Q.If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that

Concept: De Morgan's Laws for Sets

We verify the two De Morgan's laws by computing both sides independently and checking equality.

Given:

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

(i) Verify $(A \cup B)' = A' \cap B'$

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

Right side: $A' = \{1, 3, 5, 7, 9\}$ and $B' = \{1, 4, 6, 8, 9\}$, so $A' \cap B' = \{1, 9\}$.

Both sides equal $\{1, 9\}$. ✓

(ii) Verify $(A \cap B)' = A' \cup B'$

Left side: $A \cap B = \{2\}$, so $(A \cap B)' = \{1, 3, 4, 5, 6, 7, 8, 9\}$. …

Understanding De Morgan's Laws

De Morgan's Laws are fundamental identities in set theory that reveal a beautiful duality between union and intersection under complementation. When you take the complement of a union, it "flips" to become an intersection of complements—and vice versa. These laws appear everywhere: logic, probability, computer science, and of course set theory problems in your exams.

The intuition is straightforward. An element is not in $A \cup B$ precisely when it's in neither $A$ nor $B$—that is, when it's in both $A'$ and $B'$ simultaneously. Similarly, an element is not in $A \cap B$ when it fails to be in at least one of them—meaning it's in $A'$ or in $B'$ (or both).

Let's verify both laws with the given sets.

Given:

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

(i) Verifying $(A \cup B)' = A' \cap B'$

Left-hand side: $(A \cup B)'$

1. Find $A \cup B$: Combine all elements that appear in either set.

2. Take the complement: Elements in $U$ but not in $A \cup B$.

Right-hand side: $A' \cap B'$

3. Find $A'$: Elements in $U$ but not in $A$.

4. Find $B'$: Elements in $U$ but not in $B$.

5. Find $A' \cap B'$: Elements common to both complements.

Comparison:

(ii) Verifying $(A \cap B)' = A' \cup B'$

Left-hand side: $(A \cap B)'$

1. Find $A \cap B$: Elements common to both sets.

(Only $2$ appears in both $A$ and $B$.)

2. Take the complement: Elements in $U$ but not in $A \cap B$. $(A \cap B)' = U - \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$ …

Concept: Set Membership & Complement — The complement $X'$ of a set $X$ contains all elements of the universal set $U$ that are not in $X$.

Step 1: Recall $U = \{1,2,3,4,5,6,7,8,9\}$.

Step 2: For each part, list elements of $U$ missing from the given set.

• (i) $A' = U - A = \{5,6,7,8,9\}$
• (ii) $B' = U - B = \{1,3,5,7,9\}$
• (iii) $A \cup C = \{1,2,3,4,5,6\}$, so $(A \cup C)' = \{7,8,9\}$
• (iv) $A \cup B = \{1,2,3,4,6,8\}$, so $(A \cup B)' = \{5,7,9\}$
• (v) $(A')' = A = \{1,2,3,4\}$ (complement of complement is the original set) …

The Correct Answer

If $A \subset B$, then $A \cup B = B$.

Why?

Because every element of $A$ is already inside $B$. So when you take the union (all elements in $A$ or in $B$), you don’t add anything new beyond what $B$ already has. The union just gives back $B$.

Common Mistakes & How to Avoid Them

Mistake 1: Writing $A \cup B = A$

• What students think: “Since $A$ is inside $B$, the union is just the smaller set $A$.”
• Why it’s wrong: The union must include everything from both sets. $B$ has extra elements that $A$ doesn’t have — those must be included.
• How to avoid: Draw a Venn diagram. Shade $A$ and $B$ separately, then shade the union. You’ll see the larger set $B$ is fully covered.

Mistake 2: Writing $A \cup B = A \cap B$

• What students think: “If one is inside the other, union and intersection are the same.”
• Why it’s wrong:
  • $A \cup B$ = all elements in either set = $B$ (the bigger one).
  • $A \cap B$ = only elements in both sets = $A$ (the smaller one). They are equal only if $A = B$.
• How to avoid: Memorise the difference:
  • Union → bigger set (or equal).
  • Intersection → smaller set (or equal).

Mistake 3: Forgetting the special case $A = B$

• What students think: “If $A \subset B$, then $A$ is strictly smaller.”
• Why it’s wrong: In many textbooks, $A \subset B$ allows $A = B$ (some use $\subseteq$ for that). If $A = B$, then $A \cup B = A = B$ — still correct, but students sometimes panic. …