Let $A$, $B$ and $C$ be sets. Then show that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

Let A, B and C be sets. Then show that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

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Concept understanding — Set Operations

The idea in plain words

Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".

The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.

Why this works

Sets are just labelled buckets that hold distinct items. The universal set $U$ is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:

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

Step by step

Let's take two concrete sets so you can see each operation in action:

A = \{1, 2, 3\}, \quad B = \{3, 4, 5\}

Step 1: Union — gather everything from both, but don't repeat anything.

A \cup B = \{1, 2, 3, 4, 5\}

Step 2: Intersection — only what's common to both.

A \cap B = \{3\}

Step 3: Difference ( $A$ minus $B$ ) — start with $A$ , remove anything that's also in $B$ .

A \setminus B = \{1, 2\}

Step 4: Complement — needs a universal set. Let $U = \{1, 2, 3, 4, 5\}$ . Then:

A^c = \{4, 5\}

Step 5: Symmetric Difference — combine the two differences:

A \triangle B = (A \setminus B) \cup (B \setminus A) = \{1, 2\} \cup \{4, 5\} = \{1, 2, 4, 5\}

A common slip …

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Intersection distributes over union: an element belongs to $A \cap (B \cup C)$ if and only if it belongs to both $A$ and at least one of $B$ or $C$ , which is precisely the condition for membership in $(A \cap B) \cup (A \cap C)$ .

The distributive law for set operations mirrors the distributive property in algebra, where multiplication distributes over addition. Here, intersection plays the role of "multiplication" and union plays the role of "addition." To prove two sets are equal, we show each is a subset of the other: every element of the left-hand side belongs to the right-hand side, and vice versa.

The strategy is to track what it means for an arbitrary element $x$ to belong to each side, translating set operations into logical statements about membership.

Proof that $A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$ :

Let $x \in A \cap (B \cup C)$ be arbitrary.
By definition of intersection, $x \in A$ and $x \in B \cup C$ .
Since $x \in B \cup C$ , by definition of union we have $x \in B$ or $x \in C$ (or both).
Case 1: If $x \in B$ , then since we already know $x \in A$ , we have $x \in A \cap B$ . Therefore $x \in (A \cap B) \cup (A \cap C)$ .
Case 2: If $x \in C$ , then since $x \in A$ , we have $x \in A \cap C$ . Therefore $x \in (A \cap B) \cup (A \cap C)$ .
In both cases, $x \in (A \cap B) \cup (A \cap C)$ .

This establishes the first inclusion.

Proof that $(A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$ :

Let $x \in (A \cap B) \cup (A \cap C)$ be arbitrary.
By definition of union, $x \in A \cap B$ or $x \in A \cap C$ (or both).
Case 1: If $x \in A \cap B$ , then $x \in A$ and $x \in B$ . Since $x \in B$ , we have $x \in B \cup C$ . Combined with $x \in A$ , this gives $x \in A \cap (B \cup C)$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 52 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The shaded region in the given Venn-diagram represents:$
(a) A ∪ B
(b) A ∩ B
(c) (A ∪ B)'
(d) (A ∩ B)'
›Reveal solutionSolution

The shaded region is everything in the universal set except A and B combined, which is exactly $(A \cup B)'$ .

The rectangle is the universal set $U$ , and the two overlapping circles are sets $A$ and $B$ . The description tells us the shading covers the rectangle except the two circles — i.e. every point that lies outside both $A$ and $B$ .

…

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CBSE 2026Set ANNUAL1 mark
$Q. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 6, 7}, then (A \cup B)' = ..............$
›Reveal solutionSolution

Find A∪B first, then take its complement in U.

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

First find $A\cup B$ (all elements in A or B or both):

$A\cup B = \{2,3,4,6,7,8\}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$X = \{1, 3, 5\}$
(a) $\{1, 2, 3, 4, 5\}$
(b) $\{1, 2, 3, 5\}$
(c) $\{1, 3\}$
(d) $\phi$
›Reveal solutionSolution

$X \cap Y$ consists of elements present in both $X$ and $Y$ , which gives $\{1,3\}$ .

Given $X = \{1,3,5\}$ and $Y=\{1,2,3\}$ . The intersection $X \cap Y$ contains only those elements that belong to BOTH sets.

…

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CBSE 2026Set ANNUAL1 mark
$\{2, 6, 10\}$
›Reveal solutionSolution

Sets are disjoint when their intersection is empty; comparing the elements of $\{2,6,10\}$ and $\{3,7,11\}$ shows no overlap.

Set $A=\{2,6,10\}$ and set $B=\{3,7,11\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{1, 2, 3\}, B = \{2, 3, 7\} \Rightarrow A \cup B =$
(a) $\{1, 2, 3\}$
(b) $\{1, 3, 7\}$
(c) $\{1, 2, 3, 7\}$
(d) $\{1, 2, 7\}$
›Reveal solutionSolution

$A \cup B = \{1, 2, 3, 7\}$ : the union lists every element that is in $A$ or in $B$ (or both), each written once.

