not $(A \cup C)'$.

How to avoid: Remember De Morgan's Law:

But the safest method is:

1. Find $A \cup C$ first
2. Then take complement from $U$

✓ Correct path:

$A \cup C = \{1,2,3,4,5,6\}$

$(A \cup C)' = U - \{1,2,3,4,5,6\} = \{7,8,9\}$

Mistake #3: Errors in Set Difference $(B - C)$

The error: Students compute $B - C$ as $\{2,4,6,8\} - \{3,4,5,6\}$ and either:

• Remove too many elements (like removing 2)
• Keep elements that are in C (like keeping 4 or 6)

How to avoid: $B - C$ means elements in B that are NOT in C. Cross out common elements from B only.

✓ Correct:

$B = \{2, 4, 6, 8\}$, $C = \{3, 4, 5, 6\}$

Common: $\{4, 6\}$

$B - C = \{2, 8\}$

Then $(B - C)' = U - \{2, 8\} = \{1, 3, 4, 5, 6, 7, 9\}$

Mistake #4: Thinking $(A')'$ is Something Complicated

The error: Students try to compute $A'$ first (correctly), then get confused taking complement again — sometimes writing an empty set or a wrong result.

How to avoid: Remember the double complement law:

It's a basic property — the complement of the complement brings you back to the original set.

✓ Correct: $(A')' = A = \{1, 2, 3, 4\}$

Mistake #5: Misreading the Notation

The error: Confusing $'$ (complement) with other symbols like:

• Prime notation in derivatives
• "Not" in logic (which works differently)
• Thinking $A'$ means "A dash" as a different set

How to avoid: In set theory, $A'$ always means complement with respect to the universal set unless stated otherwise. Read the problem statement carefully.

Quick Summary Table

Final Tip

Always write $U$ at the top of your page. For each part:

1. Compute the inner set first (union, difference, etc.)
2. Then subtract from $U$ to get the complement
3. Double-check that every element in your answer is actually in $U$