Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′

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> [!TLDR]
> The complement of a set contains everything in the universal set that is not in the set. We find $A' = \{1,4,5,6\}$, $B' = \{1,2,6\}$, $A' \cap B' = \{1,6\}$, and $A \cup B = \{2,3,4,5\}$. Since $(A \cup B)' = \{1,6\}$ matches $A' \cap B'$, we have shown $(A \cup B)' = A' \cap B'$ — one of De Morgan's laws.

## The core idea: set membership and complements

The question asks you to verify a fundamental law of sets — De Morgan's law for the complement of a union. But before we dive into the mechanics, let's understand *why* this law makes sense.

Think of the universal set $U$ as the "whole world" of elements we care about. A set $A$ picks out some of those elements. Its complement $A'$ picks out everything *else* in $U$. So an element belongs to $A'$ precisely when it does **not** belong to $A$.

Now, what does $(A \cup B)'$ mean? It's the set of elements that are **not** in $A \cup B$. An element is in $A \cup B$ if it's in $A$ **or** in $B$ (or both). So being outside $A \cup B$ means the element is **neither** in $A$ **nor** in $B$ — it's outside both.

That's exactly the same as saying: the element is in $A'$ **and** in $B'$ — i.e., in $A' \cap B'$. So the law is almost obvious once you think about membership conditions. Let's verify it concretely.

## Step-by-step verification

**1. Find $A'$ — the complement of $A$**

$A = \{2, 3\}$. The universal set is $U = \{1, 2, 3, 4, 5, 6\}$. So $A'$ contains every element of $U$ that is **not** in $A$.

Removing $2$ and $3$ from $U$ leaves $\{1, 4, 5, 6\}$.

$$A' = \{1, 4, 5, 6\}$$

**2. Find $B'$ — the complement of $B$**

$B = \{3, 4, 5\}$. Removing these from $U$ leaves $\{1, 2, 6\}$.

$$B' = \{1, 2, 6\}$$

**3. Find $A' \cap B'$ — the intersection of the complements**

Intersection means elements that are in **both** $A'$ and $B'$.

$A' = \{1, 4, 5, 6\}$ and $B' = \{1, 2, 6\}$. The common elements are $1$ and $6$.

$$A' \cap B' = \{1, 6\}$$

**4. Find $A \cup B$ — the union of the original sets**

Union

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Q.Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

The core idea: set membership and complements

Step-by-step verification

Method: Set Difference (Subtraction) Method

Steps for $A - B$

Steps for $B - A$

🧠 The Concept: Complement of a Set

✗ Common Mistake #1: Forgetting the Universal Set

✗ Common Mistake #2: Including Elements of $A$ in $A'$

✗ Common Mistake #3: Confusing Complement with Difference

Q.Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3} and B = {3, 4, 5}. Find A′, B′ , A′ ∩ B′, A ∪ B and hence show that ( A ∪ B )′ = A′ ∩ B′

The core idea: set membership and complements

Step-by-step verification

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