For sets $A$ and $B$ , the union is $A \cup B = \{x : x \in A \text{ or } x \in B\}$ — combine both sets and remove duplicate entries.

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{3, 5, 7\}, Y = \{2, 3, 5\} \Rightarrow X \cap Y =$
(a) $\{3, 2\}$
(b) $\{3, 7\}$
(c) $\{5, 7\}$
(d) $\{3, 5\}$
›Reveal solutionSolution

$X \cap Y = \{3, 5\}$ : the intersection keeps only elements that belong to both sets.

For sets $X$ and $Y$ , $X \cap Y = \{x : x \in X \text{ and } x \in Y\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{1, 2\}, Y = \{2, 3, 5\}, Z = \{4, 6\} \Rightarrow X \cup Y \cup Z =$
(a) $\{1, 2, 3, 5, 6\}$
(b) $\{2, 3, 4, 5, 6\}$
(c) $\{1, 2, 3, 4, 5, 6\}$
(d) $\{1, 2\}$
›Reveal solutionSolution

$X \cup Y \cup Z = \{1, 2, 3, 4, 5, 6\}$ : list every element appearing in at least one of the three sets, once each.

Given $X = \{1, 2\}$ , $Y = \{2, 3, 5\}$ , $Z = \{4, 6\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{1, 2, 3, 6\}, Y = \{4, 5, 6\}, Z = \{4, 2, 3, 6\} \Rightarrow (X \cup Y) \cap Z =$
(a) $\{2, 3, 4, 6\}$
(b) $\{1, 5\}$
(c) $\{1, 2, 5\}$
(d) $\{1, 2, 3, 5\}$
›Reveal solutionSolution

$(X \cup Y) \cap Z = \{2, 3, 4, 6\}$ , found by first taking the union, then intersecting with $Z$ .

Given $X = \{1, 2, 3, 6\}$ , $Y = \{4, 5, 6\}$ , $Z = \{4, 2, 3, 6\}$ .

Step 1: $X \cup Y = \{1, 2, 3, 4, 5, 6\}$ (combine both, 6 counted once).

…

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CBSE 2025Set ANNUAL1 markMCQ
$X = \{a, b, c, d\}, Y = \{c, a, r\}, Z = \{r, o, b\} \Rightarrow (X \cap Y) \cup Z =$
(a) $\{a, b, c, o, r\}$
(b) $\{c, a, r, b\}$
(c) $\{r, o, b, c\}$
(d) $\phi$
›Reveal solutionSolution

$(X \cap Y) \cup Z = \{a, b, c, o, r\}$ , found by first taking the intersection, then the union with $Z$ .

Given $X = \{a, b, c, d\}$ , $Y = \{c, a, r\}$ , $Z = \{r, o, b\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{x : x - 2 = 0\}, B = \{x : 2x = 6\} \Rightarrow A \cup B =$
(a) $\{2, 6\}$
(b) $\{-2, 6\}$
(c) $\{2, 3\}$
(d) $\{2, -3\}$
›Reveal solutionSolution

$A \cup B = \{2, 3\}$ .

$A = \{x : x - 2 = 0\} = \{2\}$ . $B = \{x : 2x = 6\} = \{3\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$A = \{x : x^2 + 5x + 6 = 0\}, B = \{x : x^2 + 8x + 15 = 0\} \Rightarrow$
(a) $A \subset B$
(b) $B \subset A$
(c) $A = B$
(d) $A \cap B = \{-3\}$
›Reveal solutionSolution

$A = \{-2, -3\}$ , $B = \{-3, -5\}$ , and their only common element is $-3$ , so $A \cap B = \{-3\}$ .

$A = \{x : x^2 + 5x + 6 = 0\}$ : factorising, $(x+2)(x+3) = 0 \Rightarrow x = -2, -3$ , so $A = \{-2, -3\}$ .

$B = \{x : x^2 + 8x + 15 = 0\}$ : factorising, $(x+3)(x+5) = 0 \Rightarrow x = -3, -5$ , so $B = \{-3, -5\}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$U = \{1, 2, \ldots, 15\}$
(a) $\{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$
(b) $\{4, 6, 8, 10, 12, 14\}$
(c) $\{8, 10, 12, 14\}$
(d) $\phi$
›Reveal solutionSolution

$A' = U - A = \{4, 6, 7, 8, 9, 10, 11, 12, 13, 14\}$ .

Given $U = \{1, 2, \ldots, 15\}$ and $A = \{1, 2, 3, 5, 15\}$ . The complement $A' = U - A$ consists of every element of $U$ not in $A$ .

…

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Q.Let AAA, BBB and CCC be sets. Then show that A∩(B∪C)=(A∩B)∪(A∩C)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)A∩(B∪C)=(A∩B)∪(A∩C).

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Q.Let $A$ , $B$ and $C$ be sets. Then show that $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ .

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